Following reaction occurs at `25^(@)` :
`2NO(g, 1xx10^(-5)atm) + CI_(2)(g, 1 xx 10^(-2)atm)hArr 2NOCI(g,1 xx 10^(-2)atm)`
Calculate `DeltaG^(@) [R = 8 J//"mole" K]`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the given reaction at 25°C, we will follow these steps:

### Step 1: Write the Reaction and Identify the Given Values
The reaction is:
\[ 2 \text{NO}(g, 1 \times 10^{-5} \text{ atm}) + \text{Cl}_2(g, 1 \times 10^{-2} \text{ atm}) \rightleftharpoons 2 \text{NOCl}(g, 1 \times 10^{-2} \text{ atm}) \]

Given:
- Pressure of NO = \( 1 \times 10^{-5} \) atm
- Pressure of Cl₂ = \( 1 \times 10^{-2} \) atm
- Pressure of NOCl = \( 1 \times 10^{-2} \) atm
- Temperature (T) = 25°C = 298 K
- Gas constant (R) = 8 J/(mol K)

### Step 2: Calculate the Equilibrium Constant (K)
The equilibrium constant (K) for the reaction can be expressed in terms of the partial pressures:
\[ K = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 \cdot (P_{\text{Cl}_2})} \]

Substituting the values:
\[ K = \frac{(1 \times 10^{-2})^2}{(1 \times 10^{-5})^2 \cdot (1 \times 10^{-2})} \]
\[ K = \frac{1 \times 10^{-4}}{1 \times 10^{-10} \cdot 1 \times 10^{-2}} = \frac{1 \times 10^{-4}}{1 \times 10^{-12}} = 1 \times 10^{8} \]

### Step 3: Use the Gibbs Free Energy Equation
The relationship between ΔG° and K is given by:
\[ \Delta G° = -2.303 R T \log K \]

### Step 4: Substitute the Values into the Equation
Now substituting the values we have:
\[ \Delta G° = -2.303 \times 8 \, \text{J/(mol K)} \times 298 \, \text{K} \times \log(1 \times 10^{8}) \]

### Step 5: Calculate the Logarithm
Calculating the logarithm:
\[ \log(1 \times 10^{8}) = 8 \]

### Step 6: Final Calculation
Now substituting this back into the equation:
\[ \Delta G° = -2.303 \times 8 \times 298 \times 8 \]
Calculating:
\[ \Delta G° = -2.303 \times 8 \times 298 \times 8 = -43.92 \, \text{kJ/mol} \]

### Final Answer
Thus, the standard Gibbs free energy change (ΔG°) is:
\[ \Delta G° = -43.92 \, \text{kJ/mol} \]

---