The heat of solution of anhydrous `CuSO_(4)` and `CuSO_(4).5H_(2)O` are -15.89 and 2.80 Kcal `mol^(-1)` respectively. What will be the heat of hydration of anhydrous `CuSO_(4)` ?

Options

A. `-18.69" Kcal"`
B. `18.69" Kcal"`
C. `-28.96" Kcal"`
D. `28.96" Kcal"`

Answer

To find the heat of hydration of anhydrous CuSO₄, we can use the given heats of solution for both anhydrous CuSO₄ and CuSO₄·5H₂O. Here’s a step-by-step solution:

### Step 1: Understand the given data
We have:
- Heat of solution of anhydrous CuSO₄ (ΔH₁) = -15.89 kcal/mol
- Heat of solution of CuSO₄·5H₂O (ΔH₂) = +2.80 kcal/mol

### Step 2: Write the dissolution reactions
1. For anhydrous CuSO₄:
\[ \text{CuSO}_4 (s) \rightarrow \text{CuSO}_4 (aq) \quad \text{with} \quad \Delta H_1 = -15.89 \text{ kcal/mol} \]

2. For hydrated CuSO₄·5H₂O:
\[ \text{CuSO}_4 \cdot 5\text{H}_2\text{O} (s) \rightarrow \text{CuSO}_4 (aq) + 5\text{H}_2\text{O} (l) \quad \text{with} \quad \Delta H_2 = +2.80 \text{ kcal/mol} \]

### Step 3: Write the hydration reaction for anhydrous CuSO₄
The heat of hydration (ΔH_hydration) can be expressed as:
\[ \text{CuSO}_4 (s) + 5\text{H}_2\text{O} (l) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O} (s) \]

### Step 4: Relate the heats of reaction
Using Hess's law, we can relate the heats of the reactions:
\[ \Delta H_hydration = \Delta H_1 + \Delta H_2 \]
Substituting the values:
\[ \Delta H_hydration = (-15.89 \text{ kcal/mol}) + (2.80 \text{ kcal/mol}) \]

### Step 5: Calculate the heat of hydration
\[ \Delta H_hydration = -15.89 + 2.80 = -13.09 \text{ kcal/mol} \]

### Conclusion
The heat of hydration of anhydrous CuSO₄ is -13.09 kcal/mol.