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Chemistry · PYQ NEET UG 2024
हिन्दीConsider the following reaction in a sealed vessel at equilibrium with concentrations of \( N_2 = 3.0 \times 10^{-3} \) M, \( O_2 = 4.2 \times 10^{-3} \) M and \( NO = 2.8 \times 10^{-3} \) M. \( 2\text{NO}_{(g)} \rightleftharpoons \text{N}_{2(g)} + \text{O}_{2(g)} \). If 0.1 mol L\( ^{-1} \) of \( \text{NO}_{(g)} \) is taken in a closed vessel, what will be degree of dissociation (\( \alpha \)) of NO at equilibrium?
Correct answer · Option 2
0.717
Explanation
First, calculate \( K_c = \frac{[N_2][O_2]}{[NO]^2} = \frac{3.0 \times 10^{-3} \times 4.2 \times 10^{-3}}{(2.8 \times 10^{-3})^2} = 1.607 \). For initial concentration C of NO, \( K_c = \frac{\alpha^2}{4(1-\alpha)^2} \Rightarrow \sqrt{1.607} = \frac{\alpha}{2(1-\alpha)} \Rightarrow 1.2676 = \frac{\alpha}{2-2\alpha} \Rightarrow \alpha \approx 0.717 \).
