Verified Solution • 2025–26 Session moderate

QJEEM23Ch44fb: JEE Main 2023 Chemistry - Class 11 JEEM23Ch44fb Solution: At 298 K Based on above equilibria, then equilibrium constant…

At 298 K

N2 (g)+3H2 (g)⇌ 2NH3 (g), K1=4×105\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}

N2 (g)+O2 (g)⇌ 2NO (g), K2=1.6×1012\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}

H2 (g)+12O2 (g)⇌ H2O (g), K3=1.0×10−13\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}

Based on above equilibria, then equilibrium constant of the reaction, 2NH3(g)+52O2 (g)⇌ 2NO (g)+3H2O (g)\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)} is ____________ ×10−33\times10^{-33} (Nearest integer).

Step-by-Step Verified Solution

2NH3(g)52O2(g)⇌2NO(g)+3H2O(g)\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}

Clearly, Keq=1k1×k2×k33\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}

=1.6×1012×10−394×105= {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}

=0.4×10−32= 0.4 \times {10^{ - 32}}

=4×10−33= 4 \times {10^{ - 33}}

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