2.0 mole of`PCl_(5)` were nttoducedd in a vessel of 5.0 L capacity of a particular temperature At equilibrium, `PCl_(5)` was found to be 35 % dissociated into `PCl_(3)and Cl_(2)` the value of `K_(c)` for the reaction
`PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)`

Options

A. `1.89`
B. `0.377`
C. `1.33`
D. `13.3`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction:

\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]

Given that 2.0 moles of \( PCl_5 \) are introduced into a 5.0 L vessel and that 35% of \( PCl_5 \) is dissociated at equilibrium, we can follow these steps:

### Step 1: Determine the amount of \( PCl_5 \) dissociated
- Initial moles of \( PCl_5 \) = 2.0 moles
- Percentage dissociation = 35%
- Moles of \( PCl_5 \) dissociated = \( 35\% \) of \( 2.0 \) moles

\[ \text{Moles dissociated} = \frac{35}{100} \times 2.0 = 0.7 \text{ moles} \]

### Step 2: Calculate the moles at equilibrium
- Moles of \( PCl_5 \) at equilibrium = Initial moles - Moles dissociated
\[ \text{Moles of } PCl_5 = 2.0 - 0.7 = 1.3 \text{ moles} \]
- Moles of \( PCl_3 \) at equilibrium = Moles dissociated = 0.7 moles
- Moles of \( Cl_2 \) at equilibrium = Moles dissociated = 0.7 moles

### Step 3: Calculate the concentrations at equilibrium
- Volume of the vessel = 5.0 L
- Concentration of \( PCl_5 \):
\[ \text{Concentration of } PCl_5 = \frac{1.3 \text{ moles}}{5.0 \text{ L}} = 0.26 \text{ M} \]
- Concentration of \( PCl_3 \):
\[ \text{Concentration of } PCl_3 = \frac{0.7 \text{ moles}}{5.0 \text{ L}} = 0.14 \text{ M} \]
- Concentration of \( Cl_2 \):
\[ \text{Concentration of } Cl_2 = \frac{0.7 \text{ moles}}{5.0 \text{ L}} = 0.14 \text{ M} \]

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction \( PCl_3 + Cl_2 \rightleftharpoons PCl_5 \) is given by:

\[ K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} \]

### Step 5: Substitute the equilibrium concentrations into the \( K_c \) expression
\[ K_c = \frac{0.26}{(0.14)(0.14)} = \frac{0.26}{0.0196} \approx 13.27 \]

### Step 6: Calculate \( K_c \) for the reverse reaction
Since we need \( K_c \) for the reaction \( PCl_3 + Cl_2 \rightleftharpoons PCl_5 \), we take the inverse of the \( K_c \) we just calculated:

\[ K_c' = \frac{1}{K_c} = \frac{1}{13.27} \approx 0.075 \]

### Final Result
Thus, the value of \( K_c \) for the reaction \( PCl_3 + Cl_2 \rightleftharpoons PCl_5 \) is approximately:

\[ K_c' \approx 13.3 \]