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: Understanding the Concept:
This question relates the standard Gibbs free energy change (\(\Delta_r G^\circ\)) of a reaction to its equilibrium constant (\(K_p\)). A negative \(\Delta_r G^\circ\) indicates a spontaneous reaction, which corresponds to an equilibrium constant greater than 1.
Step 2: Key Formula or Approach:
The relationship between \(\Delta_r G^\circ\) and the equilibrium constant \(K_p\) is given by the equation:
\[ \Delta_r G^\circ = -RT \ln K_p \] To work with base-10 logarithm (\(\log_{10}\)), we use the conversion \(\ln x = 2.303 \log_{10} x\):
\[ \Delta_r G^\circ = -2.303 RT \log_{10} K_p \] Step 3: Detailed Explanation:
First, ensure all units are consistent. We should use Joules for energy.
- \(\Delta_r G^\circ = -115 \text{ kJ} = -115 \times 10^3 \text{ J}\).
- Temperature, \(T = 298 \text{ K}\).
- Gas constant, \(R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}\).
Rearrange the formula to solve for \(\log_{10} K_p\):
\[ \log_{10} K_p = \frac{-\Delta_r G^\circ}{2.303 RT} \] Substitute the given values into the equation:
\[ \log_{10} K_p = \frac{-(-115 \times 10^3 \text{ J})}{2.303 \times (8.314 \text{ J K}^{-1}\text{mol}^{-1}) \times (298 \text{ K})} \] \[ \log_{10} K_p = \frac{115000}{2.303 \times 8.314 \times 298} \] Calculate the value of the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5705.84 \] Now, calculate the final value:
\[ \log_{10} K_p = \frac{115000}{5705.84} \approx 20.1548 \] This value is approximately +20.15.
Step 4: Final Answer:
The value of \(\log_{10} K_p\) is +20.15. Therefore, option (A) is correct.