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Question:
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(1) $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \mathrm{K}_1$
(2) $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_2$
(3) $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}), \mathrm{K}_3$
The equation for the equilibrium constant of the reaction
$$
2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {, }
$$
$\left(K_4\right)$ in terms of $\mathrm{K}_1, \mathrm{~K}_2$ and $\mathrm{K}_3$ is :
Options:
(2) $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_2$
(3) $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}), \mathrm{K}_3$
The equation for the equilibrium constant of the reaction
$$
2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {, }
$$
$\left(K_4\right)$ in terms of $\mathrm{K}_1, \mathrm{~K}_2$ and $\mathrm{K}_3$ is :
Solution:
1079 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{K}_2 \cdot \mathrm{K}_3^3}{\mathrm{~K}_1}$
$\frac{\mathrm{K}_2 \cdot \mathrm{K}_3^3}{\mathrm{~K}_1}$
To calculate the value of $\mathrm{K}_4$ in the given equation we should apply :
eqn. (2) $+$ eqn. $(3) \times 3-$ eqn. (1)
hence $\mathrm{K}_4=\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}$
eqn. (2) $+$ eqn. $(3) \times 3-$ eqn. (1)
hence $\mathrm{K}_4=\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}$
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