For a reaction `A(g) hArr B(g)` at equilibrium . The partial pressure of B is found to be one fourth of the partial pressure of A . The value of `DeltaG^(@)` of the reaction `A rarrB` is

Options

A. RT ln `4`
B. `-"RT" ln 4`
C. `"RT" log 4`
D. `-"RT" log 4`

Explanation

To find the value of \( \Delta G^\circ \) for the reaction \( A(g) \rightleftharpoons B(g) \) at equilibrium, we can follow these steps:

### Step 1: Understand the relationship between partial pressures and equilibrium constant
At equilibrium, the relationship between the partial pressures of the reactants and products can be expressed in terms of the equilibrium constant \( K \). For the reaction \( A(g) \rightleftharpoons B(g) \), the equilibrium constant \( K \) is given by:

\[ K = \frac{P_B}{P_A} \]

where \( P_A \) is the partial pressure of A and \( P_B \) is the partial pressure of B.

### Step 2: Substitute the given values
From the problem, we know that the partial pressure of B is one fourth of the partial pressure of A:

\[ P_B = \frac{1}{4} P_A \]

Substituting this into the expression for \( K \):

\[ K = \frac{\frac{1}{4} P_A}{P_A} = \frac{1}{4} \]

### Step 3: Relate \( \Delta G^\circ \) to \( K \)
The standard Gibbs free energy change \( \Delta G^\circ \) is related to the equilibrium constant \( K \) by the equation:

\[ \Delta G^\circ = -RT \ln K \]

### Step 4: Substitute the value of \( K \)
Now we substitute the value of \( K \) we found into the equation:

\[ \Delta G^\circ = -RT \ln\left(\frac{1}{4}\right) \]

### Step 5: Simplify the logarithm
Using the property of logarithms, we can simplify \( \ln\left(\frac{1}{4}\right) \):

\[ \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = 0 - \ln(4) = -\ln(4) \]

### Step 6: Substitute back into the equation
Now substituting this back into the equation for \( \Delta G^\circ \):

\[ \Delta G^\circ = -RT (-\ln(4)) = RT \ln(4) \]

### Conclusion
Thus, the value of \( \Delta G^\circ \) for the reaction \( A(g) \rightleftharpoons B(g) \) is:

\[ \Delta G^\circ = RT \ln(4) \]

The correct answer is:

\[ \Delta G^\circ = -RT \ln(4) \]