Question:
At a given temperature T in a 10.0 L flask, 2.0 moles of N$_2$O$_4$(g) is heated. At equilibrium, 20% of N$_2$O$_4$(g) dissociates into NO$_2$(g). The value of K$_C$ for the reaction N$_2$O$_4$(g) $\rightleftharpoons$ 2NO$_2$(g) is:
At a given temperature T in a 10.0 L flask, 2.0 moles of N$_2$O$_4$(g) is heated. At equilibrium, 20% of N$_2$O$_4$(g) dissociates into NO$_2$(g). The value of K$_C$ for the reaction N$_2$O$_4$(g) $\rightleftharpoons$ 2NO$_2$(g) is:
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For dissociation problems, always express equilibrium concentrations in terms of degree of dissociation before substituting into K$_C$.
Updated On: Jun 10, 2026
- 2 10^-2
- 4 10^-2
- 3 10^-2
- 6 10^-2
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The Correct Option is B
Solution and Explanation
Concept:
For a dissociation equilibrium, the equilibrium constant is determined using equilibrium concentrations:
\[
K_C = \frac{[NO_2]^2}{[N_2O_4]}
\]
Step 1: Set up dissociation \[ N_2O_4 \rightleftharpoons 2NO_2 \] Initial moles = 2.0 mol Degree of dissociation = 20% = 0.2
Step 2: Calculate equilibrium moles Dissociated moles: \[ 2.0 \times 0.2 = 0.4 \text{ mol} \] Equilibrium moles: \[ N_2O_4 = 2.0 - 0.4 = 1.6 \text{ mol} \] \[ NO_2 = 2 \times 0.4 = 0.8 \text{ mol} \]
Step 3: Convert to concentration (V = 10 L) \[ [N_2O_4] = \frac{1.6}{10} = 0.16 \, M \quad,\quad [NO_2] = \frac{0.8}{10} = 0.08 \, M \]
Step 4: Calculate K$_C$ \[ K_C = \frac{(0.08)^2}{0.16} = \frac{0.0064}{0.16} = 0.04 = 4 \times 10^{-2} \]
Step 1: Set up dissociation \[ N_2O_4 \rightleftharpoons 2NO_2 \] Initial moles = 2.0 mol Degree of dissociation = 20% = 0.2
Step 2: Calculate equilibrium moles Dissociated moles: \[ 2.0 \times 0.2 = 0.4 \text{ mol} \] Equilibrium moles: \[ N_2O_4 = 2.0 - 0.4 = 1.6 \text{ mol} \] \[ NO_2 = 2 \times 0.4 = 0.8 \text{ mol} \]
Step 3: Convert to concentration (V = 10 L) \[ [N_2O_4] = \frac{1.6}{10} = 0.16 \, M \quad,\quad [NO_2] = \frac{0.8}{10} = 0.08 \, M \]
Step 4: Calculate K$_C$ \[ K_C = \frac{(0.08)^2}{0.16} = \frac{0.0064}{0.16} = 0.04 = 4 \times 10^{-2} \]
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