A LevelChemistryOCRRevision Notes5. Physical Chemistry & Transition ElementsEquilibriaKc & Kp

Kc & Kp (OCR A Level Chemistry A): Revision Note

Exam code: H432

Richard Boole

Written by: Richard Boole

Reviewed by: Caroline Carroll

Updated on

Kc & Kp Expressions

  • As previously discussed, the equilibrium expression links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account

  • So, for the general reaction:

aA + bB ⇌ cC + dD

  • The Kc expression is:

Kc = [C]c [D]d[A]a [B]b

Units of Kc

  • The units of Kc are deduced from the expression, in the same way as for Kp

  • Each concentration term carries the unit mol dm-3, raised to its stoichiometric power

  • Count the terms above and below the line, then cancel

  • For the contact process equilibrium:

2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

Kc = [SO3]2[SO2]2 [O2]

  • There are two concentration terms above the line and three below:

(mol dm-3)2 ÷ (mol dm-3)3 = (mol dm-3)-1

  • So the units of Kc are dm3 mol-1

  • Where the number of terms above and below the line is equal, the units cancel completely and Kc has no units

  • For example, the esterification equilibrium has two terms on each side of the expression

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Kc = [CH3COOC2H5] [H2O][CH3COOH] [C2H5OH]

Examiner Tips and Tricks

The units of Kc are worth a separate mark at A Level, so always work them out from your own expression.

Write the units as dm3 mol-1 or mol-1 dm3 as both are accepted. Do not write them as a fraction such as dm3/mol, which is not accepted.

If your Kc expression is wrong, you can still gain the units mark for deducing units correctly from the expression you wrote.

Homogeneous and heterogeneous equilibria

  • An equilibrium is homogeneous when all the species are in the same phase

  • An equilibrium is heterogeneous when they are not all in the same phase

  • In a heterogeneous equilibrium, the concentrations of solids and pure liquids are constant, so they are omitted from the equilibrium expression

  • This applies to Kc in exactly the same way as it applies to Kp

  • For the thermal decomposition of calcium carbonate

CaCO3 (s) ⇌ CaO (s) + CO2 (g)

  • Calcium carbonate and calcium oxide are both solids, so both are omitted from each expression:

Kc = [CO2]

Kp = p(CO2)

  • Each expression contains only the carbon dioxide term

  • The units follow as usual, so here Kc has units of mol dm-3

Examiner Tips and Tricks

Do not include a solid or a pure liquid term in a heterogeneous equilibrium expression. Leaving it in is a common error and loses the mark, and it changes the units as well as the expression.

Gaseous Equilibria

  • This section covers how we manage gases in calculating the equilibrium constant and how an equilibrium yield is affected by the partial pressures of reactants and products

  • Many industrial process involve reactions between gases so this application has important consequences for controlling reaction conditions

  • In the generic example above, if all the substances are gases, we can show the equation with that state symbol

aA (g) + bB (g) ⇌ cC (g) + dD (g)

  • We can write a different equilibrium expression in terms of the partial pressure of the gases

  • This equilibrium constant is called Kp and is defined as follows

Kp expression for aA(g) + bB(g) <=> cC(g) + dD(g): Kp = p^c C(g) eqm times p^d D(g) eqm, divided by p^a A(g) eqm times p^b B(g) eqm. p is the equilibrium partial pressure in kPa; a, b, c and d are the coefficients in the balanced equation.

Equilibrium expression linking the partial pressures of reactants and products at equilibrium

Kp = pc(C) × pd(D) / [pa(A) × pb(B)], using equilibrium partial pressures in kPa; a, b, c and d are the coefficients in the balanced equation.

(Partial pressures are written p(C), p(D), p(A), p(B); no square brackets.)

  • In a heterogeneous equilibrium, solids and pure liquids are ignored in both Kc and Kp equilibrium expressions

  • The Kp of a reaction is constant and only changes if the temperature of the reaction changes

Examiner Tips and Tricks

There are a variety of ways to represent the partial pressure terms in a Kp expression

The only key point is do not use square brackets as these represent concentration and, therefore, imply a KC expression

Worked Example

Write a Kp expression for the following equilibria and deduce the units of Kp :

  1. N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

  2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

Answer 1

Worked answer 1: Kp for N2(g) + 3H2(g) <=> 2NH3(g) equals p^2 NH3 divided by (pN2 times p^3 H2), with units kPa^2 / (kPa x kPa^3) = kPa^-2.

Kp = p2(NH3) / [p(N2) × p3(H2)] = kPa2 / (kPa × kPa3) = kPa-2

Answer 2

Worked answer 2: Kp for 2SO2(g) + O2(g) <=> 2SO3(g) equals p^2 SO3 divided by (p^2 SO2 times pO2), with units kPa^2 / (kPa^2 x kPa) = kPa^-1.

Kp = p2(SO3) / [p2(SO2) × p(O2)] = kPa2 / (kPa2 × kPa) = kPa-1

Kc & Kp Calculations

Kp calculations

  • Kp calculations are a step-by-step process in which you need to find

    • the mole fractions of the gases present

    • their partial pressures

    • the Kp expression

    • the value of Kp

Worked Example

Finding Kp

Hydrogen and bromine were mixed in a flask in a 1: 1 ratio and allowed to reach equilibrium at 450 K. When equilibrium had been achieved the total pressure in the flask was 140 kPa and the mole fraction of bromine was 0.35.

The equation for the reaction is

H2 (g) + Br2 (g)  ⇌ 2HBr (g)

Determine the partial pressures for each gas at equilibrium and the value of Kp

Worked answer: mole fractions H2 0.35, Br2 0.35, HBr 0.30; partial pressures 49 kPa, 49 kPa and 42 kPa; Kp = 42^2 / (49 x 49) = 0.735, no units.

Mole fraction of H2 = mole fraction of Br2 = 0.35 (1:1 ratio); mole fraction of HBr = 1 - (2 × 0.35) = 0.30

p(H2) = p(Br2) = 0.35 × 140 = 49 kPa; p(HBr) = 0.30 × 140 = 42 kPa

Kp = p2(HBr) / [p(H2) × p(Br2)] = 422 / (49 × 49) = 0.735 (no units)

  • Another style of Kp calculation involves being given the value of Kp and working backwards to deduce the partial pressure of one of the gases

Worked Example

Finding Partial Pressure

Kp for the dissociation equilibrium reaction of nitrogen monoxide is 0.0021.

2NO (g)  ⇌ N2 (g) + O2 (g)

If pure NO is introduced into a reaction flask at an initial pressure of 100 kPa, what is the equilibrium partial pressure of nitrogen? 

Worked answer step 1: let x be the equilibrium partial pressure of N2; initially NO is 100 kPa and N2 and O2 are 0; at equilibrium NO is (100 - 2x) kPa and N2 and O2 are x kPa each.

Let x = equilibrium partial pressure of N2. Initially: NO 100 kPa, N2 0, O2 0. At equilibrium: NO (100 - 2x) kPa, N2 x kPa, O2 x kPa.

Worked answer step 2: Kp = pN2 x pO2 / p^2 NO = x^2 / (100 - 2x)^2 = 0.0021; square roots give x = 4.2 kPa (2 s.f.) so N2 and O2 are 4.2 kPa and NO is 91.6 kPa.

Kp = p(N2) × p(O2) / p2(NO) = x2 / (100 - 2x)2 = 0.0021

Taking square roots: x / (100 - 2x) = 0.04583, so x = 0.04583 × (100 - 2x), 1.09166x = 4.583, x = 4.198

Therefore p(N2) = 4.2 kPa (2 s.f.); p(O2) = 4.2 kPa; p(NO) = 100 - (2 × 4.2) = 91.6 kPa

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Richard Boole

Author: Richard Boole

Expertise: Curriculum Expert

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.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.