Calculate `Delta_(r)G^(c-)` of the reaction `:`
`Ag^(o+)(aq)+Cl^(c-)(aq) rarr AgCl(s)`
Given `:Delta_(f)G^(c-)._(AgCl)=-109kJ mol ^(-1)`
`Delta_(f)G^(c-)_((Cl^(c-)))=-129k J mol ^(-1)`
`Delta_(f)G^(c-)._((Ag^(o+)))=-77 kJ mol ^(-1)`

Options

A. `-97 kJ mol^(-1)`
B. `-57kJ mol^(-1)`
C. `57kJmol^(-1)`
D. `97kJ mol^(-1)`

Explanation

To calculate the standard Gibbs free energy change (Δ_rG°) for the reaction:

\[ \text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s) \]

we can use the Gibbs free energy of formation (Δ_fG°) values provided for the reactants and products. The formula to calculate Δ_rG° is:

\[ \Delta_rG^\circ = \Delta_fG^\circ(\text{products}) - \Delta_fG^\circ(\text{reactants}) \]

### Step-by-Step Solution:

1. **Identify the Gibbs free energy of formation values:**
- For AgCl(s): \( \Delta_fG^\circ(\text{AgCl}) = -109 \, \text{kJ/mol} \)
- For Ag\(^+(aq)\): \( \Delta_fG^\circ(\text{Ag}^+) = -77 \, \text{kJ/mol} \)
- For Cl\(^-(aq)\): \( \Delta_fG^\circ(\text{Cl}^-) = -129 \, \text{kJ/mol} \)

2. **Calculate the total Gibbs free energy of formation for the reactants:**
\[ \Delta_fG^\circ(\text{reactants}) = \Delta_fG^\circ(\text{Ag}^+) + \Delta_fG^\circ(\text{Cl}^-) \]
\[ = (-77 \, \text{kJ/mol}) + (-129 \, \text{kJ/mol}) \]
\[ = -77 - 129 = -206 \, \text{kJ/mol} \]

3. **Calculate the total Gibbs free energy of formation for the products:**
\[ \Delta_fG^\circ(\text{products}) = \Delta_fG^\circ(\text{AgCl}) \]
\[ = -109 \, \text{kJ/mol} \]

4. **Substitute the values into the Δ_rG° formula:**
\[ \Delta_rG^\circ = \Delta_fG^\circ(\text{products}) - \Delta_fG^\circ(\text{reactants}) \]
\[ = (-109 \, \text{kJ/mol}) - (-206 \, \text{kJ/mol}) \]
\[ = -109 + 206 = 97 \, \text{kJ/mol} \]

5. **Final Result:**
\[ \Delta_rG^\circ = 97 \, \text{kJ/mol} \]