Question

Predict the direction in which $△G^0$ for the equilibrium $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right);△G_{298}^0=-33.32kJ$
will change with increase in temperature. Calculate $△G^0$ at ${500}^{\circ }C$, assuming the $△H$ and $△S$ do not change with temperature, that is, $△H_{298}^0=△H_{793}$ and $△S_{298}^0=△S_{793}$
$△H^0=-92.38kJ$ and $△S^0=-198.2J/K$
(The eqn. shifts to left with increase in temp., $+60.83kJ$).

Answer 1

When given in this reaction $\Delta G$ is negative and the enthalpy change and entropy change are negative so when we increase the temperature in the reaction the equilibrium shift in the left side