If the equilibrium concentration of the components in a reaction
`A+B hArr C+D` are 3,5,10 and 15 mol `L^(-1)` respectively then what is `Delta G^(@)` for the reactiona at 300K ?

Options

A. `13.81`
B. `-1381.8`
C. `-138.18`
D. `1391.6`

Answer

To find the standard Gibbs free energy change (ΔG°) for the reaction at 300 K, we can follow these steps:

### Step 1: Write the reaction and identify the components
The reaction is given as:
\[ A + B \rightleftharpoons C + D \]

The equilibrium concentrations are:
- \([A] = 3 \, \text{mol L}^{-1}\)
- \([B] = 5 \, \text{mol L}^{-1}\)
- \([C] = 10 \, \text{mol L}^{-1}\)
- \([D] = 15 \, \text{mol L}^{-1}\)

### Step 2: Calculate the equilibrium constant (K)
The equilibrium constant \( K \) is calculated using the formula:
\[ K = \frac{[C][D]}{[A][B]} \]

Substituting the equilibrium concentrations:
\[ K = \frac{(10)(15)}{(3)(5)} \]

Calculating the values:
\[ K = \frac{150}{15} = 10 \]

### Step 3: Use the Gibbs free energy formula
The standard Gibbs free energy change is given by the equation:
\[ \Delta G° = -2.303 \, RT \, \log K \]

Where:
- \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \) (gas constant)
- \( T = 300 \, \text{K} \)
- \( K = 10 \)

### Step 4: Substitute the values into the formula
Substituting the values into the Gibbs free energy equation:
\[ \Delta G° = -2.303 \times (8.314) \times (300) \times \log(10) \]

Since \( \log(10) = 1 \):
\[ \Delta G° = -2.303 \times 8.314 \times 300 \times 1 \]

### Step 5: Calculate ΔG°
Calculating the value:
\[ \Delta G° = -2.303 \times 8.314 \times 300 \]
\[ \Delta G° = -2.303 \times 2494.2 \approx -5748.5 \, \text{J mol}^{-1} \]

### Final Answer
\[ \Delta G° \approx -5748.5 \, \text{J mol}^{-1} \text{ or } -5.75 \, \text{kJ mol}^{-1} \]

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