Question

1. The water-gas-shift reaction, CO(g) + H2O(g)  CO2(g) + H2(g) is carried out with the reactants consists of 2 mol of H2O vapor and 1 mol of CO. The temperature is 1,100 K and the pressure is 1 bar. 1). Find out the reaction constant K from the Figure13.2 on the textbook based on the condition given. (5 points) 2). Calculate the fraction of steam reacted. Assume the mixture behaves as an ideal gas. (10 points)

Answer 1

Answer:

Answer (a) K=1

            (b) 1/3  fraction of steam reacted

Explanation:

CO  + H20 → CO2 + H2

t=0   (1/v)   (2/v)

t= teq   (1-a/v)   (2-a/v)   (a/v)  (a/v)

k= (a/v)²/ (1-a/v)(2-a/v)

1 = a²/(1-a)(2-a)

2-a - 2a + a²= a²

2 = 3a

a = 2/3

Moles of H2O reacted ; a=2/3

Fraction of H2O reacted ; 2/3

                                             2                  = 1/3

1/3 fraction of H20 reacted

Answer 2

The equilibrium constant of this reaction is K = 1 and the fraction of moles that reacted is 1/3 of the steam that participated in the reaction.

Equilibrium Constant

From the equation of reaction;

CO + H$_2$O ⇒ CO$_2$ + H$_2$

but

t = 0   (1/v)   (2/v)

and

t = t$_e_q$ = $(\frac{1-x}{v})(\frac{2-x}{v})(\frac{x}{v})(\frac{x}{v})$

K =

$\frac{(\frac{x}{v}^2 )}{(\frac{1-x}{v})(\frac{2-x}{v} ) } \\1 = \frac{x^2}{(1-x)(2-x)}$

This shows that the equilibrium constant k = 1

b)

The fraction of steam that reacted is

$1 = \frac{x^2}{(1-x)(2-x)} \\2-x-2x+x^2=x^2\\2x=3x\\x= 2/3$

The number of moles that reacted is 2/3;

The fraction of H$_2$O that reacted is

$\frac{2}{3}/2 =\frac{1}{3}$

The fraction of moles that reacted is 1/3 of the steam in the reaction.

From the calculations above, we come to the conclusions that

• Equilibrium Constant K = 1
• Fraction of moles of steam that reacted = 1/3

Learn more on equilibrium constant here;

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