The equilibrium constant for the reaction
`A+BhArrC+D`
is 10. `DeltaG^(@)` for the reaction at 300 K is

Options

A. `-0.6` kcal
B. `-116` kcal
C. `-691` kcal
D. `-1.382` kcal

Answer

To find the standard Gibbs free energy change (ΔG°) for the reaction at 300 K, we can use the relationship between ΔG° and the equilibrium constant (K) given by the equation:

\[ \Delta G^\circ = -RT \ln K \]

### Step 1: Identify the values needed for the equation.
- **R** (universal gas constant) = 1.987 cal/(mol·K) (or 0.001987 kcal/(mol·K) if we want to convert to kilocalories)
- **T** (temperature) = 300 K
- **K** (equilibrium constant) = 10

### Step 2: Convert R to the appropriate units.
Since the answer is required in kilocalories, we will use:
\[ R = 0.001987 \text{ kcal/(mol·K)} \]

### Step 3: Substitute the values into the equation.
\[ \Delta G^\circ = - (0.001987 \text{ kcal/(mol·K)}) \times (300 \text{ K}) \times \ln(10) \]

### Step 4: Calculate \(\ln(10)\).
Using a calculator or logarithm table:
\[ \ln(10) \approx 2.303 \]

### Step 5: Substitute \(\ln(10)\) into the equation.
\[ \Delta G^\circ = - (0.001987) \times (300) \times (2.303) \]

### Step 6: Perform the multiplication.
First, calculate:
\[ 0.001987 \times 300 \approx 0.5961 \]
Then multiply by \(2.303\):
\[ 0.5961 \times 2.303 \approx 1.372 \]

### Step 7: Apply the negative sign.
\[ \Delta G^\circ \approx -1.372 \text{ kcal/mol} \]

### Step 8: Round the answer.
The final value of \(\Delta G^\circ\) is approximately:
\[ \Delta G^\circ \approx -1.38 \text{ kcal/mol} \]

### Final Answer:
The value of \(\Delta G^\circ\) for the reaction at 300 K is approximately \(-1.38 \text{ kcal/mol}\).

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