CHE 214 Thermodynamics II
Chemical-Reaction Equilibria
Example 12 Solution
March 28th, 2025
Equilibrium Conversions for Single Reactions
Suppose a single reaction occurs in a homogeneous system, and suppose the equilibrium
constant is known. In this event, the calculation of the phase composition at equilibrium is
straight-forward if the phase is assumed an ideal gas or an ideal solution. When an
assumption of ideality is not reasonable, the problem is still tractable for gas-phase reactions
through application of an equation of state and solution by computer. At equilibrium, there
can be no tendency for change to occur, either by mass transfer between phases or by
chemical reaction.
Example 13.5
The water gas-shift reaction
𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑔) → 𝐶𝑂2 (𝑔) + 𝐻2 (𝑔)
is carried out under the different sets of conditions described below. Calculate the fraction
of steam reacted in each case. Assume the mixture behaves as an ideal gas.
(a) The reactants consist of 1 mol of 𝐻2 𝑂 vapor and 1 mol of 𝐶𝑂. The temperature is 1,100 𝐾
and the pressure is 1 bar.
(b) Same as (a) except that the pressure is 10 bar.
(c) Same as (a) except that 2 mol of 𝑁2 is included in the reactants.
(d) The reactants are 2 mol of 𝐻2 𝑂 and 1 mol of 𝐶𝑂. Other conditions are the same as in (a).
(e) The reactants are 1 mol of 𝐻2 𝑂 and 2 mol of 𝐶𝑂. Other conditions are the same as in (a).
(f) The initial mixture consists of 1 mol of 𝐻2 𝑂, 1 mol of 𝐶𝑂 and 1 mol of 𝐶𝑂2. Other
conditions are the same as in (a).
(g) Same as (a) except that the temperature is 1,650 𝐾.
Solution
(a) Because the reaction mixture is an ideal gas,
∏(𝑦𝑖
) 𝜈𝑖
𝑖
𝑦𝐻 𝑦𝐶𝑂2
𝑃 −𝜈
1 0
= ( °) 𝐾 = ( ) 𝐾 = 𝐾 = 2
𝑃
1
𝑦𝐶𝑂 𝑦𝐻2 𝑂
104
At 𝑇 = 1,100 𝐾, 𝑇 = 9.05
ln 𝐾 = 0, 𝐾 = 1
𝑦𝐻2 𝑦𝐶𝑂2
=1
𝑦𝐶𝑂 𝑦𝐻2 𝑂
𝑦𝐶𝑂2 =
𝜀𝑒
2
;
𝑦𝐻2 =
𝜀𝑒
2
;
𝑦𝐻2 𝑂 =
1 − 𝜀𝑒
2
;
𝑦𝐶𝑂 =
1 − 𝜀𝑒
2
Page 1 of 2
CHE 214 Thermodynamics II
Chemical-Reaction Equilibria
Example 12 Solution
March 28th, 2025
𝜀𝑒 𝜀𝑒
=1
(1 − 𝜀𝑒 )(1 − 𝜀𝑒 )
𝜀𝑒 = 0.5
(b) The increase in pressure has no effect on the ideal-gas reaction.
(c) 𝑁2 doesn’t participate in the reaction. Hence, no effects on the equilibrium reaction
coordinate.
(d) In this case,
𝑦𝐶𝑂2 =
𝜀𝑒
3
;
𝑦𝐻2 =
𝜀𝑒
3
;
𝑦𝐻2 𝑂 =
2 − 𝜀𝑒
3
;
𝑦𝐶𝑂 =
1 − 𝜀𝑒
3
𝜀𝑒 𝜀𝑒
=1
(2 − 𝜀𝑒 )(1 − 𝜀𝑒 )
𝜀𝑒 = 0.667
(e) Same as (d).
(f) In this case
𝜀𝑒 (1 + 𝜀𝑒 )
=1
(1 − 𝜀𝑒 )(1 − 𝜀𝑒 )
𝜀𝑒 = 0.333
104
(g) At 𝑇 = 1,650 𝐾, 𝑇 = 6.06
ln 𝐾 = −1.15,
𝐾 = 0.316
𝜀𝑒 𝜀𝑒
= 0.316
(1 − 𝜀𝑒 )(1 − 𝜀𝑒 )
𝜀𝑒 = 0.36
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