37.8 g of was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K The total pressure at equilibrium was found to be 18.65 bar. Then, = ______ [nearest integer] Assume to behave ideally under these conditions. G

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Solution

Q.
37.8 g of $N_{2} O_{5}$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

$2 N_{2} O_{5} (g) ⇌ 2 N_{2} O_{4} (g) + O_{2} (g)$

The total pressure at equilibrium was found to be 18.65 bar.

Then, $K_{p}$ = ______ $\times 10^{- 2}$ [nearest integer]

Assume $N_{2} O_{5}$ to behave ideally under these conditions.

Given: R = 0.082 bar L ${mol}^{- 1}$ $K^{- 1}$ [2025]

Ans.
(962)

Initial moles of $N_{2} O_{5} (n_{N_{2} O_{5}})$ $= \frac{Given mass of N_{2} O_{5}}{Molar mass of N_{2} O_{5}} = \frac{37.8}{108} mol = 0.35 mol$

Initial pressure of $N_{2} O_{5} = \frac{n_{N_{2} O_{5}} R T}{V}$

$= \frac{0.35 \times 0.082 \times 500}{1} bar = 14.35 bar$

$\begin{matrix} 2 N_{2} O_{5} (g) & ⇌ & 2 N_{2} O_{4} (g) & + & O_{2} (g) \\ t = 0 & 14.35 bar & 0 & 0 \\ t = \infty & 14.35 - 2 p & 2 p & p \end{matrix} \begin{matrix} total = 14.35 + p \end{matrix}$

Total pressure at equilibrium = 14.35 + p = 18.65, p = 4.3 bar

$K_{p} = \frac{P_{O_{2}} \times P_{N_{2} O_{4}}^{2}}{P_{N_{2} O_{5}}^{2}} = \frac{p \times {(2 p)}^{2}}{{(14.35 - 2 p)}^{2}}$

$= \frac{4.3 \times {(2 \times 4.3)}^{2}}{{(14.35 - 2 \times 4.3)}^{2}} = \frac{4.3 \times {(2 \times 4.3)}^{2}}{{(5.75)}^{2}} ≃ 9.62 = 962 \times 10^{- 2}$

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