JEE MAIN - Chemistry (2023 - 29th January Evening Shift - No. 18)
At 298 K
$$\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}$$
$$\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}$$
$$\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}$$
Based on above equilibria, then equilibrium constant of the reaction, $$\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)}$$ is ____________ $$\times10^{-33}$$ (Nearest integer).
Answer
4
Explanation
$$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$$
Clearly, $$\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$$
$$ = {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$$
$$ = 0.4 \times {10^{ - 32}}$$
$$ = 4 \times {10^{ - 33}}$$
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