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NVAMediumJEE 2025Equilibrium Basics

JEE Chemistry 2025 Question with Solution

The equilibrium constant for decomposition of H2OH_2O (g)\text{(g)}

H2O(g)⇌H2(g)+12O2(g)(ΔG∘=92.34 kJ mol−1)H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, kJ \ mol^{-1})

is 8.0×10−38.0 \times 10^{-3} at 2300 K2300 \, \text{K} and total pressure at equilibrium is 1 bar1 \, \text{bar}. Under this condition, the degree of dissociation (α\alpha) of water is _____ ×10−2\times 10^{-2} (nearest integer value).

[Assume α\alpha is negligible with respect to 11]

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: The reaction is

H2O(g)⇌H2(g)+12O2(g)H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)

with Kp=8.0×10−3K_p = 8.0 \times 10^{-3} and total pressure P=1 barP = 1 \, \text{bar}.

Find: The degree of dissociation α\alpha of water.

Assume initially there is 11 mole of water vapour. At equilibrium:

  • H2O=1−αH_2O = 1-\alpha
  • H2=αH_2 = \alpha
  • O2=α2O_2 = \frac{\alpha}{2}

Hence total moles are

nT=1+α2≈1n_T = 1 + \frac{\alpha}{2} \approx 1

since α≪1\alpha \ll 1.

Now write the equilibrium constant expression:

Kp=PH2 PO21/2PH2OK_p = \frac{P_{H_2} \, P_{O_2}^{1/2}}{P_{H_2O}}

Using the approximation and total pressure 1 bar1 \, \text{bar},

PH2=αP,P_{H_2} = \alpha P, PO2=α2P,P_{O_2} = \frac{\alpha}{2} P, PH2O=(1−α)PP_{H_2O} = (1-\alpha)P

So,

Kp=(αP)(α2P)1/2(1−α)PK_p = \frac{(\alpha P) \left(\frac{\alpha}{2}P\right)^{1/2}}{(1-\alpha)P}

Detailed Calculation

With P=1P = 1, the expression becomes

8.0×10−3=α3/228.0 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}

Therefore,

α3/2=82×10−3\alpha^{3/2} = 8\sqrt{2} \times 10^{-3}

Squaring both sides appropriately gives

α3=128×10−6\alpha^3 = 128 \times 10^{-6}

Taking cube root,

α=128×10−63=5.03×10−2\alpha = \sqrt[3]{128 \times 10^{-6}} = 5.03 \times 10^{-2}

So the degree of dissociation is approximately 5×10−25 \times 10^{-2}. Therefore, the required nearest integer value is 55.

Common mistakes

  • Using mole numbers directly in KpK_p without converting to partial pressures is incorrect because KpK_p is defined in terms of equilibrium partial pressures. First express each partial pressure in terms of α\alpha and total pressure.

  • Forgetting that total moles become 1+α21 + \frac{\alpha}{2} is a conceptual mistake. Even though it is later approximated as 11, this change must be identified before applying the approximation α≪1\alpha \ll 1.

  • Ignoring the stoichiometric coefficient of oxygen leads to writing O2=αO_2 = \alpha instead of α2\frac{\alpha}{2}. This changes the power and numerical factor in KpK_p, so use the balanced reaction carefully.

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