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Chemical Equilibrium question

2025 · 8 Apr · Shift 2 · Q23
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Chemical Equilibrium question

2025 · 8 Apr · Shift 2 · Q23

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The equilibrium constant for decomposition of H2O(g)H2O(g)⇌H2(g)+12O2(g)(ΔG∘=92.34 kJ mol−1)\text{H}_2\text{O(g)} \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1})H2​O(g)H2​O(g)⇌H2​(g)+21​O2​(g)(ΔG∘=92.34kJ mol−1) is 8.0×10−38.0 \times 10^{-3}8.0×10−3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α\alphaα) of water is ‾×10−2\underline{\hspace{2cm}}\times 10^{-2}​×10−2(nearest integer value). [Assume α\alphaα is negligible with respect to 1]
Numerical answer
View written solutionFree

Correct answer: 5

  1. Reaction and equilibrium constant

The reaction is

H2O(g)⇌H2(g)+12O2(g)\text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}H2​O(g)⇌H2​(g)+21​O2​(g)

Given:

  • K=8.0×10−3K = 8.0 \times 10^{-3}K=8.0×10−3 at 2300 K2300\,\text{K}2300K
  • Total pressure at equilibrium, P=1 barP = 1\,\text{bar}P=1bar
  • Degree of dissociation of water = α\alphaα
  • Assume α≪1\alpha \ll 1α≪1

We need to find α×102\alpha \times 10^2α×102.

  1. Start with 1 mole of water

Initially:

  • moles of H2O=1\text{H}_2\text{O} = 1H2​O=1
  • moles of H2=0\text{H}_2 = 0H2​=0
  • moles of O2=0\text{O}_2 = 0O2​=0

If degree of dissociation is α\alphaα, then at equilibrium:

  • H2O=1−α\text{H}_2\text{O} = 1-\alphaH2​O=1−α
  • H2=α\text{H}_2 = \alphaH2​=α
  • O2=α2\text{O}_2 = \dfrac{\alpha}{2}O2​=2α​

Total moles at equilibrium:

ntot=1−α+α+α2=1+α2n_{\text{tot}} = 1-\alpha+\alpha+\frac{\alpha}{2} = 1+\frac{\alpha}{2}ntot​=1−α+α+2α​=1+2α​

Since α\alphaα is very small,

1+α2≈11+\frac{\alpha}{2} \approx 11+2α​≈1
  1. Write partial pressures

At total pressure P=1 barP=1\,\text{bar}P=1bar,

pi=yiPp_i = y_i Ppi​=yi​P

So,

pH2O=1−α1+α/2⋅1p_{\text{H}_2\text{O}} = \frac{1-\alpha}{1+\alpha/2} \cdot 1pH2​O​=1+α/21−α​⋅1 pH2=α1+α/2⋅1p_{\text{H}_2} = \frac{\alpha}{1+\alpha/2} \cdot 1pH2​​=1+α/2α​⋅1 pO2=α/21+α/2⋅1p_{\text{O}_2} = \frac{\alpha/2}{1+\alpha/2} \cdot 1pO2​​=1+α/2α/2​⋅1

Using α≪1\alpha \ll 1α≪1,

pH2O≈1,pH2≈α,pO2≈α2p_{\text{H}_2\text{O}} \approx 1, \qquad p_{\text{H}_2} \approx \alpha, \qquad p_{\text{O}_2} \approx \frac{\alpha}{2}pH2​O​≈1,pH2​​≈α,pO2​​≈2α​
  1. Expression for equilibrium constant

For the reaction,

K=pH2 pO21/2pH2OK = \frac{p_{\text{H}_2}\, p_{\text{O}_2}^{1/2}}{p_{\text{H}_2\text{O}}}K=pH2​O​pH2​​pO2​1/2​​

Substitute the approximated partial pressures:

K≈α(α2)1/2K \approx \alpha \left(\frac{\alpha}{2}\right)^{1/2}K≈α(2α​)1/2

So,

8.0×10−3=αα28.0\times 10^{-3} = \alpha \sqrt{\frac{\alpha}{2}}8.0×10−3=α2α​​ 8.0×10−3=α3/228.0\times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}8.0×10−3=2​α3/2​

Therefore,

α3/2=8.0×10−32\alpha^{3/2} = 8.0\times 10^{-3}\sqrt{2}α3/2=8.0×10−32​

Using 2≈1.414\sqrt{2} \approx 1.4142​≈1.414,

α3/2≈8.0×10−3×1.414=1.131×10−2\alpha^{3/2} \approx 8.0\times 10^{-3} \times 1.414 = 1.131\times 10^{-2}α3/2≈8.0×10−3×1.414=1.131×10−2

Now,

α=(1.131×10−2)2/3\alpha = \left(1.131\times 10^{-2}\right)^{2/3}α=(1.131×10−2)2/3

Compute it:

α≈0.050\alpha \approx 0.050α≈0.050

Thus,

α≈5.0×10−2\alpha \approx 5.0 \times 10^{-2}α≈5.0×10−2

So the required nearest integer is

5\boxed{5}5​
  1. Comparison with stored answer

Stored correct answer = 5

Our derived answer also = 5, so they agree.

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