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NVAMediumJEE 2023Equilibrium Basics

JEE Chemistry 2023 Question with Solution

Water decomposes at 2300 K2300 \, \text{K}:

2 H2O(g)⇌2 H2(g)+O2(g)2 \, \text{H}_2\text{O}(g) \rightleftharpoons 2 \, \text{H}_2(g) + \text{O}_2(g)

The percent of water decomposing at 2300 K2300 \, \text{K} and 1 bar1 \, \text{bar} is _____ (Nearest integer). Equilibrium constant for the reaction is 2×10−32 \times 10^{-3} at 2300 K2300 \, \text{K}.

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: Water decomposes at 2300 K2300 \, \text{K} and 1 bar1 \, \text{bar} with equilibrium constant Kp=2×10−3K_p = 2 \times 10^{-3}.

Find: The percent of water decomposed.

Write the reaction per mole of water as

H2O(g)⇌H2(g)+12 O2(g)\text{H}_2\text{O}(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2} \, \text{O}_2(g)

Let the degree of decomposition be α\alpha.

At equilibrium, the partial pressures are:

P0(1−α) for H2O,P0α for H2,P0α2 for O2P_0(1-\alpha) \text{ for } \text{H}_2\text{O}, \quad P_0\alpha \text{ for } \text{H}_2, \quad \frac{P_0\alpha}{2} \text{ for } \text{O}_2

The total pressure at equilibrium is

P0(1+α2)=1P_0\left(1 + \frac{\alpha}{2}\right) = 1

The equilibrium constant expression is

Kp=(P0α)(P0α2)1/2P0(1−α)K_p = \frac{(P_0\alpha)\left(\frac{P_0\alpha}{2}\right)^{1/2}}{P_0(1-\alpha)}

Substituting Kp=2×10−3K_p = 2 \times 10^{-3},

αα21−α=2×10−3\frac{\alpha \sqrt{\frac{\alpha}{2}}}{1-\alpha} = 2 \times 10^{-3}

Since α\alpha is very small, use 1−α≈11-\alpha \approx 1. Then

αα2=2×10−3\alpha \sqrt{\frac{\alpha}{2}} = 2 \times 10^{-3}

So,

α3/2⋅12=2×10−3\alpha^{3/2} \cdot \frac{1}{\sqrt{2}} = 2 \times 10^{-3} α3/2=22×10−3\alpha^{3/2} = 2\sqrt{2} \times 10^{-3}

Raising both sides to the power 23\frac{2}{3},

α=(22×10−3)2/3\alpha = \left(2\sqrt{2} \times 10^{-3}\right)^{2/3} α=2×10−2\alpha = 2 \times 10^{-2}

Therefore, percentage decomposition is

%α=2%\%\alpha = 2\%

So the required nearest integer is 22.

Common mistakes

  • Using the original reaction 2 H2O⇌2 H2+O22 \, \text{H}_2\text{O} \rightleftharpoons 2 \, \text{H}_2 + \text{O}_2 but writing the equilibrium partial pressures without dividing consistently by stoichiometric ratio. This gives a wrong KpK_p expression. Write the reaction basis and the corresponding equilibrium pressures consistently.

  • Ignoring the denominator term 1−α1-\alpha without first checking that KpK_p is small. The approximation works here because Kp=2×10−3K_p = 2 \times 10^{-3} implies small decomposition. Always justify the small-α\alpha assumption.

  • Confusing degree of decomposition α\alpha with percentage decomposition. Here α=2×10−2\alpha = 2 \times 10^{-2}, so the percentage is α×100=2%\alpha \times 100 = 2\%. Convert to percent only at the final step.

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