1. thermochemistry

2. INTRODUCTION
Thermochemistry is the study of heat (the
transfer of thermal energy) in chemical
reactions.
Heat is the transfer of thermal energy.
Heat is either absorbed or released during a
process.

3. Chemical Reactions
• All chemical reactions involve bond breaking
and bond forming.
• Energy is needed to break bonds and released
when bonds are formed.
• Chemical reactions are accompanied by a
change in energy, mainly in the form of heat.

4. HEAT REACTION=Exothermic and
Endothermic
• A reaction in which heat is given out is
exothermic.
2H2(g) + O2(g) 2H2O(l) + energy
• A reaction in which heat is taken in is
endothermic.
energy + 2HgO(s) 2Hg(l) + O2(g)

5. Heat of:
• Formation- is the heat change in kilojoules, when
one mole of a substance is formed from its
elements in their standard state.
• Of reaction- heat in kilojoules released or
absorbed when the number of moles of reactants
indicated, in the balanced equation describing
the reaction, react completely.
• Combustion- is the heat change in kilojoules
when one mole of the substance is completely
burned in excess oxygen

6. cont
• Heat of Neutralization-? The enthalpy of neutralization
(ΔHn) is the change in enthalpy that occurs when one
equivalent of an acid and one equivalent of a base undergo a
neutralization reaction to form water and a salt.
• Heat of vaporization:is the enthalpy change required to
transform a given quantity of a substance from a liquid into a
gas at a given pressure (often atmospheric pressure, as in STP
).

7. TThhee FFiirrsstt LLaaww ooff TThheerrmmooddyynnaammiiccss
The first law of thermodynamics states that energy can be converted from one form to
another, but cannot be created or destroyed./
The law of conservation of energy states that energy cannot be created or
destroyed, but only changed from one form to another.
ΔUsys + ΔUsurr = 0
ΔU is the change in the internal energy.
“sys” and “surr” denote system and surroundings, respectively.
ΔU = Uf – Ui; the difference in the energies of the initial and final states.
ΔUsys = –ΔUsurr

8. WWoorrkk aanndd HHeeaatt
The overall change in the system’s internal energy is given by:
ΔU = q + w
q is heat
q is positive for an endothermic process (heat absorbed by the system)
q is negative for an exothermic process (heat released by the system)
w is work
w is positive for work done on the system
w is negative for work done by the system
Calculate the overall change in internal energy, ΔU, (in joules) for a system that absorbs
188 J of heat and does 141 J of work on its surroundings.
Solution The system absorbs heat, so q is positive. The system does work on the
surroundings, so w is negative.
ΔU = q + w = 188 J + (-141 J) = 47 J

9. Enthalpy changes of exothermic and
endothermic reactions

10. Heat of:
• HEAT Of reaction- heat in kilojoules released or
absorbed when the number of moles of reactants
indicated, in the balanced equation describing the
reaction, react completely.
• For an exothermic reaction ΔH is negative(-)
• For an endothermic reaction ΔH is positive(+)
• HEAT OF Combustion- is the heat change in kilojoules
when one mole of the substance is completely burned
in excess oxygen
• HEAT OF Formation- is the heat change in kilojoules,
when one mole of a substance is formed from its
elements in their standard state

11. • HEAT OF Neutralization-?

12. Heat of Combustion of Different Fuels
• Methane (natural gas) -890 kj mol-1
• Propane (LPG) -2219 kj mol-1
• Hydrogen -286 kj mol-1
• Petrol (octane) -5470 kj mol-1

13. Bond Energy
• Bond energy is the amount of energy in
kilojoules needed to break one mole of bonds
of the same type, all species being in the
gaseous state.
• The average C-H bond energy in methane is
412kj mol-1 i.e. E(C-H) = 412kj mol-1 .
• The energy of a particular bond type can vary.
• It is usual to quote the average bond energies.

14. At Constant Pressure
Recall, by definition a change in energy equals heat
transferred (q) plus work (w):
DE = q + w
Consider a process carried out at constant pressure.
At constant pressure, work involves only a change
in volume. We can then substitute -PDV for w.
DE = qp - PDV
Then if we want to solve for the heat transferred, qp,
at constant pressure, we simply rearrange the
equation.
qp = DE + PDV

15. EEnntthhaallppyy
Sodium azide detonates to give a large quantity of nitrogen gas.
2NaN3(s) 2Na(s) + 3N2(g)
Under constant volume conditions, pressure increases:

16. EEnntthhaallppyy
Sodium azide detonates to give a large quantity of nitrogen gas.
2NaN3(s) 2Na(s) + 3N2(g)
Under constant volume conditions, pressure increases:

17. EEnntthhaallppyy
Pressure-volume, or PV work, is done when there is a volume change under constant
pressure.
w = −PΔV
P is the external opposing pressure.
ΔV is the change in the volume of the
container.

18. Worked Example 10.2
Determine the work done (in joules) when a sample of gas extends from 552 mL
to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm,
(b) against a constant pressure of 1.00 atm, and (c) against a vacuum
(1 L∙atm = 101.3 J).
Strategy Determine change in volume (ΔV), identify the external pressure (P),
and use w = −PΔV to calculate w. The result will be in L∙atm; use the equality 1
L∙atm = 101.3 J to convert to joules.
Solution ΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm,
(c) P = 0 atm.
(a) w = -(1.25 atm)(339 mL)
(b) w = -(1.00 atm)(339 mL)
1 L
1000 mL
101.3 J
1 L∙atm = -42.9 J
1 L
1000 mL
101.3 J
1 L∙atm = -34.3 J

19. Worked Example 10.2 (cont.)
Solution
(c) w = -(0 atm)(339 mL) 1 L
1000 mL
101.3 J
1 L∙atm = 0 J
Think About It Remember that the negative sign in the answers to part (a) and
(b) indicate that the system does work on the surroundings. When an expansion
happens against a vacuum, no work is done. This example illustrates that work is
not a state function. For an equivalent change in volume, the work varies
depending on external pressure against which the expansion must occur.

20. EEnntthhaallppyy
Pressure-volume, or PV work, is done when there is a volume change under constant
pressure.
w = −PΔV
ΔU = q + w
ΔU = q − PΔV
When a change occurs at constant volume, ΔV = 0 and no work is done.
qV = ΔU
substitute

21. EEnntthhaallppyy
Under conditions of constant pressure:
ΔU = q + w
ΔU = q − PΔV
qP = ΔU + PΔV

22. EEnntthhaallppyy
The thermodynamic function of a system called enthalpy (H) is defined by the equation:
H = U + PV
A note about SI units:
Pressure: pascal; 1Pa = 1 kg/(m . s2)
Volume: cubic meters; m3
PV: 1kg/(m . s2) x m3 = 1(kg . m2)/s2 =
1 J
Enthalpy:joules
U, P, V, and H are all state functions.

23. Enthalpy aanndd EEnntthhaallppyy CChhaannggeess
For any process, the change in enthalpy is:
ΔH = ΔU + Δ(PV) (1)
ΔH = ΔU + PΔV
If pressure is constant:
(2)
ΔU = ΔH + PΔV
Rearrange to solve for ΔU:
(3)
qp = ΔU + ΔV
Remember, qp:
(4)
Substitute equation (3) into equation (4) and solve:
qp = (ΔH − PΔV) + PΔV
(5)
qp = ΔH for a constant-pressure process

24. Enthalpy = Heat Transferred
Recall our original definition of enthalpy:
H = E + PV
Then for a change in enthalpy:
DH = D E + D(PV)
If we set P constant, then:
DH = D E + P D V
Since
qp = DE + PDV
Then
DH = qp
The change in enthalpy, D H, is then equal to the
heat transferred at constant pressure, qp.

25. Enthalpy aanndd EEnntthhaallppyy CChhaannggeess
The enthalpy of reaction (ΔH) is the difference between the enthalpies of the products and
the enthalpies of the reactants:
ΔH = H(products) – H(reactants)
Assumes reactions in the lab occur at constant pressure
ΔH > 0 (positive) endothermic process
ΔH < 0 (negative) exothermic process

26. TThheerrmmoocchheemmiiccaall EEqquuaattiioonnss
H2O(s) H2O(l) ΔH = +6.01 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
ΔH > 0 endothermic

27. TThheerrmmoocchheemmiiccaall EEqquuaattiioonnss
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
ΔH < 0 exothermic

28. TThheerrmmoocchheemmiiccaall EEqquuaattiioonnss
Enthalpy is an extensive property.
Extensive properties are dependent on the amount of matter involved.
H2O(l) → H2O(g) ΔH = +44 kJ/mol
Double the amount of matter Double the enthalpy
2H2O(l) → 2H2O(g) ΔH =
+88 kJ/mol
Units refer to
mole of reaction as written

29. TThheerrmmoocchheemmiiccaall EEqquuaattiioonnss
The following guidelines are useful when considering thermochemical equations:
1) Always specify the physical states of reactants and products because they help
determine the actual enthapy changes.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol
different states different enthalpies
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = +890.4 kJ/mol

30. TThheerrmmoocchheemmiiccaall EEqquuaattiioonnss
The following guidelines are useful when considering thermochemical equations:
2) When multiplying an equation by a factor (n), multiply the ΔH value by same factor.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = − 802.4 kJ/mol
2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) ΔH = − 1604.8 kJ/mol
3) Reversing an equation changes the sign but not the magnitude of ΔH.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = − 802.4 kJ/mol
CO2(g) + 2H2O(g) CH4(g) + 2O2(g) ΔH = +802.4 kJ/mol

31. Worked Example 10.3
Given the thermochemical equation for photosynthesis,
6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g) ΔH = +2803 kJ/mol
calculate the solar energy required to produce 75.0 g of C6H12O6.
Strategy The thermochemical equation shows that for every mole of C6H12O6
produced, 2803 kJ is absorbed. We need to find out how much energy is
absorbed for the production of 75.0 g of C6H12O6. We must first find out how
many moles there are in 75.0 g of C6H12O6.
The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is
1 mol C6H12O6
180.2 g C6H12O6
75.0 g C6H12O6 ×
= 0.416 mol C6H12O6
We will multiply the thermochemical equation, including the enthalpy change, by
0.416, in order to write the equation in terms of the appropriate amount of
C6H12O6.

32. Worked Example 10.3 (cont.)
Solution
(0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]
and (0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives
2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g) ΔH = +1.17×103 kJ
Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the
production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are
canceled when we multiply the thermochemical equation by the number of moles
of C6H12O6.
Think About It The specified amount of C6H12O6 is less than half a mole.
Therefore, we should expect the associated enthalpy change to be less than half
that specified in the thermochemical equation for the production of 1 mole of
C6H12O6.

33. Heat Capacity, C
“C” is an extensive property; so a large object has a larger
heat capacity than a small object made of the same material.
q
T
……..of a substance is the amount of heat required to raise the temperature of 1 g
of the substance by 1°C.
Using the Equation:
C heat absorbed
increase in temperature
D
= =
Looking at the figures on the
left, it can be seen that the
temperature change is
constant, but the heat absorbed
by the larger object is greater.
This results in a larger
heat capacity for the larger
object because more heat is
absorbed.

34. Specific heat capacity: The energy (joules) required to
raise the temperature of 1 gram of substance by 1°C
Unit: J g-1K-1 or J g-1°C-1
C C s =
Molar heat capacity: The energy (joules) required to
raise the temperature of 1 mol of substance by 1°C
Unit: J mol-1 K-1 or J mol-1°C-1
m
C C m =
n

35. Substance
Specific Heat, Cs
(cal/gram°C) (J/kg °C)
Pure water 1.00 4,186*
Wet mud 0.60 2,512
Ice (0 °C) 0.50 2,093
Sandy clay 0.33 1,381
Dry air (sea level) 0.24 1,005
Quartz sand 0.19 295
Granite 0.19 294
1 calorie = 4.186 joules
*The high heat capacity of water makes it ideal for storing heat
in solar heating systems.

36. Neutralization
The reaction between an acid and a base
which results in a salt plus water.
HClaq + NaOHaq ® NaClaq + H2O
Another example, cyanic
acid and a hydroxide ion.
For example, hydrochloric acid
and sodium hydroxide:
acid + base ® salt + water

37. Heat of Neutralization
Net ionic equation for neutralization:
H+(aq) + OH-(aq) ® H2O(l)
Energy released by reaction = Energy absorbed by solution
Specific heat capacity, Cs, is defined as the quantity of
heat transferred, q, divided by the mass of the substance
times the change in temperature. A value of Cs is specific to
the given substance.
Cs = q / [(mass) (Tfinal-Tinitial)]
This can then be rearranged to solve for the heat transferred.
q = Cs (mass) (Tfinal-Tinitial)

38. Enthalpy of Fusion (Melting)
Enthalpy of Fusion is defined as the heat that is absorbed
when the melting occurs at constant pressure. If the
substance freezes, the reaction is reversed, and an equal
amount of heat is given off to the surroundings; i.e.,
ΔHfreez = - ΔHfus
solid liquid
Melting (fusion) is an endothermic process
H H (liquid) H (solid) fus m m D = -

39. Heat Capacity of Calorimeter
• To determine the heat capacity of coffee cup
calorimeter:
– A 25.0-g sample of warm water at 40.0oC was
added to a 25.0-g sample of water in a Styrofoam
coffee cup calorimeter initially at 20.0oC. The final
temperature of the mixed water and calorimeter
was 29.5oC and the specific heat capacity of water
is 4.184 J/g.oC. Calculate the heat capacity, Ch, of
the calorimeter.

40. Specific Heat Capacity
• To determine the specific heat capacity of a
metal using coffee cup calorimeter:
– A 55.0-g sample of hot metal initially at 99.5oC
was added to 40.0 g of water in a Styrofoam coffee
cup calorimeter. The water and calorimeter were
initially at 21.0oC. If the final temperature of
mixture was 30.5oC, calculate the total heat lost by
metal and the specific heat capacity of the metal.
The specific heat of water is 4.184 J/(g.oC) and
heat capacity of calorimeter is 10.0 J/oC.

41. Heat of Neutralization
• To determine the molar enthalpy of acid-base reaction using
coffee cup calorimeter.
– 50.0 mL of 2.0 M HCl was reacted with 50.0 mL of 2.0 M NaOH
in a coffee cup calorimeter. The reaction was exothermic, which
caused the temperature of the solution to increase from 22.0oC to
35.6oC. Assume the density of solution as 1.0 g/mL, its specific
heat capacity as 4.18 J/g.oC, and the heat capacity of calorimeter
as 10.J/oC. Calculate the total amount of heat produced by the
reaction. Calculate the enthalpy change (DH, in kJ/mol) for the
following reaction:
• HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

42. Calculation the enthalpy of reaction
using Styrofoam cup calorimeter
• To determine the enthalpy of reaction using coffee cup
calorimeter.
– Suppose 100. mL of 1.0 M HCl solution is placed in a Styrofoam
coffee cup calorimeter. The initial temperature of HCl solution is
22.5oC. A 0.255-g sample of magnesium ribbon is cut to short pieces
and added to the acid solution in which the following exothermic
reaction occurred.
• Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
– The heat produced by the above reaction is completely absorbed by the
solution and calorimeter, which attained the highest temperature of
34.2oC. Assume the acid solution has a density of 1.0 g/mL and its
specific heat capacity as 4.0 J/g.oC, and the calorimeter has a heat
capacity of 10. J/oC. Calculate the molar enthalpy change (DH, in
kJ/mol) for the above reaction.

43. Calculating the enthalpy of reaction using
Styrofoam cup calorimeter
Reaction: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g);
Calculations:

44. Heat Capacity of Calorimeter
• To determine the heat capacity of bomb
calorimeter:
– When a 1.200-g sample of glucose, C6H12O6, was
completely combusted in a bomb calorimeter, the
temperature of the calorimeter assembly increased
by 4.48oC. If the combustion of glucose produces
14.0 kJ/g of energy, how much heat energy is
absorbed by the calorimeter. Calculate the heat
capacity, Ch, of the calorimeter. (Assume that all of
the heat produced by the combustion of glucose is
absorbed by the calorimeter.)

45. Heat of Combustion
• To calculate the enthalpy of combustion using bomb
calorimeter
– When a 1.010-g sample of sucrose (cane sugar) is
completely combusted in a bomb calorimeter, the
temperature of the calorimeter was increased by
4.50oC. If the heat capacity of calorimeter is 3.75
kJ/oC, how much heat was absorbed by the
calorimeter? Calculate the molar enthalpy of
combustion of sucrose according to the following
equation:
• C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(l)

46. Calorimetry
• Heat capacity- the quantity of heat needed to change
the temperature of the system 1K
– Cp = q/ DT (units are J/k or J/oc)
• Specific Heat capacity – the quantity of heat need to
raise the temperature of 1 gram of a substance 1 oC
– q = C m DT

47. Specific Heat
• The quantity of heat required to raise 1 gram
of water 1oC.
• The higher the specific heat the harder it is to
change the temperature of the substance.
C (J/goC)
– Al .092
– Cu. .385
– Ethanol 2.46
– Water 4.182

48. Calorimetry
• Measures the amount of heat generated from a chemical
reaction by letting the heat generated flow into a mass
of cooler water.
Calorimetry is the measurement of heat changes.
Heat changes are measured in a device called a calorimeter
q= m DT C
q = heat DT = Tf – Ti
m = mass C = specific heat
(J/g oC)

49. Specific HHeeaatt aanndd HHeeaatt CCaappaacciittyy
The heat capacity (C) is the amount of heat required to raise the temperature of an object
by 1°C.
The “object” may be a given quantity of a particular substance.
heat capacity of 1 kg of water = 4.184 J 1000 g = 4184 J/ C
Specific heat capacity has units of J/(g • °C)
Heat capacity has units of J/°C
1 g C
´ °
· °
Specific heat capacity of water heat capacity of
1 kg water

50. Coffee Cup Calorimeter
• A coffee cup calorimeter is essentially a polystyrene
(Styrofoam) cup with a lid.
• The cup is partially filled with a known volume of water
and a thermometer is inserted through the lid of the cup
so that its bulb is below the water surface.
• When a chemical reaction occurs in the coffee cup
calorimeter, the heat of the reaction if absorbed by the
water.
• The change in the water temperature is used to calculate
the amount of heat that has been absorbed or evolved in
the reaction.

51. For example,
a chemical reaction which occurs in 200 grams of water with an
initial temperature of 25.0°C. The reaction is allowed to proceed
in the coffee cup calorimeter. As a result of the reaction, the
temperature of the water changes to 31.0°C. The heat flow is
calculated:
qwater = 4.18 J/(g·°C) x 200 g x (31.0°C - 25.0°C)
qwater = +5.0 x 103 J
ΔHreaction = -(qwater)

52. CCaalloorriimmeettrryy
Calculate the amount of heat required to heat 1.01 kg of water from 0.05°C to 35.81°C.
Solution:
Step 1: Use the equation q = msΔT to calculate q.
1.01 kg 1000 g 4.184 J [35.81 C 0.05 C] = 151000 J = 151 kJ
q = ´ ´ ´ ° - °
1 kg g · °
C

53. CCaalloorriimmeettrryy
A coffee-cup calorimeter may be used to measure the heat exchange for a
variety of reactions at constant pressure:
Heat of neutralization:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Heat of ionization:
H2O(l) → H+(aq) + OH‒(aq)
Heat of fusion:
H2O(s) → H2O(l)
Heat of vaporization:
H2O(l) → H2O(g)

54. CCaalloorriimmeettrryy
Concepts to consider for coffee-cup calorimetry:
qP = ΔH
System: reactants and products (the reaction)
Surroundings: water in the calorimeter
For an exothermic reaction:
 the system loses heat
 the surroundings gain (absorb) heat
qsys = −msΔT
qsurr = msΔT qsys = −qsurr
The minus sign is used to keep
sign conventions consistent.

55. Worked Example 10.5
A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125 g
of water originally at 25.1°C. The final temperature of both pellet and the water is
31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.
Strategy Water constitutes the surroundings; the pellet is the system. Use qsurr =
msΔT to determine the heat absorbed by the water; then use q = CΔT to
determine the heat capacity of the metal pellet.
mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C. The
heat absorbed by the water must be released by the pellet: qwater = -qpellet, mpellet =
100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.

56. Worked Example 10.5 (cont.)
Solution
4.184 qg∙°C =
J
water × 125 g × 6.2°C = 3242.6 J
Thus,
qpellet = -3242.6 J
From q = CΔT we have
-3242.6 J = Cpellet × (-57.1°C)
Thus,
Cpellet = 57 J/°C
Think About It The units cancel properly to give appropriate units for heat
capacity. Moreover, ΔTpellet is a negative number because the temperature of the
pellet decreases.

57. CCoonnssttaanntt--VVoolluummee CCaalloorriimmeettrryy
Constant volume calorimetry is carried out in a device known as a constant-volume bomb.
A constant-volume calorimeter is an
isolated system.
Bomb calorimeters are typically used to
determine heats of combustion.
qcal = −qrxn

58. CCoonnssttaanntt--VVoolluummee CCaalloorriimmeettrryy
To calculate qcal, the heat capacity of the calorimeter must be known.
qcal = CcalΔT
qrxn = −qcal
qrxn = −CcalΔT

59. Worked Example 10.6
A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a
bomb calorimeter to determine its energy content. The heat capacity of the
calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water
in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of
the cookie.
Think About It According to the label on the cookie package, a
service size is four cookies, or 29 g, and each serving contains 150
Cal. Convert the energy per gram to Calories per serving to verify
the result.
Strategy Use qrxn = -CcalΔT to calculate the heat released by the combustion of
the cookie. Divide the heat released by the mass of the cookie to determine its
energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C.
Solution
21.5 kJ
g ×
1 Cal
4.184 kJ ×
29 g
serving = 1.5×102 Cal/serving
qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ
Because energy content is a positive quantity, we write
1.559×102 kJ
7.25 g = 21.5 kJ/g
energy content per gram =

60. HHeessss’’ss LLaaww
Hess’s law states that the change in enthalpy for a stepwise process is the sum of the
enthalpy changes for each of the steps.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol
2H2O(l) 2H2O(g) ΔH = +88.0 kJ/mol
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol
CH4(g) + 2O2(g)
ΔH = −890.4 kJ
CO2(g) + 2H2O(l)
ΔH = −802.4 kJ
CO2(g) + 2H2O(g)
ΔH = +88.0 kJ
10.5

61. Bomb calorimetry is used to determine the enthalpy of
combustion, DH, for hydrocarbons:

62. Bomb Calorimeter
• In a coffee cup calorimeter, the reaction takes place in the
water.
• In a bomb calorimeter, the reaction takes place in a sealed
metal container, which is placed in the water in an insulated
container.
• Heat flow from the reaction crosses the walls of the sealed
container to the water.
• The temperature difference of the water is measured, just as
it was for a coffee cup calorimeter.

63. Bomb Calorimeter
• Analysis of the heat flow is more complex
than for the coffee cup calorimeter because
the heat flow into the metal parts of the
calorimeter must be accounted for (heat
capacity, Cp):
• qreaction = - (qwater + qbomb)
• where qwater = 4.18 J/(g·°C) x mwater x Δt
• qbomb = Cp x Δt

64. Calorimetry
• A 1.5886 g sample of glucose (C6H12O6) was
ignited in a bomb calorimeter. The
temperature increased by 3.682oc. The heat
capacity of the calorimeter was 3.56 kJ/oc, and
the calorimeter contained 1.00 kg of water.
Find the molar heat in kJ/molrxn
C6H12O6 + 6 O2  6 CO2 + 6 H2O

65. Calorimetry
qbomb = 3.562 kJ/oc (3.682oc) = 13.12 kJ
qwater = 1000g (4.184 J/goc) (3.682oc) = 15400 J
Qtotal = qbomb + qwater
qtotal = - 28.52 kJ (exothermic)
per molrxn
1.5886 g / 180.16g/mol = .0088177 mol
-28.52 kJ/ .0088177 mol = -3234 kJ/molrxn

66. Enthalpy
• Enthalpy is a measure of the total energy of
a thermodynamic system.
• Most chemistry reactions take place at
constant pressure, open to the atmosphere.
So, Enthalpy (H) is used to describes these types
of reactions.
H = U + PV

67. Enthalpy (H) –
The Heat of the Reaction
• The change in enthalpy (DH) is equal to the
difference in the heat of the reaction.
DH = DHfinal – DHinitial
= DHproducts – DHreactants
• Heat of formation (DHo
f) - the change in
enthalpy when a compound forms from its
pure elements. (table in back of text book)
DHo
f of a pure element = 0

68. Enthalpy and Internal Energy
DH = DU + DVP
• DVP = DngasRT @ const. T and P
DH = DU + DngasRT
DU and DH are very close to the same value
and are the same when no gas is generated by
the reaction.

69. Hess’s Law
Hess’s law states that the change in enthalpy for a stepwise process is the sum of the
enthalpy changes for each of the steps.
• .
Example:
1.What is the DH for C + ½ O2  CO?
CO + ½ O2  CO2 DH = -283.0 kJ
C + O2  CO2 DH = -393.5 kJ

70. What is the DH for
C + ½ O2  CO?
CO2  CO + ½ O2 DH = +283.0 kJ
C + O2  CO2 DH = -393.5 kJ
C + ½ O2  CO DH = -110.5 kJ

71. HHeessss’’ss LLaaww
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol
2H2O(l) 2H2O(g) ΔH = +88.0 kJ/mol
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol
CH4(g) + 2O2(g)
ΔH = −890.4 kJ
CO2(g) + 2H2O(l)
ΔH = −802.4 kJ
CO2(g) + 2H2O(g)
ΔH = +88.0 kJ
10.5
2

72. Worked Example 10.7
Given the following thermochemical equations,
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol
32
O3(g) → O2(g) ΔH = –142.3 kJ/mol
O2(g) → 2O(g) ΔH = +495 kJ/mol
determine the enthalpy change for the reaction
NO(g) + O(g) → NO2(g)
Strategy Arrange the given thermochemical equations so that they sum to the
desired equation. Make the corresponding changes to the enthalpy changes, and
add them to get the desired enthalpy change.

73. Worked Example 10.7 (cont.)
Solution The first equation has NO as a reactant with the correct coefficients,
so we will use it as is.
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol
The second equation must be reversed so that the O3 introduced by the first
equation will cancel (O3 is not part of the overall chemical equation). We also
must change the sign on the corresponding ΔH value.
32 NO(g) + O3(g) → NO2(g) + O2(g)
O2(g) → O3(g) ΔH = +142.3 kJ/mol
These two steps sum to give:
O2(g) → O3(g) 32 ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
+ O2(g) 12
NO(g) + O2(g) → NO2(g) ΔH = –56.6 kJ/mol 12

74. Worked Example 10.7 (cont.)
Solution We then replace O2 on the left with O by incorporating the last
equation. To do so, we divide the third equation by 2 and reverse its direction. As
a result, we must also divide ΔH value by 2 and change its sign.
12
12
Finally, we sum all the steps and add their enthalpy changes.
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol
O2(g) → O3(g) ΔH = +142.3 kJ/mol
32
+
O(g) → O2(g) ΔH = –247.5 kJ/mol
12
O(g) → O2(g) ΔH = –247.5 kJ/mol
NO(g) + O(g) → NO2(g) ΔH = –304 kJ/mol
Think About It Double-check the cancellation of identical items–especially
where fractions are involved.

75. Standard EEnntthhaallppiieess ooff FFoorrmmaattiioonn
The standard enthalpy of formation (ΔH f°) is defined as the heat change that results when
1 mole of a compound is formed from its constituent elements in their standard states.
C(graphite) + O2(g) CO2(g) ΔH f° = −393.5 kJ/mol
Elements in standard
states
1 mole of product
10.6

76. Standard EEnntthhaallppiieess ooff FFoorrmmaattiioonn
The standard enthalpy of formation (ΔH f°) is defined as the heat change that results when
1 mole of a compound is formed from its constituent elements in their standard states.
The superscripted degree sign denotes standard conditions.
1 atm pressure for gases
1 M concentration for solutions
“f” stands for formation.
ΔH f° for an element in its most stable form is zero.
ΔH f° for many substances are tabulated in Appendix 2 of the textbook.

77. Standard EEnntthhaallppiieess ooff FFoorrmmaattiioonn
The standard enthalpy of reaction (ΔH °rxn) is defined as the enthalpy of a reaction carried
out under standard conditions.
aA + bB → cC + dD
ΔH °rxn = [cΔH f°(C) + dΔH f°(D) ] – [aΔH f°(A) + bΔH f°(B)]
n and m are the stoichΔioHmetric coefficients for the reactants and products. °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants)

78. Worked Example 10.8
Using data from Appendix 2, calculate ΔH °rxn for Ag+(aq) + Cl-(aq) → AgCl(s).
Strategy Use ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants) and ΔH f° values
from Appendix 2 to calculate ΔH °rxn. The ΔH f° values for Ag+(aq), Cl-(aq), and
AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.
Solution
ΔH °rxn = ΔH f°(AgCl) – [ΔH f°(Ag+) + ΔH f°(Cl-)]
= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2
kJ/mol)]
= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol
Think About It Watch out for misplaced or missing minus signs. This is an
easy place to lose track of them.

79. Worked Example 10.9
Given the following information, calculate the standard enthalpy of formation of
acetylene (C2H2) from its constituent elements:
C(graphite) + O2(g) → CO2(g) ΔH °rxn = –393.5 kJ/mol
H2(g) + O2(g) → H2O(l) ΔH °rxn = –285.5 kJ/mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ΔH °rxn = –2598.8 kJ/mol
Strategy Arrange the equations that are provided so that they will sum to the
desired equation. This may require reversing or multiplying one or more of the
equations. For any such change, the corresponding change must also be made to
the ΔH °rxn value. The desired equation, corresponding to the standard enthalpy
of formation of acetylene, is
2C(graphite) + H2(g) → C2H2(g)
12
(1)
(2)
(3)

80. Worked Example 10.9 (cont.)
Solution We multiply Equation (1) and its ΔH °rxn value by 2:
2C(graphite) + 2O2(g) → 2CO2(g) ΔH °rxn = –787.0 kJ/mol
We include Equation (2) and its ΔH °rxn value as is:
12
H2(g) + O2(g) → H2O(l) ΔH °rxn = –285.5 kJ/mol
We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):
Think About It Remember that a ΔH °rxn is only a ΔH °f when there
is just one product, just one mole produced, and all the reactants are
elements in their standard states.
52
2CO2(g) + H2O(l) → C2H2(g) + O2(g) ΔH °rxn = +1299.4 kJ/mol
Summing the resulting equations and the corresponding ΔH °rxn values:
2C(graphite) + 2O2(g) → 2CO2(g) ΔH °rxn = –787.0 kJ/mol
12
H2(g) + O2(g) → H2O(l) ΔH °rxn = –285.5 kJ/mol
2CO2(g) + H2O(l) → C2H2(g) + O2(g) ΔH °rxn = +1299.4 kJ/mol
52
2C(graphite) + H2(g) → C2H2(g) ΔH °f = +226.6 kJ/mol