Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at `298 K ,2A + B rarr C` `DeltaH =100 kcal` and `DeltaS=0.050 kcal K^(-1)`. If `DeltaH` and `DeltaS` are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?A. 1000 KB. 1500 KC. 2000 KD. 2500 K

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1.	Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)` The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous . For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at `298 K ,2A + B rarr C` `DeltaH =100 kcal` and `DeltaS=0.050 kcal K^(-1)`. If `DeltaH` and `DeltaS` are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?A. 1000 KB. 1500 KC. 2000 KD. 2500 K
Answer» Correct Answer - C Reaction to be spontaneous `DeltaG lt 0 implies DeltaH-TDeltaS lt0 implies DeltaT gt (DeltaH)/(DeltaS) implies T gt (100xx10^(3))/(05xx10^(3)) implies T gt 2000` K.