The equilibrium constant `K_(p)` for the homogeneous reaction is `10^(-3)`. The standard Gibbs free energy change `DeltaG^(ɵ)` for the reaction at `27^(@)C ("using" R=2 cal K^(-1) mol^(-1))` is

Options

A. Zero
B. `-1.8 kcal`
C. `-4.145 kcal`
D. `+4.145 kcal`

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the reaction at equilibrium, we can use the following relationship:

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

Where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step 1: Identify the values
- Given \( K_p = 10^{-3} \)
- Temperature \( T = 27°C = 27 + 273 = 300 K \)
- \( R = 2 \, \text{cal K}^{-1} \text{mol}^{-1} \)

### Step 2: Substitute the values into the equation
\[ \Delta G° = -2.303 \cdot (2 \, \text{cal K}^{-1} \text{mol}^{-1}) \cdot (300 \, \text{K}) \cdot \log(10^{-3}) \]

### Step 3: Calculate \( \log(10^{-3}) \)
Using the logarithmic identity:
\[ \log(10^{-3}) = -3 \cdot \log(10) = -3 \cdot 1 = -3 \]

### Step 4: Substitute \( \log(10^{-3}) \) back into the equation
\[ \Delta G° = -2.303 \cdot (2) \cdot (300) \cdot (-3) \]

### Step 5: Calculate the product
\[ \Delta G° = -2.303 \cdot 2 \cdot 300 \cdot (-3) \]
\[ = 2.303 \cdot 2 \cdot 300 \cdot 3 \]
\[ = 2.303 \cdot 6 \cdot 300 \]
\[ = 2.303 \cdot 1800 \]
\[ = 4145.4 \, \text{calories} \]

### Step 6: Convert calories to kilocalories
Since the answer is required in kilocalories:
\[ \Delta G° = \frac{4145.4}{1000} = 4.145 \, \text{kcal} \]

### Final Answer
\[ \Delta G° = 4.145 \, \text{kcal} \]