The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction `PCl_(5)(g)rarrPCl_(3)(g)+Cl_(2)(g)`.
Assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.3)

Options

A. `4.30`
B. `3.54`
C. `2.64`
D. `1.56`

Answer

To solve the problem, we need to calculate the density of the equilibrium mixture of the reaction:

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

Given:
- Degree of dissociation (\( \alpha \)) = 0.4
- Temperature (T) = 420 K
- Pressure (P) = 1.0 atm
- Molar mass of P = 31 g/mol
- Molar mass of Cl = 35.3 g/mol

### Step 1: Calculate the total number of moles at equilibrium.

Initially, we have 1 mole of \( PCl_5 \).

At equilibrium:
- Moles of \( PCl_5 \) = \( 1 - \alpha = 1 - 0.4 = 0.6 \)
- Moles of \( PCl_3 \) = \( \alpha = 0.4 \)
- Moles of \( Cl_2 \) = \( \alpha = 0.4 \)

Total moles at equilibrium (\( n \)):
\[ n = (1 - \alpha) + \alpha + \alpha = 1 - 0.4 + 0.4 + 0.4 = 1 + 0.4 = 1.4 \]

### Step 2: Calculate the molar mass of the equilibrium mixture.

The molar mass of the components is:
- Molar mass of \( PCl_5 \) = \( 31 + 5 \times 35.3 = 31 + 176.5 = 207.5 \) g/mol
- Molar mass of \( PCl_3 \) = \( 31 + 3 \times 35.3 = 31 + 105.9 = 136.9 \) g/mol
- Molar mass of \( Cl_2 \) = \( 2 \times 35.3 = 70.6 \) g/mol

Now, we calculate the average molar mass of the mixture:
\[ \text{Average Molar Mass} = \frac{(0.6 \times 207.5) + (0.4 \times 136.9) + (0.4 \times 70.6)}{1.4} \]

Calculating each term:
- Contribution from \( PCl_5 \): \( 0.6 \times 207.5 = 124.5 \)
- Contribution from \( PCl_3 \): \( 0.4 \times 136.9 = 54.76 \)
- Contribution from \( Cl_2 \): \( 0.4 \times 70.6 = 28.24 \)

Total mass:
\[ \text{Total Mass} = 124.5 + 54.76 + 28.24 = 207.5 \text{ g} \]

Now, average molar mass:
\[ \text{Average Molar Mass} = \frac{207.5}{1.4} = 148.2143 \text{ g/mol} \]

### Step 3: Calculate the volume using the ideal gas equation.

Using the ideal gas equation:
\[ PV = nRT \]
Where:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)

Rearranging for volume \( V \):
\[ V = \frac{nRT}{P} = \frac{1.4 \times 0.0821 \times 420}{1.0} \]

Calculating:
\[ V = \frac{1.4 \times 0.0821 \times 420}{1} = 48.275 \text{ L} \]

### Step 4: Calculate the density of the equilibrium mixture.

Density (\( \rho \)) is given by:
\[ \rho = \frac{\text{mass}}{\text{volume}} = \frac{207.5 \text{ g}}{48.275 \text{ L}} \approx 4.3 \text{ g/L} \]

### Final Answer:
The density of the equilibrium mixture at 420 K and 1.0 atm is approximately **4.3 g/L**.

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