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)` respectively. 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 will be in equilibrium, we can use the Gibbs free energy equation:

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

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

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

Rearranging this gives us:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Convert Enthalpy from kJ to J
We are given:
- \(\Delta H = 40 \, \text{kJ/mol}\)
- \(\Delta S = 110 \, \text{J/K/mol}\)

Since \(\Delta H\) is in kJ, we need to convert it to J:

\[ \Delta H = 40 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40000 \, \text{J/mol} \]

### Step 2: Substitute Values into the Temperature Equation
Now we can substitute the values of \(\Delta H\) and \(\Delta S\) into the equation for temperature:

\[ T = \frac{40000 \, \text{J/mol}}{110 \, \text{J/K/mol}} \]

### Step 3: Calculate the Temperature
Now we perform the calculation:

\[ T = \frac{40000}{110} \approx 363.64 \, \text{K} \]

### Final Answer
The temperature at which the reaction will be in equilibrium is approximately:

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

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