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23 Nov 2019

At a certain temperature, Kc for the reaction, PCl5 = PCl3 + Cl2, is equal to 3.33. After 0.20 mole of PCl5 and 1.0 mole of PCl are introduced into a 2.00-L evacuated chamber, calculate the equilibrium concentration of PCl5. (Hint: use ICE table and solve the quadratic equation)

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Irving Heathcote
Irving HeathcoteLv2
22 Mar 2019

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Related questions

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.

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.

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.

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

Calculate the equilibrium concentrations of reactant and products when 0.309 moles of PCl3 and 0.309 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.

[PCl3] = M

[Cl2] = M

[PCl5] = M

(please show work especially quadratic equation)

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.

PCl3(g) + Cl2(g) <-----> PCl5(g)

Calculate the equilibrium concentrations of reactant and products when 0.358 moles of PCl3 and 0.358 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.

[PCl3] = ____ M
[Cl2] = ____ M
[PCl5] = ____ M

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