$(i)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$ at $298 \ K$ has $\Delta G^{\Theta} = -16.5 \ kJ \ mol^{-1}$. Find $K_p$.
$(ii)$ At $298 \ K$,for $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,calculate $K_p$ and $\Delta G^{\Theta}$.
$(ii)$ At $298 \ K$,for $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,calculate $K_p$ and $\Delta G^{\Theta}$.
(N/A) $(i)$ Using the relation $\Delta G^{\Theta} = -RT \ln K_p$:
$-16500 \ J \ mol^{-1} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln K_p$
$\ln K_p = \frac{16500}{8.314 \times 298} \approx 6.658$
$K_p = e^{6.658} \approx 779.4$
$(ii)$ The reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is twice the first reaction.
Therefore,$\Delta G^{\Theta}_{new} = 2 \times (-16.5 \ kJ \ mol^{-1}) = -33 \ kJ \ mol^{-1}$.
$K_{p, new} = (K_p)^2 = (779.4)^2 \approx 6.07 \times 10^5$.
$-16500 \ J \ mol^{-1} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln K_p$
$\ln K_p = \frac{16500}{8.314 \times 298} \approx 6.658$
$K_p = e^{6.658} \approx 779.4$
$(ii)$ The reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is twice the first reaction.
Therefore,$\Delta G^{\Theta}_{new} = 2 \times (-16.5 \ kJ \ mol^{-1}) = -33 \ kJ \ mol^{-1}$.
$K_{p, new} = (K_p)^2 = (779.4)^2 \approx 6.07 \times 10^5$.
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