CCEA A-Level Chemistry 1110
Formulae, Equations and Amounts of Substance
Atomic Structure
Bonding
Intermolecular Forces and Structure
Shapes of Molecules and Ions
Redox
The Halogens
Acid-Base Titrations
Qualitative Tests
Formulae and Amounts of a Substance
Nomenclature and Isomerism in Organic Chemistry
Alkanes
Alkenes
Halogenoalkanes
Alcohols
Infrared Spectroscopy
Energetics
Kinetics
Equilibrium
Group II Elements and Their Compounds
Lattice Enthalpy, Entropy and Free Energy
Rates of Reaction
Equilibrium and Acid-Base Equilibria
Isomerism
Aldehydes and Ketones
Carboxylic Acids
Derivatives of Carboxylic Acids
Aromatic Chemistry
Spectroscopy and Chromatography
Volumetric Analysis
Transition Metals
Electrode Potentials
Amines
Amides and Polymers
Amino Acids and Proteins
Equilibrium and Acid-Base Equilibria
Equilibrium concentrations from initial amounts
- 3 things to know
- Unit A2 1
Overview
Recall questions on the initial, change and equilibrium layout, why the changes follow the balancing numbers, worked examples on hydrogen with iodine, an esterification, the dissociation of dinitrogen tetroxide and of phosphorus pentachloride, converting moles to concentrations, and what a negative equilibrium amount shows.
What you need to know
Answer each one before you reveal it.
Why must the change in moles for each species follow the balancing numbers in the equation?
One mole of ethanoic acid and one mole of ethanol reach equilibrium with Kc = 4. State the amount of water present at equilibrium
Questions from this subtopic
Each one with the model answer underneath.
- 1
Describe the layout used to work out equilibrium amounts from initial amounts.
Model answer
A table is drawn with a row for the initial moles, a row for the change in moles, and a row for the equilibrium moles of each species.
- 2
Why must the change in moles for each species follow the balancing numbers in the equation?
Model answer
The stoichiometry fixes the ratio in which species are used up and formed, so if one mole of a reactant reacts the others change by the amounts the equation requires.
- 3
In the reaction H₂(g) + I₂(g) ⇌ 2HI(g), 1.00 mol of each reactant is placed in a 1.00 dm³ vessel and 1.56 mol of HI is present at equilibrium. Calculate the equilibrium moles of hydrogen.
Model answer
Forming 1.56 mol of HI uses 0.78 mol of hydrogen, so 1.00 − 0.78 = 0.22 mol of hydrogen remains.
- 4
1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed and 0.667 mol of ester is present at equilibrium. Calculate the equilibrium moles of ethanoic acid.
Model answer
Forming 0.667 mol of ester consumes 0.667 mol of acid, so 1.00 − 0.667 = 0.333 mol of ethanoic acid remains.
- 5
In that esterification mixture, state the amount of water present at equilibrium.
Model answer
Water is formed in a one to one ratio with the ester, so 0.667 mol of water is present.
- 6
1.00 mol of N₂O₄ is placed in a 2.00 dm³ vessel and 0.60 mol remains at equilibrium in the reaction N₂O₄(g) ⇌ 2NO₂(g). Calculate the equilibrium concentration of NO₂.
Model answer
0.40 mol of N₂O₄ has dissociated, producing 0.80 mol of NO₂, so the concentration is 0.80 ÷ 2.00 = 0.40 mol dm⁻³.
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