Question

Question 7.14 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass)
reacts with CO according to the equation,

H2O(g) + CO(g) <---> H2(g) + CO2(g)

Calculate the equilibrium constant for the reaction.

Class XI Equilibrium Page 225

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: H_2O(g)+CO(g) < = > H_2(g)+ CO_2(g) \\ at \: \: t = 0 \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ at \: eqlm \: (1 - x) \: \: \: \: \: \: \: (1 - x) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x$
A/C to question,
H2O reacted = 40% of 1 mol of H2O = 0.4 mol
x = 0.4 mol
(1 - x ) = 1 - 0.4 = 0.6 mol

Kc = [H2][CO2]/[H2O][CO]
= x . x/(1 - x ).(1 - x )
= x²/(1 - x)²
= (0.4)²/(0.6)²
= 0.444

Hence, equilibrium constant = 0.444

Answer 2

Hi

Here is your answer,

             H₂O9(g) + CO ⇆ H₂(g) + CO₂(g)

Initial conc. 1          1         0            0

Equilibrium conc. (1-x)   (1-x)   x   x  

                  H₂O reacted = 40% of 1 molecular of H₂O == 0.4 mol


      (1-x) = 1 - 0.4 = 0.6 mol 

                Kc = [H₂] [CO₂] /[H₂O] [CO] 

                 = 0.4 × 0.4 / 0.6 × 0.6 

                = 0.444 



Hope it helps you !