The reaction, `PCI_(5)hArrPCI_(3)+CI_(2)` is started in a five litre container by taking one mole of `PCI_(5)`. If `0.3` mol `PCI_(5)` is there at equilibrium, concentration of `PCI_(3) "and" K_(c)` will respectively be:

Options

A. `0.14,(49)/(150)`
B. `0.12,(23)/(100)`
C. `0.07,(23)/(100)`
D. `20,(49)/(150)`

Explanation

To solve the problem, we need to find the equilibrium concentrations of \( PCl_3 \) and the equilibrium constant \( K_c \) for the reaction:

\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step 1: Write the initial conditions
Initially, we have:
- Moles of \( PCl_5 = 1 \) mole
- Moles of \( PCl_3 = 0 \)
- Moles of \( Cl_2 = 0 \)

### Step 2: Set up the change in moles at equilibrium
At equilibrium, we are given that there are \( 0.3 \) moles of \( PCl_5 \). Therefore, we can calculate the change in moles:

\[ \text{Change in moles of } PCl_5 = 1 - 0.3 = 0.7 \text{ moles} \]

This means \( 0.7 \) moles of \( PCl_5 \) have dissociated to form \( PCl_3 \) and \( Cl_2 \).

### Step 3: Calculate moles of products at equilibrium
Since the stoichiometry of the reaction shows that 1 mole of \( PCl_5 \) produces 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \), we have:

- Moles of \( PCl_3 = 0 + 0.7 = 0.7 \) moles
- Moles of \( Cl_2 = 0 + 0.7 = 0.7 \) moles

### Step 4: Calculate concentrations at equilibrium
To find the concentrations, we use the formula:

\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \]

Given that the volume of the container is \( 5 \) liters:

- Concentration of \( PCl_5 \):

\[ \text{Concentration of } PCl_5 = \frac{0.3 \text{ moles}}{5 \text{ L}} = 0.06 \text{ M} \]

- Concentration of \( PCl_3 \):

\[ \text{Concentration of } PCl_3 = \frac{0.7 \text{ moles}}{5 \text{ L}} = 0.14 \text{ M} \]

- Concentration of \( Cl_2 \):

\[ \text{Concentration of } Cl_2 = \frac{0.7 \text{ moles}}{5 \text{ L}} = 0.14 \text{ M} \]

### Step 5: Calculate the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

Substituting the concentrations we found:

\[ K_c = \frac{(0.14)(0.14)}{0.06} \]

Calculating this gives:

\[ K_c = \frac{0.0196}{0.06} = 0.3267 \approx \frac{49}{150} \]

### Final Answer
The concentrations of \( PCl_3 \) and \( K_c \) at equilibrium are:

- Concentration of \( PCl_3 = 0.14 \, \text{M} \)
- \( K_c = \frac{49}{150} \)

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