Question

The value of \( \log_{10} K \) for a reaction \( A \rightleftharpoons B \) is (Given: \( \Delta H^\circ_{298K} = -54.07 \, \text{kJ mol}^{-1} \), \( \Delta S^\circ_{298K} = 10 \, \text{JK}^{-1} \, \text{mol}^{-1} \) and \( R = 8.314 \, \text{JK}^{-1} \, \text{mol}^{-1} \)) \[ 2.303 \times 8.314 \times 298 = 5705 \]

• A. 5
• B. 10
• C. 95
• D. 100

Hint:

To find \( \log_{10} K \), use the relationship between Gibbs free energy, enthalpy, and entropy, along with the gas constant \( R \).

Answer 1

The Correct Option is B

To calculate the value of \( \log_{10} K \), we use the equation: \[ \log_{10} K = \frac{-\Delta G^\circ}{2.303RT} \] Using the relation \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \), we can substitute the values and simplify the expression, resulting in \( \log_{10} K = 10 \).