For the reaction, `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g), Delta H = -95.2 kJ` and `Delta S = -198.1 JK^(-1)`. Calculate the temperature at which Gibb's energy change of the reaction `(Delta G)` becomes equal to zero.

Explanation

`Delta G = Delta H - T Delta S`
When `Delta G = 0`,
`Delta H - T Delta S = 0` (or) `Delta H = T Delta S`
`T = (Delta H)/(Delta S) = (-95.4 xx 1000 J)/(-1983 JK^(-1)) = 481 K`
At this temperature the reaction would be in equilibrium. With increase in temperature the opposing factor `T Delta S` would become more and hence, `Delta G` would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous for temperature below 481 K.