Question

Given the reaction at 300 K: 

with \(\Delta G^\circ = -15 \, kJ \, mol^{-1}\). Calculate the value of log \(K\) for the reaction at the same temperature (R = 8.3 J \(K^{-1} \, mol^{-1}\)).

• A. 13.1
• B. 1.31
• C. 26.2
• D. 2.62

Hint:

To convert \(\ln\) to \(\log_{10}\), divide the \(\ln\) value by \(\ln 10\) which is approximately \(2.3026\).

Answer 1

The Correct Option is D

Step 1: Convert \(\Delta G^\circ\) from \(kJ/mol\) to \(J/mol\). \[ \Delta G^\circ = -15 \, kJ/mol \times 1000 = -15000 \, J/mol \] Step 2: Use the relationship between \(\Delta G^\circ\) and \(K\). The relationship is given by the equation: \[ \Delta G^\circ = -RT \ln K \] Where \( R = 8.3 \, J \, K^{-1} \, mol^{-1} \) and \( T = 300 \, K \). 
Step 3: Solve for \(\ln K\). \[ \ln K = -\frac{\Delta G^\circ}{RT} = -\frac{-15000}{8.3 \times 300} \approx 6.02 \] 
Step 4: Convert \(\ln K\) to \(\log_{10} K\) (common logarithm). \[ \log_{10} K = \frac{\ln K}{\ln 10} \approx \frac{6.02}{2.3026} \approx 2.62 \]