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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.
The enthalpy change for a certain rection at 300 K is `-15.0` K cal `mol^(-1)` . The entropy change under these conditions is `-7.2` cal `K^(-1)mol^(-1)` . The free energy change for the reaction and its spontaneous/ non-spontaneous character will be
A. `-12.84` Kcal`"mol"^(-1)`, spontaneous
B. `-12.84` Kcal`"mol"^(-1)`,non-spontaneous
C. `-17.16` Kcal`"mol"^(-1)`, spontaneous
D. None of these

1 Answer

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by (81.1k points)
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Best answer
Correct Answer - A
Use `DeltaG = DeltaH - TDeltaS`
`DeltaG =- 15 -(33xx(-7.2))/(1000) =-12.84 "Kcal mol"^(-1)`

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