Question

a reaction mixture of 0.00623 M H2, 0.00414 M I2 and 0.0224 M HI is placed...

a reaction mixture of 0.00623 M H2, 0.00414 M I2 and 0.0224 M HI is placed in a steel container of 1.0 L. The reaction that takes place is:

H2 (g) + I2 (g) ⇌ 2 HI (g).

Kc= 54.3 at 430 °C

a) Calculate the reaction quotient, Qc, for the initial mixture.

b) Will the reaction proceed toward making HI?

c) Calculate the concentration of these species at equilibrium.

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Homework Answers

Answer #1

Solution:- (a) Qc = [HI]2/[H2][I2]

Qc = (0.0224)2/(0.00623)(0.00414) = 19.5

(b Since Qc < Kc , So, the reaction would proceed towards the products and more HI is formed.

(c) Let's say the charge is X to the concentrations. Since the reaction proceed to the foward direction, The equilibrium concentrations would be..

[HI] = (0.0224 + 2X)

[H2] = (0.00623 - X)

[I2] = (0.00414 - X)

54.3 = [(0.0224 + 2X)2]/[(0.00623 - X) (0.00414 - X)]

On solving this complex math's equation

X = 0.00157

So, [HI] = 0.0224 + 2(0.00157) = 0.0255 M

[H2] = 0.00623 - 0.00157 = 0.00466 M

[I2] = 0.00414 - 0.00157 = 0.00257 M

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