Question

For the equilibrium what is the value of \(\log_{10}K\) at 298 K? \((\Delta_r H^\circ = -54.07 \text{ kJ mol}^{-1}, \Delta_r S^\circ = 10 \text{ J K}^{-1}\) and \(2.303RT = 5705)\)}

• A. 10
• B. 5
• C. 90
• D. 95
• E. 100

Hint:

Use: \[ \Delta G^\circ = -2.303RT\log K \] and always convert \(\Delta H^\circ\), \(\Delta S^\circ\), and \(RT\) into consistent units first.

Answer 1

The Correct Option is A

First calculate \(\Delta G^\circ\): \[ \Delta G^\circ=\Delta H^\circ - T\Delta S^\circ \] Given: \[ \Delta H^\circ=-54.07\text{ kJ mol}^{-1} \] \[ \Delta S^\circ=10\text{ J K}^{-1}\text{mol}^{-1}=0.01\text{ kJ K}^{-1}\text{mol}^{-1} \] So, \[ \Delta G^\circ=-54.07-(298)(0.01) \] \[ \Delta G^\circ=-54.07-2.98=-57.05\text{ kJ mol}^{-1} \] Now use: \[ \Delta G^\circ = -2.303RT\log K \] Given: \[ 2.303RT=5705\text{ J mol}^{-1}=5.705\text{ kJ mol}^{-1} \] Thus, \[ -57.05=-5.705\log K \] \[ \log K = \frac{57.05}{5.705}=10 \]
Hence, the correct answer is: \[ \boxed{(A)\ 10} \]