Chemistry 3411 Solutions Worksheet - Georgia Institute Of Technology - 2010 Page 6
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12CHEM 3411, Fall 2010
Solution Set 2
3
Exercise 2.24a pg 87
Given
⊖
(g) → 2CO
1
For the reaction C
H
OH(l) + 3O
(g) + 3H
O(g), ∆
U
=
1373 kJ mol
at T = 298 K.
2
5
2
2
2
r
In terms of given variables, this is written:
⊖
1
∆
U
=
1373 kJ mol
r
T = 298 K
Find
⊖
Calculate ∆
H
of this reaction.
r
Strategy
The following equation relates the enthalpy change of a reaction to the internal energy change (Equation 2.21 on pg
58 in your text book).
∆H = ∆U + ∆n
RT
g
where ∆n
RT is the change in quantity of gas molecules of the reaction.
g
In analyzing this reaction
(g) → 2CO
C
H
OH(l) + 3O
(g) + 3H
O(g)
2
5
2
2
2
we see that 5 moles of gas are produced (2 moles of CO
(g) and 3 of H
O(g)) while 3 are consumed (O
(g)). Therefore
2
2
2
⊖
∆n
= 2 and ∆
H
is easily calculated as
g
r
⊖
⊖
∆
H
= ∆
U
+ ∆n
RT
r
r
g
2 $ $ $ $
$
$ $ $
× 8.314 J ¨ ¨ ¨
mol gas
1373 × 10
× 298 & & K
3
1
1
1
=
J mol
+
K
mol
1 mol reaction
1.36804 × 10
6
1
=
J mol
1
=
1368.04 kJ mol
Solution
⊖
1
∆
H
= 1368 kJ mol
r
6
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