Question

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,
$H_2O (g) + CO (g) ⇋ H_2 (g) + CO_2 (g)$
Calculate the equilibrium constant for the reaction.

Answer 1

The given reaction is:
$H_2O(g) + CO(g) ↔ H_2(g) + CO_2(g)$

Initial conc. $\frac {1}{10} M$ $\frac {1}{10} M$ 0 0

At equilibrium $\frac {1 - 0.4}{10} M$ $\frac {1 - 0.4}{10} M$ $\frac {0.4}{10}\ M$ $\frac {0.4}{10}\ M$

                       $= 0.06\ M$      $= 0.06\ M$  $= 0.06\ M$  $= 0.06\ M$  

Therefore, the equilibrium constant for the reaction,

$K_c = \frac {[H_2] [CO_2]}{[H_2O][CO]}$

$K_c= \frac {0.04 × 0.04}{0.06 × 0.06}$

$K_c= 0.444$ (approximately)