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Chapter 19- Chemical
     Thermodynamics
Sections 19.1 - 19.3
  Spontaneous processes & entropy
First Law of Thermodynamics




                            Chemical
                         Thermodynamics
First Law of Thermodynamics

• You will recall from Chapter 5 that
  energy cannot be created nor
  destroyed.




                                           Chemical
                                        Thermodynamics
First Law of Thermodynamics

• You will recall from Chapter 5 that
  energy cannot be created nor
  destroyed.
• Therefore, the total energy of the
  universe is a constant.



                                           Chemical
                                        Thermodynamics
First Law of Thermodynamics

• You will recall from Chapter 5 that
  energy cannot be created nor
  destroyed.
• Therefore, the total energy of the
  universe is a constant.
• Energy can, however, be converted
  from one form to another or transferred
  from a system to the surroundings or
  vice versa.                            Chemical
                                           Thermodynamics
Spontaneous Processes




                           Chemical
                        Thermodynamics
Spontaneous Processes
• Spontaneous processes
  are those that can
  proceed without any
  outside intervention.




                                 Chemical
                              Thermodynamics
Spontaneous Processes
• Spontaneous processes
  are those that can
  proceed without any
  outside intervention.
• The gas in vessel B will
  spontaneously effuse into
  vessel A, but once the
  gas is in both vessels, it
  will not spontaneously
  return
                                  Chemical
                               Thermodynamics
Spontaneous Processes
 • Spontaneous processes
   are those that can
   proceed without any
   outside intervention.
 • The gas in vessel B will
   spontaneously effuse into
   vessel A, but once the
   gas is in both vessels, it
   will not spontaneously
   return
A process is either spontaneous,         Chemical
non-spontaneous, or at equilibrium.   Thermodynamics
Spontaneous Processes

            Processes that are
            spontaneous in one
            direction are
            nonspontaneous in
            the reverse
            direction.



                              Chemical
                           Thermodynamics
Melting a cube of water ice at −10°C
and atmospheric pressure is a
________ process.
1. Spontaneous

2. Nonspontaneous




                                      Chemical
                                   Thermodynamics
Correct Answer:

                    Melting of an ice cube at
1. Spontaneous      1 atm pressure will only
                    be spontaneous above
2. Nonspontaneous   the freezing point (0°C).




                                           Chemical
                                        Thermodynamics
Spontaneous Processes




                           Chemical
                        Thermodynamics
Spontaneous Processes
• Processes that are spontaneous at one
  temperature may be nonspontaneous at other
  temperatures.




                                            Chemical
                                         Thermodynamics
Spontaneous Processes
• Processes that are spontaneous at one
  temperature may be nonspontaneous at other
  temperatures.
• Above 0°C it is spontaneous for ice to melt.




                                             Chemical
                                          Thermodynamics
Spontaneous Processes
• Processes that are spontaneous at one
  temperature may be nonspontaneous at other
  temperatures.
• Above 0°C it is spontaneous for ice to melt.
• Below 0°C the reverse process is spontaneous.




                                             Chemical
                                          Thermodynamics
Chemical
Thermodynamics
No. Nonspontaneous processes can occur with
    some external assistance.




                                            Chemical
                                         Thermodynamics
SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes

   Predict whether these processes are spontaneous as
   described, spontaneous in the reverse direction, or in
   equilibrium:
   (a) When a piece of metal heated to 150°C is added
          to water at 40°C, the water gets hotter.
   (b) Water at room temperature decomposes into
          H2(g) and O2(g).
   (c) Benzene vapor, C6H6(g), at a pressure of 1 atm
          condenses to liquid benzene at the normal
          boiling point of benzene, 80.1°C.
SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes

   Predict whether these processes are spontaneous as
   described, spontaneous in the reverse direction, or in
   equilibrium:
   (a) When a piece of metal heated to 150°C is added
          to water at 40°C, the water gets hotter.
   (b) Water at room temperature decomposes into
          H2(g) and O2(g).
   (c) Benzene vapor, C6H6(g), at a pressure of 1 atm
          condenses to liquid benzene at the normal
          boiling point of benzene, 80.1°C.
Analyze: We are asked to judge whether each process will proceed spontaneously in the direction indicated, in the reverse
direction, or in neither direction.
Plan: We need to think about whether each process is consistent with our experience about the natural direction of events or
whether it is the reverse process that we expect to occur.
Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is
transferred from the hotter object to the colder one. In this instance heat is transferred from the hot metal to the cooler water.
The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere
between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we
certainly have never seen hydrogen and oxygen gas bubbling up out of water! Rather, the reverse process—the reaction of
H2 and O2 to form H2O—is spontaneous once initiated by a spark or flame. (c) By definition, the normal boiling point is the
temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. Neither the
condensation of benzene vapor nor the reverse process is spontaneous. If the temperature were below 80.1°C,
condensation would be spontaneous.
PRACTICE EXERCISE
Under 1 atm pressure CO2(s) sublimes at
–78°C. Is the transformation of CO2(s) to
CO2(g) a spontaneous process at –100ºC
and 1 atm pressure?




                                       Chemical
                                    Thermodynamics
PRACTICE EXERCISE
Under 1 atm pressure CO2(s) sublimes at
–78°C. Is the transformation of CO2(s) to
CO2(g) a spontaneous process at –100ºC
and 1 atm pressure?


 Answer: No, the reverse process is
 spontaneous at this temperature.
                                         Chemical
                                      Thermodynamics
Reversible Processes

            In a reversible
            process the system
            changes in such a
            way that the system
            and surroundings
            can be put back in
            their original states
            by exactly reversing
            the process.

                                 Chemical
                              Thermodynamics
Irreversible Processes




                            Chemical
                         Thermodynamics
Irreversible Processes




• Irreversible processes cannot be undone by
  exactly reversing the change to the system.


                                              Chemical
                                           Thermodynamics
Irreversible Processes




• Irreversible processes cannot be undone by
  exactly reversing the change to the system.
• Spontaneous processes are irreversible.

                                              Chemical
                                           Thermodynamics
Chemical
Thermodynamics
No




        Chemical
     Thermodynamics
Entropy




             Chemical
          Thermodynamics
Entropy

• Entropy (S) is a term coined by Rudolph
  Clausius in the 19th century.




                                         Chemical
                                      Thermodynamics
Entropy

• Entropy (S) is a term coined by Rudolph
  Clausius in the 19th century.
• Clausius was convinced of the
  significance of the ratio of heat
  delivered and the temperature at which
  it is delivered, q
                   T

                                         Chemical
                                      Thermodynamics
Entropy




             Chemical
          Thermodynamics
Entropy

• Entropy can be thought of as a measure
  of the randomness of a system.




                                        Chemical
                                     Thermodynamics
Entropy

• Entropy can be thought of as a measure
  of the randomness of a system.
• It is related to the various modes of
  motion in molecules.




                                        Chemical
                                     Thermodynamics
Entropy




             Chemical
          Thermodynamics
Entropy

• Like total energy, E, and enthalpy, H,
  entropy is a state function.




                                              Chemical
                                           Thermodynamics
Entropy

• Like total energy, E, and enthalpy, H,
  entropy is a state function.
• Therefore,




                                              Chemical
                                           Thermodynamics
Entropy

• Like total energy, E, and enthalpy, H,
  entropy is a state function.
• Therefore,
              ΔS = Sfinal − Sinitial




                                              Chemical
                                           Thermodynamics
Entropy

• For a process occurring at constant
  temperature (an isothermal process), the
  change in entropy is equal to the heat that
  would be transferred if the process were
  reversible divided by the temperature:




                                                   Chemical
                                                Thermodynamics
Entropy

• For a process occurring at constant
  temperature (an isothermal process), the
  change in entropy is equal to the heat that
  would be transferred if the process were
  reversible divided by the temperature:

                      qrev
                 ΔS =
                       T
                                                   Chemical
                                                Thermodynamics
Chemical
Thermodynamics
•        ΔS depends not merely on q but on qrev.
    There is only one reversible isothermal path
    between two states regardless of the number of
    possible paths.




                                                 Chemical
                                              Thermodynamics
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change


The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 2.29 kJ/mol. What is the
entropy change of the system when 50.0 g of
Hg(l) freezes at the normal freezing point?




                                                            Chemical
                                                         Thermodynamics
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change


The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 2.29 kJ/mol. What is the
entropy change of the system when 50.0 g of
Hg(l) freezes at the normal freezing point?




                                                            Chemical
                                                         Thermodynamics
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change


The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point
of mercury is –38.9ºC, and its molar enthalpy
of fusion is ΔΗfusion = 2.29 kJ/mol. What is the
entropy change of the system when 50.0 g of
Hg(l) freezes at the normal freezing point?




                                                            Chemical
                                                         Thermodynamics
SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change


The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and
its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l)
freezes at the normal freezing point?

Solution
Analyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the
surroundings when a liquid freezes (q < 0). The enthalpy of fusion is ΔΗ for the melting process. Because freezing is
the reverse of melting, the enthalpy change that accompanies the freezing of
1 mol of Hg is –ΔΗfusion = 2.29 kJ/mol.


Plan: We can use –ΔΗfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg:




We can use this value of q as qrev in Equation 19.2. We must first, however, convert the temperature to K:



Solve: We can now calculate the value of ΔSsys:




Check: The entropy change is negative because heat flows from the system making qrev negative.
Comment: The procedure we have used here can be used to calculate ΔS for other isothermal phase changes, such
as the vaporization of a liquid at its boiling point.
                                                                                                                 Chemical
                                                                                                              Thermodynamics
SAMPLE EXERCISE 19.2 continued


PRACTICE EXERCISE
The normal boiling point of ethanol,
C2H5OH, is 78.3°C (Figure 11.24), and
its molar enthalpy of vaporization is
38.56 kJ/mol. What is the change in
entropy in the system when 68.3 g of
C2H5OH(g) at 1 atm condenses to
liquid at the normal boiling point?

                                     Chemical
                                  Thermodynamics
SAMPLE EXERCISE 19.2 continued


PRACTICE EXERCISE
The normal boiling point of ethanol,
C2H5OH, is 78.3°C (Figure 11.24), and
its molar enthalpy of vaporization is
38.56 kJ/mol. What is the change in
entropy in the system when 68.3 g of
C2H5OH(g) at 1 atm condenses to
liquid at the normal boiling point?
Answer: –163 J/K
                                     Chemical
                                  Thermodynamics
Second Law of Thermodynamics

  The second law of thermodynamics
  states that the entropy of the universe
  increases for spontaneous processes,
  and the entropy of the universe does
  not change for reversible processes.



                                           Chemical
                                        Thermodynamics
Second Law of Thermodynamics




                            Chemical
                         Thermodynamics
Second Law of Thermodynamics

In other words:
For reversible processes:
         ΔSuniv = ΔSsystem + ΔSsurroundings = 0
For irreversible processes:
         ΔSuniv = ΔSsystem + ΔSsurroundings > 0


                                              Chemical
                                           Thermodynamics
Second Law of Thermodynamics

These last truths mean that as a result of all
spontaneous processes

     the entropy of the
     universe increases.

                                              Chemical
                                           Thermodynamics
Chemical
Thermodynamics
The entropy of the universe for the process must
increase by a greater amount than the entropy of the
system decreases.




                                               Chemical
                                            Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
  entropy on the molecular level.




                                            Chemical
                                         Thermodynamics
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
  entropy on the molecular level.
• Temperature is a measure of the average
  kinetic energy of the molecules in a sample.




                                               Chemical
                                            Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale
• Molecules exhibit several types of motion:
   Translational: Movement of the entire molecule from
    one place to another.
   Vibrational: Periodic motion of atoms within a molecule.
   Rotational: Rotation of the molecule on about an axis or
    rotation about σ bonds.




                                                        Chemical
                                                     Thermodynamics
Chemical
Thermodynamics
A molecule can vibrate (atoms moving relative to
one another) and rotate (tumble); a single atom can
only rotate.




                                                 Chemical
                                              Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale
• Boltzmann envisioned the motions of a sample of
  molecules at a particular instant in time.
   This would be akin to taking a snapshot of all the
    molecules.
• He referred to this sampling as a microstate of the
  thermodynamic system.




                                                            Chemical
                                                         Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
  microstates, W, associated with it.




                                               Chemical
                                            Thermodynamics
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
  microstates, W, associated with it.
• Entropy is




                                               Chemical
                                            Thermodynamics
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
  microstates, W, associated with it.
• Entropy is
                      S = k lnW




                                               Chemical
                                            Thermodynamics
Entropy on the Molecular Scale
• Each thermodynamic state has a specific number of
  microstates, W, associated with it.
• Entropy is
                      S = k lnW
  where k is the Boltzmann constant, 1.38 × 10−23 J/K.




                                                 Chemical
                                              Thermodynamics
S = k lnW




               Chemical
            Thermodynamics
S = k lnW




S = 0




                       Chemical
                    Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale

• The change in entropy for a process,
  then, is
     ΔS = k lnWfinal − k lnWinitial

                lnWfinal

    ΔS = k ln
               lnWinitial

                                            Chemical
                                         Thermodynamics
Entropy on the Molecular Scale

• The change in entropy for a process,
  then, is
     ΔS = k lnWfinal − k lnWinitial

                 lnWfinal

     ΔS = k ln
                lnWinitial

• Entropy increases with the number of
  microstates in the system.              Chemical
                                       Thermodynamics
Entropy on the Molecular Scale




                              Chemical
                           Thermodynamics
Entropy on the Molecular Scale
• The number of microstates and,
  therefore, the entropy tends to increase
  with increases in
  Temperature.
  Volume.
  The number of independently moving
   molecules.


                                           Chemical
                                        Thermodynamics
Entropy and Physical States




                             Chemical
                          Thermodynamics
Entropy and Physical States

• Entropy increases with
  the freedom of motion
  of molecules.




                                 Chemical
                              Thermodynamics
Entropy and Physical States

• Entropy increases with
  the freedom of motion
  of molecules.
• Therefore,




                                 Chemical
                              Thermodynamics
Entropy and Physical States

• Entropy increases with
  the freedom of motion
  of molecules.
• Therefore,
  S(g) > S(l) > S(s)




                                 Chemical
                              Thermodynamics
Solutions

      Generally, when
      a solid is
      dissolved in a
      solvent, entropy
      increases.



                       Chemical
                    Thermodynamics
Entropy Changes




                     Chemical
                  Thermodynamics
Entropy Changes

• In general, entropy
  increases when
   Gases are formed from
    liquids and solids.
   Liquids or solutions are
    formed from solids.
   The number of gas
    molecules increases.
   The number of moles
    increases.
                                  Chemical
                               Thermodynamics
The sign of entropy change, ΔS,
associated with the boiling of water is
_______.
1. Positive

2. Negative

3. Zero




                                       Chemical
                                    Thermodynamics
Correct Answer:

1. Positive       Vaporization of a liquid
                  to a gas leads to a large
2. Negative       increase in volume and
                  hence entropy; ΔS
3. Zero           must be positive.




                                          Chemical
                                       Thermodynamics
SAMPLE EXERCISE 19.3 Predicting the Sign of ΔS


Predict whether ΔS is positive or
negative for each of the following
processes, assuming each occurs at
constant temperature:




                                                    Chemical
                                                 Thermodynamics
SAMPLE EXERCISE 19.3 Predicting the Sign of ΔS


Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant
temperature:




Solution
Analyze: We are given four equations and asked to predict the sign of ΔS for each chemical reaction.
Plan: The sign of ΔS will be positive if there is an increase in temperature, an increase in the volume in which the
molecules move, or an increase in the number of gas particles in the reaction. The question states that the temperature
is constant. Thus, we need to evaluate each equation with the other two factors in mind.


Solve: (a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water
(18 g) occupies about 18 mL as a liquid and 22.4 L as a gas at STP. Because the molecules are distributed throughout
a much larger volume in the gaseous state than in the liquid state, an increase in motional freedom accompanies
vaporization. Therefore, ΔS is positive.

     (b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in which
they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ΔS is negative.


     (c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than
do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ΔS is negative.

    (d) The number of moles of gases is the same on both sides of the equation, and so the entropy
change will be small. The sign of ΔS is impossible to predict based on our discussions thus far, but we
can predict that ΔS will be close to zero.
PRACTICE EXERCISE
Indicate whether each of the following
processes produces an increase or
decrease in the entropy of the system:




                                      Chemical
                                   Thermodynamics
PRACTICE EXERCISE
Indicate whether each of the following
processes produces an increase or
decrease in the entropy of the system:




Answers: (a) increase, (b) decrease,
(c) decrease, (d) decrease                Chemical
                                       Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has
the Higher Entropy



Choose the sample of matter that
has greater entropy in each pair,
and explain your choice:
(a) 1 mol of NaCl(s) or 1 mol of
     HCl(g) at 25°C,
(b) 2 mol of HCl(g) or 1 mol of
     HCl(g) at 25°C,
(c) 1 mol of HCl(g) or 1 mol of Ar(g)
     at 298 K.                                           Chemical
                                                      Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy


Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1
mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.




                                                                                                                 Chemical
                                                                                                              Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy


Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1
mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.


Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.




                                                                                                                 Chemical
                                                                                                              Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy


Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1
mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.


Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.


Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The
sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol
sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because
the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms
cannot.




                                                                                                                 Chemical
                                                                                                              Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy


Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1
mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.


Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.


Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The
sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol
sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because
the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms
cannot.


PRACTICE EXERCISE
Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and
0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1
mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.




                                                                                                                  Chemical
                                                                                                               Thermodynamics
SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy


Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1
mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.


Solution
Analyze: We need to select the system in each pair that has the greater entropy.
Plan: To do this, we examine the state of the system and the complexity of the molecules it contains.


Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The
sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol
sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because
the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms
cannot.


PRACTICE EXERCISE
Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and
0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1
mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.

Answers: (a) 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP, (d) 2 mol
NO2(g) of at STP




                                                                                                                  Chemical
                                                                                                               Thermodynamics
PRACTICE EXERCISE
Choose the substance with the greater
entropy in each case:
(a) 1 mol of H2(g) at STP or 1 mol of
     H2(g) at 100°C and 0.5 atm,
(b) 1 mol of H2O(s) at 0°C or 1 mol of
     H2O(l) at 25°C,
(c) 1 mol of H2(g) at STP or 1 mol of
     SO2(g) at STP,
(d) 1 mol of N2O4(g) at STP or 2 mol of
     NO2(g) at STP.
PRACTICE EXERCISE
     Choose the substance with the greater
     entropy in each case:
     (a) 1 mol of H2(g) at STP or 1 mol of
           H2(g) at 100°C and 0.5 atm,
     (b) 1 mol of H2O(s) at 0°C or 1 mol of
           H2O(l) at 25°C,
     (c) 1 mol of H2(g) at STP or 1 mol of
           SO2(g) at STP,
     (d) 1 mol of N2O4(g) at STP or 2 mol of
           NO2(g) at STP.
Answers: (a) 1 mol of H (g) at 100°C and 0.5 atm,
                         2
(b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP,
(d) 2 mol NO (g) of at STP
Which of the following would be
 expected to have the highest entropy?
1.   1 mole N2(l)
2.   1 mole N2(g)
3.   1 mole Ar(g)
4.   1 mole Ar(s)




                                      Chemical
                                   Thermodynamics
Correct Answer:
1.   1 mole N2(l)   Given the same amounts, gases
2.   1 mole N2(g)   have higher entropy than
                    liquids, which have higher
3.   1 mole Ar(g)   entropy than solids. Of the
4.   1 mole Ar(s)   gases, N2 has greater entropy
                    than Ar because N2 is a
                    molecule that can store energy
                    in ways not possible for Ar.




                                              Chemical
                                           Thermodynamics
Third Law of Thermodynamics

       The entropy of a pure crystalline
       substance at absolute zero is 0.




                                          Chemical
                                       Thermodynamics
Chemical
Thermodynamics
It must be a perfect crystal at 0K (absolute zero).




                                                  Chemical
                                               Thermodynamics
Chapter 19- Chemical
     Thermodynamics
Sections 19.4 - 19.6
All about Free Energy
Standard Entropies




                        Chemical
                     Thermodynamics
Standard Entropies

• These are molar
  entropy values of
  substances in their
  standard states.




                                   Chemical
                                Thermodynamics
Standard Entropies

• These are molar
  entropy values of
  substances in their
  standard states.
• Standard entropies tend
  to increase with
  increasing molar mass.

                                  Chemical
                               Thermodynamics
Standard Entropies

Larger and more complex molecules have
greater entropies.




                                      Chemical
                                   Thermodynamics
Which substance will
 have the highest
     entropy?
         1. Butane
         2. Isobutane
         3. Pentane
         4. Isopentane
         5. Neopentane


                   Chemical
                Thermodynamics
Which substance will
 have the highest
     entropy?
         1. Butane
         2. Isobutane
         3. Pentane
         4. Isopentane
         5. Neopentane


                   Chemical
                Thermodynamics
Predict whether entropy increases,
decreases, or stays constant for:
       2 NO2(g) → 2 NO(g) + O2(g)


1. Increases
2. Decreases
3. Stays constant




                                        Chemical
                                     Thermodynamics
Predict whether entropy increases,
  decreases, or stays constant for:
         2 NO2(g) → 2 NO(g) + O2(g)


  1. Increases
  2. Decreases
  3. Stays constant

The entropy increases because two molecules of gas
react to form three molecules of gas, leading to more
disorder.

                                                     Chemical
                                                  Thermodynamics
Which process will have a
          negative ΔS?
1. Dissolving a substance
2. Melting
3. Falling leaves
4. Oxidation of a metal
5. Decomposition of water
   into H2 and O2           Dissolution of NaCl

                                             Chemical
                                          Thermodynamics
Which process will have a
          negative ΔS?
1. Dissolving a substance
2. Melting
3. Falling leaves
4. Oxidation of a metal
5. Decomposition of water
                            Dissolution of NaCl
   into H2 and O2

                                               Chemical
                                            Thermodynamics
Entropy Changes




                     Chemical
                  Thermodynamics
Entropy Changes

Entropy changes for a reaction can be
estimated in a manner analogous to that by
which ΔH is estimated:

 ΔS° = ΣnΔS°(products) - ΣmΔS°(reactants)

where n and m are the coefficients in the
balanced chemical equation.                Chemical
                                        Thermodynamics
SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies



Calculate ΔS° for the
synthesis of ammonia
from N2(g) and H2(g) at
298 K:




                                                                  Chemical
                                                               Thermodynamics
SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies


  Calculate ΔS° for the synthesis of ammonia from N2(g) and H2(g) at 298 K:



  Solution
  Analyze: We are asked to calculate the entropy change for the synthesis of NH3(g)
  from its constituent elements.
  Plan: We can make this calculation using Equation 19.8 and the standard molar
  entropy values for the reactants and the products that are given in Table 19.2 and in
  Appendix C.

  Solve: Using Equation 19.8, we have




Substituting the appropriate S° values from Table 19.2 yields




                                                                                             Chemical
                                                                                          Thermodynamics
     Check: The value for ΔS° is negative, in agreement with our qualitative prediction
     based on the decrease in the number of molecules of gas during the reaction.
PRACTICE EXERCISE
Using the standard
entropies in Appendix C,
calculate the standard
entropy change, ΔS°, for
the following reaction at
298 K:



                               Chemical
                            Thermodynamics
PRACTICE EXERCISE
Using the standard
entropies in Appendix C,
calculate the standard
entropy change, ΔS°, for
the following reaction at
298 K:



Answer: 180.39 J/K             Chemical
                            Thermodynamics
Calculate ΔS° for the equation below using the
standard entropy data given:
             2 NO(g) + O2(g) → 2 NO2(g)
ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211,
O2(g) = 205.

     1. +176 J/mol        3. −147 J/mol
     2. +147 J/mol        4. −76 J/mol




                                             Chemical
                                          Thermodynamics
Correct Answer:
1. +176 J/mol             3. −147 J/mol
2. +147 J/mol             4. −76 J/mol


ΔS° = Σ ΔS ° (products) − Σ ΔS ° (reactants)
    ΔS° = 2(240) − [2(211) + 205]
         ΔS° = 480 − [627]
             ΔS° = − 147



                                                  Chemical
                                               Thermodynamics
Entropy Changes in Surroundings




                              Chemical
                           Thermodynamics
Entropy Changes in Surroundings
 • Heat that flows into or out of the
   system changes the entropy of the
   surroundings.




                                      Chemical
                                   Thermodynamics
Entropy Changes in Surroundings
 • Heat that flows into or out of the
   system changes the entropy of the
   surroundings.
 • For an isothermal process:
                       −qsys
              ΔSsurr =
                        T

                                      Chemical
                                   Thermodynamics
Entropy Changes in Surroundings
 • Heat that flows into or out of the
   system changes the entropy of the
   surroundings.
 • For an isothermal process:
                         −qsys
                ΔSsurr =
                          T
 • At constant pressure, qsys is simply
   ΔH° for the system.                       Chemical
                                          Thermodynamics
−qsys
               ΔSsurr =
                         T
• At constant pressure, qsys is simply
  ΔH° for the system.                       Chemical
                                         Thermodynamics
always increase



                             −qsys
                    ΔSsurr =
                              T
• At constant pressure, qsys is simply
  ΔH° for the system.                       Chemical
                                         Thermodynamics
Entropy Change in the Universe




                              Chemical
                           Thermodynamics
Entropy Change in the Universe

• The universe is composed of the system
  and the surroundings.




                                        Chemical
                                     Thermodynamics
Entropy Change in the Universe

• The universe is composed of the system
  and the surroundings.
• Therefore,




                                        Chemical
                                     Thermodynamics
Entropy Change in the Universe

• The universe is composed of the system
  and the surroundings.
• Therefore,
     ΔSuniverse = ΔSsystem + ΔSsurroundings




                                          Chemical
                                       Thermodynamics
Entropy Change in the Universe

• The universe is composed of the system
  and the surroundings.
• Therefore,
     ΔSuniverse = ΔSsystem + ΔSsurroundings
• For spontaneous processes


                                          Chemical
                                       Thermodynamics
Entropy Change in the Universe

• The universe is composed of the system
  and the surroundings.
• Therefore,
     ΔSuniverse = ΔSsystem + ΔSsurroundings
• For spontaneous processes
                ΔSuniverse > 0
                                          Chemical
                                       Thermodynamics
Entropy Change in the Universe




                              Chemical
                           Thermodynamics
Entropy Change in the Universe

• This becomes:
                             −ΔHsystem
     ΔSuniverse = ΔSsystem +
                                T

 Multiplying both sides by −T,
   −TΔSuniverse = ΔHsystem − TΔSsystem


                                            Chemical
                                         Thermodynamics
Gibbs Free Energy




                       Chemical
                    Thermodynamics
Gibbs Free Energy

• −TΔSuniverse is defined as the Gibbs free
  energy, ΔG.




                                                 Chemical
                                              Thermodynamics
Gibbs Free Energy

• −TΔSuniverse is defined as the Gibbs free
  energy, ΔG.
• When ΔSuniverse is positive, ΔG is negative.




                                               Chemical
                                            Thermodynamics
Gibbs Free Energy

• −TΔSuniverse is defined as the Gibbs free
  energy, ΔG.
• When ΔSuniverse is positive, ΔG is negative.
• Therefore, when
              ΔG is negative,
 a process is spontaneous.
                                               Chemical
                                            Thermodynamics
Chemical
Thermodynamics
Entropy of universe increases and free energy of
the system decreases.




                                                 Chemical
                                              Thermodynamics
Entropy of universe increases and free energy of
the system decreases.




                                                 Chemical
                                              Thermodynamics
Gibbs Free Energy




                       Chemical
                    Thermodynamics
Gibbs Free Energy
         1. If ΔG is negative, the
            forward reaction is
            spontaneous.




                              Chemical
                           Thermodynamics
Gibbs Free Energy
         1. If ΔG is negative, the
            forward reaction is
            spontaneous.
         2. If ΔG is 0, the system
            is at equilibrium.




                              Chemical
                           Thermodynamics
Gibbs Free Energy
         1. If ΔG is negative, the
            forward reaction is
            spontaneous.
         2. If ΔG is 0, the system
            is at equilibrium.
         3. If ΔG is positive, the
            reaction is
            spontaneous in the
            reverse direction.

                              Chemical
                           Thermodynamics
Standard Free Energy Changes

 Analogous to standard enthalpies of
 formation are standard free energies of
 formation, ΔG°.
               f




                                         Chemical
                                      Thermodynamics
Standard Free Energy Changes

 Analogous to standard enthalpies of
 formation are standard free energies of
 formation, ΔG°.
                 f


 ΔG° = ΣnΔG°(products) − ΣmΔG°(reactants)
             f                 f

where n and m are the stoichiometric
coefficients.
                                          Chemical
                                       Thermodynamics
Calculate ΔG° for the equation below using
the Gibbs free energy data given:
           2 SO2(g) + O2(g)  2 SO3(g)
ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4

   1.   +70 kJ
   2.   +140 kJ
   3.   −140 kJ
   4.   −70 kJ




                                             Chemical
                                          Thermodynamics
Correct Answer:
   1.    +70 kJ
   2.    +140 kJ
   3.    −140 kJ
   4.    −70 kJ

ΔG ° =    ∑ΔGf° (products) − ∑ ΔGf° (reactants)
    ΔG° = (2 × −370.4) − [(2 × −300.4) + 0]
      ΔG° = −740.8 − [−600.8] = −140



                                                 Chemical
                                              Thermodynamics
Chemical
Thermodynamics
It indicates the process has taken place under standard
      conditions.




                                                    Chemical
                                                 Thermodynamics
It indicates the process has taken place under standard
      conditions.




                                                    Chemical
                                                 Thermodynamics
It indicates the process has taken place under standard
      conditions.


298K = 25°C      this is NOT STP (273K)




                                                    Chemical
                                                 Thermodynamics
It indicates the process has taken place under standard
      conditions.


298K = 25°C       this is NOT STP (273K)
1 atm for gases




                                                    Chemical
                                                 Thermodynamics
It indicates the process has taken place under standard
      conditions.


298K = 25°C        this is NOT STP (273K)
1 atm for gases
1M for solutions



                                                    Chemical
                                                 Thermodynamics
Calculate ΔG° for the equation below using
the thermodynamic data given:
           N2(g) + 3 H2(g) → 2 NH3(g)
ΔH° (NH3) = −46 kJ; ΔS° values (J/mol-K) are

NH3 = 192.5, N2 = 191.5, H2 = 130.6.

        1. +33 kJ          3. −66 kJ
        2. +66 kJ          4. −33 kJ




                                           Chemical
                                        Thermodynamics
Correct Answer:

1.   +33 kJ
                         ΔH° = −92 kJ
2.   +66 kJ         ΔS° = −198.7J/mol − K
3.   −66 kJ            ΔG° = ΔH° − TΔS°
4.   −33 kJ       ΔG° = (−92) − (298)(−0.1987)
                   ΔG° = (−92) + (59.2) = −33




                                             Chemical
                                          Thermodynamics
SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy
Change from Free Energies of Formation


(a) By using data from Appendix C,
calculate the standard free-energy change
for the following reaction at 298 K:



(b) What is ΔG° for the reverse of the
above reaction?

                                                           Chemical
                                                        Thermodynamics
SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change
                                         from Free Energies of Formation
 (a) By using data from Appendix C, calculate the standard free-energy change for the following reaction at
 298 K:


 (b) What is ΔG° for the reverse of the above reaction?


 Solution
 Analyze: We are asked to calculate the free-energy change for the indicated reaction, and then to determine the free-
 energy change of its reverse.
 Plan: To accomplish our task, we look up the free-energy values for the products and reactants and use Equation
 19.13: We multiply the molar quantities by the coefficients in the balanced equation, and subtract the total for the
 reactants from that for the products.

Solve: (a) Cl2(g) is in its standard state, so      is zero for this reactant. P4(g), however, is not in its standard
state, so      is not zero for this reactant. From the balanced equation and using Appendix C, we have:




 The fact that ΔG° is negative tells us that a mixture of P4(g), Cl2(g), and PCl3(g), at 25°C, each present at a partial
 pressure of 1 atm, would react spontaneously in the forward direction to form more PCl3. Remember, however, that the
 value of ΔG° tells us nothing about the rate at which the reaction occurs.

    (b) Remember that ΔG = G (products) – G (reactants). If we reverse the reaction, we
reverse the roles of the reactants and products. Thus, reversing the reaction changes the
sign of ΔG, just as reversing the reaction changes the sign of ΔH. • (Section 5.4) Hence,                                  Chemical
using the result from part (a):                                                                                         Thermodynamics
PRACTICE EXERCISE
By using data from Appendix C,
calculate ΔG° at 298 K for the
combustion of methane:




                                    Chemical
                                 Thermodynamics
PRACTICE EXERCISE
By using data from Appendix C,
calculate ΔG° at 298 K for the
combustion of methane:



 Answer: –800.7 kJ

                                    Chemical
                                 Thermodynamics
SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°

In Section 5.7 we used Hess’s law to
calculate ΔH° for the combustion of
propane gas at 298 K:

(a) Without using data from Appendix C,
predict whether ΔG° for this reaction is
more negative or less negative than ΔH°.
(b) Use data from Appendix C to calculate
the standard free-energy change for the
reaction at 298 K. Is your prediction from
part (a) correct?
SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG°
 In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K:



(a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH
°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your
prediction from part (a) correct?

Solution
Analyze: In part (a) we must predict the value for ΔG° relative to that for ΔH° on the basis of the balanced equation for
the reaction. In part (b) we must calculate the value for ΔG° and compare with our qualitative prediction.
Plan: The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction
(Equation 19.11), so under standard conditions:
                                                    ΔG° = ΔH° – TΔS°
To determine whether ΔG° is more negative or less negative than ΔH° we need to determine the sign of the term TΔS°.
T is the absolute temperature, 298 K, so it is a positive number. We can predict the sign of ΔS° by looking at the
reaction.
Solve: (a) We see that the reactants consist of six molecules of gas, and the products consist of three molecules of gas
and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By
using the general rules we discussed in Section 19.3, we would expect a decrease in the number of gas molecules to
lead to a decrease in the entropy of the system—the products have fewer accessible microstates than the reactants.
We therefore expect ΔS° and TΔS° to be negative numbers. Because we are subtracting TΔS°, which is a negative
number, we would predict that ΔG° is less negative than ΔH°.

       (b) Using Equation 19.13 and values from Appendix C, we can calculate the value of ΔG°:




                                                                                                                     Chemical
Notice that we have been careful to use the value of                  as in the calculation of ΔH values, the     Thermodynamics
phases of the reactants and products are important. As we predicted, ΔG° is less negative than ΔH° because
of the decrease in entropy during the reaction.
PRACTICE EXERCISE
Consider the combustion of propane to
form CO2(g) and H2O(g) at 298 K:

Would you expect ΔG° to be more
negative or less negative than ΔH°?



                                         Chemical
                                      Thermodynamics
PRACTICE EXERCISE
Consider the combustion of propane to
form CO2(g) and H2O(g) at 298 K:

Would you expect ΔG° to be more
negative or less negative than ΔH°?



 Answer: more negative
                                         Chemical
                                      Thermodynamics
Free Energy Changes




                         Chemical
                      Thermodynamics
Free Energy Changes

At temperatures other than 25°C,

          ΔG° = ΔH° − TΔS°

How does ΔG° change with temperature?


                                      Chemical
                                   Thermodynamics
Free Energy and Temperature




                            Chemical
                         Thermodynamics
Free Energy and Temperature

• There are two parts to the free energy
  equation:
  ΔH°— the enthalpy term
  TΔS° — the entropy term
• The temperature dependence of free
  energy, then comes from the entropy
  term.
                                           Chemical
                                        Thermodynamics
Upon heating, limestone (CaCO3) decomposes to
CaO and CO2. Speculate on the sign of ΔH and
ΔS for this process.
   CaCO3(s)            CaO(s) + CO2(g)




                                 Limestone Cave
                          Carlsbad Caverns, New Mexico


                                                    Chemical
                                                 Thermodynamics
Upon heating, limestone (CaCO3) decomposes to
CaO and CO2. Speculate on the sign of ΔH and
ΔS for this process.
   CaCO3(s)            CaO(s) + CO2(g)




                                 Limestone Cave
                          Carlsbad Caverns, New Mexico


                                                    Chemical
                                                 Thermodynamics
Free Energy and Temperature




                            Chemical
                         Thermodynamics
Chemical
Thermodynamics
ΔH < TΔS




              Chemical
           Thermodynamics
ΔH < TΔS
ΔG(Negative/spontaneous)=




                               Chemical
                            Thermodynamics
ΔH < TΔS
ΔG(Negative/spontaneous)=
 ΔH (Positive/vaporization) - TΔS (positive/liquid to gas)



                                                     Chemical
                                                  Thermodynamics
ΔH < TΔS
ΔG(Negative/spontaneous)=
 ΔH (Positive/vaporization) - TΔS (positive/liquid to gas)
                                ↑must be larger

                                                     Chemical
                                                  Thermodynamics
Both the entropy, ΔS° and the enthalpy ΔH°
have positive values. Near absolute zero (0 K),
will the process be spontaneous or
nonspontaneous?

  1. Spontaneous

  2. Nonspontaneous




                                           Chemical
                                        Thermodynamics
Correct Answer:
     1. Spontaneous

     2. Nonspontaneous

At low temperatures, the enthalpy
term dominates the entropy term,
leading to a positive ΔG and a
nonspontaneous process.




                                       Chemical
                                    Thermodynamics
A reaction will be spontaneous at
lower temperatures when the sign
          of ΔH and ΔS is




                                  Chemical
                               Thermodynamics
A reaction will be spontaneous at
lower temperatures when the sign
          of ΔH and ΔS is




                                  Chemical
                               Thermodynamics
SAMPLE EXERCISE 19.8 Determining the Effect of Temperature
on Spontaneity

The Haber process for the production
of ammonia involves the equilibrium

Assume that ΔH° and ΔS° for this
reaction do not change with temperature.
(a) Predict the direction in which ΔG° for
this reaction changes with increasing
temperature.
(b) Calculate the values of ΔG° for the
reaction at 25°C and 500°C.
SAMPLE EXERCISE 19.8 continued

    Solution
    Analyze: In part (a) we are asked to predict the direction in which ΔG° for the ammonia synthesis reaction changes as
    temperature increases. In part (b) we need to determine ΔG° for the reaction at two different temperatures.
    Plan: In part (a) we can make this prediction by determining the sign of ΔS° for the reaction and then using that
    information to analyze Equation 19.20. In part (b) we need to calculate ΔH° and ΔS° for the reaction by using the data
    in Appendix C. We can then use Equation 19.20 to calculate ΔG°.
    Solve: (a) Equation 19.20 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –TΔS°. The
    temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because
    the number of molecules of gas is smaller in the products. Because ΔS° is negative, the term
    –TΔS° is positive and grows larger with increasing temperature. As a result, ΔG° becomes less negative (or more
    positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing
    temperature.
     (b) We calculated the value of ΔH° in Sample
Exercise 15.14, and the value of ΔS° was
determined in Sample Exercise 19.5: ΔH° = –92.38
kJ and ΔS° = –198.4 J/K. If we assume that these
values don’t change with temperature, we can
calculate ΔG° at any temperature by using
Equation 19.20. At T = 298 K we have:




    Notice that we have been careful to convert –TΔS° into units of kJ so that it can be added to ΔH°, which has units of kJ.

Comment: Increasing the temperature from 298 K to 773 K changes ΔG° from –33.3 kJ to +61 kJ. Of course, the result at 773 K
depends on the assumption that ΔH° and ΔS° do not change with temperature. In fact, these values do change slightly with
temperature. Nevertheless, the result at 773 K should be a reasonable approximation. The positive increase in ΔG° with
increasing T agrees with our prediction in part (a) of this exercise. Our result indicates that a mixture of N2(g), H2(g), and NH3(g),
each present at a partial pressure of 1 atm, will react spontaneously at 298 K to form more NH3(g). In contrast, at 773 K the
positive value of ΔG° tells us that the reverse reaction is spontaneous. Thus, when the mixture of three gases, each at a partial
pressure of 1 atm, is heated to 773 K, some of the NH (g) spontaneously decomposes into N (g) and H (g).
PRACTICE EXERCISE
(a) Using standard enthalpies of
     formation and standard entropies in
    Appendix C, calculate ΔH° and ΔS° at
     298 K for the following reaction:


(b) Using the values obtained in part (a),
    estimate ΔG° at 400 K.
PRACTICE EXERCISE
 (a) Using standard enthalpies of
      formation and standard entropies in
     Appendix C, calculate ΔH° and ΔS° at
      298 K for the following reaction:


 (b) Using the values obtained in part (a),
     estimate ΔG° at 400 K.

Answer: (a) ΔH° = –196.6 kJ, ΔS° = –189.6 J/K;
(b) ΔG° = –120.8 kJ
Chapter 19- Chemical
     Thermodynamics
Section 19.7
  Free Energy & Keq
Free Energy and Equilibrium




                            Chemical
                         Thermodynamics
Free Energy and Equilibrium
 Under any conditions, standard or
 nonstandard, the free energy change
 can be found this way:

             ΔG = ΔG° + RT lnQ

(Under standard conditions, all concentrations are 1 M,
        so Q = 1 and lnQ = 0; the last term drops out.)
                                                     Chemical
                                                  Thermodynamics
SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium

The normal boiling point is the temperature
at which a pure liquid is in equilibrium with
its vapor at a pressure of 1 atm.
(a) Write the chemical equation that
defines the normal boiling point of liquid
     carbon tetrachloride, CCl4(l).
(b) What is the value of ΔG° for the
     equilibrium in part (a)?
(c) Use thermodynamic data in Appendix C
     and Equation 19.20 to estimate the
     normal boiling point of CCl4.
SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium


As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its
vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon
tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in
Appendix C and Equation 19.20 to estimate the normal boiling point of CCl4.

Solution
Analyze: (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous
CCl4 at the normal boiling point. (b) We must determine the value of ΔG° for CCl4 in equilibrium with its vapor at the
normal boiling point. (c) We must estimate the normal boiling point of CCl4, based on available thermodynamic data.
Plan: (a) The chemical equation will merely show the change of state of CCl4 from liquid to solid. (b) We need to
analyze Equation 19.21 at equilibrium (ΔG = 0). (c) We can use Equation 19.20 to calculate T
when ΔG = 0.



Solve: (a) The normal boiling point of CCl4 is the temperature at which pure liquid CCl4 is in equilibrium with its vapor at
a pressure of 1 atm:



(b) At equilibrium ΔG = 0. In any normal boiling-point equilibrium both the liquid and the vapor are in their standard
states (Table 19.2). As a consequence, Q = 1, ln Q = 0, and ΔG = ΔG° for this process. Thus, we conclude that ΔG° = 0
for the equilibrium involved in the normal boiling point of any liquid. We would also find that ΔG° = 0 for the equilibria
relevant to normal melting points and normal sublimation points of solids.

(c) Combining Equation 19.20 with the result from part (b), we see that the equality at the normal boiling point, Tb, of
CCl4(l) or any other pure liquid is
                                                                                                                    Chemical
                                                                                                                 Thermodynamics
SAMPLE EXERCISE 19.9 continued

Solving the equation for Tb, we obtain


Strictly speaking, we would need the values of ΔH° and ΔS° for the equilibrium between CCl4(l) and
CCl4(g) at the normal boiling point to do this calculation. However, we can estimate the boiling point by
using the values of ΔH° and ΔS° for CCl4 at 298 K, which we can obtain from the data in Appendix C
and Equations 5.31 and 19.8:




Notice that, as expected, the process is endothermic (ΔH > 0) and produces a gas in which energy
can be more spread out (ΔS > 0). We can now use these values to estimate Tb for CCl4(l):




Note also that we have used the conversion factor between J and kJ to make sure that the units of ΔH
° and ΔS° match.
Check: The experimental normal boiling point of CCl (l) is 76.5°C. The small deviation of our
                                                           4
estimate from the experimental value is due to the assumption that ΔH° and ΔS° do not change with
temperature.


                                                                                                   Chemical
                                                                                                Thermodynamics
PRACTICE EXERCISE
Use data in Appendix C to estimate the
normal boiling point, in K, for elemental
bromine, Br2(l). (The experimental
value is given in Table 11.3.)




                                         Chemical
                                      Thermodynamics
PRACTICE EXERCISE
Use data in Appendix C to estimate the
normal boiling point, in K, for elemental
bromine, Br2(l). (The experimental
value is given in Table 11.3.)




Answer: 330 K                            Chemical
                                      Thermodynamics
SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change
Under Nonstandard Conditions


We will continue to explore the Haber
process for the synthesis of ammonia:


Calculate ΔG at 298 K for a reaction
mixture that consists of 1.0 atm N2,
3.0 atm H2, and 0.50 atm NH3.


                                                         Chemical
                                                      Thermodynamics
SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions
   We will continue to explore the Haber process for the synthesis of ammonia:
   Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.

   Solution
   Analyze: We are asked to calculate ΔG under nonstandard conditions.
   Plan: We can use Equation 19.21 to calculate ΔG. Doing so requires that we calculate the value of the reaction
   quotient Q for the specified partial pressures of the gases and evaluate ΔG°, using a table of standard free energies of
   formation.
   Solve: Solving for the reaction quotient gives:




  In Sample Exercise 19.8 we calculated ΔG° = 33.3 kJ for this reaction. We will have to change the units of this quantity
  in applying Equation 19.21, however. In order for the units in Equation 19.21 to work out, we will use kJ/mol as our
  units for ΔG°, where “per mole” means “per mole of the reaction as written.” Thus,
  ΔG° = –33.3kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.

       We can now use Equation 19.21 to calculate ΔG for these nonstandard conditions:




Comment: We see that ΔG becomes more negative, changing from –33.3 kJ/mol to –44.9 kJ/mol as the pressures of N2, H2,
and NH3 are changed from 1.0 atm each (standard conditions, ΔG° ) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger
negative value for ΔG indicates a larger “driving force” to produce NH3.
    We would have made the same prediction on the basis of Le Châtelier’s principle. • (Section 15.6) Relative to standard
conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s
principle predicts that both of these changes should shift the reaction more to the product side, thereby forming more NH3.
PRACTICE EXERCISE
Calculate ΔG at 298 K for the reaction
of nitrogen and hydrogen to form
ammonia if the reaction mixture
consists of 0.50 atm N2, 0.75 atm H2,
and 2.0 atm NH3.




                                       Chemical
                                    Thermodynamics
PRACTICE EXERCISE
Calculate ΔG at 298 K for the reaction
of nitrogen and hydrogen to form
ammonia if the reaction mixture
consists of 0.50 atm N2, 0.75 atm H2,
and 2.0 atm NH3.



Answer: –26.0 kJ/mol
                                       Chemical
                                    Thermodynamics
Decomposition of CaCO3 has a ΔH° =
178.3 kJ/mol and ΔS° = 159 J/mol • K.
At what temperature does this become
spontaneous?

 1.   121°C
 2.   395°C
 3.   848°C
 4.   1121°C




                                     Chemical
                                  Thermodynamics
Correct Answer:

1.   121°C               T = ΔH°/ΔS°
2.   395°C
3.   848°C    T = 178.3 kJ/mol/0.159 kJ/mol • K
4.   1121°C
                          T = 1121 K
                  T (°C) = 1121 – 273 = 848




                                             Chemical
                                          Thermodynamics
Free Energy and Equilibrium




                            Chemical
                         Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.




                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes




                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
              0 = ΔG° + RT lnK




                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
              0 = ΔG° + RT lnK
• Rearranging, this becomes




                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
              0 = ΔG° + RT lnK
• Rearranging, this becomes
               ΔG° = −RT lnK



                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
              0 = ΔG° + RT lnK
• Rearranging, this becomes
               ΔG° = −RT lnK
  or,

                                          Chemical
                                       Thermodynamics
Free Energy and Equilibrium

• At equilibrium, Q = K, and ΔG = 0.
• The equation becomes
              0 = ΔG° + RT lnK
• Rearranging, this becomes
               ΔG° = −RT lnK
  or,
                 K = e−ΔG°/RT
                                          Chemical
                                       Thermodynamics
Suppose ΔG° is a large, positive value.
What then will be the value of the
equilibrium constant, K?

1.   K=0
2.   K=1
3.   0<K<1
4.   K>1




                                       Chemical
                                    Thermodynamics
Correct Answer:

                       ΔG° = −RTlnK
1.   K=0
             Thus, large positive values of ΔG°
2.   K=1
             lead to large negative values of lnK.
3.   0<K<1   The value of K itself, then, is very
4.   K>1     small, approaching zero.




                                                Chemical
                                             Thermodynamics
N2(g) + 3 H2(g)  2 NH3(g)      ΔH° = -92.4 kJ/mol
If the temperature is decreased, what is the effect
        on the equilibrium constant and ΔG?


 1. K increases, ΔG increases
 2. K decreases, ΔG decreases
 3. K increases, ΔG decreases
 4. K decreases, ΔG increases
 5. Temperature does not affect
                                   Josiah Willard Gibbs
    equilibrium or ΔG                  (1839-1903)
                                                      Chemical
                                                   Thermodynamics
N2(g) + 3 H2(g)  2 NH3(g)      ΔH° = -92.4 kJ/mol
If the temperature is decreased, what is the effect
        on the equilibrium constant and ΔG?


 1. K increases, ΔG increases
 2. K decreases, ΔG decreases
 3. K increases, ΔG decreases
 4. K decreases, ΔG increases
 5. Temperature does not affect
                                   Josiah Willard Gibbs
    equilibrium or ΔG                  (1839-1903)
                                                      Chemical
                                                   Thermodynamics
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°

 Use standard free energies of
 formation to calculate the equilibrium
 constant, K, at 25°C for the reaction
 involved in the Haber process:


  The standard free-energy change for
  this reaction was calculated in Sample
  Exercise 19.8:
  ΔG° = –33.3 kJ/mol = –33,300 J/mol.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
 Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
 Haber process:

 The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.




Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
 Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
 Haber process:

 The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.


  Solution
  Analyze: We are asked to calculate K for a reaction, given ΔG°.
  Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
 Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
 Haber process:

 The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.


  Solution
  Analyze: We are asked to calculate K for a reaction, given ΔG°.
  Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




  In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
  when using Equations 19.21, 19.22, or 19.23).




Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
 Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
 Haber process:

 The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.


  Solution
  Analyze: We are asked to calculate K for a reaction, given ΔG°.
  Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




  In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
  when using Equations 19.21, 19.22, or 19.23).

  Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have




Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
 Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
 Haber process:

 The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.


  Solution
  Analyze: We are asked to calculate K for a reaction, given ΔG°.
  Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




  In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
  when using Equations 19.21, 19.22, or 19.23).

  Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have




  We insert this value into Equation 19.23 to obtain K:



Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG°
    Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the
    Haber process:

    The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.


     Solution
     Analyze: We are asked to calculate K for a reaction, given ΔG°.
     Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form




     In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG°
     when using Equations 19.21, 19.22, or 19.23).

     Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have




     We insert this value into Equation 19.23 to obtain K:



   Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture
   at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than
   the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one.
   Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.

Thermodynamics can tell us the direction and extent of a
reaction, but tells us nothing about the rate at which it occurs.
If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be
needed to force the equilibrium toward NH3.
PRACTICE EXERCISE
Use data from Appendix C to calculate
the standard free-energy change, ΔG°,
and the equilibrium constant, K, at 298
K for the reaction




                                        Chemical
                                     Thermodynamics
PRACTICE EXERCISE
Use data from Appendix C to calculate
the standard free-energy change, ΔG°,
and the equilibrium constant, K, at 298
K for the reaction




Answer: ΔG° = –106.4 kJ/mol, K = 5 × 1018
                                        Chemical
                                     Thermodynamics
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together
Consider the equilibria in which these salts dissolve in water to form aqueous
solutions of ions:



  (a) Calculate the value of ΔG° at 298 K for each of the
    preceding reactions.
  (b) The two values from part (a) are very different. Is this
    difference primarily due to the enthalpy term or the
    entropy term of the standard free-energy change?
  (c) Use the values of ΔG° to calculate the Ksp values for
    the two salts at 298 K.
  (d) Sodium chloride is considered a soluble salt, whereas
    silver chloride is considered insoluble. Are these
    descriptions consistent with the answers to part (c)?
  (e) How will ΔG° for the solution process of these salts
    change with increasing T? What effect should this
    change have on the solubility of the salts?
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together


Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to
form aqueous solutions of ions:




(a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very
different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change?
(c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a
soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part
(c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change
have on the solubility of the salts?


Solution (a) We will use Equation 19.13 along with         values from Appendix C to calculate the
values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to indicate that these are
thermodynamic quantities for the formation of a solution.) We find




                                                                                                                     Chemical
                                                                                                                  Thermodynamics
SAMPLE INTEGRATIVE EXERCISE continued

   (b) We can write                 as the sum of an enthalpy term,                          and an entropy term,

We can calculate the values of                       and              by using Equations 5.31 and 19.8.

We can then calculate                                         at T = 298 K. All these
calculations are now familiar to us. The results are summarized in the following table:




  The entropy terms for the solution of the two salts are very similar. That seems sensible because each
  solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions.
   • (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two
  salts. The difference in the values of         is dominated by the difference in the values of


      (c) The solubility product, Ksp, is the equilibrium constant for the solution process. • (Section 17.4) As
  such, we can relate Ksp directly to          by using Equation 19.23:




  We can calculate the Ksp values in the same way we applied Equation 19.23 in Sample Exercise 19.13.
  We use the         values we obtained in part (a), remembering to convert them from to J/mol:



                                                                                                                      Chemical
                                                                                                                   Thermodynamics
  The value calculated for the Ksp of AgCl is very close to that listed in Appendix D.
SAMPLE INTEGRATIVE EXERCISE continued


     (d) A soluble salt is one that dissolves appreciably in water. • (Section 4.2) The Ksp value for NaCl
is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small,
indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble
salt.




    (e) As we expect, the solution process has a positive value of ΔS for both salts (see the table in part b).
As such, the entropy term of the free-energy change,                   is negative. If we assume that
and         do not change much with temperature, then an increase in T will serve to make                more
negative. Thus, the driving force for dissolution of the salts will increase with increasing T, and we therefore
expect the solubility of the salts to increase with increasing T. In Figure 13.17 we see that the solubility of
NaCl (and that of nearly any salt) increases with increasing temperature. • (Section 13.3)




                                                                                                    Chemical
                                                                                                 Thermodynamics

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Chapter 19 Lecture- Thermodynamics

  • 1. Chapter 19- Chemical Thermodynamics Sections 19.1 - 19.3 Spontaneous processes & entropy
  • 2. First Law of Thermodynamics Chemical Thermodynamics
  • 3. First Law of Thermodynamics • You will recall from Chapter 5 that energy cannot be created nor destroyed. Chemical Thermodynamics
  • 4. First Law of Thermodynamics • You will recall from Chapter 5 that energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. Chemical Thermodynamics
  • 5. First Law of Thermodynamics • You will recall from Chapter 5 that energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. • Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. Chemical Thermodynamics
  • 6. Spontaneous Processes Chemical Thermodynamics
  • 7. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. Chemical Thermodynamics
  • 8. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return Chemical Thermodynamics
  • 9. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return A process is either spontaneous, Chemical non-spontaneous, or at equilibrium. Thermodynamics
  • 10. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. Chemical Thermodynamics
  • 11. Melting a cube of water ice at −10°C and atmospheric pressure is a ________ process. 1. Spontaneous 2. Nonspontaneous Chemical Thermodynamics
  • 12. Correct Answer: Melting of an ice cube at 1. Spontaneous 1 atm pressure will only be spontaneous above 2. Nonspontaneous the freezing point (0°C). Chemical Thermodynamics
  • 13. Spontaneous Processes Chemical Thermodynamics
  • 14. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Chemical Thermodynamics
  • 15. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. • Above 0°C it is spontaneous for ice to melt. Chemical Thermodynamics
  • 16. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. • Above 0°C it is spontaneous for ice to melt. • Below 0°C the reverse process is spontaneous. Chemical Thermodynamics
  • 18. No. Nonspontaneous processes can occur with some external assistance. Chemical Thermodynamics
  • 19. SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes Predict whether these processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150°C is added to water at 40°C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g). (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1°C.
  • 20. SAMPLE EXERCISE 19.1 Identifying Spontaneous Processes Predict whether these processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150°C is added to water at 40°C, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and O2(g). (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1°C. Analyze: We are asked to judge whether each process will proceed spontaneously in the direction indicated, in the reverse direction, or in neither direction. Plan: We need to think about whether each process is consistent with our experience about the natural direction of events or whether it is the reverse process that we expect to occur. Solve: (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. In this instance heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gas bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is spontaneous once initiated by a spark or flame. (c) By definition, the normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. Neither the condensation of benzene vapor nor the reverse process is spontaneous. If the temperature were below 80.1°C, condensation would be spontaneous.
  • 21. PRACTICE EXERCISE Under 1 atm pressure CO2(s) sublimes at –78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100ºC and 1 atm pressure? Chemical Thermodynamics
  • 22. PRACTICE EXERCISE Under 1 atm pressure CO2(s) sublimes at –78°C. Is the transformation of CO2(s) to CO2(g) a spontaneous process at –100ºC and 1 atm pressure? Answer: No, the reverse process is spontaneous at this temperature. Chemical Thermodynamics
  • 23. Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. Chemical Thermodynamics
  • 24. Irreversible Processes Chemical Thermodynamics
  • 25. Irreversible Processes • Irreversible processes cannot be undone by exactly reversing the change to the system. Chemical Thermodynamics
  • 26. Irreversible Processes • Irreversible processes cannot be undone by exactly reversing the change to the system. • Spontaneous processes are irreversible. Chemical Thermodynamics
  • 28. No Chemical Thermodynamics
  • 29. Entropy Chemical Thermodynamics
  • 30. Entropy • Entropy (S) is a term coined by Rudolph Clausius in the 19th century. Chemical Thermodynamics
  • 31. Entropy • Entropy (S) is a term coined by Rudolph Clausius in the 19th century. • Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q T Chemical Thermodynamics
  • 32. Entropy Chemical Thermodynamics
  • 33. Entropy • Entropy can be thought of as a measure of the randomness of a system. Chemical Thermodynamics
  • 34. Entropy • Entropy can be thought of as a measure of the randomness of a system. • It is related to the various modes of motion in molecules. Chemical Thermodynamics
  • 35. Entropy Chemical Thermodynamics
  • 36. Entropy • Like total energy, E, and enthalpy, H, entropy is a state function. Chemical Thermodynamics
  • 37. Entropy • Like total energy, E, and enthalpy, H, entropy is a state function. • Therefore, Chemical Thermodynamics
  • 38. Entropy • Like total energy, E, and enthalpy, H, entropy is a state function. • Therefore, ΔS = Sfinal − Sinitial Chemical Thermodynamics
  • 39. Entropy • For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: Chemical Thermodynamics
  • 40. Entropy • For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: qrev ΔS = T Chemical Thermodynamics
  • 42. ΔS depends not merely on q but on qrev. There is only one reversible isothermal path between two states regardless of the number of possible paths. Chemical Thermodynamics
  • 43. SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? Chemical Thermodynamics
  • 44. SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? Chemical Thermodynamics
  • 45. SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? Chemical Thermodynamics
  • 46. SAMPLE EXERCISE 19.2 Calculating ΔS for a Phase Change The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is –38.9ºC, and its molar enthalpy of fusion is ΔΗfusion = 2.29 kJ/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? Solution Analyze: We first recognize that freezing is an exothermic process; heat is transferred from the system to the surroundings when a liquid freezes (q < 0). The enthalpy of fusion is ΔΗ for the melting process. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is –ΔΗfusion = 2.29 kJ/mol. Plan: We can use –ΔΗfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg: We can use this value of q as qrev in Equation 19.2. We must first, however, convert the temperature to K: Solve: We can now calculate the value of ΔSsys: Check: The entropy change is negative because heat flows from the system making qrev negative. Comment: The procedure we have used here can be used to calculate ΔS for other isothermal phase changes, such as the vaporization of a liquid at its boiling point. Chemical Thermodynamics
  • 47. SAMPLE EXERCISE 19.2 continued PRACTICE EXERCISE The normal boiling point of ethanol, C2H5OH, is 78.3°C (Figure 11.24), and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point? Chemical Thermodynamics
  • 48. SAMPLE EXERCISE 19.2 continued PRACTICE EXERCISE The normal boiling point of ethanol, C2H5OH, is 78.3°C (Figure 11.24), and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point? Answer: –163 J/K Chemical Thermodynamics
  • 49. Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes. Chemical Thermodynamics
  • 50. Second Law of Thermodynamics Chemical Thermodynamics
  • 51. Second Law of Thermodynamics In other words: For reversible processes: ΔSuniv = ΔSsystem + ΔSsurroundings = 0 For irreversible processes: ΔSuniv = ΔSsystem + ΔSsurroundings > 0 Chemical Thermodynamics
  • 52. Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases. Chemical Thermodynamics
  • 54. The entropy of the universe for the process must increase by a greater amount than the entropy of the system decreases. Chemical Thermodynamics
  • 55. Entropy on the Molecular Scale Chemical Thermodynamics
  • 56. Entropy on the Molecular Scale • Ludwig Boltzmann described the concept of entropy on the molecular level. Chemical Thermodynamics
  • 57. Entropy on the Molecular Scale • Ludwig Boltzmann described the concept of entropy on the molecular level. • Temperature is a measure of the average kinetic energy of the molecules in a sample. Chemical Thermodynamics
  • 58. Entropy on the Molecular Scale Chemical Thermodynamics
  • 59. Entropy on the Molecular Scale • Molecules exhibit several types of motion:  Translational: Movement of the entire molecule from one place to another.  Vibrational: Periodic motion of atoms within a molecule.  Rotational: Rotation of the molecule on about an axis or rotation about σ bonds. Chemical Thermodynamics
  • 61. A molecule can vibrate (atoms moving relative to one another) and rotate (tumble); a single atom can only rotate. Chemical Thermodynamics
  • 62. Entropy on the Molecular Scale Chemical Thermodynamics
  • 63. Entropy on the Molecular Scale • Boltzmann envisioned the motions of a sample of molecules at a particular instant in time.  This would be akin to taking a snapshot of all the molecules. • He referred to this sampling as a microstate of the thermodynamic system. Chemical Thermodynamics
  • 64. Entropy on the Molecular Scale Chemical Thermodynamics
  • 65. Entropy on the Molecular Scale • Each thermodynamic state has a specific number of microstates, W, associated with it. Chemical Thermodynamics
  • 66. Entropy on the Molecular Scale • Each thermodynamic state has a specific number of microstates, W, associated with it. • Entropy is Chemical Thermodynamics
  • 67. Entropy on the Molecular Scale • Each thermodynamic state has a specific number of microstates, W, associated with it. • Entropy is S = k lnW Chemical Thermodynamics
  • 68. Entropy on the Molecular Scale • Each thermodynamic state has a specific number of microstates, W, associated with it. • Entropy is S = k lnW where k is the Boltzmann constant, 1.38 × 10−23 J/K. Chemical Thermodynamics
  • 69. S = k lnW Chemical Thermodynamics
  • 70. S = k lnW S = 0 Chemical Thermodynamics
  • 71. Entropy on the Molecular Scale Chemical Thermodynamics
  • 72. Entropy on the Molecular Scale • The change in entropy for a process, then, is ΔS = k lnWfinal − k lnWinitial lnWfinal ΔS = k ln lnWinitial Chemical Thermodynamics
  • 73. Entropy on the Molecular Scale • The change in entropy for a process, then, is ΔS = k lnWfinal − k lnWinitial lnWfinal ΔS = k ln lnWinitial • Entropy increases with the number of microstates in the system. Chemical Thermodynamics
  • 74. Entropy on the Molecular Scale Chemical Thermodynamics
  • 75. Entropy on the Molecular Scale • The number of microstates and, therefore, the entropy tends to increase with increases in Temperature. Volume. The number of independently moving molecules. Chemical Thermodynamics
  • 76. Entropy and Physical States Chemical Thermodynamics
  • 77. Entropy and Physical States • Entropy increases with the freedom of motion of molecules. Chemical Thermodynamics
  • 78. Entropy and Physical States • Entropy increases with the freedom of motion of molecules. • Therefore, Chemical Thermodynamics
  • 79. Entropy and Physical States • Entropy increases with the freedom of motion of molecules. • Therefore, S(g) > S(l) > S(s) Chemical Thermodynamics
  • 80. Solutions Generally, when a solid is dissolved in a solvent, entropy increases. Chemical Thermodynamics
  • 81. Entropy Changes Chemical Thermodynamics
  • 82. Entropy Changes • In general, entropy increases when  Gases are formed from liquids and solids.  Liquids or solutions are formed from solids.  The number of gas molecules increases.  The number of moles increases. Chemical Thermodynamics
  • 83. The sign of entropy change, ΔS, associated with the boiling of water is _______. 1. Positive 2. Negative 3. Zero Chemical Thermodynamics
  • 84. Correct Answer: 1. Positive Vaporization of a liquid to a gas leads to a large 2. Negative increase in volume and hence entropy; ΔS 3. Zero must be positive. Chemical Thermodynamics
  • 85. SAMPLE EXERCISE 19.3 Predicting the Sign of ΔS Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature: Chemical Thermodynamics
  • 86. SAMPLE EXERCISE 19.3 Predicting the Sign of ΔS Predict whether ΔS is positive or negative for each of the following processes, assuming each occurs at constant temperature: Solution Analyze: We are given four equations and asked to predict the sign of ΔS for each chemical reaction. Plan: The sign of ΔS will be positive if there is an increase in temperature, an increase in the volume in which the molecules move, or an increase in the number of gas particles in the reaction. The question states that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind. Solve: (a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g) occupies about 18 mL as a liquid and 22.4 L as a gas at STP. Because the molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an increase in motional freedom accompanies vaporization. Therefore, ΔS is positive. (b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ΔS is negative. (c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ΔS is negative. (d) The number of moles of gases is the same on both sides of the equation, and so the entropy change will be small. The sign of ΔS is impossible to predict based on our discussions thus far, but we can predict that ΔS will be close to zero.
  • 87. PRACTICE EXERCISE Indicate whether each of the following processes produces an increase or decrease in the entropy of the system: Chemical Thermodynamics
  • 88. PRACTICE EXERCISE Indicate whether each of the following processes produces an increase or decrease in the entropy of the system: Answers: (a) increase, (b) decrease, (c) decrease, (d) decrease Chemical Thermodynamics
  • 89. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Chemical Thermodynamics
  • 90. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Chemical Thermodynamics
  • 91. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Solution Analyze: We need to select the system in each pair that has the greater entropy. Plan: To do this, we examine the state of the system and the complexity of the molecules it contains. Chemical Thermodynamics
  • 92. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Solution Analyze: We need to select the system in each pair that has the greater entropy. Plan: To do this, we examine the state of the system and the complexity of the molecules it contains. Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot. Chemical Thermodynamics
  • 93. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Solution Analyze: We need to select the system in each pair that has the greater entropy. Plan: To do this, we examine the state of the system and the complexity of the molecules it contains. Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot. PRACTICE EXERCISE Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. Chemical Thermodynamics
  • 94. SAMPLE EXERCISE 19.4 Predicting Which Sample of Matter Has the Higher Entropy Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25°C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25°C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. Solution Analyze: We need to select the system in each pair that has the greater entropy. Plan: To do this, we examine the state of the system and the complexity of the molecules it contains. Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing energy in more ways than is Ar. HCl molecules can rotate and vibrate; Ar atoms cannot. PRACTICE EXERCISE Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. Answers: (a) 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP, (d) 2 mol NO2(g) of at STP Chemical Thermodynamics
  • 95. PRACTICE EXERCISE Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP.
  • 96. PRACTICE EXERCISE Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100°C and 0.5 atm, (b) 1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at 25°C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. Answers: (a) 1 mol of H (g) at 100°C and 0.5 atm, 2 (b) 1 mol of H2O(l) at 25°C, (c) 1 mol of SO2(g) at STP, (d) 2 mol NO (g) of at STP
  • 97. Which of the following would be expected to have the highest entropy? 1. 1 mole N2(l) 2. 1 mole N2(g) 3. 1 mole Ar(g) 4. 1 mole Ar(s) Chemical Thermodynamics
  • 98. Correct Answer: 1. 1 mole N2(l) Given the same amounts, gases 2. 1 mole N2(g) have higher entropy than liquids, which have higher 3. 1 mole Ar(g) entropy than solids. Of the 4. 1 mole Ar(s) gases, N2 has greater entropy than Ar because N2 is a molecule that can store energy in ways not possible for Ar. Chemical Thermodynamics
  • 99. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. Chemical Thermodynamics
  • 101. It must be a perfect crystal at 0K (absolute zero). Chemical Thermodynamics
  • 102. Chapter 19- Chemical Thermodynamics Sections 19.4 - 19.6 All about Free Energy
  • 103. Standard Entropies Chemical Thermodynamics
  • 104. Standard Entropies • These are molar entropy values of substances in their standard states. Chemical Thermodynamics
  • 105. Standard Entropies • These are molar entropy values of substances in their standard states. • Standard entropies tend to increase with increasing molar mass. Chemical Thermodynamics
  • 106. Standard Entropies Larger and more complex molecules have greater entropies. Chemical Thermodynamics
  • 107. Which substance will have the highest entropy? 1. Butane 2. Isobutane 3. Pentane 4. Isopentane 5. Neopentane Chemical Thermodynamics
  • 108. Which substance will have the highest entropy? 1. Butane 2. Isobutane 3. Pentane 4. Isopentane 5. Neopentane Chemical Thermodynamics
  • 109. Predict whether entropy increases, decreases, or stays constant for: 2 NO2(g) → 2 NO(g) + O2(g) 1. Increases 2. Decreases 3. Stays constant Chemical Thermodynamics
  • 110. Predict whether entropy increases, decreases, or stays constant for: 2 NO2(g) → 2 NO(g) + O2(g) 1. Increases 2. Decreases 3. Stays constant The entropy increases because two molecules of gas react to form three molecules of gas, leading to more disorder. Chemical Thermodynamics
  • 111. Which process will have a negative ΔS? 1. Dissolving a substance 2. Melting 3. Falling leaves 4. Oxidation of a metal 5. Decomposition of water into H2 and O2 Dissolution of NaCl Chemical Thermodynamics
  • 112. Which process will have a negative ΔS? 1. Dissolving a substance 2. Melting 3. Falling leaves 4. Oxidation of a metal 5. Decomposition of water Dissolution of NaCl into H2 and O2 Chemical Thermodynamics
  • 113. Entropy Changes Chemical Thermodynamics
  • 114. Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which ΔH is estimated: ΔS° = ΣnΔS°(products) - ΣmΔS°(reactants) where n and m are the coefficients in the balanced chemical equation. Chemical Thermodynamics
  • 115. SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔS° for the synthesis of ammonia from N2(g) and H2(g) at 298 K: Chemical Thermodynamics
  • 116. SAMPLE EXERCISE 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔS° for the synthesis of ammonia from N2(g) and H2(g) at 298 K: Solution Analyze: We are asked to calculate the entropy change for the synthesis of NH3(g) from its constituent elements. Plan: We can make this calculation using Equation 19.8 and the standard molar entropy values for the reactants and the products that are given in Table 19.2 and in Appendix C. Solve: Using Equation 19.8, we have Substituting the appropriate S° values from Table 19.2 yields Chemical Thermodynamics Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction.
  • 117. PRACTICE EXERCISE Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K: Chemical Thermodynamics
  • 118. PRACTICE EXERCISE Using the standard entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K: Answer: 180.39 J/K Chemical Thermodynamics
  • 119. Calculate ΔS° for the equation below using the standard entropy data given: 2 NO(g) + O2(g) → 2 NO2(g) ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205. 1. +176 J/mol 3. −147 J/mol 2. +147 J/mol 4. −76 J/mol Chemical Thermodynamics
  • 120. Correct Answer: 1. +176 J/mol 3. −147 J/mol 2. +147 J/mol 4. −76 J/mol ΔS° = Σ ΔS ° (products) − Σ ΔS ° (reactants) ΔS° = 2(240) − [2(211) + 205] ΔS° = 480 − [627] ΔS° = − 147 Chemical Thermodynamics
  • 121. Entropy Changes in Surroundings Chemical Thermodynamics
  • 122. Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. Chemical Thermodynamics
  • 123. Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process: −qsys ΔSsurr = T Chemical Thermodynamics
  • 124. Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process: −qsys ΔSsurr = T • At constant pressure, qsys is simply ΔH° for the system. Chemical Thermodynamics
  • 125. −qsys ΔSsurr = T • At constant pressure, qsys is simply ΔH° for the system. Chemical Thermodynamics
  • 126. always increase −qsys ΔSsurr = T • At constant pressure, qsys is simply ΔH° for the system. Chemical Thermodynamics
  • 127. Entropy Change in the Universe Chemical Thermodynamics
  • 128. Entropy Change in the Universe • The universe is composed of the system and the surroundings. Chemical Thermodynamics
  • 129. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, Chemical Thermodynamics
  • 130. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, ΔSuniverse = ΔSsystem + ΔSsurroundings Chemical Thermodynamics
  • 131. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, ΔSuniverse = ΔSsystem + ΔSsurroundings • For spontaneous processes Chemical Thermodynamics
  • 132. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, ΔSuniverse = ΔSsystem + ΔSsurroundings • For spontaneous processes ΔSuniverse > 0 Chemical Thermodynamics
  • 133. Entropy Change in the Universe Chemical Thermodynamics
  • 134. Entropy Change in the Universe • This becomes: −ΔHsystem ΔSuniverse = ΔSsystem + T Multiplying both sides by −T, −TΔSuniverse = ΔHsystem − TΔSsystem Chemical Thermodynamics
  • 135. Gibbs Free Energy Chemical Thermodynamics
  • 136. Gibbs Free Energy • −TΔSuniverse is defined as the Gibbs free energy, ΔG. Chemical Thermodynamics
  • 137. Gibbs Free Energy • −TΔSuniverse is defined as the Gibbs free energy, ΔG. • When ΔSuniverse is positive, ΔG is negative. Chemical Thermodynamics
  • 138. Gibbs Free Energy • −TΔSuniverse is defined as the Gibbs free energy, ΔG. • When ΔSuniverse is positive, ΔG is negative. • Therefore, when ΔG is negative, a process is spontaneous. Chemical Thermodynamics
  • 140. Entropy of universe increases and free energy of the system decreases. Chemical Thermodynamics
  • 141. Entropy of universe increases and free energy of the system decreases. Chemical Thermodynamics
  • 142. Gibbs Free Energy Chemical Thermodynamics
  • 143. Gibbs Free Energy 1. If ΔG is negative, the forward reaction is spontaneous. Chemical Thermodynamics
  • 144. Gibbs Free Energy 1. If ΔG is negative, the forward reaction is spontaneous. 2. If ΔG is 0, the system is at equilibrium. Chemical Thermodynamics
  • 145. Gibbs Free Energy 1. If ΔG is negative, the forward reaction is spontaneous. 2. If ΔG is 0, the system is at equilibrium. 3. If ΔG is positive, the reaction is spontaneous in the reverse direction. Chemical Thermodynamics
  • 146. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, ΔG°. f Chemical Thermodynamics
  • 147. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, ΔG°. f ΔG° = ΣnΔG°(products) − ΣmΔG°(reactants) f f where n and m are the stoichiometric coefficients. Chemical Thermodynamics
  • 148. Calculate ΔG° for the equation below using the Gibbs free energy data given: 2 SO2(g) + O2(g)  2 SO3(g) ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4 1. +70 kJ 2. +140 kJ 3. −140 kJ 4. −70 kJ Chemical Thermodynamics
  • 149. Correct Answer: 1. +70 kJ 2. +140 kJ 3. −140 kJ 4. −70 kJ ΔG ° = ∑ΔGf° (products) − ∑ ΔGf° (reactants) ΔG° = (2 × −370.4) − [(2 × −300.4) + 0] ΔG° = −740.8 − [−600.8] = −140 Chemical Thermodynamics
  • 151. It indicates the process has taken place under standard conditions. Chemical Thermodynamics
  • 152. It indicates the process has taken place under standard conditions. Chemical Thermodynamics
  • 153. It indicates the process has taken place under standard conditions. 298K = 25°C this is NOT STP (273K) Chemical Thermodynamics
  • 154. It indicates the process has taken place under standard conditions. 298K = 25°C this is NOT STP (273K) 1 atm for gases Chemical Thermodynamics
  • 155. It indicates the process has taken place under standard conditions. 298K = 25°C this is NOT STP (273K) 1 atm for gases 1M for solutions Chemical Thermodynamics
  • 156. Calculate ΔG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g) → 2 NH3(g) ΔH° (NH3) = −46 kJ; ΔS° values (J/mol-K) are NH3 = 192.5, N2 = 191.5, H2 = 130.6. 1. +33 kJ 3. −66 kJ 2. +66 kJ 4. −33 kJ Chemical Thermodynamics
  • 157. Correct Answer: 1. +33 kJ ΔH° = −92 kJ 2. +66 kJ ΔS° = −198.7J/mol − K 3. −66 kJ ΔG° = ΔH° − TΔS° 4. −33 kJ ΔG° = (−92) − (298)(−0.1987) ΔG° = (−92) + (59.2) = −33 Chemical Thermodynamics
  • 158. SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change from Free Energies of Formation (a) By using data from Appendix C, calculate the standard free-energy change for the following reaction at 298 K: (b) What is ΔG° for the reverse of the above reaction? Chemical Thermodynamics
  • 159. SAMPLE EXERCISE 19.6 Calculating Standard Free-Energy Change from Free Energies of Formation (a) By using data from Appendix C, calculate the standard free-energy change for the following reaction at 298 K: (b) What is ΔG° for the reverse of the above reaction? Solution Analyze: We are asked to calculate the free-energy change for the indicated reaction, and then to determine the free- energy change of its reverse. Plan: To accomplish our task, we look up the free-energy values for the products and reactants and use Equation 19.13: We multiply the molar quantities by the coefficients in the balanced equation, and subtract the total for the reactants from that for the products. Solve: (a) Cl2(g) is in its standard state, so is zero for this reactant. P4(g), however, is not in its standard state, so is not zero for this reactant. From the balanced equation and using Appendix C, we have: The fact that ΔG° is negative tells us that a mixture of P4(g), Cl2(g), and PCl3(g), at 25°C, each present at a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl3. Remember, however, that the value of ΔG° tells us nothing about the rate at which the reaction occurs. (b) Remember that ΔG = G (products) – G (reactants). If we reverse the reaction, we reverse the roles of the reactants and products. Thus, reversing the reaction changes the sign of ΔG, just as reversing the reaction changes the sign of ΔH. • (Section 5.4) Hence, Chemical using the result from part (a): Thermodynamics
  • 160. PRACTICE EXERCISE By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane: Chemical Thermodynamics
  • 161. PRACTICE EXERCISE By using data from Appendix C, calculate ΔG° at 298 K for the combustion of methane: Answer: –800.7 kJ Chemical Thermodynamics
  • 162. SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG° In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K: (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH°. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct?
  • 163. SAMPLE EXERCISE 19.7 Estimating and Calculating ΔG° In Section 5.7 we used Hess’s law to calculate ΔH° for the combustion of propane gas at 298 K: (a) Without using data from Appendix C, predict whether ΔG° for this reaction is more negative or less negative than ΔH °. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? Solution Analyze: In part (a) we must predict the value for ΔG° relative to that for ΔH° on the basis of the balanced equation for the reaction. In part (b) we must calculate the value for ΔG° and compare with our qualitative prediction. Plan: The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction (Equation 19.11), so under standard conditions: ΔG° = ΔH° – TΔS° To determine whether ΔG° is more negative or less negative than ΔH° we need to determine the sign of the term TΔS°. T is the absolute temperature, 298 K, so it is a positive number. We can predict the sign of ΔS° by looking at the reaction. Solve: (a) We see that the reactants consist of six molecules of gas, and the products consist of three molecules of gas and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By using the general rules we discussed in Section 19.3, we would expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system—the products have fewer accessible microstates than the reactants. We therefore expect ΔS° and TΔS° to be negative numbers. Because we are subtracting TΔS°, which is a negative number, we would predict that ΔG° is less negative than ΔH°. (b) Using Equation 19.13 and values from Appendix C, we can calculate the value of ΔG°: Chemical Notice that we have been careful to use the value of as in the calculation of ΔH values, the Thermodynamics phases of the reactants and products are important. As we predicted, ΔG° is less negative than ΔH° because of the decrease in entropy during the reaction.
  • 164. PRACTICE EXERCISE Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K: Would you expect ΔG° to be more negative or less negative than ΔH°? Chemical Thermodynamics
  • 165. PRACTICE EXERCISE Consider the combustion of propane to form CO2(g) and H2O(g) at 298 K: Would you expect ΔG° to be more negative or less negative than ΔH°? Answer: more negative Chemical Thermodynamics
  • 166. Free Energy Changes Chemical Thermodynamics
  • 167. Free Energy Changes At temperatures other than 25°C, ΔG° = ΔH° − TΔS° How does ΔG° change with temperature? Chemical Thermodynamics
  • 168. Free Energy and Temperature Chemical Thermodynamics
  • 169. Free Energy and Temperature • There are two parts to the free energy equation: ΔH°— the enthalpy term TΔS° — the entropy term • The temperature dependence of free energy, then comes from the entropy term. Chemical Thermodynamics
  • 170. Upon heating, limestone (CaCO3) decomposes to CaO and CO2. Speculate on the sign of ΔH and ΔS for this process. CaCO3(s) CaO(s) + CO2(g) Limestone Cave Carlsbad Caverns, New Mexico Chemical Thermodynamics
  • 171. Upon heating, limestone (CaCO3) decomposes to CaO and CO2. Speculate on the sign of ΔH and ΔS for this process. CaCO3(s) CaO(s) + CO2(g) Limestone Cave Carlsbad Caverns, New Mexico Chemical Thermodynamics
  • 172. Free Energy and Temperature Chemical Thermodynamics
  • 174. ΔH < TΔS Chemical Thermodynamics
  • 175. ΔH < TΔS ΔG(Negative/spontaneous)= Chemical Thermodynamics
  • 176. ΔH < TΔS ΔG(Negative/spontaneous)= ΔH (Positive/vaporization) - TΔS (positive/liquid to gas) Chemical Thermodynamics
  • 177. ΔH < TΔS ΔG(Negative/spontaneous)= ΔH (Positive/vaporization) - TΔS (positive/liquid to gas) ↑must be larger Chemical Thermodynamics
  • 178. Both the entropy, ΔS° and the enthalpy ΔH° have positive values. Near absolute zero (0 K), will the process be spontaneous or nonspontaneous? 1. Spontaneous 2. Nonspontaneous Chemical Thermodynamics
  • 179. Correct Answer: 1. Spontaneous 2. Nonspontaneous At low temperatures, the enthalpy term dominates the entropy term, leading to a positive ΔG and a nonspontaneous process. Chemical Thermodynamics
  • 180. A reaction will be spontaneous at lower temperatures when the sign of ΔH and ΔS is Chemical Thermodynamics
  • 181. A reaction will be spontaneous at lower temperatures when the sign of ΔH and ΔS is Chemical Thermodynamics
  • 182. SAMPLE EXERCISE 19.8 Determining the Effect of Temperature on Spontaneity The Haber process for the production of ammonia involves the equilibrium Assume that ΔH° and ΔS° for this reaction do not change with temperature. (a) Predict the direction in which ΔG° for this reaction changes with increasing temperature. (b) Calculate the values of ΔG° for the reaction at 25°C and 500°C.
  • 183. SAMPLE EXERCISE 19.8 continued Solution Analyze: In part (a) we are asked to predict the direction in which ΔG° for the ammonia synthesis reaction changes as temperature increases. In part (b) we need to determine ΔG° for the reaction at two different temperatures. Plan: In part (a) we can make this prediction by determining the sign of ΔS° for the reaction and then using that information to analyze Equation 19.20. In part (b) we need to calculate ΔH° and ΔS° for the reaction by using the data in Appendix C. We can then use Equation 19.20 to calculate ΔG°. Solve: (a) Equation 19.20 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –TΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negative, the term –TΔS° is positive and grows larger with increasing temperature. As a result, ΔG° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature. (b) We calculated the value of ΔH° in Sample Exercise 15.14, and the value of ΔS° was determined in Sample Exercise 19.5: ΔH° = –92.38 kJ and ΔS° = –198.4 J/K. If we assume that these values don’t change with temperature, we can calculate ΔG° at any temperature by using Equation 19.20. At T = 298 K we have: Notice that we have been careful to convert –TΔS° into units of kJ so that it can be added to ΔH°, which has units of kJ. Comment: Increasing the temperature from 298 K to 773 K changes ΔG° from –33.3 kJ to +61 kJ. Of course, the result at 773 K depends on the assumption that ΔH° and ΔS° do not change with temperature. In fact, these values do change slightly with temperature. Nevertheless, the result at 773 K should be a reasonable approximation. The positive increase in ΔG° with increasing T agrees with our prediction in part (a) of this exercise. Our result indicates that a mixture of N2(g), H2(g), and NH3(g), each present at a partial pressure of 1 atm, will react spontaneously at 298 K to form more NH3(g). In contrast, at 773 K the positive value of ΔG° tells us that the reverse reaction is spontaneous. Thus, when the mixture of three gases, each at a partial pressure of 1 atm, is heated to 773 K, some of the NH (g) spontaneously decomposes into N (g) and H (g).
  • 184. PRACTICE EXERCISE (a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: (b) Using the values obtained in part (a), estimate ΔG° at 400 K.
  • 185. PRACTICE EXERCISE (a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔH° and ΔS° at 298 K for the following reaction: (b) Using the values obtained in part (a), estimate ΔG° at 400 K. Answer: (a) ΔH° = –196.6 kJ, ΔS° = –189.6 J/K; (b) ΔG° = –120.8 kJ
  • 186. Chapter 19- Chemical Thermodynamics Section 19.7 Free Energy & Keq
  • 187. Free Energy and Equilibrium Chemical Thermodynamics
  • 188. Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way: ΔG = ΔG° + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.) Chemical Thermodynamics
  • 189. SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.20 to estimate the normal boiling point of CCl4.
  • 190. SAMPLE EXERCISE 19.9 Relating ΔG to a Phase Change at Equilibrium As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ΔG° for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.20 to estimate the normal boiling point of CCl4. Solution Analyze: (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous CCl4 at the normal boiling point. (b) We must determine the value of ΔG° for CCl4 in equilibrium with its vapor at the normal boiling point. (c) We must estimate the normal boiling point of CCl4, based on available thermodynamic data. Plan: (a) The chemical equation will merely show the change of state of CCl4 from liquid to solid. (b) We need to analyze Equation 19.21 at equilibrium (ΔG = 0). (c) We can use Equation 19.20 to calculate T when ΔG = 0. Solve: (a) The normal boiling point of CCl4 is the temperature at which pure liquid CCl4 is in equilibrium with its vapor at a pressure of 1 atm: (b) At equilibrium ΔG = 0. In any normal boiling-point equilibrium both the liquid and the vapor are in their standard states (Table 19.2). As a consequence, Q = 1, ln Q = 0, and ΔG = ΔG° for this process. Thus, we conclude that ΔG° = 0 for the equilibrium involved in the normal boiling point of any liquid. We would also find that ΔG° = 0 for the equilibria relevant to normal melting points and normal sublimation points of solids. (c) Combining Equation 19.20 with the result from part (b), we see that the equality at the normal boiling point, Tb, of CCl4(l) or any other pure liquid is Chemical Thermodynamics
  • 191. SAMPLE EXERCISE 19.9 continued Solving the equation for Tb, we obtain Strictly speaking, we would need the values of ΔH° and ΔS° for the equilibrium between CCl4(l) and CCl4(g) at the normal boiling point to do this calculation. However, we can estimate the boiling point by using the values of ΔH° and ΔS° for CCl4 at 298 K, which we can obtain from the data in Appendix C and Equations 5.31 and 19.8: Notice that, as expected, the process is endothermic (ΔH > 0) and produces a gas in which energy can be more spread out (ΔS > 0). We can now use these values to estimate Tb for CCl4(l): Note also that we have used the conversion factor between J and kJ to make sure that the units of ΔH ° and ΔS° match. Check: The experimental normal boiling point of CCl (l) is 76.5°C. The small deviation of our 4 estimate from the experimental value is due to the assumption that ΔH° and ΔS° do not change with temperature. Chemical Thermodynamics
  • 192. PRACTICE EXERCISE Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.) Chemical Thermodynamics
  • 193. PRACTICE EXERCISE Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Table 11.3.) Answer: 330 K Chemical Thermodynamics
  • 194. SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions We will continue to explore the Haber process for the synthesis of ammonia: Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. Chemical Thermodynamics
  • 195. SAMPLE EXERCISE 19.10 Calculating the Free-Energy Change Under Nonstandard Conditions We will continue to explore the Haber process for the synthesis of ammonia: Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. Solution Analyze: We are asked to calculate ΔG under nonstandard conditions. Plan: We can use Equation 19.21 to calculate ΔG. Doing so requires that we calculate the value of the reaction quotient Q for the specified partial pressures of the gases and evaluate ΔG°, using a table of standard free energies of formation. Solve: Solving for the reaction quotient gives: In Sample Exercise 19.8 we calculated ΔG° = 33.3 kJ for this reaction. We will have to change the units of this quantity in applying Equation 19.21, however. In order for the units in Equation 19.21 to work out, we will use kJ/mol as our units for ΔG°, where “per mole” means “per mole of the reaction as written.” Thus, ΔG° = –33.3kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3. We can now use Equation 19.21 to calculate ΔG for these nonstandard conditions: Comment: We see that ΔG becomes more negative, changing from –33.3 kJ/mol to –44.9 kJ/mol as the pressures of N2, H2, and NH3 are changed from 1.0 atm each (standard conditions, ΔG° ) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for ΔG indicates a larger “driving force” to produce NH3. We would have made the same prediction on the basis of Le Châtelier’s principle. • (Section 15.6) Relative to standard conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s principle predicts that both of these changes should shift the reaction more to the product side, thereby forming more NH3.
  • 196. PRACTICE EXERCISE Calculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. Chemical Thermodynamics
  • 197. PRACTICE EXERCISE Calculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. Answer: –26.0 kJ/mol Chemical Thermodynamics
  • 198. Decomposition of CaCO3 has a ΔH° = 178.3 kJ/mol and ΔS° = 159 J/mol • K. At what temperature does this become spontaneous? 1. 121°C 2. 395°C 3. 848°C 4. 1121°C Chemical Thermodynamics
  • 199. Correct Answer: 1. 121°C T = ΔH°/ΔS° 2. 395°C 3. 848°C T = 178.3 kJ/mol/0.159 kJ/mol • K 4. 1121°C T = 1121 K T (°C) = 1121 – 273 = 848 Chemical Thermodynamics
  • 200. Free Energy and Equilibrium Chemical Thermodynamics
  • 201. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. Chemical Thermodynamics
  • 202. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes Chemical Thermodynamics
  • 203. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes 0 = ΔG° + RT lnK Chemical Thermodynamics
  • 204. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes 0 = ΔG° + RT lnK • Rearranging, this becomes Chemical Thermodynamics
  • 205. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes 0 = ΔG° + RT lnK • Rearranging, this becomes ΔG° = −RT lnK Chemical Thermodynamics
  • 206. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes 0 = ΔG° + RT lnK • Rearranging, this becomes ΔG° = −RT lnK or, Chemical Thermodynamics
  • 207. Free Energy and Equilibrium • At equilibrium, Q = K, and ΔG = 0. • The equation becomes 0 = ΔG° + RT lnK • Rearranging, this becomes ΔG° = −RT lnK or, K = e−ΔG°/RT Chemical Thermodynamics
  • 208. Suppose ΔG° is a large, positive value. What then will be the value of the equilibrium constant, K? 1. K=0 2. K=1 3. 0<K<1 4. K>1 Chemical Thermodynamics
  • 209. Correct Answer: ΔG° = −RTlnK 1. K=0 Thus, large positive values of ΔG° 2. K=1 lead to large negative values of lnK. 3. 0<K<1 The value of K itself, then, is very 4. K>1 small, approaching zero. Chemical Thermodynamics
  • 210. N2(g) + 3 H2(g)  2 NH3(g) ΔH° = -92.4 kJ/mol If the temperature is decreased, what is the effect on the equilibrium constant and ΔG? 1. K increases, ΔG increases 2. K decreases, ΔG decreases 3. K increases, ΔG decreases 4. K decreases, ΔG increases 5. Temperature does not affect Josiah Willard Gibbs equilibrium or ΔG (1839-1903) Chemical Thermodynamics
  • 211. N2(g) + 3 H2(g)  2 NH3(g) ΔH° = -92.4 kJ/mol If the temperature is decreased, what is the effect on the equilibrium constant and ΔG? 1. K increases, ΔG increases 2. K decreases, ΔG decreases 3. K increases, ΔG decreases 4. K decreases, ΔG increases 5. Temperature does not affect Josiah Willard Gibbs equilibrium or ΔG (1839-1903) Chemical Thermodynamics
  • 212. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol.
  • 213. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
  • 214. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Solution Analyze: We are asked to calculate K for a reaction, given ΔG°. Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
  • 215. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Solution Analyze: We are asked to calculate K for a reaction, given ΔG°. Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG° when using Equations 19.21, 19.22, or 19.23). Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
  • 216. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Solution Analyze: We are asked to calculate K for a reaction, given ΔG°. Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG° when using Equations 19.21, 19.22, or 19.23). Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
  • 217. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Solution Analyze: We are asked to calculate K for a reaction, given ΔG°. Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG° when using Equations 19.21, 19.22, or 19.23). Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have We insert this value into Equation 19.23 to obtain K: Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
  • 218. SAMPLE EXERCISE 19.11 Calculating an Equilibrium Constant from ΔG° Use standard free energies of formation to calculate the equilibrium constant, K, at 25°C for the reaction involved in the Haber process: The standard free-energy change for this reaction was calculated in Sample Exercise 19.8: ΔG° = –33.3 kJ/mol = –33,300 J/mol. Solution Analyze: We are asked to calculate K for a reaction, given ΔG°. Plan: We can use Equation 19.23 to evaluate the equilibrium constant, which in this case takes the form In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of ΔG° when using Equations 19.21, 19.22, or 19.23). Solve: Solving Equation 19.22 for the exponent –ΔG°/RT, we have We insert this value into Equation 19.23 to obtain K: Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25°C. The equilibrium constants for temperatures in the range of 300°C to 600°C, given in Table 15.2, are much smaller than the value at 25°C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature. Thermodynamics can tell us the direction and extent of a reaction, but tells us nothing about the rate at which it occurs. If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be needed to force the equilibrium toward NH3.
  • 219. PRACTICE EXERCISE Use data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction Chemical Thermodynamics
  • 220. PRACTICE EXERCISE Use data from Appendix C to calculate the standard free-energy change, ΔG°, and the equilibrium constant, K, at 298 K for the reaction Answer: ΔG° = –106.4 kJ/mol, K = 5 × 1018 Chemical Thermodynamics
  • 221. SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together Consider the equilibria in which these salts dissolve in water to form aqueous solutions of ions: (a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?
  • 222. SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: (a) Calculate the value of ΔG° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of ΔG° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ΔG° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts? Solution (a) We will use Equation 19.13 along with values from Appendix C to calculate the values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to indicate that these are thermodynamic quantities for the formation of a solution.) We find Chemical Thermodynamics
  • 223. SAMPLE INTEGRATIVE EXERCISE continued (b) We can write as the sum of an enthalpy term, and an entropy term, We can calculate the values of and by using Equations 5.31 and 19.8. We can then calculate at T = 298 K. All these calculations are now familiar to us. The results are summarized in the following table: The entropy terms for the solution of the two salts are very similar. That seems sensible because each solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions. • (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two salts. The difference in the values of is dominated by the difference in the values of (c) The solubility product, Ksp, is the equilibrium constant for the solution process. • (Section 17.4) As such, we can relate Ksp directly to by using Equation 19.23: We can calculate the Ksp values in the same way we applied Equation 19.23 in Sample Exercise 19.13. We use the values we obtained in part (a), remembering to convert them from to J/mol: Chemical Thermodynamics The value calculated for the Ksp of AgCl is very close to that listed in Appendix D.
  • 224. SAMPLE INTEGRATIVE EXERCISE continued (d) A soluble salt is one that dissolves appreciably in water. • (Section 4.2) The Ksp value for NaCl is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small, indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble salt. (e) As we expect, the solution process has a positive value of ΔS for both salts (see the table in part b). As such, the entropy term of the free-energy change, is negative. If we assume that and do not change much with temperature, then an increase in T will serve to make more negative. Thus, the driving force for dissolution of the salts will increase with increasing T, and we therefore expect the solubility of the salts to increase with increasing T. In Figure 13.17 we see that the solubility of NaCl (and that of nearly any salt) increases with increasing temperature. • (Section 13.3) Chemical Thermodynamics