For the gas phase reaction SO3(g) SO2 (g) + ½ O2 (g) if the gaseous mixture in a closed container is allowed to come at equilibrium and the degree of dissociation (α) at equilibrium is found to be 2/3 at 400 K and 1 atm pressure then calculate the vapour density of the equilibrium mixture.

To calculate the vapor density of the equilibrium mixture for the reaction SO3(g) ⇌ SO2(g) + ½ O2(g), we first need to determine the number of moles of each component at equilibrium based on the degree of dissociation (α).

Step 1: Initial Moles

Assume we start with 1 mole of SO3. At equilibrium, the dissociation can be represented as follows:

• Initial moles of SO3 = 1
• Change in moles of SO3 = -α = -2/3
• Equilibrium moles of SO3 = 1 - 2/3 = 1/3

Step 2: Moles of Products

From the stoichiometry of the reaction:

• Moles of SO2 produced = α = 2/3
• Moles of O2 produced = ½ * α = 1/3

Step 3: Total Moles at Equilibrium

Now, we can calculate the total moles at equilibrium:

• Total moles = Moles of SO3 + Moles of SO2 + Moles of O2
• Total moles = (1/3) + (2/3) + (1/3) = 4/3

Step 4: Calculate Vapor Density

The vapor density (VD) is defined as the mass of the gas mixture divided by the volume. To find the mass, we need the molar masses:

• Molar mass of SO3 = 80 g/mol
• Molar mass of SO2 = 64 g/mol
• Molar mass of O2 = 32 g/mol

Now, calculate the total mass of the mixture:

• Mass of SO3 = (1/3) mol × 80 g/mol = 26.67 g
• Mass of SO2 = (2/3) mol × 64 g/mol = 42.67 g
• Mass of O2 = (1/3) mol × 32 g/mol = 10.67 g

Total mass = 26.67 g + 42.67 g + 10.67 g = 80 g

Final Calculation of Vapor Density

Using the ideal gas law, at 1 atm and 400 K, the volume (V) can be calculated using:

PV = nRT, where R = 0.0821 L·atm/(K·mol).

V = (4/3 mol) × (0.0821 L·atm/(K·mol) × 400 K) / 1 atm = 10.94 L

Finally, the vapor density (VD) is:

VD = Total mass / Volume = 80 g / 10.94 L = 7.31 g/L

Result

The vapor density of the equilibrium mixture is approximately 7.31 g/L.