Question

At 298K, if the standard Gibbs energy change \(\Delta_r G^\circ\) of a reaction is - 115 kJ, the value of \(\log_{10} K_p\) will be (\(R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}\))

• A. +20.15
• B. -20.15
• C. -10.30
• D. +10.30

Hint:

A negative value for \(\Delta_r G^\circ\) means the reaction is spontaneous and the equilibrium lies far to the right, so \(K_p>1\). This implies that \(\log_{10} K_p\) must be positive. This quick check can help you eliminate negative options immediately.

Answer 1

The Correct Option is A

Step 1: Formula: \[ \Delta G^\ominus = -2.303 RT \log K_p \]
Step 2: Substitution: \( -115000 \text{ J} = -2.303 \times 8.314 \times 298 \times \log K_p \) \[ 115000 = 5705.8 \times \log K_p \] \[ \log K_p = \frac{115000}{5705.8} \approx 20.15 \]