If a given quantity of phosphorus pentachloride is heated to 250°C and allowed to come to
equilibrium at atmospheric pressure it is found to be dissociated to the extent of 80% into
phosphorous trichloride and chlorine. If now, the pressure on this mixture is increased so that
finally the equilibrium mixture occupies only one half of its original volume (temperature
maintained constant).
(a) What will be the % dissociation at the new pressure?
(b) What will be the effect of adding He gas keeping the volume fixed at V/2.
Answers
a) The percent dissociation at the new pressure is 69 %.
b) When we add He gas keeping the volume fixed at V/2 the equilibrium constant did not change because temperature is constant.
Given-
- Temperature = 250° C
- Dissociation = 80 %
- Pressure = 1 atm
Reaction is -
PCl₅ ↔ PCl₃ + Cl₂
Let us assume that the initial mole of PCl₅ = 1 mol
Total volume of the mixture at equilibrium = V liters
Since percent dissociation is 80 %, so
[PCl₅] = 0.2/V mol/lit
[PCl₃] = 0.8/V mol/lit
[Cl₂] = 0.8/V mol/lit
From the above reaction, Kc can be given by -
Kc = [PCl₃] [Cl₂] / [PCl₅]
By substituting the values we get -
Kc =
Kc = 3.2/ V
If volume is changes to V/2, let x be the fraction dissociated.
So, Concentrations are -
[PCl₅] = 1-x / V/2 = 2 (1-x)/V
[PCl₃] = x / V/2 = 2x/V
[Cl₂] = x / V/2 = 2x/V
Equilibrium constant will be same because temperature is constant, so
Kc = 3.2/V = 2x² / (1-x)V
By solving the above equation we get x = 0.69. So the percentage dissociation is 69 %
Answer:
a) The percent dissociation at the new pressure is 69 %.
b) When we add He gas keeping the volume fixed at V/2 the equilibrium constant did not change because temperature is constant.
Given-
Temperature = 250° C
Dissociation = 80 %
Pressure = 1 atm
Reaction is -
PCl₅ ↔ PCl₃ + Cl₂
Let us assume that the initial mole of PCl₅ = 1 mol
Total volume of the mixture at equilibrium = V liters
Since percent dissociation is 80 %, so
[PCl₅] = 0.2/V mol/lit
[PCl₃] = 0.8/V mol/lit
[Cl₂] = 0.8/V mol/lit
From the above reaction, Kc can be given by -
Kc = [PCl₃] [Cl₂] / [PCl₅]
By substituting the values we get -
Kc =
Kc = 3.2/ V
If volume is changes to V/2, let x be the fraction dissociated.
So, Concentrations are -
[PCl₅] = 1-x / V/2 = 2 (1-x)/V
[PCl₃] = x / V/2 = 2x/V
[Cl₂] = x / V/2 = 2x/V
Equilibrium constant will be same because temperature is constant, so
Kc = 3.2/V = 2x² / (1-x)V
By solving the above equation we get x = 0.69. So the percentage dissociation is 69 %
