`triangleG^@ and triangleH^(@)` for a reaction at 300K are -66.9 kJ/mole and 41.8 kJ/mole respectively. `triangleG^(@)` for the same reaction at 330K is

Options

A. `-25.1kJ`
B. `+25.1kJ`
C. `118.7kJ`
D. `-69.4kJ`

Answer

To find the standard Gibbs free energy change (ΔG°) for the reaction at 330 K, we can follow these steps:

### Step 1: Calculate the Entropy Change (ΔS°) at 300 K
The relationship between Gibbs free energy change (ΔG°), enthalpy change (ΔH°), and entropy change (ΔS°) is given by the equation:
\[ \Delta G° = \Delta H° - T \Delta S° \]
We can rearrange this equation to find ΔS°:
\[ \Delta S° = \frac{\Delta H° - \Delta G°}{T} \]

### Step 2: Substitute the Given Values
We have the following values:
- ΔG° = -66.9 kJ/mol
- ΔH° = 41.8 kJ/mol
- T = 300 K

Substituting these values into the equation:
\[ \Delta S° = \frac{41.8 \, \text{kJ/mol} - (-66.9 \, \text{kJ/mol})}{300 \, \text{K}} \]
\[ \Delta S° = \frac{41.8 + 66.9}{300} \]
\[ \Delta S° = \frac{108.7}{300} \approx 0.3623 \, \text{kJ/K·mol} \]
To convert this to kJ, we multiply by 1000:
\[ \Delta S° \approx 0.3623 \, \text{kJ/K·mol} = 0.3623 \, \text{kJ/K·mol} \times 1000 = 0.3623 \, \text{kJ/K·mol} \]

### Step 3: Calculate ΔG° at 330 K
Now we will use the same equation to calculate ΔG° at 330 K:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Substituting the values:
- ΔH° = 41.8 kJ/mol
- T = 330 K
- ΔS° = 0.3623 kJ/K·mol

\[ \Delta G° = 41.8 \, \text{kJ/mol} - (330 \, \text{K} \times 0.3623 \, \text{kJ/K·mol}) \]
\[ \Delta G° = 41.8 \, \text{kJ/mol} - 119.56 \, \text{kJ/mol} \]
\[ \Delta G° = 41.8 - 119.56 \approx -77.76 \, \text{kJ/mol} \]

### Step 4: Final Result
Thus, the standard Gibbs free energy change (ΔG°) for the reaction at 330 K is approximately:
\[ \Delta G° \approx -77.76 \, \text{kJ/mol} \]