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Dependence of Spontaneity on Temperature...

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.
A reaction has a value of `DeltaH =-40 Kcal` at 400 k cal `mol^(-1)`. The reaction is spontaneous, below this temperature , it is not . The values fo `DeltaG` and `DeltaS` at 400 k are respectively

A

`0,-0.1 cal K^(-1)`

B

`0,100 cal K^(-1)`

C

`-10` Kcal, `-100"cal"K^(-1)`

D

`0,-100 cal K^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D

Clearly at `400` K reaction is in equilibrium `implies DeltaG=0 implies DeltaS =(DeltaH)/(T) =- (40000)/(400) =- 100 cal K^(-1)`
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