For a given reaction, $$\Delta H = 35.5\,kJ\,mo{l^{ - 1}}$$    and $$\Delta S = 83.6\,J{K^{ - 1}}mo{l^{ - 1}}.$$     The reaction is spontaneous at : ( Assume that $$\Delta H$$  ans $$\Delta S$$  do not vary with temperature )

Solution :
According to Gibbs-Helmholtz equation,
Gibbs energy $$\left( {\Delta G} \right) = \Delta H - T\Delta S$$
Where, $$\Delta H$$ = Enthalpy change
$$\Delta S$$  = Entropy change
$$T$$ = Temperature
For a reaction to be spontaneous
$$\Delta G < 0.$$
∴ Gibbs -Helmholtz equation becomes,
$$\eqalign{ & \Delta G = \Delta H - T\Delta S < 0 \cr & {\text{or,}}\,\Delta H < T\Delta S \cr & {\text{or}},\,\,T > \frac{{\Delta H}}{{\Delta S}} \cr & = \frac{{35.5\,kJ\,mo{l^{ - 1}}}}{{83.6\,J{K^{ - 1}}mo{l^{ - 1}}}} \cr & = \frac{{35.5 \times 1000}}{{83.6}} \cr & = 425K \cr & T > 425K \cr} $$