Using the Gibbs energy change, `Delta G^(@)=+ 63.3 kJ`, for the following reaction,
`Ag_(2)CO_(3)hArr2Ag^(+)(aq)+CO_(3)^(2-)`
the `K_(sp)` of `Ag_(2)CO_(3)(s)` in water at `25^(@)C` is
`(R=8.314 JK^(-1)mol^(-1))`

Options

A. `3.2xx10^(-26)`
B. `8.0xx10^(-12)`
C. `2.9xx10^(-3)`
D. `7.9xx10^(-2)`

Explanation

To find the solubility product constant \( K_{sp} \) of \( Ag_2CO_3(s) \) using the Gibbs energy change \( \Delta G^\circ = +63.3 \, \text{kJ} \), we can follow these steps:

### Step 1: Convert \( \Delta G^\circ \) to Joules
Since the value of \( \Delta G^\circ \) is given in kilojoules, we need to convert it to joules for consistency with the gas constant \( R \).

\[ \Delta G^\circ = 63.3 \, \text{kJ} = 63.3 \times 1000 \, \text{J} = 63300 \, \text{J} \]

### Step 2: Use the Gibbs Free Energy Equation
The relationship between Gibbs free energy change and the equilibrium constant is given by the equation:

\[ \Delta G^\circ = -RT \ln K_{sp} \]

Where:
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin (for \( 25^\circ C \), \( T = 298 \, \text{K} \))

### Step 3: Rearrange the Equation to Solve for \( K_{sp} \)
Rearranging the equation to solve for \( K_{sp} \):

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

### Step 4: Substitute the Values
Now, substitute the values into the equation:

\[ \ln K_{sp} = -\frac{63300 \, \text{J}}{(8.314 \, \text{J K}^{-1} \text{mol}^{-1})(298 \, \text{K})} \]

Calculating the denominator:

\[ RT = 8.314 \times 298 = 2477.572 \, \text{J} \]

Now substituting back:

\[ \ln K_{sp} = -\frac{63300}{2477.572} \approx -25.53 \]

### Step 5: Calculate \( K_{sp} \)
To find \( K_{sp} \), we take the exponential of both sides:

\[ K_{sp} = e^{-25.53} \]

Calculating \( K_{sp} \):

\[ K_{sp} \approx 1.12 \times 10^{-11} \]

### Final Answer
Thus, the solubility product constant \( K_{sp} \) of \( Ag_2CO_3(s) \) at \( 25^\circ C \) is approximately:

\[ K_{sp} \approx 1.12 \times 10^{-11} \]

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