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A student ran the following reaction in the laboratory at 293 K: 2CH2Cl2(g) CH4(g) + CCl4(g)...

A student ran the following reaction in the laboratory at 293 K:

2CH2Cl2(g) CH4(g) + CCl4(g)

When she introduced 6.91×10-2 moles of CH2Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of CCl4(g) to be 3.19×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

Kc=?

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Answer #1

ICE Table:

[CH2C12] [CH4] [CC14] initial 0.0691 change -2x +1x +1x equilibrium 0.0691-2x +1x +1x

Given at equilibrium,

[CCl4] = 0.0319

+1x = 0.0319

x = 0.0319

Equilibrium constant expression is

Kc = [CH4][CCl4]/[CH2Cl2]^2

Kc = (+1x)(+1x)/(0.0691-2x)^2

Kc = (+1*0.0319)(+1*0.0319)/(0.0691-2*0.0319)^2

Kc = 36.2

Answer: 36.2

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