The following concentrations were observed at 500 K for the formation of NH 3 from N 2 and H 2 . At equilibrium : [ N 2 ]=2 × 10 -2 M , [ H 2 ]=3 × 10 -2 M and [ NH 3 ]=1.5 × 10 -2 M . Equilibrium constant for the reaction is .

Chemistry · Chemical Equilibrium

JEE Main 2024 — 29 January, Shift 2 — Question 77

The following concentrations were observed at 500 K for the formation of $\mathrm{NH}_{3}$ from $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ . At equilibrium :

$\left[\mathrm{N}_{2}\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_{2}\right]=3 \times 10^{-2} \mathrm{M}$ and $\left[\mathrm{NH}_{3}\right]=1.5 \times 10^{-2} \mathrm{M}$ .

Equilibrium constant for the reaction is $\qquad$ .

Answer: 417

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}$

$K_{C}=\frac{\left(1.5 \times 10^{-2}\right)^{2}}{\left(2 \times 10^{-2}\right) \times\left(3 \times 10^{-2}\right)^{3}}$

$\mathrm{K}_{\mathrm{C}}=417$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient