Question

37.8 g \( N_2O_5 \) was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K: \[ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) \]
The total pressure at equilibrium was found to be 18.65 bar. Then, \( K_p \) is: Given: \[ R = 0.082 \, \text{bar L mol}^{-1} \, \text{K}^{-1} \]

Hint:

For the equilibrium constant in terms of pressure, use the partial pressures of the gases involved in the reaction, and apply the ideal gas law to calculate the total moles at equilibrium.

Answer 1

Step 1: Initial Moles of N\(_2\)O\(_5\)

Molar mass of N\(_2\)O\(_5\): \[ \text{Molar mass of N}_2\text{O}_5 = 2(14) + 5(16) = 108 \, \text{g/mol} \] Initial moles of N\(_2\)O\(_5\): \[ n_0 = \frac{37.8}{108} = 0.35 \, \text{mol} \]

Step 2: ICE Table for Partial Pressures

Initial pressure \( P_0 \) via ideal gas law: \[ P_0 = \frac{n_0RT}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35 \, \text{bar} \] Total pressure at equilibrium: \[ P_T = (P_0 - 2x) + 2x + x = P_0 + x \] With \( P_T = 18.65 \, \text{bar} \), solving for \( x \): \[ 18.65 = 14.35 + x \quad \Rightarrow \quad x = 4.3 \, \text{bar} \]

Step 3: Equilibrium Partial Pressures

Equilibrium partial pressures: \[ P_{\text{N}_2\text{O}_5} = P_0 - 2x = 14.35 - 2(4.3) = 5.75 \, \text{bar} \] \[ P_{\text{N}_2\text{O}_4} = 2x = 2(4.3) = 8.6 \, \text{bar} \] \[ P_{\text{O}_2} = x = 4.3 \, \text{bar} \]

Step 4: Calculate \( K_p \)

Equilibrium constant \( K_p \): \[ K_p = \frac{P_{\text{N}_2\text{O}_4}^2 \cdot P_{\text{O}_2}}{P_{\text{N}_2\text{O}_5}^2} \] Substituting values: \[ K_p = \frac{(8.6)^2 \cdot (4.3)}{(5.75)^2} = \frac{318.028}{33.0625} \approx 9.619 \]

Conclusion

\( K_p \) is approximately \( 9.62 \). The final answer is \( \boxed{962} \).