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 can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

For the reaction to be spontaneous, \( \Delta G \) must be less than zero:

\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]

This can be rearranged to find the critical temperature:

\[ \Delta H < T \Delta S \implies T > \frac{\Delta H}{\Delta S} \]

### Step 1: Convert Enthalpy Change to Appropriate Units
The enthalpy change \( \Delta H \) is given as \( 7.1 \, \text{kcal/mol} \). We need to convert this to calories:

\[ \Delta H = 7.1 \, \text{kcal/mol} = 7.1 \times 1000 \, \text{cal/mol} = 7100 \, \text{cal/mol} \]

### Step 2: Use the Given Entropy Change
The entropy change \( \Delta S \) is given as \( 25 \, \text{cal/K mol} \).

### Step 3: Calculate the Critical Temperature
Now we can substitute the values of \( \Delta H \) and \( \Delta S \) into the equation:

\[ T > \frac{\Delta H}{\Delta S} = \frac{7100 \, \text{cal/mol}}{25 \, \text{cal/K mol}} = 284 \, \text{K} \]

### Step 4: Conclusion
The reaction will be spontaneous at temperatures greater than \( 284 \, \text{K} \). Therefore, the minimum temperature for spontaneity is approximately \( 290 \, \text{K} \).

### Final Answer
The reaction will be spontaneous at temperatures greater than \( 284 \, \text{K} \), and specifically, we can say it will be spontaneous at \( 290 \, \text{K} \).

---