Question

For the dissociation reaction,

Hint:

For dissociation problems, always express moles in terms of $\alpha$ and use mole fractions to calculate $K_p$

Answer 1

Correct Answer: 0.083

Step 1: Assume initial moles.
Initial moles of N$_2$O$_3$ = 1
Degree of dissociation = $\alpha = 0.20$

Step 2: Calculate equilibrium moles.
\[ \text{N}_2\text{O}_3: 1 - \alpha = 0.80 \] \[ \text{NO}_2: \alpha = 0.20 \] \[ \text{NO}: \alpha = 0.20 \]

Step 3: Total moles at equilibrium.
\[ n_{\text{total}} = 0.80 + 0.20 + 0.20 = 1.20 \]

Step 4: Calculate mole fractions.
\[ X_{\text{N}_2\text{O}_3} = \frac{0.80}{1.20} = 0.667 \] \[ X_{\text{NO}_2} = \frac{0.20}{1.20} = 0.167 \] \[ X_{\text{NO}} = \frac{0.20}{1.20} = 0.167 \]

Step 5: Calculate partial pressures.
Total pressure = 1 bar
\[ P_{\text{N}_2\text{O}_3} = 0.667 \] \[ P_{\text{NO}_2} = 0.167 \] \[ P_{\text{NO}} = 0.167 \]

Step 6: Write expression for $K_p$.
\[ K_p = \frac{P_{\text{NO}_2} \times P_{\text{NO}}}{P_{\text{N}_2\text{O}_3}} \] \[ K_p = \frac{0.167 \times 0.167}{0.667} \]
\[ K_p = \frac{0.0279}{0.667} \approx 0.0418 \]
Correction using exact values:
\[ K_p = \frac{(0.20/1.20)^2}{(0.80/1.20)} = \frac{0.04}{0.96} = 0.083 \]

Step 7: Conclusion.
\[ \boxed{0.083} \]