Consider the following reaction: PCl5(g)PCl3(g)+Cl2(g) Part B Initially, 0.62 mo

Question

Consider the following reaction:

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

Part B

Initially, 0.62 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.14 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5?

Express your answer to two significant figures and include the appropriate units

Part C

What is the equilibrium concentration of Cl2?

Express your answer to two significant figures and include the appropriate units.

Part D

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

Express your answer using two significant figures.

Part E

If 0.18 mol of Cl2 is added to the equilibrium mixture, will the concentration of PCl5 increase or decrease?

Explanation / Answer

B)

since volume is 1L, number of mol will be equal to molarity

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

0.62 0 0 (initial)

0.62-x       x          x (at equilibrium)

at equilibrium,

[PCl3] = x = 0.14 M

so,

x = 0.14 M

[PCl5] = 0.62-x = 0.62 - 0.14 = 0.48 M

C)

[Cl2] = x = 0.14 M

D)

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

= 0.14 * 0.14 / (0.48)

= 0.0408

E)

we are adding a product

According to LeChattelier's Principle,

Adding product will shift reaction towards reactant side

Equilibrium moves to reactant side

so,

concentration of PCl5 will increase

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