Question

Calculate the value of $\log K_p$ (nearest integer) for the reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at ${25}^{\circ }C$. The standard enthalpy of formation of $N{H}_3\left(g\right)$ is -46 kJ and standard entropies of $N_2\left(g\right),H_2\left(g\right),N{H}_3\left(g\right)$ are $191,130,and192J{K}^{-1}mo{l}^{-1}$ respectively. $(R=8.3J{K}^{-1}mo{l}^{-1})$

Answer 1

At equilibrium, the relationship between the Gibbs standard free energy change and the equilibrium constant is
$-\Delta G^{\circ }=2.303RT\log K_p$

Also for the reaction,
$N_2\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$;

The standard enthalpy change $\Delta H^{\circ }=-46\times 2KJ=-92kJ$
$(\Delta H_{formation}^{\circ }ofN{H}_3=-46kJ)$

Also, the standard entropy change $\Delta S_{reaction}^{\circ }=2\times S_{N{H}_3}^{\circ }-S_{N_2}^{\circ }-3\times S_{H_2}^{\circ }$
$=2\times 192-191-3\times 130=-197J$
$T=273+25=298K$
The relationship between the standard Gibbs free energy change, the standard enthalpy change, and the standard entropy change is as given below.
$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
$=-46\times 2\times {10}^3-298\times (-197)$
$=-33294J$
Thus, from (i)
$33294=2.303\times 298\times 8.3\times \log K_p$
$\therefore \log K_p=5.845\approx 6$