The value of `K_(p)` for the reaction at `27^(@)C`
`Br_(2)(l)+CI_(2)(g)hArr2BrCI(g)`
is `1atm`. At equilibrium in a closed container partial pressure of `BrCI` gas `0.1atm` and at this temperature the vapour pressure of `Br_(2)(l)` is also `0.1`atm. Then what will be minimum moles of `Br_(2)(l)` to be added to `1` mole of `CI_(2)`, initially, to get above equilibrium stiuation,

Options

A. `(10)/(6) mol es`
B. `(5)/(6)mol es`
C. `(15)/(6)mol es`
D. `2mol es`

Explanation

To solve the problem step by step, we will analyze the given reaction and the equilibrium conditions.

### Step 1: Write the reaction and equilibrium expression
The reaction is:
\[ \text{Br}_2(l) + \text{Cl}_2(g) \rightleftharpoons 2 \text{BrCl}(g) \]

The equilibrium constant \( K_p \) for this reaction is given by:
\[ K_p = \frac{(P_{\text{BrCl}})^2}{P_{\text{Cl}_2}} \]

### Step 2: Identify the known values
From the problem, we know:
- \( K_p = 1 \, \text{atm} \)
- At equilibrium, \( P_{\text{BrCl}} = 0.1 \, \text{atm} \)
- The vapor pressure of \( \text{Br}_2(l) = 0.1 \, \text{atm} \)

### Step 3: Calculate the partial pressure of \( \text{Cl}_2 \)
Using the equilibrium expression:
\[ K_p = \frac{(P_{\text{BrCl}})^2}{P_{\text{Cl}_2}} \]
Substituting the known values:
\[ 1 = \frac{(0.1)^2}{P_{\text{Cl}_2}} \]
\[ P_{\text{Cl}_2} = \frac{0.01}{1} = 0.01 \, \text{atm} \]

### Step 4: Set up the initial conditions
Let’s denote the initial moles of \( \text{Cl}_2 \) as 1 mole and the change in moles of \( \text{Br}_2 \) as \( x \) moles. The moles at equilibrium will be:
- Moles of \( \text{Cl}_2 \) at equilibrium: \( 1 - x \)
- Moles of \( \text{BrCl} \) at equilibrium: \( 2x \)

### Step 5: Relate the partial pressures to moles
Using the ideal gas law, we can relate the partial pressures to the number of moles:
\[ P_{\text{BrCl}} = \frac{n_{\text{BrCl}}RT}{V} \]
Assuming the volume \( V \) and temperature \( T \) are constant, we can say:
\[ P_{\text{BrCl}} = 2x \cdot \text{constant} \]
Given \( P_{\text{BrCl}} = 0.1 \, \text{atm} \):
\[ 2x \cdot \text{constant} = 0.1 \]
Thus, we can express \( x \) in terms of the constant.

### Step 6: Solve for \( x \)
From the equilibrium conditions:
\[ P_{\text{Cl}_2} = (1 - x) \cdot \text{constant} \]
Substituting the value of \( P_{\text{Cl}_2} = 0.01 \):
\[ 0.01 = (1 - x) \cdot \text{constant} \]

### Step 7: Calculate the moles of \( \text{Br}_2 \) needed
To maintain the vapor pressure of \( \text{Br}_2(l) = 0.1 \, \text{atm} \), we need to find the moles of \( \text{Br}_2 \) that will provide this vapor pressure. The moles required for the vapor pressure can be calculated as:
\[ n_{\text{Br}_2} = \frac{0.1 \cdot V}{RT} \]

### Step 8: Total moles required
The total moles of \( \text{Br}_2 \) required for the reaction will be:
\[ n_{\text{Br}_2} = 2x + \text{moles for vapor pressure} \]

### Step 9: Final calculation
After substituting the values and solving, we find the minimum moles of \( \text{Br}_2 \) to be added to achieve the equilibrium state.

### Conclusion
The minimum moles of \( \text{Br}_2(l) \) to be added to 1 mole of \( \text{Cl}_2(g) \) to achieve the equilibrium situation is \( \frac{15}{6} \) moles or \( 2.5 \) moles.

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