JEE Main 2025 Jan 24 Shift 1, Chemistry Q21: Law of Chemical Equilibrium And Equilibrium Constant

JEE Main2025Jan 24, Shift 1Chemistry

37.8 g N $_{2}$ O $_{5}$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

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

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

Then, $K_{p}$ = _______ $\times 1 0^{- 2}$ [nearest integer]

Assume N $_{2}$ O $_{5}$ to behave ideally under these conditions

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

Your answer

Solution

Step 1 $-$ initial pressure of N $_{2}$ O $_{5}$ from ideal gas law:

Moles $= \frac{37.8}{108} = 0.35$ mol; $P_{0} = \frac{n R T}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35$ bar.

Step 2 $-$ set up the ICE table in pressures. Let $2 x$ bar of N $_{2}$ O $_{5}$ react:

$P_{N_{2} O_{5}} = 14.35 - 2 x$ , $P_{N_{2} O_{4}} = 2 x$ , $P_{O_{2}} = x$ .

Total: $P_{t o t} = 14.35 - 2 x + 2 x + x = 14.35 + x = 18.65$ bar, so $x = 4.30$ bar.

Step 3 $-$ equilibrium partial pressures:

$P_{N_{2} O_{5}} = 14.35 - 8.60 = 5.75$ bar, $P_{N_{2} O_{4}} = 8.60$ bar, $P_{O_{2}} = 4.30$ bar.

Step 4 $-$ assemble $K_{p}$ for $2 N_{2} O_{5} \to 2 N_{2} O_{4} + O_{2}$ :

$K_{p} = \frac{( P _{N_{2} O_{4}} ) ^{2} ( P _{O_{2}} )}{( P _{N_{2} O_{5}} ) ^{2}} = \frac{( 8.60 ) ^{2} \times 4.30}{( 5.75 ) ^{2}} = 9.619$ bar.

Expressed as $x \times 1 0^{- 2}$ : $x \approx 961.9 \approx 962$ .

$∴$ Answer: $962$ .

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