Please Help I Hate Chem !!!!!!!!!!!!! Consider the following reaction: PCl5( g )

Question

Please Help I Hate Chem !!!!!!!!!!!!!

Consider the following reaction:
PCl5(g)PCl3(g)+Cl2(g)

Part A

Choose the correct equilibrium constant expression for the reaction.

D) Kc=[PCl3][Cl2][PCl5]

E) Kc=1[PCl3][Cl2]

Part B

Initially 0.62 mole of PCl5 is placed in a 2.00 L flask. At equilibrium, there is 0.16 mole of PCl3 in the flask. What is the equilibrium concentration of the PCl5?

Express your answer using two significant figures.

[PCl5] =          M

Part C

What is the equilibrium concentration of the Cl2?

Express your answer using two significant figures.

[Cl2] =      M

Part D

What is the value of the equilibrium constant, Kc, for the reaction?

Express your answer using two significant figures.

Kc =

Part E

If 0.20 mole of Cl2 is added to the equilibrium mixture, will [PCl5] increase or decrease?

A) Kc=[PCl5][PCl3][Cl2] B) Kc=[PCl3][PCl5] C) Kc=[PCl3][Cl2]

D) Kc=[PCl3][Cl2][PCl5]

E) Kc=1[PCl3][Cl2]

Part B

Initially 0.62 mole of PCl5 is placed in a 2.00 L flask. At equilibrium, there is 0.16 mole of PCl3 in the flask. What is the equilibrium concentration of the PCl5?

Express your answer using two significant figures.

[PCl5] =          M

Part C

What is the equilibrium concentration of the Cl2?

Express your answer using two significant figures.

[Cl2] =      M

Part D

What is the value of the equilibrium constant, Kc, for the reaction?

Express your answer using two significant figures.

Kc =

Part E

If 0.20 mole of Cl2 is added to the equilibrium mixture, will [PCl5] increase or decrease?

A) [PCl5] will increase. B) [PCl5] will decrease.

Explanation / Answer

Part A :

PCl5(g)PCl3(g)+Cl2(g)

answer: equilibrium constant expression = [products]/[reactants]

D) Kc=[PCl3][Cl2][PCl5]

Part B

concentration of PCl5 = 0.62/2 = 0.31 M

            PCl5(g)PCl3(g)+Cl2(g)

initial     0.31 M      0      0

at equi     0.31-0.08 M   0.08 M   0.08 M

equilibrium concentration of the PCl5 = 0.31-0.08 = 0.23 M

Part C

equilibrium concentration of the Cl2 = 0.08 M

Part D

Kc = (0.08*0.08)/0.23 = 0.278 M

part E

A) [PCl5] will increase.

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