If `Delta H_(f)^(@)` for `Ag^(+)` (infinitely diluted), `NO_(3)^(-)` (infinitely diluted), `Cl^(-)` (infinitely diluted) and `AgCl(s)` are `10.5 .579-207.36, -67.159` and `-127.068` respectively. Calculate the enthalpy change for the reaction
`AgNO_(3)(aq.)rarr+HCl(aq.)rarrAgCl(s)+HNO_(3)(aq.)`

Options

A. `21.471 KJ//mol`
B. `145.688 KJ//mol`
C. `-65.488KJ//mol`
D. None

Explanation

To calculate the enthalpy change for the reaction:

\[ \text{AgNO}_3(aq) + \text{HCl}(aq) \rightarrow \text{AgCl}(s) + \text{HNO}_3(aq) \]

we will use the standard enthalpy of formation values provided for each species involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Enthalpy of Formation Values**:
- \( \Delta H_f^\circ (\text{Ag}^+) = 105.579 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{NO}_3^-) = -207.36 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{Cl}^-) = -67.159 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{AgCl}(s)) = -127.068 \, \text{kJ/mol} \)

2. **Write the Enthalpy Change Equation**:
The enthalpy change (\( \Delta H^\circ \)) for the reaction can be calculated using the formula:
\[ \Delta H^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \]

3. **Identify Products and Reactants**:
- **Products**: \( \text{AgCl}(s) \) and \( \text{HNO}_3(aq) \)
- **Reactants**: \( \text{AgNO}_3(aq) \) and \( \text{HCl}(aq) \)

4. **Calculate the Enthalpy of Formation for the Products**:
- For \( \text{AgCl}(s) \): \( \Delta H_f^\circ = -127.068 \, \text{kJ/mol} \)
- For \( \text{HNO}_3(aq) \): Since we don't have a direct value, we can use the enthalpy of formation of \( \text{NO}_3^- \) and \( \text{H}^+ \) (which is 0 for elemental state):
\[ \Delta H_f^\circ (\text{HNO}_3) = \Delta H_f^\circ (\text{NO}_3^-) + \Delta H_f^\circ (\text{H}^+) = -207.36 + 0 = -207.36 \, \text{kJ/mol} \]

5. **Calculate the Enthalpy of Formation for the Reactants**:
- For \( \text{AgNO}_3(aq) \): We can express it as:
\[ \Delta H_f^\circ (\text{AgNO}_3) = \Delta H_f^\circ (\text{Ag}^+) + \Delta H_f^\circ (\text{NO}_3^-) = 105.579 - 207.36 = -101.781 \, \text{kJ/mol} \]
- For \( \text{HCl}(aq) \): Similar to \( \text{HNO}_3 \):
\[ \Delta H_f^\circ (\text{HCl}) = \Delta H_f^\circ (\text{H}^+) + \Delta H_f^\circ (\text{Cl}^-) = 0 - 67.159 = -67.159 \, \text{kJ/mol} \]

6. **Substituting Values into the Enthalpy Change Equation**:
\[ \Delta H^\circ = \left[ -127.068 + (-207.36) \right] - \left[ -101.781 + (-67.159) \right] \]
\[ \Delta H^\circ = (-334.428) - (-168.94) \]
\[ \Delta H^\circ = -334.428 + 168.94 = -165.488 \, \text{kJ/mol} \]

7. **Final Calculation**:
\[ \Delta H^\circ = -65.488 \, \text{kJ/mol} \]

### Conclusion:
The enthalpy change for the reaction is \( \Delta H^\circ = -65.488 \, \text{kJ/mol} \).