JEE Main 2025 Apr 8 Shift 2, Chemistry Q24: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2025Apr 8, Shift 2Chemistry

The equilibrium constant for decomposition of H $_{2}$ O(g)

$H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g)$ ( $Δ G^{\circ} = 92.34$ kJ mol $^{- 1}$ )

is $8.0 \times 1 0^{- 3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ( $α$ ) of water is _____ $\times 1 0^{- 2}$ (nearest integer value).

[Assume $α$ is negligible with respect to 1]

Your answer

Solution

Starting with 1 mole of H $_{2}$ O and degree of dissociation $α$ , at equilibrium the moles are: H $_{2}$ O $= 1 - α$ , H $_{2}$ $= α$ , O $_{2}$ $= α /2$ , and total moles $\approx 1$ since $α ≪ 1$ .

With total pressure $P = 1$ bar, the partial pressures approximate to $P_{H_{2}} \approx α$ , $P_{O_{2}} \approx α /2$ , and $P_{H_{2} O} \approx 1$ .

The equilibrium constant is $K_{p} = \frac{P _{H_{2}} \cdot P _{O_{2}}^{1/2}}{P _{H_{2} O}} = \frac{α \cdot ( α /2 ) ^{1/2}}{1} = \frac{α ^{3/2}}{2}$ .

Substituting $K_{p} = 8.0 \times 1 0^{- 3}$ : $α^{3/2} = 82 \times 1 0^{- 3}$ , so $α^{3} = (82 \times 1 0^{- 3})^{2} = 128 \times 1 0^{- 6}$ .

Taking the cube root: $α = (128 \times 1 0^{- 6})^{1/3} = 12 8^{1/3} \times 1 0^{- 2} \approx 5.03 \times 1 0^{- 2}$ .

Rounded to the nearest integer, the coefficient is 5.

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q24: Law of Chemical Equilibrium And Equilibrium Constant

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