First law of thermodynamics as taught in introductory physical chemistry (includes general chemistry material). Covers concepts such as internal energy, heat, work, heat capacity, enthalpy, bomb calorimetry, Hess's law, thermochemical equations, bond energy, and heat of formations.

1. 3. First Law of
Thermodynamics
Dr. Yujung Dong
1

2. Energy
When energy is transferred,
it can perform work or heat an object.

3. Kinetic Energy
• The energy from the movement of an object
𝐾𝐸 =
1
2
𝑚𝑣2

4. Potential Energy
• Stored energy
• Energy due to the position within a field (gravity, electric potential)
• Chemical Energy is also a form of potential energy.
potential energy
within a gravity field
𝑃𝐸 = 𝑚𝑔ℎ
+
−
d
+Q
-q
potential energy
within an electric field
𝑃𝐸 = −
𝑄𝑞
4𝜋𝜀0 𝑟
chemical energy

5. Potential-Kinetic Energy Transition
• http://ca.pbslearningmedia.org/resource/hew06.sci.phys.maf.rollerco
aster/energy-in-a-roller-coaster-ride/

6. • The absolute internal energy of the system is difficult to determine.
• But the change in the energy can be measured by the work they do or
heat they release/absorb.
Heat and work are vehicles of energy transfer.

7. Units of Energy
• Energy, heat, and work have the same units.
• Joule (J)
• The SI unit of energy
• 𝐽 =
𝑘𝑔∙𝑚2
𝑠2 = 𝑁 ∙ 𝑚
• Calorie (cal)
• Amount of heat necessary to raise the temperature of 1 g of water by 1˚C
• 4.184 J = 1 cal
• 101.32 J = 1 L·atm

8. First Law of Thermodynamics
• Law of Conservation of Energy
• Energy cannot be created or destroyed
• It can be converted from one form to another
• Potential energy → kinetic energy
• Chemical energy → heat
• etc.
• Total energy of the universe is constant (ΔEuniv=0).

9. First Law of Thermodynamics
• System: The part of the universe that is the focus.
• Surroundings: Everything in the universe that is not part
of the system
• Universe = System + Surroundings
• First law of thermodynamics:
ΔEuniv = 0
ΔEuniv = Δ Esys + ΔEsurr = 0
Δ Esys = -ΔEsurr
• Whatever energy the system loses (ΔEsys < 0), the
surrounding gains (ΔEsurr < 0), and vice versa.
universe
system
surrounding

10. Internal Energy (U)
• Energy contained within the system
• Includes
• energy due to the microscopic movements of the particles (kinetic energy)
• potential energy of the particles
• chemical energy
• Excludes KE and PE of the system as a whole.
• Esys,total = KE + PE + U
• In most cases, we are interested in systems that are at rest, and no external fields
will be present (KE=0, PE=0). We will be mostly considering internal energy.
• Esys,total = U

11. 11
• Types of molecular motions that contributes to the internal energy of a system
• a) translational
• b) rotational
• c) vibrational

12. ΔU = q + w
• The change in the energy can be measured by the work they do or
heat they release/absorb.
• ΔU = q + w q: heat w: work
q > 0 heat flows into system
w > 0 surroundings do work on system
q < 0 heat flows out of system
w < 0 system does work on surroundings

13. • Calculate ∆E for the combustion of a gas that releases 210.0 kJ of heat
to its surroundings and does 65.5 kJ of work on its surroundings.

14. • Calculate ∆E when the system absorbs 85.0 kJ of heat and the
surroundings do 33.5 kJ of work on the system.

15. Internal energy is a state function
• Internal energy is a state function:
• It only depends on the energy the system at the initial and final stage
• It does NOT depend on the pathway or sequence of events
• ΔU = Ufinal – Uinitial
Whether the skiers at the base of
the mountain
1) ride a ski lift
2) hike up a mountain trail,
their increase in altitude is the
same
= “Altitude” is a state function

16. Same ΔE for both paths:
ΔE = Edead battery – Echarged battery
w = 0
q ≠ 0
w ≠ 0
q ≠ 0
Work and heat are not state
functions.
They depend on the path.
They are “path functions.”

17. PV work (Expansion of Gases)
1) Constant external pressure (Pex)
w = -Pex ΔV
• When volume increases, work is negative (ΔU<0).
 The system does work on the surroundings.
• When volume decreases, work is positive (ΔU>0).
 The surroundings do work on the system.
2) Variable external pressure (Pex)
𝑤 = −
𝑉1
𝑉2
𝑃𝒆𝒙 𝑑𝑉
ΔV
P

18. PV work (Expansion of Gases)
• For a reversible system,
• 𝑤 = − 𝑉1
𝑉2
𝑃𝒆𝒙 𝑑𝑉 = − 𝑉1
𝑉2
𝑃𝒊𝒏 𝑑𝑉
• For ideal gases, PV = nRT
• 𝑃𝑖𝑛 =
𝑛𝑅𝑇
𝑉
• 𝑤 = − 𝑉1
𝑉2
𝑃𝒊𝒏 𝑑𝑉 = − 𝑉1
𝑉2 𝑛𝑅𝑇
𝑉
𝑑𝑉 = −𝑛𝑅𝑇 ln
𝑉2
𝑉1
= −𝑛𝑅𝑇 ln
𝑃1
𝑃2
18

19. Reversible Process
• In a reversible process, the changes of the system happens in infinitesimal
amounts.
• The system and the properties of the system are always close to equilibrium.
• When reversed, properties are reversed to same value.
• When a system is reversible, Pex and Pin are in equilibrium.
• Work calculated assuming reversibility
= maximum amount of work
• (when not reversible, loss of energy)
• In reality, processes are not reversible. However, calculations based on
reversibility gives the limit for maximum work possible.
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20. 20

21. 21

22. 22

23. Heat
• Energy transferred between objects with different temperature.
• Energy is transferred from the hotter object to the colder object.

24. Heat Capacity
• Energy needed to raise the energy of an object by 1˚C.
• If the heat capacity is higher, you need more energy to raise the
temperature of the object.
The heat capacity of
the beach is lower
than the water, so the
beach gets hotter and
colder faster.

25. Heat Capacity
• Heat capacity (C, J K-1)
• Energy needed to raise the temperature of the whole object by 1°C
• q = C ΔT
• Specific heat (s, J g-1 K-1)
• Energy required to raise the temperature of 1 g of a substance by 1°C
• q = ms ΔT
• Molar heat capacity ( 𝑪 , J mol-1 K-1)
• Quantity of energy required to raise the temperature of 1 mol of a substance
by 1°C
• q = n 𝐂 ΔT
• C = ms C = n C

26. Water: high heat capacity
“heat sink”
Due to hydrogen boning

27. • If a piece of copper wire with a mass of 20.9 g changes in
temperature from 25.00 to 28.00˚C, how much heat has it absorbed?
(specific heat of copper = 0.387 J g-1 K-1)

28. • Calculate the heat needed to heat 2 g of water from 10˚C to 30˚C.
• Specific heat of water = 4.18 J/(g∙K)

29. Enthalpy (H)
• @ constant P
• ΔU = q + w = qP – PΔV (From PV work, w = – PexΔV)
• qP = ΔU + PΔV
• Define Enthalpy H ≡ U + PV
• ΔH = ΔU + Δ(PV) = ΔU + PΔV = qP
• Enthalpy change refers to heat at constant pressure!
• Many chemical reactions occur in constant pressure.
• We use ΔH to indicate heat of reactions @ constant P.
(Appendix 2 p651)
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30. ΔU vs. ΔH
ΔH = ΔU + Δ(PV)
ΔU = ΔH − Δ(PV)
• Internal energy includes the work.
• At constant T, Δ(PV) = RTΔn can be used.
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31. 31
#1 method

32. 32
#2 method

33. @ constant V?
• ΔU = qV + w = qV − PΔV = qV
• Internal Energy Change refers to heat at constant volume.
33

34. Heat Capacities
q = C ΔT 𝑪 =
𝒒
∆𝑻
𝑪 =
𝑪
𝒏
=
𝒒
𝒏∆𝑻
34
@ constant V
ΔU = qV
𝐶 𝑉 =
𝑞 𝑉
∆𝑇
=
∆𝑈
∆𝑇
infinitesimal: 𝐶 𝑉 =
𝜕𝑈
𝜕𝑇 𝑉
𝑑𝑈 = 𝐶 𝑉 𝑑𝑇
∆𝑈 =
𝑇1
𝑇2
𝐶 𝑉 𝑑𝑇
@ constant P
ΔH = qP
𝐶 𝑃 =
𝑞 𝑃
∆𝑇
=
∆𝐻
∆𝑇
infinitesimal: 𝐶 𝑃 =
𝜕𝐻
𝜕𝑇 𝑃
𝑑𝐻 = 𝐶 𝑃 𝑑𝑇
∆𝐻 =
𝑇1
𝑇2
𝐶 𝑃 𝑑𝑇

35. 35

36. Exothermic vs. Endothermic
H
Hi
Hf
ΔH = Hf - Hi < 0
Enthalpy of system decreases
while heat is released
to surroundings
Exothermic
progress of change
H
Hi
Hf
ΔH = Hf - Hi > 0
Enthalpy of system increases
by absorbing heat
from surroundings
Endothermic
progress of change

37. Enthalpy Change for Phase Changes
ΔHfusion
H
Hi
Hf
ΔH = Hf - Hi > 0
endothermic
progress of change
solid
liquid
fusion: solid → liquid
ΔHvaporization
H
Hi
Hf
ΔH = Hf - Hi > 0
endothermic
progress of change
liquid
gas
vaporization: liquid → gas
ΔHsublimation
H
Hi
Hf
ΔH = Hf - Hi > 0
endothermic
progress of change
solid
gas
sublimation: solid → gas
ΔHfusion
ΔHvaporization
ΔHsublimation

38. Enthalpy Change for Chemical Reactions
H
6 CO2(g), 6 H2O (g)
Hf
ΔH = Hf - Hi > 0
endothermic
Photosynthesis
6 CO2(g) + 6 H2O (g) → C6H12O6 (g) + 6 O2(g)
progress of change
Hi
C6H12O6 (g) + 6 O2(g)
ΔHrxn
H
N2(g), 3 H2 (g)
Hf
ΔH = Hf - Hi < 0
exothermic
N2(g) + 3 H2 (g) → 2 NH3 (g)
progress of change
Hi
2 NH3 (g)
ΔHrxn

39. Combustion
• Reaction of fuel with oxygen (O2) to form CO2 and H2O to generate
energy.
• Exothermic
• CxHy + O2 → a CO2 + b H2O
• When hydrogen is fuel, H2 + ½ O2 → H2O

40. Constant Volume Calorimetry
• Bomb calorimeter:
• a constant-volume device used to measure
the energy released during a combustion
reaction
• Measure Ti and Tf of water before and after the
reaction
• Heat capacity of water and all the other
insulated components of the calorimeter = Ccal
determined from known compound
insulated

41. Bomb Calorimetry
qrxn = - qcalorimeter
qcalorimeter = Ccal ΔT
(system = reactants and products)
• at constant V,
qrxn = ΔU
• Combining,
ΔU = - Ccal ΔT
insulated(1)
(2)
(3)

42. Bomb Calorimetry
• From ΔU = - Ccal ΔT ,
• H ≡ U + PV
• ΔH = ΔU + Δ(PV)
• or ΔH = ΔU + RT Δn
• Therefore, ΔH can be estimated.
• Also, ΔU = n ΔUmolar and ΔH = n ΔHmolar
insulated

43. 43
qrxn = - qcalorimeter
qcalorimeter = Ccal ΔT
qrxn = ΔU
ΔH = ΔU + RT Δn
ΔU = n ΔUmolar
ΔH = n ΔHmolar

44. 44

45. Constant-Pressure Calorimetry
• Many physical and chemical processes
take place under atmospheric
conditions (constant pressure).
• At constant pressure, the enthalpy
change corresponds to heat.
qrxn = - qsolution
qsolution = mcsolution ΔT
qrxn = ΔH
45
A simple constant-pressure
calorimeter known as the
coffee-cup calorimeter.

46. Thermochemical Equations
• Enthalpy of Reaction (ΔHrxn)
• Energy/heat absorbed/released during a chemical reaction for a given
amount of reactants
H
Hreactants
Hproducts
progress of change
ΔHrxn
E.g., thermochemical equation of :
complete combustion of butane
=> means that when 2 moles of butane reacts with 13 moles of
oxygen, 5754 kJ of heat is released.
=> half the amount

47. Thermochemical Equations
• “Standard” Enthalpy of Reaction (ΔHrxn˚ )
• Enthalpy of reaction at 1 bar (generally) or 1 M for solutions

48. Thermochemical Equations
If the coefficients of a reaction are multiplied by an integer, H is
multiplied by the same integer.
2NO(g) → N2(g) + O2(g) H = –180 kJ
6NO(g) → 3N2(g) + 3O2(g) H = 3(–180 kJ) = –540 kJ

49. Thermochemical Equations
If a reaction is reversed, the sign of H changes.
N2(g) + O2(g) → 2NO(g) H = 180 kJ
2NO(g) → N2(g) + O2(g) H = –180 kJ

50. Hess’s Law
• The Hrxn for a reaction that is the sum of two or more reactions is
equal to the sum of the Hrxn values of the constituent reactions.
x + a → b ΔH1
b + y → a + z ΔH2
----------------------
x + y → z ΔH = ΔH1 + ΔH2

51. Hess’s Law
C (s) + ½ O2 (g) → CO (g) ΔH1˚ = ?
C (s) + ½ O2 (g) → CO (g) ΔH1˚
CO (g) + ½ O2 (g) → CO2 (g) ΔH2˚ = -283.0 kJ
-------------------------------------------------------------------------
C (s) + O2 (g) → CO2 (g) ΔH3˚ = -393.5 kJ
ΔH3˚ = ΔH1˚ + ΔH2˚
Therefore, ΔH1˚ = ΔH3˚ − ΔH2˚ = -393.5 kJ – (-283.0 kJ) = -110.5 kJ

52. Hess’s Law
(1) C (s) + ½ O2 (g) → CO (g) ΔH1˚ = ?
rearrange the following equations to obtain eqn (1)
(2) CO (g) + ½ O2 (g) → CO2 (g) ΔH2˚ = -283.0 kJ
(3) C (s) + O2 (g) → CO2 (g) ΔH3˚ = -393.5 kJ
-1x(2): CO2 (g) → CO (g) + ½ O2 (g) - ΔH2˚ = 283.0 kJ
(3): C (s) + O2 (g) → CO2 (g) ΔH3˚ = -393.5 kJ
-------------------------------------------------------------------------
C (s) + ½ O2 (g) → CO (g) ΔH1˚ = - ΔH2˚ + ΔH3˚

53. 53

54. Standard Enthalpy of Formation Hf
o
The enthalpy change for the process of
forming 1 mol of a substance
from its constituent elements
in their standard states (1 bar, 25˚C, in its most stable form)
• e.g., formation reaction for NO:
• N2(g) + O2(g)  2 NO(g) Hrxn ≠ Hf
o (NO) not formation reaction
• ½ N2(g) + ½ O2(g)  NO(g) Hrxn = Hf
o (NO) formation reaction
• Hf
o values are listed in the appendix of the book.

55. Standard Enthalpy of Formation Hf
o
not formation reactions
formation reaction

56. Standard Enthalpy of Formation Hf
o
The standard enthalpy of formation
for an element
at its standard state
in its most stable form
is zero.
• e.g., Hf
o = 0 for O2(g), H2(g), Fe(s), S(s), I2(s) …

57. Hf
o and Hrxn
o
Formation reaction of H2O2
Formation reaction of H2O
Applying Hess’s law to the formation reaction
Hrxn
o = Hf
o(H2O) + ½ Hf
o(O2) - Hf
o(H2O2)

58. Hf
o and Hrxn
o
• Absolute values of enthalpy of substances are unknown.
• Standard enthalpies of formation are useful because they provide a
convenient method for applying Hess’s law without having to
manipulate thermochemical equations.
Hrxn
o = [sum of Hf
o of all the products]
– [sum of Hf
o of all the reactants]
ΔHrxn = ∑npΔHf (products) - ∑ nrΔHf (reactants)

60. Bond Energy
Potential energy curve of H2
60
Attraction
As the bond is formed
with shorter distance,
PE decreases
Repulsion
As the distance gets
shorter, coulombic
repulsion results in
higher PE
Bond
Energy
Bond
Distance

61. Bond Energy
• Bond Energy
• Energy change that occurs when 1 mol of bonds in the gas phase are broken
• A measure of the bond strength: the larger the bond energy, the stronger the
bond.
• Breaking a bond is a highly endothermic process (ΔH>0)
• Bond formation is a highly exothermic process (ΔH<0)
61
O2

62. Bond Energy
• Bond “enthalpies” are not exactly the
same as bond “energies”. However, bond
enthalpies are used for practical reasons.
• Bond enthalpies are different for the
bonds in different molecules and
environment. Therefore, average values
are used for bonds between different
atoms.
62

63. Estimating Hrxn from Bond Energies
63
Bond
Breaking
Bond
Forming
Overall
Hrxn = ∑ Hbond breaking - ∑ Hbond forming
= 4xH(C-H) + 2xH(O=O)
– [2xH(C=O) + 4xH(O-H)]
Hrxn = ∑ Hbond breaking - ∑ Hbond forming

64. Estimating Hrxn at Different Temperatures
64
at T1
at T2
H
Reactants Products
ΔH = CprΔT
ΔH = CppΔT
ΔrxnH1
ΔrxnH2
H1r
H1p
H2r
H2p
ΔrxnH1 = H1p – H1r
ΔrxnH2 = H2p – H2r
= (H1p + CppΔT ) – (H1r + CprΔT )
= H1p – H1r +(Cpp - Cpr) ΔT
= ΔrxnH1 + ΔCp ΔT
ΔrxnH2 - ΔrxnH1 = ΔCp (T2 – T1) when Cp const
or ΔrxnH2 - ΔrxnH1 = 𝑇1
𝑇2
∆𝐶 𝑝 𝑑𝑇
Kirchhoff’s law

65. 65