The enthalpy and entropy change for the reaction
`Br_(2)(l) + Cl_(2)(g) rightarrow 2BrCl(g)` are ` 40 kJ mol^(-1)` and `110 JK^(-1) mol ^(-1)` repectively. The temperature at which the reaction will be in equilibrium is

Options

A. `363.64 K`
B. `273 K`
C. `450 K`
D. `300 K`

Answer

To find the temperature at which the reaction \( \text{Br}_2(l) + \text{Cl}_2(g) \rightarrow 2\text{BrCl}(g) \) will be in equilibrium, we can use the Gibbs free energy equation:

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

At equilibrium, the change in Gibbs free energy (\( \Delta G \)) is zero. Therefore, we can set the equation to zero:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this gives us:

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

From the question, we know:

- \( \Delta H = 40 \, \text{kJ/mol} = 40 \times 10^3 \, \text{J/mol} \) (converting kJ to J)
- \( \Delta S = 110 \, \text{J/K/mol} \)

Now, substituting the values into the equation:

\[ T = \frac{\Delta H}{\Delta S} = \frac{40 \times 10^3 \, \text{J/mol}}{110 \, \text{J/K/mol}} \]

Calculating this:

\[ T = \frac{40,000}{110} \approx 363.636 \, \text{K} \]

Rounding this to two decimal places, we get:

\[ T \approx 363.64 \, \text{K} \]

Thus, the temperature at which the reaction will be in equilibrium is approximately **363.64 K**.