The degree of dissociation is `0.5 at 800K and 2` atm for the gaseous reaction
`PCl_(5)hArrPCl_(3)+Cl_(2)`
Assuming ideal behaviour of all the gases.
Calculate the density of equilibrium mixture at `800K "and" 2` atm.

Options

A. `4.232g//L`
B. `6.4g//L`
C. `8.4g//L`
D. `2.2g//L`

Answer

To solve the problem of calculating the density of the equilibrium mixture for the reaction \( \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \) at 800 K and 2 atm, given a degree of dissociation (\( \alpha \)) of 0.5, we can follow these steps:

### Step 1: Calculate the Molar Mass of \( \text{PCl}_5 \)
The molar mass of \( \text{PCl}_5 \) can be calculated as follows:
- Molar mass of Phosphorus (P) = 31 g/mol
- Molar mass of Chlorine (Cl) = 35.5 g/mol

\[ \text{Molar mass of } \text{PCl}_5 = 31 + (5 \times 35.5) = 31 + 177.5 = 208.5 \text{ g/mol} \]

### Step 2: Calculate the Density of Pure \( \text{PCl}_5 \) (Capital D)
Using the ideal gas equation \( PM = DRT \), we can rearrange it to find the density (D) of pure \( \text{PCl}_5 \):

\[ D = \frac{PM}{RT} \]

Where:
- \( P = 2 \) atm
- \( M = 208.5 \) g/mol
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 800 \) K

Substituting the values:

\[ D = \frac{2 \times 208.5}{0.0821 \times 800} \]

Calculating the denominator:

\[ 0.0821 \times 800 = 65.68 \]

Now, substituting back into the equation:

\[ D = \frac{417}{65.68} \approx 6.34 \text{ g/L} \]

### Step 3: Use the Degree of Dissociation to Find Density of the Mixture (small d)
The degree of dissociation \( \alpha \) is given by the formula:

\[ \alpha = \frac{D - d}{d} \times (n - 1) \]

Where:
- \( n \) is the number of moles of products per mole of reactant. In this case, \( n = 2 \) (1 mole of \( \text{PCl}_5 \) produces 1 mole of \( \text{PCl}_3 \) and 1 mole of \( \text{Cl}_2 \)).
- \( \alpha = 0.5 \)

Substituting the known values:

\[ 0.5 = \frac{6.34 - d}{d} \times (2 - 1) \]

This simplifies to:

\[ 0.5 = \frac{6.34 - d}{d} \]

### Step 4: Solve for \( d \)
Cross-multiplying gives:

\[ 0.5d = 6.34 - d \]

Rearranging gives:

\[ 0.5d + d = 6.34 \]

\[ 1.5d = 6.34 \]

Now, solving for \( d \):

\[ d = \frac{6.34}{1.5} \approx 4.23 \text{ g/L} \]

### Conclusion
The density of the equilibrium mixture at 800 K and 2 atm is approximately **4.23 g/L**.