Question

A 6.19g sample of PCl5 is placed in an evacuated 2.00L flask and is completely vaporized at 252C. 1. Calculate the pressure in the flask if no chemical reaction were to occur. P=.6406 ATM 2.Actually at 252C the PCl5 is partially dissociated according to the following equation: PCl5<->PCl3+Cl2 (all gass states) The observed pressure is found to be 1.00 ATM. In view of this observation, calculate the partial pressure of PCl5 and PCl3 in the flask at 252C.

Answer 1

Answer:

The answer to the question is

• The pressure in the flask if no chemical reaction were to occur is 0.6406 atm

• The partial pressure of PCl₅ and PCl₃ in the flask at 252 °C is 0.3594 atm each

Explanation:

To solve the question we list the knowns so as to plug in the values when required

Mass of PCl₅ = 6.19g

Volume of flask = 2.00 L

Temperature  252 °C = ‪525.15‬ K

Molar mass of PCl = 208.24 g/mol

Therefore the number of moles of PCl₅ = (6.19 g)/(208.24 g/mol) =moles2.973×10⁻²

PV = nRT therefore P = nRT/V = (2.973×10⁻² moles×8.314 J/(gmol·K)× ‪525.15‬ K)/(0.002 m³) = 64906.01 Pa = 0.6406 ATM

(2) At 252 °C the PCl₅ dissociates partially as follows

PCl₅↔PCl₃+Cl₂

New pressure is observed to be 1.00 atm

Dalton's law states that  $P_{Total}$ = P₁ + P₂ +...+Pₙ where P₁,  P₂, Pₙ are the partial pressures of the constituent gases

That means  $P_{Total}$ = P(PCl₅) + P(PCl₃) + P(Cl₂)

The difference in pressure due to the dissociation = 1 - 0.6406 = 0.3594 ATM, However 1 PCl dissociates into 2 moles of gases

That is Y-X +2X = 1 ATM where Y = 0.6406 ATM then X = 0.3594 ATM

Therefore the final partial pressure of PCl₅ = 0.6406-0.3594 = 0.2812 atm

and P(PCl₃) and P(Cl₂) = 0.3594 ATM each