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2018-07-27 12:40 pm

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

1. Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.16 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5?

2. What is the equilibrium concentration of Cl2?

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

回答 (1)

胡雪8°

2018-07-27 5:32 pm

✔ 最佳答案

1.
Initial concentration of PCl₅, [PCl₅]ₒ = (0.65 mol) / (1.0 L) = 0.65 mol/L

___________________ PCl₅(g) ___ ⇌ ___ PCl₃(g) ___ + ___ Cl₂(g) ______ Kc
INitial (mol/L): _______ 0.65 ___________ 0 _____________ 0
Change (mol/L): _______ -y ____________ +y ____________ +y
Equilibrium (mol/L) _ 0.65 - y ___________ y _____________ y

Equilibrium concentration of PCl₃, [PCl₃] = y mol/L = 0.16 mol/L
Hence, y = 0.16
Equilibrium concentration of PCl₅, [PCl₅] = (0.65 - y) mol/L = (0.65 - 0.16) mol/L = 0.49 mol/L

2.
Equilibrium concentration of Cl₂, [Cl₂] = y mol/L = 0.16 mol/L

3.
Equilibrium constant, Kc
= [PCl₃] [Cl₂] / [PCl₅]
= 0.16 × 0.16 / 0.49
= 0.052 (to 2 sig. fig.)

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