Enthalpies of solution of `BaCI_(2). 2H_(2)O` and `BaCI_(2)` are 8.8 and `-20.6 kJ mol^(-1)` respectively.Calculate the heat of hydration of `BaCI_(2). 2H_(2)O`

Explanation

To calculate the heat of hydration of `BaCl2·2H2O`, we can use the enthalpies of solution of `BaCl2·2H2O` and `BaCl2`. The process can be broken down into the following steps:

### Step 1: Write the equations for the dissolution of `BaCl2·2H2O` and `BaCl2`.

1. The dissolution of `BaCl2·2H2O` in water can be represented as:
\[ \text{BaCl}_2\cdot2\text{H}_2\text{O (s)} \rightarrow \text{BaCl}_2\text{ (aq)} + 2\text{H}_2\text{O (l)} \]
The enthalpy change for this process is given as:
\[ \Delta H_1 = +8.8 \, \text{kJ/mol} \]

2. The dissolution of `BaCl2` in water can be represented as:
\[ \text{BaCl}_2\text{ (s)} \rightarrow \text{BaCl}_2\text{ (aq)} \]
The enthalpy change for this process is given as:
\[ \Delta H_2 = -20.6 \, \text{kJ/mol} \]

### Step 2: Reverse the equation for the dissolution of `BaCl2`.

To find the heat of hydration of `BaCl2·2H2O`, we need to consider the reverse process of the dissolution of `BaCl2`:
\[ \text{BaCl}_2\text{ (aq)} \rightarrow \text{BaCl}_2\text{ (s)} \]
The enthalpy change for this reversed process will be:
\[ \Delta H_2' = +20.6 \, \text{kJ/mol} \]

### Step 3: Combine the equations.

Now, we can combine the two equations:
1. The dissolution of `BaCl2·2H2O`:
\[ \text{BaCl}_2\cdot2\text{H}_2\text{O (s)} \rightarrow \text{BaCl}_2\text{ (aq)} + 2\text{H}_2\text{O (l)} \quad (\Delta H_1 = +8.8 \, \text{kJ/mol}) \]
2. The reverse dissolution of `BaCl2`:
\[ \text{BaCl}_2\text{ (aq)} \rightarrow \text{BaCl}_2\text{ (s)} \quad (\Delta H_2' = +20.6 \, \text{kJ/mol}) \]

When we add these two reactions, we get:
\[ \text{BaCl}_2\cdot2\text{H}_2\text{O (s)} \rightarrow \text{BaCl}_2\text{ (s)} + 2\text{H}_2\text{O (l)} + \text{BaCl}_2\text{ (aq)} \]

### Step 4: Calculate the total enthalpy change.

The total enthalpy change for the process is:
\[ \Delta H_{hydration} = \Delta H_1 + \Delta H_2' = 8.8 \, \text{kJ/mol} + 20.6 \, \text{kJ/mol} = 29.4 \, \text{kJ/mol} \]

### Step 5: Determine the heat of hydration.

Since the heat of hydration is the energy released when `BaCl2·2H2O` is formed from `BaCl2` and water, we take the negative of the total enthalpy change:
\[ \Delta H_{hydration} = -29.4 \, \text{kJ/mol} \]

### Final Answer:
The heat of hydration of `BaCl2·2H2O` is:
\[ \Delta H_{hydration} = -29.4 \, \text{kJ/mol} \]

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