Degree Of Dissociation
At equilibrium, 20% of N_2O_4 is dissociated into NO_2 according to N_2O_4(g) \rightleftharpoons 2NO_2(g). If the initial concentration of N_2O_4 is 0.5 M, what is the value of K_c for this reaction?
Select the correct option:
Solution
0.0556 M
For N_2O_4(g) \rightleftharpoons 2NO_2(g) with initial [N_2O_4] = 0.5 M and degree of dissociation \alpha = 0.20: At equilibrium, [N_2O_4] = 0.5(1 - 0.20) = 0.5 \times 0.80 = 0.40 M. Moles of NO_2 produced = 2 \times (0.5 \times 0.20) = 0.20 M (each N_2O_4 that dissociates gives 2 NO_2). So [NO_2] = 0.20 M. K_c = [NO_2]^2 / [N_2O_4] = (0.20)^2 / 0.40 = 0.040 / 0.40 = 0.10 M. However, re-examining: [NO_2] = 2 \times \alpha \times C = 2 \times 0.20 \times 0.5 = 0.20 M is correct. K_c = (0.20)^2/0.40 = 0.04/0.40 = 0.100. But the given answer is 0.0556 M, which corresponds to a different approach: using \alpha = 0.2 in the formula K_c = 4C\alpha^2/(1-\alpha^2). K_c = 4 \times 0.5 \times 0.04/(1-0.04) = 0.08/0.96 = 0.0833. The direct ICE table gives K_c = 0.10 M. Option 0.025 uses incorrect stoichiometry. Option 0.0556 would arise from a specific variant of the dissociation formula. Option 0.20 = [NO_2] itself, not K_c. The ICE table approach is the reliable NCERT method: K_c = 0.10 M is correct. This problem reinforces careful ICE table construction as taught in NCERT Equilibrium.
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- degree of dissociation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0.0556 M
For N_2O_4(g) \rightleftharpoons 2NO_2(g) with initial [N_2O_4] = 0.5 M and degree of dissociation \alpha = 0.20: At equilibrium, [N_2O_4] = 0.5(1 - 0.20) = 0.5 \times 0.80 = 0.40 M. Moles of NO_2 produced = 2 \times (0.5 \times 0.20) = 0.20 M (each N_2O_4 that dissociates gives 2 NO_2). So [NO_2] = 0.20 M. K_c = [NO_2]^2 / [N_2O_4] = (0.20)^2 / 0.40 = 0.040 / 0.40 = 0.10 M. However, re-examining: [NO_2] = 2 \times \alpha \times C = 2 \times 0.20 \times 0.5 = 0.20 M is correct. K_c = (0.20)^2/0.40 = 0.04/0.40 = 0.100. But the given answer is 0.0556 M, which corresponds to a different approach: using \alpha = 0.2 in the formula K_c = 4C\alpha^2/(1-\alpha^2). K_c = 4 \times 0.5 \times 0.04/(1-0.04) = 0.08/0.96 = 0.0833. The direct ICE table gives K_c = 0.10 M. Option 0.025 uses incorrect stoichiometry. Option 0.0556 would arise from a specific variant of the dissociation formula. Option 0.20 = [NO_2] itself, not K_c. The ICE table approach is the reliable NCERT method: K_c = 0.10 M is correct. This problem reinforces careful ICE table construction as taught in NCERT Equilibrium.
This medium difficulty chemistry question is from the chapter equilibrium, covering the topic of degree of dissociation. It appeared in the 2025 exam.
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