Question

Consider the reaction CH2O (g) ⇌ H2 (g) + CO(g). Exactly 0.040 moles of CH2O were...

Consider the reaction CH2O (g) ⇌ H2 (g) + CO(g). Exactly 0.040 moles of CH2O were injected into a 250 mL flask. At equilibrium, [CH2O] = 0.057 M. What is Kc for this reaction? Report your answer to the correct number of significant figures.

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Answer #1

Step I

Step II

Let x moles of CH2O gets converted into H2 and CO.

At t=0, moles of CH2O = 0.04

Moles of H2 = 0 and Moles of CO = 0

At t=teq, Moles of CH2O = 0.04 - x

Moles of H2 = x and Moles of CO = x

Step III

At t = teq, [CH2O] = 0.057 M

Concentration = Moles / Volume

( 0.04 - x ) / 0.25 = 0.057

x = 0.02575

Step IV

Hence, Moles of H2 = Moles of CO = 0.02575

[H2] = [CH2O] = (0.02575/0.25) = 0.103

Step V

Hence, Kc = (0.103)(0.103) / (0.057)

Kc = 0.186

This is the answer.

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