Difficult
For the reaction ${N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$,the equilibrium constant ${K_p = 35}$ at a given temperature. Calculate the values of ${K_p}$ for the following reactions at the same temperature:
$(i) \ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
$(ii) \ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$
$(i) \ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
$(ii) \ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$
(A) Given reaction: ${N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ with ${K_p = 35}$.
$(i)$ For the reaction ${2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)}$,this is the reverse of the given reaction. Therefore,${K_{p1} = \frac{1}{K_p} = \frac{1}{35} \approx 0.0286}$.
$(ii)$ For the reaction ${\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)}$,this is the original reaction multiplied by a factor of $\frac{1}{2}$. Therefore,${K_{p2} = (K_p)^{1/2} = \sqrt{35} \approx 5.916}$.
$(i)$ For the reaction ${2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)}$,this is the reverse of the given reaction. Therefore,${K_{p1} = \frac{1}{K_p} = \frac{1}{35} \approx 0.0286}$.
$(ii)$ For the reaction ${\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)}$,this is the original reaction multiplied by a factor of $\frac{1}{2}$. Therefore,${K_{p2} = (K_p)^{1/2} = \sqrt{35} \approx 5.916}$.
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