37.8 g N₂O₅ 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, Kp = _______ × 10^–2 [nearest integer]
Assume N₂O₅ to behave ideally under these conditions
Given : R = 0.082 bar L mol^–1 K^–1

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 962.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} Initial pressure of N_{2} O_{5} \\ = \frac{\frac{37.8}{108} \times 0.082 \times 500}{1} = 14.35 bar \end{array}$

$\begin{array}{l} P_{Total} at eqb = 14.35 + P = 18.65 \\ P = 4.3 \\ P_{N_{2} O_{5}} = 5.75 bar \\ P_{N_{2} O_{4}} = 8.6 bar \\ P_{O_{2}} = 4.3 bar \\ k_{p} = \frac{(8.6)^{2} \times (4.3)}{(5.75)^{2}} = 9.619 = x \times 10^{- 2} \\ x = 961.9 \approx 962 \end{array}$