Using the Gibb's energy change , `DeltaG^@`=+63.3 kJ , for the following reaction,
`Ag_2CO_3 (s) hArr 2Ag^(+)(aq) +CO_3^(2-)(aq)` the `K_(sp)` of `Ag_2CO_3(s)` in water at `25^@C` is `("R=8.314 K"^(-1) "mol"^(-1))`

Options

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

Answer

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

### Step 1: Convert Gibbs Energy Change to Joules
Since the value of \( R \) (the gas constant) is given in \( \text{J K}^{-1} \text{mol}^{-1} \), we need to convert \( \Delta G^\circ \) from kJ to J.

\[ \Delta G^\circ = 63.3 \, \text{kJ} = 63.3 \times 10^3 \, \text{J} = 63300 \, \text{J} \]

### Step 2: Write the Relationship Between Gibbs Energy and \( K_{sp} \)
The relationship between Gibbs free energy change and the solubility product \( K_{sp} \) is given by the equation:

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

Where:
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( T = 298 \, \text{K} \) (temperature in Kelvin)

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

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

### Step 4: Substitute the Values into the Equation
Substituting the values we have:

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

Calculating the denominator:

\[ RT = 8.314 \times 298 = 2477.572 \, \text{J mol}^{-1} \]

Now substituting back:

\[ \ln K_{sp} = -\frac{63300}{2477.572} \approx -25.5 \]

### Step 5: Exponentiate to Find \( K_{sp} \)
To find \( K_{sp} \), we exponentiate both sides:

\[ K_{sp} = e^{-25.5} \]

Calculating \( K_{sp} \):

\[ K_{sp} \approx 8.2 \times 10^{-12} \]

### Final Answer
The solubility product \( K_{sp} \) of \( Ag_2CO_3 \) at \( 25^\circ C \) is approximately:

\[ K_{sp} \approx 8.2 \times 10^{-12} \]

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