The enthalpy and entropy change for the reaction,
`Br_(2)(l)+Cl_(2)(g)rarr2BrCl(g)`
are `30KJmol^(-1)` and `105JK^(-1)mol^(-1)` respectively. The temperature at which the raction will be in equilibrium is:

Options

A. 285.7 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, \( \Delta G = 0 \). Therefore, we can set up the equation:

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

Rearranging this gives:

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

### Step 1: Convert the units of \(\Delta H\)

Given:
- \(\Delta H = 30 \, \text{kJ/mol}\)

We need to convert this into joules because \(\Delta S\) is given in joules per Kelvin per mole:

\[ \Delta H = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol} = 30000 \, \text{J/mol} \]

### Step 2: Use the given \(\Delta S\)

Given:
- \(\Delta S = 105 \, \text{J/K/mol}\)

### Step 3: Substitute the values into the equation

Now we can substitute \(\Delta H\) and \(\Delta S\) into the equation for \(T\):

\[ T = \frac{30000 \, \text{J/mol}}{105 \, \text{J/K/mol}} \]

### Step 4: Calculate \(T\)

Now, perform the division:

\[ T = \frac{30000}{105} \approx 285.71 \, \text{K} \]

### Step 5: Round the answer

Rounding to one decimal place gives:

\[ T \approx 285.7 \, \text{K} \]

### Final Answer

The temperature at which the reaction will be in equilibrium is approximately **285.7 K**.