two moles of PCl5 are placed in a 5.0L container. Dissociation takes place accor

Question

two moles of PCl5 are placed in a 5.0L container. Dissociation takes place according to the equation PCl5(g) +Cl2(g). At equilibrium, 0.40 moll of Cl2 are present. calculate the equilibrium constant (Kc) for this reaction under the conditions of this experiment. two moles of PCl5 are placed in a 5.0L container. Dissociation takes place according to the equation PCl5(g) +Cl2(g). At equilibrium, 0.40 moll of Cl2 are present. calculate the equilibrium constant (Kc) for this reaction under the conditions of this experiment. two moles of PCl5 are placed in a 5.0L container. Dissociation takes place according to the equation PCl5(g) +Cl2(g). At equilibrium, 0.40 moll of Cl2 are present. calculate the equilibrium constant (Kc) for this reaction under the conditions of this experiment.

Explanation / Answer

given

two moles of PCl5 in 5 L container

we know that

conc = moles / volume (L)

so

conc of PCl5 = 2 / 5

conc of PCl5 = 0.4 M

now

the reaction is

PCl5 ----> PCl3 + Cl2

using ICE table

we get

initial conc of PCl5 , PCl3 and Cl2 are 0.4 , 0 , 0

change in conc of PCl5 , PCl3 and Cl2 are -x , x , x

equilibrium conc of PCl5 , PCl3 and Cl2 are 0.4-x , x , x

now

given

0.4 mol of Cl2 at equilibrium

so

conc of Cl2 at equilibrium = moles / volume (L)

= 0.4/5

= 0.08

so

conc of Cl2 at equilibrium = 0.08 M

but

we got

[Cl2]eq = x

so

x = 0.08

now

[PCl3]eq = x = 0.08

[PCl5]eq = 0.4 - x = 0.4-0.08 = 0.32

now

consider the reaction

PCl5 ---> PCl3 + Cl2

Kc = [PCl3] [Cl2] / [PCl5]

so using the obtained values

we get

Kc = 0.08 * 0.08 / 0.32

Kc = 0.02

so

the equilibrium constant of the reaction is 0.02

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