Question

A 0.055 mol sample of formaldehyde vapor, CH2O, was placed in a heated 500 mL vessel and some of it decomposed. The reaction is CH2O(g) H2(g) + CO(g) At equilibrium, the CH2O(g) concentration was 0.051 mol L-1. Calculate the value of Kc for this reaction?

Answer 1

Answer:

Kc for this reaction is 0.06825

Explanation:

Step 1: Data given

Number of moles formaldehyde CH2O = 0.055 moles

Volume = 500 mL = 0.500 L

At equilibrium, the CH2O(g) concentration = 0.051 mol

Step 2: The balanced equation

CH2O  <=>  H2 + CO

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CH2O] = 0.055 moles . 0.500 L

[CH2O] = 0.11 M

[H2] = 0M

[CO] = 0M

Step 4: The concentration at the equilibrium

[CH2O] = 0.11 - X M = 0.051 M

[H2] = XM

[CO] = XM

[CH2O] = 0.11 - X M = 0.051 M

X = 0.11 - 0.051 = 0.059

[H2] = XM = 0.059 M

[CO] = XM = 0.059 M

Step 5: Calculate Kc

Kc = [H2][CO]/[CHO]

Kc = (0.059 * 0.059) / 0.051

Kc = 0.06825

Kc for this reaction is 0.06825