19

To solve this problem, we require several pieces of data:

Δvap H−o (T = 373.15K ) = 40.656 k J/m ol [Note: this enthalpy change is at the transition T]
Δfus H−o (T = 273.15K ) = 6.008 k J/m ol
C−p,m
o
(H2O(l)) = 75.291 J K −1 m ol −1
C−p,m
o
(H2O(g)) = 33.58 J K −1 m ol −1
We will assume that the heat capacities are constant over the temperature range of interest.

Therefore, we have the following enthalpy changes for the three steps above:
Step 1: Heating liquid water from T = 273.15 K to 373.15 K
Δ1H = C−po (H2O(l))ΔT = nC−p,m
o
(H2O(l))ΔT
= (1 m ol )(75.291 J K m ol −1)(373.15K − 273.15K ) = 7529.1 J = 7.5291 k J
−1

Step 2: Vaporization at 373.15 K

Δ2 H = nΔvap H−o (T = 373.15K ) = (1 m ol )(40.656 k J/m ol ) = 40.656 k J

Step 3: Cooling gaseous water from T = 373.15 K to 273.15 K

Δ3 H = C−po (H2O(g))ΔT = nC−p,m
o
(H2O(g))ΔT
= (1 m ol )(33.58 J K m ol −1)(273.15K − 373.15K ) = − 3358 J = − 3.358 k J
−1

Thus,
Δvap H−o (T = 273.15K ) = Δ1H + Δ2 H + Δ3 H = 7.5192 k J + 40.656 k J − 3.358 k J = 44.83 k J

In general, we can summarize this process, i.e., the relationship between enthalpy change at one
temperature to the enthalpy change at another temperature, through Kirchoff’s Law:
T2

∫T
−
o −
o
ΔrC−po dT where ΔrC−po = νC−p,m
o
νC−p,m
o
∑ ∑
Δr H (T2) = Δr H (T1) + −
1 products reactants

How about sublimation at T = 273.15K?

1 mol H2O(s)
1 mol H2O(g)

273.15K, 1 bar 273.15 K, 1 bar

Step 1: Fusion at T = Step 2: Vaporization

273.15 K at T = 273.15 K
1 mol H2O(l)

273.15K, 1 bar
 20

Δsub H−o (T = 273.15K ) = Δ1H + Δ2 H

Step 1: Fusion at 273.15 K. This is the standard transition temperature, therefore

Δ1H = nΔfus H−o (T = 273.15K ) = (1 m ol )(6.008 k J/m ol ) = 6.008 k J

Step 2: Vaporization at 273.15 K. We just found this result in the previous example!
Δ2 H = nΔvap H−o (T = 273.15K ) = (1 m ol )(44.827 k J/m ol ) = 44.827 k J

Therefore, Δsub H−o (T = 273.15K ) = Δ1H + Δ2 H = 6.008 k J + 44.827 k J = 50.84 k J

For 1 mol, Δsub H−o (T = 273.15K ) = 50.84 k J m ol −1

What is the relationship between change in enthalpy, ΔH, and change in internal energy, ΔU?
See textbook 2B.1(b)
Recall: H = U + pV so U = H − pV
Therefore, if we want to know the relationships between changes, we have
ΔU = ΔH − Δ( pV ) = ΔH − pΔV (as pressure is constant)
We must determine the corresponding volume change,.

If we want to determine this relationship for the same problem above (water), we need the
volume for all three phases. Useful data, we need:
Densities: ρH2O(s) = 0.917 g cm −3 and ρH2O(l) = 0.99984 g cm −3 = 1.000 g cm −3
Mass of water: MH2O = 18.02 g m ol −1

Volume for the three phases:

(i) gas: VH2O(g) = 22.70 d m 3 (from the perfect gas law)
18.02 g m ol −1 × 1 m ol
(ii) liquid: VH2O(l) = = 18.02 cm 3 = 0.018 d m 3
1.000 g m ol −1
18.02 g m ol −1 × 1 m ol
(iii) solid: VH2O(s) = = 19.6 cm 3 = 0.020 d m 3
0.917 g m ol −1
Once we know the corresponding volumes, we can determine the change in internal energy for
each process, using the enthalpy changes we had determined previously.
(i) Vaporization at 273.15K:
ΔvapU = Δvap H − p(V(g) − V(l))
= 44.83k J − (1 bar)(22.70 − 0.018) d m 3

( bar d m 3 )
0.1k J
= 44.83k J − 22.68 bar d m 3
= 42.56 k J
(ii) Fusion at 273.15K:
ΔfusU = Δfus H − p(V(l) − V(s))
= 6.008 k J − (1 bar)(0.018 − 0.020) d m 3
 21

( bar d m 3 )
0.1k J
= 6.008 k J − 0.002 bar d m 3
= 6.008 k J
(iii) Sublimation at 273.15K:
ΔsubU = Δsub H − p(V(g) − V(s))
= 50.84 k J − (1 bar)(22.70 − 0.020) d m 3

( bar d m 3 )
0.1k J
= 50.84 k J − 22.68 bar d m 3
= 48.57 k J

Conclusion: the difference between ΔH and ΔU is significant if the number of moles of gas
changes in the process. We can quantify this difference as
ΔH = ΔU + Δ( pV ) = ΔU + Δngas RT

Problem #1 (an example of a problem that might appear on an exam): Determine Δr H−o for the
reaction:

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O(l) at 95 oC.
For compound C6H12O6 (s), Δf H−o = − 1274.0 k J m ol −1 and C− o
p,m = 218.87 J K
−1
m ol −1

Where can you find the other data? On the data sheet!
O2 (g), Δf H−o = 0 k J m ol −1 and C−
o
p,m = 29.355 J K
−1
m ol −1
CO2 (g), Δf H−o = − 393.5 k J m ol −1 and C− o
p,m = 37.11 J K
−1
m ol −1
H2O (l), Δf H−o = − 285.8 k J m ol −1 and C− o
p,m = 75.291 J K
−1
m ol −1

C6H12O6 (s) + 6O2 (g)

6CO2 (g) + 6H2O(l)

368.15 K, 1 bar 368.15 K, 1 bar

Step 1: Cool down Step 3: Heat up

reactants products
Step 2: Reaction @
298.15K

C6H12O6 (s) + 6O2 (g)

6CO2 (g) + 6H2O(l)

298.15 K, 1 bar 298.15 K, 1 bar

Δr H−o (T = 368.15K ) = Δ1H + Δ2 H + Δ3 H

Use Kirchoff’s Law:

 22

T2

∫T
−
o −
o
ΔrC−po dT where ΔrC−po = νC−p,m
o
νC−p,m
o
∑ ∑
Δr H (T2) = Δr H (T1) + −
1 products reactants
We will assume that the heat capacities are constant over the temperature range of interest.
T2 = 368.15 K and T1 = 298.15 K, therefore reaction at 298.15 K:
Δr H−o (T = 298.15K ) = νΔf H−o − νΔf H−o
∑ ∑
products reactants
= νΔf H−o (CO2(g)) + νΔf H−o (H2O(l )) − νΔf H−o (C6 H12O6(s)) − νΔf H−o (O2(g))
= (6 m ol )(−393.5 k J m ol −1) + (6 m ol )(−285.8 k J m ol −1) − (1 m ol )(−1274.0 k J m ol −1) − (6 m ol )(0 k J m ol −1)
= − 2801.8 k J

Enthalpy change due to heat capacities

ΔrC−po = νC−p,m
o
νC−p,m
o
∑ ∑
−
products reactants
= −
o
nCp,m(CO2(g)) + −
o
nCp,m(H2O(l )) − [nC−p,m
o
(C6 H12O6(s)) + nC−p,m
o
(O2(g))]
= (6 m ol )(37.11 J K −1 m ol −1) + (6 m ol )(75.291 J K −1 m ol −1) − [(1 m ol )(218.87 J K −1 m ol −1) + (6 m ol )(29.355 J K −1 m ol −1)]
= 279.406 J K −1
T2

∫T
ΔrC−po dT = ΔrC−po (T2 − T1) = 279.406 J K −1(368.15 K − 298.15 ) = 19.6 k J
1

Therefore,
T2

∫T
−
o −
o
Δr H (T2) = Δr H (T1) + ΔrC−po dT = − 2801.8 k J + 19.6 k J = − 2782.2 k J
1

Alternatively, we can explicitly consider the three steps in the process, i.e.,
Step 1: Cooling the reactants [C6H12O6 (s) + 6O2 (g)] from T = 368.15 K to 298.15 K
Δ1H = nC−p,m
o
(C6 H12O6(s))ΔT + nC−p,m
o
(O2(g))ΔT
= (1 m ol )(218.87 J K −1 m ol −1)(298.15.15K − 368.15K ) + (6 m ol )(29.355 J K −1 m ol −1)(298.15.15K − 368.15K )
= − 27.65 k J

Step 2: Reaction at the standard temperature of 298.15 K

Δ2 H−o = νΔf H−o − νΔf H−o
∑ ∑
products reactants
= νΔf H−o (CO2(g)) + νΔf H−o (H2O(l )) − νΔf H−o (C6 H12O6(s)) − νΔf H−o (O2(g))
= (6 m ol )(−393.5 k J m ol −1) + (6 m ol )(−285.8 k J m ol −1) − (1 m ol )(−1274.0 k J m ol −1) − (6 m ol )(0 k J m ol −1)
= − 2801.8 k J

Step 3: Heating the products [6CO2 (g) + 6H2O(l) ] from T = 298.15 K to 368.15 K
Δ3 H = nC−p,m
o
(CO2(g))ΔT + nC−p,m
o
(H2O(l ))ΔT
= (6 m ol )(37.11 J K −1 m ol −1)(368.15.15K − 298.15K ) + (6 m ol )(75.291 J K −1 m ol −1)(368.15.15K − 298.15K )
= 47.21 k J
 23

Thus,
Δr H−o (T = 368.15K ) = Δ1H + Δ2 H + Δ3 H = − 27.65 k J + (−2801.8) k J + 47.21 k J = − 2782.2 k J

This is a difference of less than 1% from the value at 298.15 K; hence, the approximation from
CHEM 102/105 that Δr H−o is T-independent would be pretty good for this reaction over this
temperature range.