Question

$$\triangle G^{\circ}$$ for $$\dfrac {1}{2} N_{2} + \dfrac {3}{2} H_{2}\rightleftharpoons NH_{3}$$ is $$-16.5\ kJ\ mol^{-1}$$ at $$25^{\circ}C$$. Find out $$K_{p}$$ for the reaction. Also report $$K_{p}$$ and $$\triangle G^{\circ}$$ for:

$$N_{2} + 3H_{2} \rightarrow 2NH_{3}$$ at $$25^{\circ}C$$.

Answer 1

We know,

$$-\triangle G^{\circ} = 2.303\ RT \log K_{p}$$

$$-(-16.5\times 10^{3}) = 2.303\times 8.314\times 298\log K_{p}$$

$$\log K_{p} = \dfrac {16500}{2.303\times 8.314\times 298}$$

$$\therefore K_{p} = 779.41\ atm^{-1}$$

Also,

$$K_{p_{1}}$$ for $$N_{2} + 3H_{2}\rightleftharpoons 2NH_{3}$$

$$K_{p_{1}} = (K_{p})^{2} = (779.41)^{2}$$

$$K_{p_{1}} = 6.07\times 10^{5} atm^{-2}$$

Also,

$$-\triangle G^{\circ}_{1} = 2.303\times 8.314\times 298 \log (6.07\times 10^{5})J$$

$$= 32.998\ kJ$$

$$\therefore \triangle G^{\circ}_{1} = -32.998\ kJ\ mol^{-1}$$.