JEE Main · 2020 · Shift-IIeasyCEQ-047

The value of Kc is 64 at 800 K for the reaction N2(g) + 3H2(g) 2NH3(g). The value of Kc for the following reaction is:…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

The value of $K_c$ is 64 at 800 K for the reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ . The value of $K_c$ for the following reaction is: $\mathrm{NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)}$

Options

a
$1/64$
b
$8$
c
$1/4$
d
$1/8$
✓

Correct Answerd

$1/8$

Detailed Solution

Step 1 — Relate the two reactions:

Given: $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ , $K_c = 64$

Target: $\mathrm{NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2}$

Step 2 — Manipulate:

Reverse the given reaction: $\mathrm{2NH_3 \rightleftharpoons N_2 + 3H_2}$ , $K = \frac{1}{64}$
Divide by 2 (take square root): $\mathrm{NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2} K = \left(\frac{1}{64}\right)^{1/2} = \frac{1}{8}$

Answer: Option (4) — $\frac{1}{8}$

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