Question

The value of \(log_{10}K\) for a reaction A ⇋ B is

(Given,

 \(\Delta H^{\circ}_{298K}\) = –54.67 kJ \(mol^{-1}\)

 \(\Delta H^{\circ}_{298K}\) = 10 kJ \(mol^{-1}\)

 and R = 8.314 J \(K^{-1}mol^{-1}\)

 2.303 × 8.314 × 298 = 5705) 

Answer 1

Correct Answer: 10

Calculation of \( \Delta G^\circ \) and \( \log K \)

Given Formula:

\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)

Step 1: Substitute the Values

\[ \Delta H^\circ = -54.07 \times 1000 \, \text{J} \quad (\text{Convert kJ to J}) \] \[ \Delta S^\circ = 10 \, \text{J K}^{-1} \] \[ T = 298 \, \text{K} \]

Substituting into the formula:

\[ \Delta G^\circ = -54.07 \times 1000 - 298 \times 10 \]

Simplify:

\[ \Delta G^\circ = -54070 - 2980 = -57050 \, \text{J} \]

Step 2: Relating \( \Delta G^\circ \) and \( \log K \)

The relationship between \( \Delta G^\circ \) and \( \log K \) is given by:

\[ \Delta G^\circ = -2.303RT\log K \]

Where:

• \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \) (Gas constant)
• \( T = 298 \, \text{K} \) (Temperature)

Rearranging for \( \log K \):

\[ \log K = \frac{-\Delta G^\circ}{2.303RT} \]

Substitute the values:

\[ \log K = \frac{-(-57050)}{2.303 \times 8.314 \times 298} \]

Numerical simplification:

\[ \log K = \frac{57050}{5705.96} \approx 10 \]

Conclusion:

The value of \( \log K \) is 10.