Question

37.8 g N₂O₅ 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, $K_p$ = ______ × $10^{-2}$ [nearest integer] Assume N₂O₅ to behave ideally under these conditions Given : R = 0.082 bar L mol⁻¹ K⁻¹.

Answer 1

962

Answer 2

1. Molar mass and initial moles of $N_2O_5$: Molar mass of $N_2O_5 = 2 \times 14.01 + 5 \times 16.00 = 108.02 \text{ g/mol}$. Initial moles of $N_2O_5$, $n = \frac{37.8 \text{ g}}{108.02 \text{ g/mol}} \approx 0.350 \text{ mol}$.

2. Initial pressure of $N_2O_5$: Using the ideal gas law, $P_0 = \frac{nRT}{V}$: $P_0 = \frac{0.350 \text{ mol} \times 0.082 \text{ bar L mol⁻¹ K⁻¹} \times 500 \text{ K}}{1 \text{ L}} = 14.35 \text{ bar}$.

3. ICE table (in terms of partial pressures): Reaction: $2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$

   Species	Initial Pressure (bar)	Change (bar)	Equilibrium Pressure (bar)
   $N_2O_5$	$14.35$	$-2x$	$14.35 - 2x$
   $N_2O_4$	$0$	$+2x$	$2x$
   $O_2$	$0$	$+x$	$x$

4. Total pressure at equilibrium: $P_{total} = (14.35 - 2x) + 2x + x = 14.35 + x$. Given $P_{total} = 18.65 \text{ bar}$. $18.65 = 14.35 + x \implies x = 4.30 \text{ bar}$.

5. Equilibrium partial pressures: $P_{N_2O_5} = 14.35 - 2(4.30) = 5.75 \text{ bar}$. $P_{N_2O_4} = 2(4.30) = 8.60 \text{ bar}$. $P_{O_2} = 4.30 \text{ bar}$.

6. Calculate $K_p$: $K_p = \frac{(P_{N_2O_4})^2 \cdot P_{O_2}}{(P_{N_2O_5})^2} = \frac{(8.60)^2 \times 4.30}{(5.75)^2} \approx 9.619 \text{ bar}$.

7. Format the answer: $K_p = 9.619 \text{ bar} = 961.9 \times 10^{-2} \text{ bar}$. Rounding to the nearest integer, the value is 962.