For the reaction, N_2(g)+3 H_2(g) 2 NH_3(g), the partial pressure… - Vidyadip

Question

For the reaction, $\mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})$, the partial pressures of $\mathrm{N}_2$ and $\mathrm{H}_2$ are 0.80 and 0.40 atmosphere respectively at equilibrium. The total pressure of the system is 2.80 atmosphere. What is $\mathrm{K}_{\mathrm{p}}$ for the above reaction?

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Answer

$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$ Given, at equilibrium, $\text{p}_{\text{N}_2}=0.80$ atmosphere, $\text{p}_{\text{H}_2}=0.40$ atmosphere $\text{p}_{\text{N}_2}+\text{p}_{\text{H}_2}+\text{p}_{\text{NH}_3}=2.80$ atmosphere$\therefore\text{p}_{\text{NH}_3}=2.80-(0.80+0.40)=1.60$ atmosphere
From, $\text{K}_{\text{p}}=\frac{\text{p}_{\text{NH}_3}^2}{\text{p}_{\text{N}_3}\times\text{p}^3_{\text{H}_2}}$ $=\frac{(1.60)^2}{0.80\times(0.40)^3}=50.0$

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