International A LevelChemistryOxford AQARevision NotesPhysical ChemistryChemical Equilibria, Le Chatelier’s Principle & KcEquilibrium Constant Kc for Homogeneous Systems

Equilibrium Constant Kc for Homogeneous Systems (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Kc Expressions & Calculations

Equilibrium expression and constant

  • The equilibrium expression is an expression that links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account

  • So, for a given reaction:

aA + bB ⇌ cC + dD

  • Kc is defined as:

K subscript straight c equals fraction numerator open square brackets straight C close square brackets to the power of straight c space open square brackets straight D close square brackets to the power of straight d over denominator open square brackets straight A close square brackets to the power of straight a space open square brackets straight B close square brackets to the power of straight b end fraction

  • Where:

    • [A] and [B] are the equilibrium concentrations of A and B, in mol dm-3 

    • [C] and [D] are the equilibrium concentrations of C and D, in mol dm-3

    • a, b, c and d are the respective number of moles of each reactant and product 

  • Solids are ignored in equilibrium expressions

  • The Kc of a reaction is specific and only changes if the temperature of the reaction changes

Worked Example

Deduce the equilibrium expression for the following reactions:

  1. Ag+ (aq) + Fe2+ (aq) rightwards harpoon over leftwards harpoon Ag (s) + Fe3+ (aq)

  2. N2 (g) + 3H2 (g) rightwards harpoon over leftwards harpoon 2NH3 (g)

  3. 2SO2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 2SO3 (g)

Answers:

  1. Kc = fraction numerator open square brackets Fe to the power of 3 plus end exponent space open parentheses aq close parentheses close square brackets over denominator open square brackets Fe to the power of 2 plus end exponent space open parentheses aq close parentheses close square brackets space open square brackets Ag to the power of plus space open parentheses aq close parentheses close square brackets end fraction

  2. Kc = fraction numerator open square brackets NH subscript 3 space open parentheses straight g close parentheses close square brackets squared over denominator stretchy left square bracket straight N subscript 2 space open parentheses straight g close parentheses stretchy right square bracket space stretchy left square bracket straight H subscript 2 space open parentheses straight g close parentheses stretchy right square bracket cubed end fraction

  3. Kc = fraction numerator open square brackets SO subscript 3 space open parentheses straight g close parentheses close square brackets squared over denominator stretchy left square bracket SO subscript 2 space stretchy left parenthesis straight g stretchy right parenthesis stretchy right square bracket squared space stretchy left square bracket straight O subscript 2 space stretchy left parenthesis straight g stretchy right parenthesis stretchy right square bracket end fraction

Calculations involving Kc

  • In the equilibrium expression, each figure within a square bracket represents the concentration in mol dm-3

  • The units of Kc therefore depend on the form of the equilibrium expression

  • Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture

  • The concentrations of the reactants and products can then be calculated from the number of moles and total volume using:

 

Worked Example

At equilibrium, 500 cm3 of the following reaction mixture contains 0.235 mol of ethanoic acid, 0.0350 mol of ethanol, 0.182 mol of ethyl ethanoate and 0.182 mol of water.

CH3COOH (l) + C2H5OH (l) rightwards harpoon over leftwards harpoon CH3COOC2H5 (l) + H2O (l)

Use this information to calculate a value of Kc for this reaction.

Answer:

  • Step 1: Calculate the concentrations of the reactants and products:

    • [CH3COOH (l)] = fraction numerator 0.235 over denominator 0.500 end fraction = 0.470 mol dm-3 

    • [C2H5OH (l)] = fraction numerator 0.0350 over denominator 0.500 end fraction = 0.070 mol dm-3 

    • [CH3COOC2H5 (l)] = fraction numerator 0.182 over denominator 0.500 end fraction = 0.364 mol dm-3 

    • [H2O (l)] = begin mathsize 14px style fraction numerator 0.182 over denominator 0.500 end fraction end style = 0.364 mol dm-3 

  • Step 2: Write out the balanced chemical equation with the calculated concentrations beneath each substance:

CH3COOH (l)

+

C2H5OH (l)

rightwards harpoon over leftwards harpoon

CH3COOC2H5 (l)

+

H2O (l)

0.470 mol dm-3 

 

0.070 mol dm-3 

 

0.364 mol dm-3 

 

0.364 mol dm-3 

  • Step 3: Write the equilibrium constant for this reaction in terms of concentration:

    • Kc = fraction numerator open square brackets straight H subscript 2 straight O close square brackets space open square brackets CH subscript 3 COOC subscript 2 straight H subscript 5 close square brackets over denominator open square brackets straight C subscript 2 straight H subscript 5 OH close square brackets space open square brackets CH subscript 3 COOH close square brackets end fraction

  • Step 4: Substitute the equilibrium concentrations into the expression:

    • Kc = fraction numerator 0.364 cross times 0.364 over denominator 0.070 cross times 0.470 end fraction

    • Kc = 4.03

  • Step 5: Deduce the correct units for Kc:

    • Kc = fraction numerator open parentheses mol space dm to the power of negative 3 end exponent close parentheses space open parentheses mol space dm to the power of negative 3 end exponent close parentheses over denominator open parentheses mol space dm to the power of negative 3 end exponent close parentheses space open parentheses mol space dm to the power of negative 3 end exponent close parentheses end fraction

    • All units cancel out

    • Therefore, Kc = 4.03

  • Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

  • Some questions give the initial and equilibrium concentrations of the reactants but products

  • An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked Example

Ethyl ethanoate is hydrolysed in water.

CH3COOC2H5 (l) + H2O (l) rightwards harpoon over leftwards harpoon CH3COOH (l) + C2H5OH (l) 

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture is made up to 1 dm3.

At equilibrium, 0.0654 mol of water are present.

Use this information to calculate a value of Kc for this reaction.

Answer:

  • Step 1: Write the balanced chemical equation, with the concentrations beneath each substance, into an initial, change and equilibrium (ICE) table:

 

CH3COOC2H5 (l)

+ H2O (l)

bold rightwards harpoon over leftwards harpoon

CH3COOH (l)

+ C2H5OH (l)

Initial moles

0.1000

0.1000

 

0

0

Change

–0.0346

–0.0346

 

+0.0346

+0.0346

Equilibrium moles

0.0654

0.0654

 

0.0346

0.0346

  • Step 2: Calculate the concentrations of the reactants and products:

    • [H2O (l)] = fraction numerator 0.0654 over denominator 1.000 end fraction = 0.0654 mol dm-3 

    • [CH3COOC2H5 (l)] = fraction numerator 0.0654 over denominator 1.000 end fraction = 0.0654 mol dm-3 

    • [C2H5OH (l)] = fraction numerator 0.0346 over denominator 1.000 end fraction = 0.0346 mol dm-3 

    • [CH3COOH (l)] = fraction numerator 0.0346 over denominator 1.000 end fraction = 0.0346 mol dm-3

  • Step 3: Write the equilibrium constant for this reaction in terms of concentration:

    • Kc = fraction numerator stretchy left square bracket straight C subscript 2 straight H subscript 5 OH stretchy right square bracket space stretchy left square bracket CH subscript 3 COOH stretchy right square bracket over denominator stretchy left square bracket straight H subscript 2 straight O stretchy right square bracket space stretchy left square bracket CH subscript 3 COOC subscript 2 straight H subscript 5 stretchy right square bracket end fraction

  • Step 4: Substitute the equilibrium concentrations into the expression:

    • Kc = fraction numerator 0.0346 cross times 0.0346 over denominator 0.0654 cross times 0.0654 end fraction

    • Kc = 0.28

  • Step 5: Deduce the correct units for Kc:

    • Kc = 

    • All units cancel out

    • Therefore, Kc = 0.28

Changes that Affect the Equilibrium Constant

Changes in concentration

  • If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in concentration of the reactants or products

  • For example, the decomposition of hydrogen iodide:

2HI ⇌ H2 + I2

  • The equilibrium expression is:

Kc = fraction numerator stretchy left square bracket H subscript 2 stretchy right square bracket bold space stretchy left square bracket I subscript 2 stretchy right square bracket over denominator stretchy left square bracket HI stretchy right square bracket to the power of bold 2 end fraction= 6.25 x 10–3

  • Adding more HI makes the ratio of [ products ] to [ reactants ] smaller

  • To restore equilibrium, [H2] and [I2] increase and [HI] decreases

  • Equilibrium is restored when the ratio is 6.25 x 10-3 again

Changes in pressure

  • A change in pressure only changes the position of the equilibrium (see Le Chatelier’s principle)

  • If all other conditions stay the same, the equilibrium constant Kc is not affected by any changes in the pressure of the reactants and products

Changes in temperature

  • Changes in temperature change the equilibrium constant Kc

  • For an endothermic reaction such as:

2HI (g) rightwards harpoon over leftwards harpoon H2 (g) + I2 (g)   Kc = fraction numerator open square brackets straight H subscript 2 close square brackets space open square brackets straight I subscript 2 close square brackets over denominator open square brackets HI close square brackets squared end fraction

  • With an increase in temperature:

    • [H2] and [I2] increases

    • [HI] decreases

    • Because [H2] and [I2] increase and [HI] decreases, the equilibrium constant Kc increases

  • For an exothermic reaction such as:

2SO2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 2SO3 (g)   Kc = fraction numerator open square brackets SO subscript 3 close square brackets cubed over denominator open square brackets SO subscript 2 close square brackets squared space open square brackets straight O subscript 2 close square brackets end fraction

  • With an increase in temperature:

    • [SO3] decreases

    • [SO2] and [O2] increases

    • Because [SO3] decreases and [SO2] and [O2] increase, the equilibrium constant Kc decreases

Examiner Tips and Tricks

You need to be able to predict the qualitative effects of changes of temperature on the value of Kc.

To do this:

  • Write the Kc expression

  • Use the balanced chemical equation and enthalpy change information to talk about the concentrations of each species in the expression

  • Then talk about the overall effect of a temperature change on the value of Kc.

Presence of a catalyst

  • If all other conditions stay the same, the equilibrium constant Kc is not affected by the presence of a catalyst

  • A catalyst speeds up both the forward and reverse reactions at the same rate so the ratio of [ products ] to [ reactants ] remains unchanged

Worked Example

Factors affecting Kc 

An equilibrium is established in the following reaction:

AB (aq) + CD (aq) rightwards harpoon over leftwards harpoon AC (aq) + BD (aq)   ΔH = +180 kJ mol-1

Which factors would affect the value of Kc in this equilibrium?

Answer:

  • Only a change in temperature will affect the value of Kc

  • Any other changes in conditions would result in the position of the equilibrium moving to oppose this change

  • Adding a catalyst increases the rate of reaction meaning the state of equilibrium will be reached faster but has no effect on the position of the equilibrium and, therefore, Kc is unchanged

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.