Question

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
$CH_{4}(g)+H_{2}O(g) \rightleftharpoons CO(g)+3H_{2}(g)$
(a) Write an expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure ?
(ii) increasing the temperature ?
(iii) using a catalyst ?

Answer 1

a) The given reaction is as follows,

$C{ H }_{ 4 }(g)\quad +\quad { H }_{ 2 }O(g)\quad \rightleftharpoons \quad CO(g)\quad +\quad 3{ H }_{ 2 }(g)\\ The\quad expression\quad { K }_{ P }\quad =\quad \frac { { P }_{ CO }\quad \times \quad { P }_{ { H }_{ 2 } }^{ 3 } }{ { P }_{ C{ H }_{ 4 } }\quad \times \quad { P }_{ { H }_{ 2 }O } }$

b) (i) On increasing the pressure ${ K }_{ P }$ unchanged. According to Le Chatelier's principle equilibrium will shifted to the backward direction.

(ii) Increase in the temperature, increases the value of endothermic reaction. According to Le Chatelier's principle, the equilibrium shifted will shifted to forward direction.

(iii) Equilibrium does not existed in the reaction but equilibrium is attained. If we use the catalyst, the equilibrium is attained quickly.

Answer 2

(i) On increasing the pressure { K }_{ P }KP​unchanged. According to Le Chatelier's principle equilibrium will shifted to the backward direction.

(ii) Increase in the temperature, increases the value of endothermic reaction. According to Le Chatelier's principle, the equilibrium shifted will shifted to forward direction.

(iii) Equilibrium does not existed in the reaction but equilibrium is attained. If we use the catalyst, the equilibrium is attained quickly.