The enthalpy and entropy change for the reaction: `Br_(2)(l) + Cl_(2)(g)` to `2BrCl(g)` are `30 kJ mol^(-1)` and `105 JK^(-1) mol^(-1)` respectively. The temperature at which the reaction will be in equilibrium is:-

Options

A. 300 K
B. 285.7 K
C. 273 K
D. 450 K

Answer

To find the temperature at which the reaction \( \text{Br}_2(l) + \text{Cl}_2(g) \rightarrow 2\text{BrCl}(g) \) is in equilibrium, we can use the relationship between Gibbs free energy change (\( \Delta G \)), enthalpy change (\( \Delta H \)), and entropy change (\( \Delta S \)).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Enthalpy change (\( \Delta H \)) = 30 kJ/mol
- Entropy change (\( \Delta S \)) = 105 J/K·mol

2. **Convert Enthalpy Change to Joules**:
Since \( \Delta H \) is given in kJ/mol, we need to convert it to J/mol for consistency:
\[ \Delta H = 30 \text{ kJ/mol} = 30 \times 1000 \text{ J/mol} = 30000 \text{ J/mol} \]

3. **Use the Gibbs Free Energy Equation**:
At equilibrium, the Gibbs free energy change (\( \Delta G \)) is zero:
\[ \Delta G = \Delta H - T \Delta S = 0 \]
Rearranging this gives:
\[ \Delta H = T \Delta S \]

4. **Solve for Temperature (T)**:
We can now solve for the temperature \( T \):
\[ T = \frac{\Delta H}{\Delta S} \]
Substituting the values:
\[ T = \frac{30000 \text{ J/mol}}{105 \text{ J/K·mol}} \]

5. **Calculate the Temperature**:
Performing the calculation:
\[ T = \frac{30000}{105} \approx 285.71 \text{ K} \]

6. **Final Result**:
Rounding to two decimal places, the temperature at which the reaction will be in equilibrium is approximately:
\[ T \approx 287.5 \text{ K} \]

### Answer:
The temperature at which the reaction will be in equilibrium is **287.5 K**.

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