Question:
\(9.2\mathrm{g}\ \mathrm{N}_2\mathrm{O}_4\) is heated in 1L vessel till equilibrium state is established.
\(\mathrm{N}_2\mathrm{O}_4(\mathrm{g})\rightleftharpoons 2\mathrm{NO}_2(\mathrm{g})\)
In equilibrium state, \(50%\ \mathrm{N}_2\mathrm{O}_4\) was dissociated, equilibrium constant will be (molecular wt. of \(\mathrm{N}_2\mathrm{O}_4 = 92\))
\(9.2\mathrm{g}\ \mathrm{N}_2\mathrm{O}_4\) is heated in 1L vessel till equilibrium state is established.
\(\mathrm{N}_2\mathrm{O}_4(\mathrm{g})\rightleftharpoons 2\mathrm{NO}_2(\mathrm{g})\)
In equilibrium state, \(50%\ \mathrm{N}_2\mathrm{O}_4\) was dissociated, equilibrium constant will be (molecular wt. of \(\mathrm{N}_2\mathrm{O}_4 = 92\))
\(\mathrm{N}_2\mathrm{O}_4(\mathrm{g})\rightleftharpoons 2\mathrm{NO}_2(\mathrm{g})\)
In equilibrium state, \(50%\ \mathrm{N}_2\mathrm{O}_4\) was dissociated, equilibrium constant will be (molecular wt. of \(\mathrm{N}_2\mathrm{O}_4 = 92\))
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Degree of dissociation \(\alpha = 0.5\). For N\(_2\)O\(_4\) ⇌ 2NO\(_2\), \(K_C = \frac{4\alpha^2 C}{1-\alpha}\).
Updated On: Apr 20, 2026
- 0.1
- 0.4
- 0.3
- 0.2
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
\(K_C = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\).
Step 2: Detailed Explanation:
Initial moles of N\(_2\)O\(_4\) = \(9.2 / 92 = 0.1\) mol. Volume = 1 L, so initial concentration = 0.1 M.
At equilibrium, 50% dissociated: [N\(_2\)O\(_4\)] = \(0.1 \times 0.5 = 0.05\) M,
[NO\(_2\)] = \(2 \times (0.1 \times 0.5) = 0.1\) M.
\(K_C = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2\).
Step 3: Final Answer:
0.2
\(K_C = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\).
Step 2: Detailed Explanation:
Initial moles of N\(_2\)O\(_4\) = \(9.2 / 92 = 0.1\) mol. Volume = 1 L, so initial concentration = 0.1 M.
At equilibrium, 50% dissociated: [N\(_2\)O\(_4\)] = \(0.1 \times 0.5 = 0.05\) M,
[NO\(_2\)] = \(2 \times (0.1 \times 0.5) = 0.1\) M.
\(K_C = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2\).
Step 3: Final Answer:
0.2
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