For the following reaction at `25^(@)C, DeltaH^(@) = + 115kJ and Delta S^(@) = +125J//K`. Calculate `DeltaG^(@)` for the reaction at `25^(@)C`
`SBr_(4) (g) rarr S(g) + 2Br_(2)(I)`

Options

A. `+152kJ`
B. `-56.7kJ`
C. `+77.8kJ`
D. `+37.1 kJ`

Answer

To calculate the change in Gibbs free energy (ΔG°) for the reaction at 25°C, we can follow these steps:

### Step 1: Identify the given values
- ΔH° (change in enthalpy) = +115 kJ
- ΔS° (change in entropy) = +125 J/K

### Step 2: Convert units
Since ΔH° is given in kJ and ΔS° is in J/K, we need to convert ΔS° into kJ/K for consistency:
- ΔS° = 125 J/K = 125/1000 kJ/K = 0.125 kJ/K

### Step 3: Convert temperature to Kelvin
The temperature is given in Celsius, so we convert it to Kelvin:
- T = 25°C + 273 = 298 K

### Step 4: Use the Gibbs free energy formula
The formula to calculate Gibbs free energy is:
\[ \Delta G° = \Delta H° - T \Delta S° \]

### Step 5: Substitute the values into the formula
Now we can substitute the values into the formula:
\[ \Delta G° = 115 \text{ kJ} - (298 \text{ K} \times 0.125 \text{ kJ/K}) \]

### Step 6: Calculate TΔS°
First, calculate TΔS°:
\[ T \Delta S° = 298 \times 0.125 = 37.25 \text{ kJ} \]

### Step 7: Calculate ΔG°
Now substitute this back into the ΔG° equation:
\[ \Delta G° = 115 \text{ kJ} - 37.25 \text{ kJ} = 77.75 \text{ kJ} \]

### Step 8: Round the answer
To match the options, we round the answer:
\[ \Delta G° \approx 77.8 \text{ kJ} \]

### Final Answer
Thus, the change in Gibbs free energy (ΔG°) for the reaction at 25°C is approximately **77.8 kJ**.

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