Q.Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4

Concept: Equilibrium Constant Calculation & Le Chatelier’s Principle

(a) For the reaction

$\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g)$,

the equilibrium constant in terms of partial pressures is:

(b) Effects on $K_p$ and composition:

• (i) Increasing pressure: $K_p$ is unchanged (depends only on temperature). The reaction has 4 moles of gas on the right and 2 on the left — higher pressure favours the side with fewer moles. Equilibrium shifts left, so CO and H$_2$ decrease, CH$_4$ and H$_2$O increase.

• (ii) Increasing temperature: The reaction is endothermic ($\Delta H > 0$). $K_p$ increases. Equilibrium shifts right to absorb heat, so CO and H$_2$ increase, CH$_4$ and H$_2$O decrease. …

Understanding the Reaction

We have the reaction:

It is endothermic (absorbs heat) and involves 4 moles of gas on the right vs. 2 moles on the left. This imbalance in moles and the heat effect are the keys to predicting how changes affect equilibrium.

(a) Expression for $K_p$

$K_p$ is the equilibrium constant expressed in terms of partial pressures of gases. For a general reaction:

the expression is:

Here, $P_X$ is the partial pressure of gas $X$ at equilibrium.

For our reaction:

(b) Effect of Changes on $K_p$ and Composition

We now analyse three changes. Remember: $K_p$ changes only with temperature, not with pressure or catalyst. The composition (position of equilibrium) shifts according to Le Chatelier’s principle.

1. Increasing the pressure

• Left side: 2 moles of gas (1 CH₄ + 1 H₂O)
• Right side: 4 moles of gas (1 CO + 3 H₂)

Increasing pressure favours the side with fewer moles, i.e., the left (reactants). So the equilibrium shifts backward, producing more CH₄ and H₂O, and less CO and H₂.

Effect on $K_p$: No change — $K_p$ is constant at a given temperature.

Effect on composition: The equilibrium mixture contains more reactants and less products than before.

2. Increasing the temperature

The reaction is endothermic (absorbs heat). Think of heat as a reactant:

$\text{CH}_4 + \text{H}_2\text{O} + \text{heat} \rightleftharpoons \text{CO} + 3\text{H}_2$ …