Chemistry · 2013
JEE Main · 23 April 2013, Shift 1 · Q41
N_2( g )+3 H_2( g ) ⇌ 2 NH_3( g ), K_1 (1); N_2( g )+ O_2( g ) ⇌ 2 NO ( g ), K_2 (2); H_2( g )+1/2 O_2( g ) ⇌ H_2 O ( g ), K_3 (3) The equation for…
$$\begin{align*}
& \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \mathrm{K}_1 \tag{1}\\
& \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g}), \mathrm{K}_2 \tag{2}\\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{~g}), \mathrm{K}_3 \tag{3}
\end{align*}
$$
The equation for the equilibrium constant of the reaction $\displaystyle 2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons$ $\displaystyle 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}),\left(\mathrm{K}_4\right)$ in terms of $\displaystyle \mathrm{K}_1, \mathrm{~K}_2$ and $\displaystyle \mathrm{K}_3$ is :
Official answer
(4)
$\displaystyle \frac{\mathrm{K}_2 \mathrm{~K}_3{ }^3}{\mathrm{~K}_1}$
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JEE Main 2013 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.