Gibbs Helmholtz equation relates the enthalpy, entropy and free energy change of the process at constant pressure and temperature as
`DeltaG=DeltaH-TDeltaS " (at constant P, T)"`
In General the magnitude of `DeltaH` does not change much with the change in temperature but the terms `TDeltaS` changes appreciably. Hence in some process spontaneity is very much dependent on temperature and such processes are generally known as entropy driven process.
The Dissolution of `CaCl_(2).6H_(2)O` in a large volume of water is endothermic to the extent of 3.5 kcal `"mol"^(-1)` and `DeltaH` for the reaction is -23.2 kcal `"mol"^(-1)`.
`CaCl_(2)(s)+6H_(2)O(l)rarrCaCl_(2).6H_(2)O(s)`
Select the correct statement :

Options

A. `DeltaH_("solution")` for anhydrous `CaCl_(2)` is - 19.7 kcal/mol and the process is enthalpy driven
B. `DeltaH_("solution")` for anhydrous `CaCl_(2)` is - 19.7 kcal/mol and the process is entropy driven
C. Dissolution of `CaCl_(2).6H_(2)O` in water is enthalpy driven process
D. The `Delta_(r )S` the reaction `CaCl_(2)(s)+6H_(2)O(l)rarrCaCl_(2).6H_(2)O(s)` is negative

Explanation

To solve the problem, we will analyze the dissolution of `CaCl2.6H2O` in water using the Gibbs-Helmholtz equation and the given data.

### Step-by-Step Solution:

1. **Understand the Reaction**:
The dissolution of `CaCl2.6H2O` can be represented as:
\[ \text{CaCl}_2(s) + 6\text{H}_2O(l) \rightarrow \text{CaCl}_2.6H_2O(s) \]
Here, `CaCl2` is dissolving in water to form a hydrated solid.

2. **Identify Given Values**:
- The dissolution is endothermic with a heat change of `3.5 kcal/mol`.
- The enthalpy change (`ΔH`) for the reaction is given as `-23.2 kcal/mol`.

3. **Apply the Gibbs-Helmholtz Equation**:
The Gibbs-Helmholtz equation is given by:
\[ ΔG = ΔH - TΔS \]
Where:
- `ΔG` is the change in Gibbs free energy.
- `ΔH` is the change in enthalpy.
- `T` is the temperature in Kelvin.
- `ΔS` is the change in entropy.

4. **Calculate the Change in Entropy (ΔS)**:
Since the process is endothermic, we can infer that the dissolution is driven by an increase in entropy. We can rearrange the equation to solve for `ΔS`:
\[ ΔS = \frac{ΔH - ΔG}{T} \]
However, we need to find `ΔG` first.

5. **Determine ΔG**:
In an endothermic process, if the dissolution occurs spontaneously, `ΔG` should be negative. Given that `ΔH` is `-23.2 kcal/mol`, we can assume that the dissolution is spontaneous at a certain temperature, which means:
\[ ΔG < 0 \]

6. **Analyze the Values**:
We have:
\[ ΔH = -23.2 \text{ kcal/mol} \]
And the heat absorbed during dissolution is `3.5 kcal/mol`. Therefore, we can assume:
\[ ΔH = -23.2 + 3.5 = -19.7 \text{ kcal/mol} \]

7. **Conclusion**:
The process is driven by entropy, and the correct interpretation of the enthalpy change indicates that the dissolution of `CaCl2.6H2O` is indeed an entropy-driven process.

### Final Statement:
Based on the calculations and analysis, the correct statement regarding the dissolution of `CaCl2.6H2O` in water is that it is an entropy-driven process, and the effective enthalpy change for the dissolution is `-19.7 kcal/mol`.

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