Question

Using the Gibbs energy change, \(△G° = + 63.3\  kJ\), for the following reaction,
\(Ag_2CO_3 (g) ⇋ 2Ag+ (aq) + CO_3^{2–} (aq)\)
the \(K_{sp}\) of \(Ag_2CO_3 \)(s) in water at \(25°C\) is :
\((R = 8.314\ J K^{–1} mol^{–1})\)

• A. \(8 .0 \times 10^{–12}\)
• B. \(2.9 \times 10^{–3}\)
• C. \(7.9 \times 10^{–2}\)
• D. \(3.2 \times 10^{–36}\)

Answer 1

The Correct Option is A

To determine the solubility product constant, \(K_{sp}\), for \(Ag_2CO_3\) at 25°C using the Gibbs energy change (\(△G°\)), we follow these steps:

1. We know the relationship between Gibbs free energy change (\(△G°\)) and equilibrium constant (\(K\)) is given by the equation: \(△G° = -RT \ln K\) where:
   • \(R = 8.314 \ J \, K^{-1} \, mol^{-1}\) (universal gas constant)
   • \(T = 298 \ K\) (temperature in Kelvin, since \(25°C = 298 \ K\))
2. Substitute the known values into the equation to find \(K_{sp}\):
   \(63.3 \ \text{kJ} \, mol^{-1} = - (8.314 \ J \, K^{-1} \, mol^{-1}) \times (298 \ K) \times \ln K_{sp}\)
3. Convert \(\Delta G°\) from kJ to J by multiplying by 1000:
   \(\Delta G° = 63.3 \times 10^3 \ \text{J} \, mol^{-1}\)
4. Now, solving for \(\ln K_{sp}\):
   \(63.3 \times 10^3 \ = - (2480.172) \times \ln K_{sp}\)
5. Rearranging to solve for \(\ln K_{sp}\):
   \(\ln K_{sp} = - \frac{63.3 \times 10^3}{2480.172}\)
6. Calculate \(\ln K_{sp}\):
   \(\ln K_{sp} \approx -25.5\)
7. Convert \(\ln K_{sp}\) to \(K_{sp}\) by taking the exponential:
   \(K_{sp} = e^{-25.5} \approx 8.0 \times 10^{-12}\)

The correct answer is \(\boxed{8.0 \times 10^{-12}}\). This option matches our calculated value of \(K_{sp}\), validating our procedure and ensuring the consistent use of thermodynamic principles.