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: Calculate the Initial Moles of N\(_2\)O\(_5\)

The molar mass of N\(_2\)O\(_5\) is: \[ \text{Molar mass of N}_2\text{O}_5 = 2(14) + 5(16) = 28 + 80 = 108 \, \text{g/mol} \] The initial moles of N\(_2\)O\(_5\) are given by: \[ n_0 = \frac{37.8}{108} = 0.35 \, \text{mol} \]

Step 2: Set Up the ICE Table for Partial Pressures

Using the ideal gas law, we calculate the initial pressure \( P_0 \): \[ P_0 = \frac{n_0RT}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35 \, \text{bar} \] The total pressure at equilibrium is: \[ P_T = (P_0 - 2x) + 2x + x = P_0 + x \] Given \( P_T = 18.65 \, \text{bar} \), we solve for \( x \): \[ 18.65 = 14.35 + x \quad \Rightarrow \quad x = 4.3 \, \text{bar} \]

Step 3: Calculate the Equilibrium Partial Pressures

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

Step 4: Calculate \( K_p \)

The equilibrium constant \( K_p \) is: \[ K_p = \frac{P_{\text{N}_2\text{O}_4}^2 \cdot P_{\text{O}_2}}{P_{\text{N}_2\text{O}_5}^2} \] Substituting the values: \[ K_p = \frac{(8.6)^2 \cdot (4.3)}{(5.75)^2} = \frac{73.96 \cdot 4.3}{33.0625} = \frac{318.028}{33.0625} \approx 9.619 \]

Conclusion

The value of \( K_p \) is approximately \( 962 \times 10^{-2} \), and the correct answer is \( \boxed{962} \).