Question

The equilibrium constant (Kp) for the interconversion of PCl5 and PCl3 is 0.0121 at certain temperature. A vessel is charged with PCl5 giving an initial pressure of 0.123 atm.

PCl5 (g) ⥦ PCl3 (g) + Cl2 (g)

a) Calculate equilibrium partial pressure of PCl3

b) Calculate total pressure at equilibrium

Answer 1

Equilibrium constant Kp is defined as ratio of product of partial pressures of products at equilibrium to product of partial pressures of reactants at equilibrium raised to power their stoichiometric coefficients.

Let us make the ICE chart for given reaction

     

PC15 (9) =PC13(9) + Cl2(g

Initial partial pressure (atm) 0.123 0 0

Change -y +y +y

Equilibrium partial pressure (atm) 0.123-y y y

Kp=
PPC13(g) PC12(9) PPCIs (g)
(
PPC1s(9), PPC13(g) and Polz(g)
are partial pressures of PCl₅, PCl₃ and Cl₂ respectively)

0.0121=y x y/(0.123-y)

0.0121 (0.123-y)=y²

0.0014883-0.0121 y=y²

y²+0.0121y-0.0014883=0

y=0.033 atm (we reject the negative square root as pressure cannot be negative)

a) So equilibrium partial pressure of PCl₃==y=0.033 atm

PPC13(g) = PcLz9)

And =0.123 atm-0.033 atm=0.090 atm

PPCis(9)

b) Total pressure at equilibrium=

PPCis(g) + PPC13(g) + Plz()

=0.090 atm + 0.033 atm + 0.033 atm=0.156 atm

Answer 2

Equilibrium constant Kp is defined as ratio of product of partial pressures of products at equilibrium to product of partial pressures of reactants at equilibrium raised to power their stoichiometric coefficients.

Let us make the ICE chart for given reaction

     

PC15 (9) =PC13(9) + Cl2(g

Initial partial pressure (atm) 0.123 0 0

Change -y +y +y

Equilibrium partial pressure (atm) 0.123-y y y

Kp=
PPC13(g) PC12(9) PPCIs (g)
(
PPC1s(9), PPC13(g) and Polz(g)
are partial pressures of PCl₅, PCl₃ and Cl₂ respectively)

0.0121=y x y/(0.123-y)

0.0121 (0.123-y)=y²

0.0014883-0.0121 y=y²

y²+0.0121y-0.0014883=0

y=0.033 atm (we reject the negative square root as pressure cannot be negative)

a) So equilibrium partial pressure of PCl₃==y=0.033 atm

PPC13(g) = PcLz9)

And =0.123 atm-0.033 atm=0.090 atm

PPCis(9)

b) Total pressure at equilibrium=

PPCis(g) + PPC13(g) + Plz()

=0.090 atm + 0.033 atm + 0.033 atm=0.156 atm