1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed and left to reach equilibrium, forming ethyl ethanoate and water. At equilibrium 0.25 mol of the acid remains. Calculate Kc.

0.75 mol of each reactant has been used and 0.75 mol of each product formed, so Kc = (0.75 × 0.75) ÷ (0.25 × 0.25) = 9.0, with no units.

Why does the volume of the flask not need to be known in that esterification calculation?

There are two concentration terms above and below the line, so every volume term cancels.

0.100 mol dm⁻³ of N₂O₄ is allowed to reach equilibrium as N₂O₄(g) ⇌ 2NO₂(g), and 40.0% of it dissociates. Calculate the equilibrium concentrations.

0.0400 mol dm⁻³ of N₂O₄ reacts, leaving 0.0600 mol dm⁻³ of N₂O₄ and forming 0.0800 mol dm⁻³ of NO₂.

Use those concentrations to calculate Kc, with units.

Kc = 0.0800² ÷ 0.0600 = 0.107 mol dm⁻³.

Confirm that answer by substituting it back into the expression.

0.107 × 0.0600 = 6.4 × 10⁻³, and the square root of that is 0.0800 mol dm⁻³, which is the quoted concentration of NO₂.

1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 2.00 dm³ vessel. At equilibrium 1.60 mol of HI is present. Calculate the equilibrium concentrations of all three gases.

0.800 mol of each reactant has been used, so [H₂] = [I₂] = 0.100 mol dm⁻³ and [HI] = 0.800 mol dm⁻³.

OCR A-Level Chemistry B H433

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ICE-table calculations and measuring Kc experimentally

1 thing to know
Components 01, 02 and 03

Overview

Recall questions on using an initial, change and equilibrium table, converting amounts to concentrations, and measuring Kc for an esterification.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

When Kc is measured experimentally, why must the reaction mixture be left for a long time before it is sampled?

Questions from this subtopic

Each one with the model answer underneath.

1
1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed and left to reach equilibrium, forming ethyl ethanoate and water. At equilibrium 0.25 mol of the acid remains. Calculate Kc.
Model answer
0.75 mol of each reactant has been used and 0.75 mol of each product formed, so Kc = (0.75 × 0.75) ÷ (0.25 × 0.25) = 9.0, with no units.
2
Why does the volume of the flask not need to be known in that esterification calculation?
Model answer
There are two concentration terms above and below the line, so every volume term cancels.
3
0.100 mol dm⁻³ of N₂O₄ is allowed to reach equilibrium as N₂O₄(g) ⇌ 2NO₂(g), and 40.0% of it dissociates. Calculate the equilibrium concentrations.
Model answer
0.0400 mol dm⁻³ of N₂O₄ reacts, leaving 0.0600 mol dm⁻³ of N₂O₄ and forming 0.0800 mol dm⁻³ of NO₂.
4
Use those concentrations to calculate Kc, with units.
Model answer
Kc = 0.0800² ÷ 0.0600 = 0.107 mol dm⁻³.
5
Confirm that answer by substituting it back into the expression.
Model answer
0.107 × 0.0600 = 6.4 × 10⁻³, and the square root of that is 0.0800 mol dm⁻³, which is the quoted concentration of NO₂.
6
1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 2.00 dm³ vessel. At equilibrium 1.60 mol of HI is present. Calculate the equilibrium concentrations of all three gases.
Model answer
0.800 mol of each reactant has been used, so [H₂] = [I₂] = 0.100 mol dm⁻³ and [HI] = 0.800 mol dm⁻³.

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