If the heat fo dissolution of anhydrous `CuSO_(4)` and `CuSO_(4).5H_(2)O` is `-15.89 kcal` and `2.80 kcal`, respectively, then the heat of hydration fo `CuSO_(4)` to form `CuSO_(4).5H_(2)O` is

Options

A. `-13.09 kcal`
B. `-18.69 kcal`
C. `+13.09 kcal`
D. `+18.69kcal`

Explanation

To find the heat of hydration of anhydrous \( \text{CuSO}_4 \) to form \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \), we can use the given heat of dissolution values for both compounds.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Heat of dissolution of anhydrous \( \text{CuSO}_4 \) (solid to aqueous):
\[ \Delta H_1 = -15.89 \text{ kcal} \]
- Heat of dissolution of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) (solid to aqueous):
\[ \Delta H_2 = -2.80 \text{ kcal} \]

2. **Write the Dissolution Reactions:**
- For \( \text{CuSO}_4 \):
\[ \text{CuSO}_4 (s) \rightarrow \text{CuSO}_4 (aq) \quad \Delta H_1 = -15.89 \text{ kcal} \]
- For \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \):
\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} (s) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O} (aq) \quad \Delta H_2 = -2.80 \text{ kcal} \]

3. **Set Up the Equation for Heat of Hydration:**
- The heat of hydration (\( \Delta H_{\text{hydration}} \)) can be calculated using the following relationship:
\[ \Delta H_{\text{hydration}} = \Delta H_1 - \Delta H_2 \]
- This represents the heat change when \( \text{CuSO}_4 \) dissolves to form \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).

4. **Substitute the Values:**
- Substitute the known values into the equation:
\[ \Delta H_{\text{hydration}} = (-15.89 \text{ kcal}) - (-2.80 \text{ kcal}) \]
- This simplifies to:
\[ \Delta H_{\text{hydration}} = -15.89 \text{ kcal} + 2.80 \text{ kcal} \]
\[ \Delta H_{\text{hydration}} = -15.89 + 2.80 = -13.09 \text{ kcal} \]

5. **Final Answer:**
- The heat of hydration of \( \text{CuSO}_4 \) to form \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is:
\[ \Delta H_{\text{hydration}} = -13.09 \text{ kcal} \]