For a process top be spontaneous, at constant temperature and pressure, there must be decreases in free energy of the system in the direction of the process, i.e. `DeltaG_(P.T.)lt0.Delta_(P.T.)=0` implies the equilibrium condition and `DeltaG_(P.T.)gt0` corresponding to non-spontaneity.
Gibb's Helmholtz equation relates the free energy change to the enthalpy and entropy change of the process as :
`DeltaG_(P.T.)=DeltaH-TDeltaS`......(i)
The magnitude of `Delta H` does not change much with the change in temperature but the entropy factor `TDeltaS` changes appreciably. Thus, spontaneity of a process depends very much on temperature.
For edothermic proces, both `DeltaH "and " DeltaS` are positive. The energy factor,the first factor of equation, opposes the spontaneity whereas entropy factor favours it . At low temperature, the favourable factor `TDeltaS` will be small and may be less than `Delta H, DeltaG` will have positive value indicating the non-spontaneity 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 exothermic process, both `DeltaH " and " DeltaS` would be negative. In this case, the first factor of equation(i) favours the spontaneity whereas the second factor opposes it. At high temperature, when `TDeltaSgt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity of the process. However, on decreasing temperature, the factore `TDeltaSlt 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 `25^(@)C,X_(2)O_(2)(l) to 2XO_(2)(g)" "DeltaH=2.1` kcal and `DeltaS=20` cal `K^(-1)` . The reaction would be:

Options

A. spontaneous
B. non-spontaneous
C. at equilibrium
D. unpredictable

Answer

a