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$_5$.
Molar mass of N$_2$O$_5$ = 2(14) + 5(16) = 28 + 80 = 108 g/mol
Initial moles of N$_2$O$_5$, $n_0 = \frac{37.8}{108} = 0.35$ mol
Step 2: Set up the ICE table for partial pressures.
Let the initial pressure of N$_2$O$_5$ be $P_0$. Using the ideal gas law, $P_0V = n_0RT$, so
$P_0 = \frac{n_0RT}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35$ bar
Total pressure at equilibrium, $P_T = (P_0 - 2x) + 2x + x = P_0 + x$
We are given $P_T = 18.65$ bar, so $18.65 = 14.35 + x$, which means $x = 4.3$ bar 
Step 3: Calculate the equilibrium partial pressures. $P_{{N}_2{O}_5} = P_0 - 2x = 14.35 - 2(4.3) = 14.35 - 8.6 = 5.75$ bar $P_{{N}_2{O}_4} = 2x = 2(4.3) = 8.6$ bar $P_{{O}_2} = x = 4.3$ bar 
Step 4: Calculate Kp. \[ K_p = \frac{P_{{N}_2{O}_4}^2 \cdot P_{{O}_2}}{P_{{N}_2{O}_5}^2} = \frac{(8.6)^2 \cdot (4.3)}{(5.75)^2} = \frac{73.96 \cdot 4.3}{33.0625} = \frac{318.028}{33.0625} \approx 9.619 \] Then $K_p = 9.619$ Kp = _________ $\times 10^{-2}$ \(961.9 \times 10^{-2}\) approximately \(962 \times 10^{-2}\) (nearest integer) Therefore the closest integer is 962