The enthalpy and entropy change for the reaction, `Br_2`(I) + `CI_2`(g) `rightarrow` 2BrCl(g) are 7.1 kcal `mol^-1` and 25 cal `K^-1 mol^-1` respectively. This reaction will be spontaneous at

Options

A. `10^oC`
B. 280 K
C. 290 K
D. `-5^oC`

Answer

To determine the temperature at which the reaction \( \text{Br}_2 (l) + \text{Cl}_2 (g) \rightarrow 2 \text{BrCl} (g) \) becomes spontaneous, we need to analyze the Gibbs free energy change (\( \Delta G \)) for the reaction. The reaction will be spontaneous when \( \Delta G < 0 \).

### Given Data:
- Enthalpy change (\( \Delta H \)) = 7.1 kcal/mol = 7100 cal/mol (since 1 kcal = 1000 cal)
- Entropy change (\( \Delta S \)) = 25 cal/K·mol

### Step 1: Write the Gibbs Free Energy Equation
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

### Step 2: Substitute Known Values
We need to find the temperature \( T \) at which \( \Delta G = 0 \) (the threshold for spontaneity):
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ T = \frac{\Delta H}{\Delta S} \]

### Step 3: Calculate the Temperature
Substituting the values of \( \Delta H \) and \( \Delta S \):
\[ T = \frac{7100 \text{ cal/mol}}{25 \text{ cal/K·mol}} = 284 \text{ K} \]

### Step 4: Analyze the Given Temperatures
We need to check the temperatures provided in the question to determine at which the reaction is spontaneous:
1. **10°C = 283 K**
- \( \Delta G = 7100 - 283 \times 25 = 7100 - 7075 = 25 \) (not spontaneous)

2. **280 K**
- \( \Delta G = 7100 - 280 \times 25 = 7100 - 7000 = 100 \) (not spontaneous)

3. **290 K**
- \( \Delta G = 7100 - 290 \times 25 = 7100 - 7250 = -150 \) (spontaneous)

4. **-5°C = 268 K**
- \( \Delta G = 7100 - 268 \times 25 = 7100 - 6700 = 400 \) (not spontaneous)

### Conclusion
The reaction will be spontaneous at **290 K**.