For a reversible reaction at 298 K the equilibrium constant K is 200. What is value of `DeltaG^(@)` at 298 K ?

Options

A. a. `-13.13kcal`
B. b.-0.13kcal`
C. c. `-3.158kcal`
D. d. `-0.413kcal`

Explanation

To find the value of \(\Delta G^\circ\) at 298 K for a reversible reaction with an equilibrium constant \(K = 200\), we can use the relationship between Gibbs free energy change and the equilibrium constant. The formula is:

\[ \Delta G^\circ = -2.303 \cdot R \cdot T \cdot \log K \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Temperature (\(T\)) = 298 K
- Equilibrium constant (\(K\)) = 200
- Universal gas constant (\(R\)) = 1.987 cal/(mol·K) (approximately 2 cal/(mol·K) for simplicity)

2. **Convert \(R\) to Kcal:**
- Since the final answer is required in kilocalories, we can use \(R = 0.001987\) kcal/(mol·K).

3. **Calculate \(\log K\):**
- We need to calculate \(\log 200\).
- Using the logarithmic property, we can express \(200\) as \(2 \times 100\):
\[ \log 200 = \log(2 \times 100) = \log 2 + \log 100 \]
- We know that \(\log 100 = 2\) and \(\log 2 \approx 0.30103\), so:
\[ \log 200 = 0.30103 + 2 = 2.30103 \]

4. **Substitute the Values into the Formula:**
- Now, substitute \(R\), \(T\), and \(\log K\) into the equation:
\[ \Delta G^\circ = -2.303 \cdot (0.001987 \text{ kcal/(mol·K)}) \cdot (298 \text{ K}) \cdot (2.30103) \]

5. **Perform the Calculation:**
- First, calculate the product:
\[ -2.303 \cdot 0.001987 \cdot 298 \cdot 2.30103 \]
- This results in:
\[ \Delta G^\circ \approx -3.158 \text{ kcal} \]

6. **Final Answer:**
- Therefore, the value of \(\Delta G^\circ\) at 298 K is approximately \(-3.158 \text{ kcal}\).

### Summary:
\[ \Delta G^\circ \approx -3.158 \text{ kcal} \]