At elevated temperature , `PCl_(5)` dissociates as `PCI_(5(a)) hArr PCl_(3(g)) + Cl_(2(g)) . " At " 300^(@) C, K_(p) = 11.8` and
`[PCl_(3)] = [Cl_(2)] = 0.01 "mol" L^(-1)` at equilibrium `[PCl_(5)] = x xx 10^(-4) "mol" L^(-1)` . What is the value of x ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression for the dissociation of PCl5.
The dissociation reaction is:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{\text{PCl}_3} \cdot P_{\text{Cl}_2}}{P_{\text{PCl}_5}} \]

### Step 2: Relate \( K_p \) to \( K_c \).
We know that:
\[ K_p = K_c \cdot (RT)^{\Delta n} \]
where \( \Delta n \) is the change in moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Here, the number of moles of products (PCl3 + Cl2) is 2, and the number of moles of reactants (PCl5) is 1. Thus:
\[ \Delta n = 2 - 1 = 1 \]

### Step 3: Convert temperature to Kelvin.
Given temperature \( T = 300^\circ C \):
\[ T(K) = 300 + 273 = 573 \, K \]

### Step 4: Substitute known values into the equation.
Given \( K_p = 11.8 \) and \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \):
\[ K_p = K_c \cdot (0.0821 \cdot 573)^1 \]
\[ 11.8 = K_c \cdot (0.0821 \cdot 573) \]

Calculating \( 0.0821 \cdot 573 \):
\[ 0.0821 \cdot 573 \approx 47.0523 \]

Now, substituting back:
\[ 11.8 = K_c \cdot 47.0523 \]
\[ K_c = \frac{11.8}{47.0523} \approx 0.25 \]

### Step 5: Write the expression for \( K_c \) in terms of concentrations.
At equilibrium, we have:
- \([PCl_3] = 0.01 \, mol/L\)
- \([Cl_2] = 0.01 \, mol/L\)
- \([PCl_5] = x \times 10^{-4} \, mol/L\)

Thus, the expression for \( K_c \) is:
\[ K_c = \frac{[PCl_3] \cdot [Cl_2]}{[PCl_5]} \]
Substituting the values:
\[ 0.25 = \frac{0.01 \cdot 0.01}{x \times 10^{-4}} \]

### Step 6: Solve for \( x \).
Rearranging gives:
\[ 0.25 = \frac{0.0001}{x \times 10^{-4}} \]
\[ 0.25 = \frac{0.0001}{x \cdot 10^{-4}} \]
\[ 0.25 = \frac{0.0001}{x \cdot 0.0001} \]
\[ 0.25 = \frac{1}{x} \]
\[ x = \frac{1}{0.25} = 4 \]

### Final Answer:
The value of \( x \) is \( 4 \).

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