The value of `1og_(10)` K for a reaction `AhArrB` is:
(Given, `Delta_(r)H_(298K)^(@)=-54.07kJ" "mol^(-1),Delta_(r)S_(298K)^(@)=10JK^(-1)" "mol^(-1)` and
R=8.314JK^(-1)` "mol^(-1),2.303xx8.314xx298=5705`)

Options

A. `5`
B. `10`
C. `95`
D. `100`

Explanation

To solve the problem, we need to find the value of \( \log_{10} K \) for the reaction \( A \rightleftharpoons B \). We will use the provided thermodynamic data and the relationship between Gibbs free energy, enthalpy, and entropy.

### Step-by-Step Solution:

1. **Write the Gibbs Free Energy Equation**:
The Gibbs free energy change at standard conditions is given by:
\[ \Delta G = \Delta G^\circ - T \Delta S^\circ \]

2. **Convert Given Values**:
- Given \( \Delta_r H_{298K}^\circ = -54.07 \, \text{kJ/mol} \). Convert this to joules:
\[ \Delta_r H_{298K}^\circ = -54.07 \times 10^3 \, \text{J/mol} = -54070 \, \text{J/mol} \]
- Given \( \Delta_r S_{298K}^\circ = 10 \, \text{J/(K mol)} \).
- Given \( T = 298 \, \text{K} \).

3. **Calculate \( \Delta G \)**:
Substitute the values into the Gibbs free energy equation:
\[ \Delta G = -54070 \, \text{J/mol} - (298 \, \text{K} \times 10 \, \text{J/(K mol)}) \]
\[ \Delta G = -54070 \, \text{J/mol} - 2980 \, \text{J/mol} = -57050 \, \text{J/mol} \]

4. **Use the Relationship Between \( \Delta G \) and \( K \)**:
We know that:
\[ \Delta G^\circ = -2.303 R T \log_{10} K \]
Rearranging gives:
\[ \log_{10} K = -\frac{\Delta G^\circ}{2.303 R T} \]

5. **Substitute Values**:
Substitute \( \Delta G^\circ = -57050 \, \text{J/mol} \), \( R = 8.314 \, \text{J/(K mol)} \), and \( T = 298 \, \text{K} \):
\[ \log_{10} K = -\frac{-57050}{2.303 \times 8.314 \times 298} \]
We know from the problem statement that:
\[ 2.303 \times 8.314 \times 298 = 5705 \]
Therefore:
\[ \log_{10} K = \frac{57050}{5705} = 10 \]

6. **Final Answer**:
Thus, the value of \( \log_{10} K \) for the reaction \( A \rightleftharpoons B \) is:
\[ \log_{10} K = 10 \]