At -20 ° C and 1 atm pressure, a cylinder is filled with equal number of H 2 . I 2 and HI molecules for the reaction H 2 ( g )+ I 2 ( g ) rightleftharpoons 2 HI ( g ) , the K P for the process is x × 10 -1 × x = . [Given : R =0.082 L atm K -1 mol -1 ]

Chemistry · Chemical Equilibrium

JEE Main 2024 — 6 April, Shift 1 — Question 73

At $-20^{\circ} \mathrm{C}$ and 1 atm pressure, a cylinder is filled with equal number of $\mathrm{H}_{2} . \mathrm{I}_{2}$ and HI molecules for the reaction

$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ , the $\mathrm{K}_{\mathrm{P}}$ for the process is $\mathrm{x} \times 10^{-1} \cdot \mathrm{x}=$ $\qquad$ .

[Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

Option A:
2
Option B:
1
Option C:
10
Correct
Option D:
0.01

Answer: C

Step-by-step solution

$\Delta \mathrm{ng}=0$

$\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{n}_{\mathrm{HI}}\right)^{2}}{\mathrm{n}_{\mathrm{H}_{2}} \mathrm{n}_{\mathrm{I}_{2}}}\left(\frac{\mathrm{P}_{\mathrm{T}}}{\mathrm{n}_{\mathrm{T}}}\right)^{\Delta \mathrm{n}_{\mathrm{g}}}$ $\mathrm{n}_{\mathrm{HI}}=\mathrm{n}_{\mathrm{H}_{2}}=\mathrm{n}_{\mathrm{I}_{2}}$ so $K_{P}=1$

$1=\mathrm{x} \times 10^{-1}$ $\mathrm{x}=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient