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.

Explanation

`DeltaG=DeltaH-TDeltaS`
For a reaction to be spontaneous, `DeltaG=-ve`
and therefore, `( :' DeltaH=-ve,DeltaS=-ve)`
`DeltaH-TDeltaS=-ve` or `DeltaH gt TDeltaS` or `(DeltaH)/(DeltaS) gt T`
or `(-95.4xx10^(+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. also at `481 K`, the reaction will be in equilibrium. an increase in temperature above `481.0` will develop non-spontaneity for the reaction.