4.0 moles of `PCI_5` dissociated at 760 K in a 2 litre flask `PCI_5 (g) hArr PCI_3 (g) +CI_2(g)` at equilibrium .
0.8 mole of `CI_2` was present in the flask .The equilibrium constant would be

Options

A. `1.0 xx 10^(-1)`
B. `1.0 xx 10^(-4)`
C. `1.0xx10^(-2)`
D. `1.0 xx 10^(-3)`

Answer

To find the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) at 760 K, we can follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of phosphorus pentachloride can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of \( PCl_5 \) = 4.0 moles
- Moles of \( PCl_3 \) = 0 moles
- Moles of \( Cl_2 \) = 0 moles

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of \( PCl_5 \) that dissociates at equilibrium. Therefore:
- Moles of \( PCl_5 \) at equilibrium = \( 4.0 - x \)
- Moles of \( PCl_3 \) at equilibrium = \( x \)
- Moles of \( Cl_2 \) at equilibrium = \( x \)

### Step 4: Use the information given
According to the problem, at equilibrium, there are 0.8 moles of \( Cl_2 \) present. Thus, we have:
\[ x = 0.8 \]

### Step 5: Calculate the moles of each species at equilibrium
Substituting \( x \) into the expressions for moles at equilibrium:
- Moles of \( PCl_5 \) = \( 4.0 - 0.8 = 3.2 \)
- Moles of \( PCl_3 \) = \( 0.8 \)
- Moles of \( Cl_2 \) = \( 0.8 \)

### Step 6: Calculate the concentrations
Since the volume of the flask is 2 liters, we can calculate the concentrations:
- Concentration of \( PCl_5 \):
\[ [C_{PCl_5}] = \frac{3.2 \text{ moles}}{2 \text{ L}} = 1.6 \text{ M} \]
- Concentration of \( PCl_3 \):
\[ [C_{PCl_3}] = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \]
- Concentration of \( Cl_2 \):
\[ [C_{Cl_2}] = \frac{0.8 \text{ moles}}{2 \text{ L}} = 0.4 \text{ M} \]

### Step 7: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 8: Substitute the concentrations into the expression
Substituting the values we calculated:
\[ K_c = \frac{(0.4)(0.4)}{1.6} \]

### Step 9: Perform the calculation
Calculating \( K_c \):
\[ K_c = \frac{0.16}{1.6} = 0.1 \]

### Final Answer
The equilibrium constant \( K_c \) is \( 0.1 \).

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