For the following reaction
Ag_((aq))^(+)+Cl_((aq))^(-)rarrAgCl_((s))`
Given : `DeltaG_(f)^(@), AgCl=-"112.44 kJ/mol," DeltaG_(f)^(@) Cl^(-)=-"130 kJ/mol", DeltaG_(f)^(@)Ag^(+)="75 kJ/mol"` Report your answer by rounding it upto nearest whole number. The `K_(sp)` of `AgCl` is `nxx10^(-10)`. The value of 'n'' is .

Answer

To solve the problem, we need to calculate the solubility product constant (Ksp) for the reaction:

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

Given the standard Gibbs free energy of formation (\( \Delta G_f^\circ \)) values:
- \( \Delta G_f^\circ \) for AgCl = -112.44 kJ/mol
- \( \Delta G_f^\circ \) for Cl^- = -130 kJ/mol
- \( \Delta G_f^\circ \) for Ag^+ = 75 kJ/mol

### Step 1: Calculate \( \Delta G^\circ \) for the reaction

The standard Gibbs free energy change for the reaction can be calculated using the formula:

\[ \Delta G^\circ_{\text{reaction}} = \Delta G_f^\circ (\text{products}) - \Delta G_f^\circ (\text{reactants}) \]

Substituting the values:

\[ \Delta G^\circ_{\text{reaction}} = \Delta G_f^\circ (\text{AgCl}) - \left( \Delta G_f^\circ (\text{Ag}^+) + \Delta G_f^\circ (\text{Cl}^-) \right) \]

\[ = (-112.44 \text{ kJ/mol}) - \left( 75 \text{ kJ/mol} - 130 \text{ kJ/mol} \right) \]

Calculating the right-hand side:

\[ = -112.44 \text{ kJ/mol} - (75 \text{ kJ/mol} - 130 \text{ kJ/mol}) \]
\[ = -112.44 \text{ kJ/mol} - (75 + 130) \text{ kJ/mol} \]
\[ = -112.44 \text{ kJ/mol} - 205 \text{ kJ/mol} \]
\[ = -112.44 + 205 \text{ kJ/mol} \]
\[ = -57.44 \text{ kJ/mol} \]

### Step 2: Convert to Joules

Convert \( \Delta G^\circ_{\text{reaction}} \) to Joules:

\[ \Delta G^\circ_{\text{reaction}} = -57.44 \text{ kJ/mol} \times 1000 \text{ J/kJ} = -57440 \text{ J/mol} \]

### Step 3: Relate \( \Delta G^\circ \) to Ksp

Using the relationship between \( \Delta G^\circ \) and the equilibrium constant \( K \):

\[ \Delta G^\circ = -RT \ln K \]

Where:
- \( R = 8.314 \text{ J/(mol K)} \)
- \( T = 300 \text{ K} \)

Rearranging gives:

\[ \ln K = -\frac{\Delta G^\circ}{RT} \]

Substituting the values:

\[ \ln K = -\frac{-57440 \text{ J/mol}}{(8.314 \text{ J/(mol K)})(300 \text{ K})} \]

Calculating the denominator:

\[ = 8.314 \times 300 = 2494.2 \text{ J/mol} \]

Now substituting back:

\[ \ln K = \frac{57440}{2494.2} \approx 23.05 \]

### Step 4: Calculate K

Exponentiating to find \( K \):

\[ K = e^{23.05} \approx 9.8 \times 10^{10} \]

### Step 5: Determine Ksp

The solubility product \( K_{sp} \) can be expressed as:

\[ K_{sp} = s^2 \]

Where \( s \) is the solubility. Since we have \( K_{sp} \approx 10^{10} \), we can express it in the form \( n \times 10^{-10} \):

\[ K_{sp} = 1 \times 10^{-10} \]

Thus, the value of \( n \) is:

\[ \boxed{1} \]