Question:
Consider the following reaction in a sealed vessel at equilibrium with concentrations of
N2 = 3.0 × 10–3 M, O2 = 4.2 × 10–3 M and NO = 2.8 × 10–3 M.
2NO(g)⇋N2(g) + O2(g)
If 0.1 mol L–1 of NO(g) is taken in a closed vessel, what will be degree of dissociation (a) of NO(g) at equilibrium?
Consider the following reaction in a sealed vessel at equilibrium with concentrations of
N2 = 3.0 × 10–3 M, O2 = 4.2 × 10–3 M and NO = 2.8 × 10–3 M.
2NO(g)⇋N2(g) + O2(g)
If 0.1 mol L–1 of NO(g) is taken in a closed vessel, what will be degree of dissociation (a) of NO(g) at equilibrium?
N2 = 3.0 × 10–3 M, O2 = 4.2 × 10–3 M and NO = 2.8 × 10–3 M.
2NO(g)⇋N2(g) + O2(g)
If 0.1 mol L–1 of NO(g) is taken in a closed vessel, what will be degree of dissociation (a) of NO(g) at equilibrium?
Updated On: Dec 9, 2024
- 0.00889
- 0.0889
- 0.8889
- 0.717
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The Correct Option is D
Solution and Explanation
Step 1: Use the stoichiometric relationship:
Initial concentration of NO: $[NO] = 0.1 \, \text{mol/L}$
Change in concentration: $\Delta[NO] = 2\alpha[NO]$
Final concentration: $[NO]_f = 0.1 - 2\alpha$
Step 2: Calculate $\alpha$ at equilibrium: Using equilibrium concentrations:
$K_c = \frac{[N_2][O_2]}{[NO]^2}$, $K_c = \frac{\alpha^2}{(1 - 2\alpha)^2}$
Simplifying for $\alpha$, we find:
$\alpha = 0.717$
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