For the chemical equilibrium
`PCl_(5)(g) hArr PCl_(3)(g) +Cl_(2) (g) " at " 298K, K_(c )=1.8xx10^(-7)`.
Calculate `Delta_(r )G^(Theta)` for the forward reaction `R=8.31JK^(-1)"mol"^(-1)`

Explanation

To calculate \(\Delta_r G^\Theta\) for the forward reaction of the equilibrium \(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\) at 298 K with \(K_c = 1.8 \times 10^{-7}\), we can use the following formula:

\[ \Delta_r G^\Theta = -2.303 \cdot R \cdot T \cdot \log K_c \]

where:
- \(R\) is the universal gas constant, given as \(8.31 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(T\) is the temperature in Kelvin, given as \(298 \, \text{K}\)
- \(K_c\) is the equilibrium constant, given as \(1.8 \times 10^{-7}\)

### Step-by-Step Solution:

1. **Identify the values**:
- \(R = 8.31 \, \text{J K}^{-1} \text{mol}^{-1}\)
- \(T = 298 \, \text{K}\)
- \(K_c = 1.8 \times 10^{-7}\)

2. **Calculate \(\log K_c\)**:
\[ \log K_c = \log(1.8 \times 10^{-7}) \]
Using a calculator or logarithm table:
\[ \log(1.8) \approx 0.2553 \quad \text{and} \quad \log(10^{-7}) = -7 \]
Therefore:
\[ \log K_c = 0.2553 - 7 = -6.7447 \]

3. **Substitute the values into the equation**:
\[ \Delta_r G^\Theta = -2.303 \cdot 8.31 \cdot 298 \cdot (-6.7447) \]

4. **Calculate the product**:
- First calculate \(2.303 \cdot 8.31 \cdot 298\):
\[ 2.303 \cdot 8.31 \cdot 298 \approx 5730.86 \]

5. **Calculate \(\Delta_r G^\Theta\)**:
\[ \Delta_r G^\Theta = -5730.86 \cdot (-6.7447) \approx 38516.52 \, \text{J} \]
Converting to kilojoules:
\[ \Delta_r G^\Theta \approx 38.516 \, \text{kJ} \]

### Final Answer:
\[ \Delta_r G^\Theta \approx 38.52 \, \text{kJ} \]