The standard free energy change `(DeltaG^(@))` for 50 % dissociation of `N_(2)O_(4)` into `NO_(2)` at `27^(@)C` and 1 atm pressure is - x J `mol^(-1)` . The value of x is _________.
[Given : `R = 8,.31 " J "K^(-1)mol^(-1)log 1.33 = 0.1239 " In " 10 = 2.3`]

Explanation

To solve the problem, we need to calculate the standard free energy change (ΔG°) for the 50% dissociation of N₂O₄ into NO₂ at 27°C and 1 atm pressure.

### Step-by-Step Solution:

1. **Write the Reaction:**
The dissociation of dinitrogen tetroxide (N₂O₄) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

2. **Define the Degree of Dissociation (α):**
For 50% dissociation, α = 0.5. This means that if we start with 1 mole of N₂O₄, 0.5 moles will dissociate to form 1 mole of NO₂.

3. **Calculate the Equilibrium Constant (Kp):**
The expression for Kp in terms of α is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2\alpha)^2}{(1 - \alpha)} \]
Substituting α = 0.5:
\[ K_p = \frac{(2 \times 0.5)^2}{(1 - 0.5)} = \frac{(1)^2}{(0.5)} = \frac{1}{0.5} = 2 \]

4. **Use the Gibbs Free Energy Equation:**
The relationship between ΔG° and Kp is given by:
\[ \Delta G° = -RT \ln K_p \]
Where:
- R = 8.31 J K⁻¹ mol⁻¹ (given)
- T = 27°C = 300 K (convert to Kelvin by adding 273)
- Kp = 2 (calculated above)

5. **Calculate ΔG°:**
First, we need to find ln Kp:
\[ \ln K_p = \ln 2 \approx 0.693 \]
Now substituting the values into the Gibbs equation:
\[ \Delta G° = - (8.31 \, \text{J K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K}) \times (0.693) \]
\[ \Delta G° = - (8.31 \times 300 \times 0.693) \]
\[ \Delta G° \approx - 1727.43 \, \text{J mol}^{-1} \]

6. **Final Result:**
The value of x is approximately:
\[ x \approx 1727 \, \text{J mol}^{-1} \]

### Final Answer:
The value of x is **1727**.