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

Options

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

Explanation

To solve the problem, we will use the Gibbs-Helmholtz equation, which relates the change in free energy (ΔG) to the changes in enthalpy (ΔH) and entropy (ΔS) of a reaction. The equation is given by:

\[ \Delta G = \Delta H - T \Delta S \tag{1} \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Enthalpy change (ΔH) = -15.0 kcal/mol
- Entropy change (ΔS) = -7.2 cal/K·mol
- Temperature (T) = 300 K

2. **Convert Units**:
- Since ΔH is given in kilocalories and ΔS in calories, we need to convert ΔH to calories for consistency:
\[ \Delta H = -15.0 \text{ kcal/mol} = -15.0 \times 1000 \text{ cal/mol} = -15000 \text{ cal/mol} \]

3. **Calculate TΔS**:
- Now, we can calculate the term \(T \Delta S\):
\[ T \Delta S = 300 \text{ K} \times (-7.2 \text{ cal/K·mol}) = -2160 \text{ cal/mol} \]

4. **Substitute Values into the Gibbs-Helmholtz Equation**:
- Substitute the values of ΔH and \(T \Delta S\) into equation (1):
\[ \Delta G = -15000 \text{ cal/mol} - (-2160 \text{ cal/mol}) \]
\[ \Delta G = -15000 \text{ cal/mol} + 2160 \text{ cal/mol} \]
\[ \Delta G = -12840 \text{ cal/mol} \]

5. **Convert ΔG to kilocalories**:
- Convert ΔG back to kilocalories:
\[ \Delta G = -12840 \text{ cal/mol} = -12.84 \text{ kcal/mol} \]

6. **Determine Spontaneity**:
- Since ΔG is negative (\(-12.84 \text{ kcal/mol} < 0\)), the reaction is spontaneous.

### Final Answer:
The free energy change for the reaction is \(-12.84 \text{ kcal/mol}\) and the reaction is spontaneous.

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