For reaction, `N_(2(g))+3H_(2(g))to2NH_(3(g))`, `DeltaH=-95.4kJ` and `DeltaS=-198.3JK^(-1)`. Calculate the maximum temperature at which the reaction will proceed in forward direction.

Answer

`T gt 481`

Explanation

`Delta G = Delta H - T Delta S`
For a reaction to be spontaneous, `Delta G = -ve`
`Delta H -T Delta S = - ve` or `Delta H gt T Delta S` or `(Delta H)/(Delta S) gt T` or `(-95.4 xx 10^(+3))/(-198.3) gt T` or `481.0 gt T`
Thus, if temperature of system is lesser than 481 K, the reaction would be spontaneous. At 481 K, the reaction will be spontaneous. An increase in temperature above 481 k will develop non-spontaneity for the reaction.