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 step by step, we will follow these procedures:

### Step 1: Determine the initial conditions
We start with 2.0 moles of \( PCl_5 \) in a 5.0 L vessel.

### Step 2: Calculate the initial concentration of \( PCl_5 \)
The initial concentration of \( PCl_5 \) can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in L}} \]
\[ \text{Initial concentration of } PCl_5 = \frac{2.0 \, \text{moles}}{5.0 \, \text{L}} = 0.4 \, \text{M} \]

### Step 3: Determine the amount dissociated
Since \( PCl_5 \) is 35% dissociated, we can calculate the moles of \( PCl_5 \) that dissociate:
\[ \text{Moles dissociated} = 0.35 \times 2.0 = 0.7 \, \text{moles} \]

### Step 4: Calculate the moles of products formed
For every mole of \( PCl_5 \) that dissociates, one mole of \( PCl_3 \) and one mole of \( Cl_2 \) are formed. Therefore, the moles of \( PCl_3 \) and \( Cl_2 \) formed are also 0.7 moles each.

### Step 5: Calculate the moles at equilibrium
At equilibrium, the moles of each species are:
- Moles of \( PCl_5 \) = Initial moles - Moles dissociated = \( 2.0 - 0.7 = 1.3 \, \text{moles} \)
- Moles of \( PCl_3 \) = 0.7 moles
- Moles of \( Cl_2 \) = 0.7 moles

### Step 6: Calculate the equilibrium concentrations
Now we can calculate the equilibrium concentrations:
- 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 7: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction:
\[ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \]
is given by:
\[ K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} \]

### Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{0.26}{(0.14)(0.14)} = \frac{0.26}{0.0196} \]
Calculating this gives:
\[ K_c \approx 13.3 \]

### Final Answer
Thus, the value of \( K_c \) for the reaction is approximately **13.3**.

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