The value of `log_(10)K` for the reaction :` A harr B` , If `Delta H^(@)` = - 55.07 kj `"mole"^(-1)` at 298K. `Delta S^(@) = 10 jK^(-1) "mole"^(-1) ` at 298 K , R = 8.314 `jK^(-1) "mole"^(-1) `

Options

A. 5.57
B. 10.174
C. 95.55
D. 100.0

Answer

To solve the problem, we need to calculate the value of \( \log_{10} K \) for the reaction \( A \rightleftharpoons B \) using the given thermodynamic data. We will follow these steps:

### Step 1: Convert Units
Given:
- \( \Delta H^\circ = -55.07 \, \text{kJ/mol} \)
- \( \Delta S^\circ = 10 \, \text{J/K/mol} \)

Convert \( \Delta H^\circ \) from kJ to J:
\[ \Delta H^\circ = -55.07 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -55070 \, \text{J/mol} \]

### Step 2: Calculate \( \Delta G^\circ \)
Using the formula:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting the values:
\[ \Delta G^\circ = -55070 \, \text{J/mol} - (298 \, \text{K} \times 10 \, \text{J/K/mol}) \]
\[ \Delta G^\circ = -55070 \, \text{J/mol} - 2980 \, \text{J/mol} \]
\[ \Delta G^\circ = -58050 \, \text{J/mol} \]

### Step 3: Relate \( \Delta G^\circ \) to \( K \)
At equilibrium, \( \Delta G^\circ \) is related to the equilibrium constant \( K \) by the equation:
\[ \Delta G^\circ = -RT \ln K \]
We can convert this to base 10 logarithm:
\[ \Delta G^\circ = -2.303 RT \log_{10} K \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \log_{10} K = -\frac{\Delta G^\circ}{2.303 RT} \]

### Step 5: Substitute Values
Substituting the values into the equation:
- \( R = 8.314 \, \text{J/K/mol} \)
- \( T = 298 \, \text{K} \)

\[ \log_{10} K = -\frac{-58050}{2.303 \times 8.314 \times 298} \]

### Step 6: Calculate the Denominator
Calculating the denominator:
\[ 2.303 \times 8.314 \times 298 \approx 5706.1 \]

### Step 7: Final Calculation
Now substituting back:
\[ \log_{10} K = \frac{58050}{5706.1} \approx 10.18 \]

### Conclusion
Thus, the value of \( \log_{10} K \) is approximately \( 10.18 \).

### Final Answer
The value of \( \log_{10} K \) for the reaction \( A \rightleftharpoons B \) is approximately \( 10.18 \).

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