Equilibrium Constant Kp for Homogeneous Systems (Oxford AQA International A Level (IAL) Chemistry): Revision Note

Exam code: 9622

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Updated on

Kp Expressions

  • 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 a given reaction:

aA + bB ⇌ cC + dD

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

Gaseous Equilibria

  • If all the substances in the general equation above are gases:

aA (g) + bB (g) ⇌ cC (g) + dD (g)

  • The equilibrium constant Kp is deduced from a gaseous reversible reaction equation

  • Kp is defined as:

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  • Where:

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

    • paA (g) and pbB (g) are the equilibrium partial pressures of A and B, kPa

    • pcC (g) and pdD (g) are the equilibrium partial pressures of C and D, in kPa

  • Equilibrium partial pressure is calculated from  the:

    • The total number of moles

    • The mole fraction

    • The total pressure

  • Solids and liquids are ignored in 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) size 16px rightwards harpoon over leftwards harpoon 2NH3 (g)

  2. N2O4 (g) size 16px rightwards harpoon over leftwards harpoon 2NO2 (g)

  3. 2SO2 (g) + O2 (g) size 16px rightwards harpoon over leftwards harpoon 2SO3 (g)

Answers:

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

    • Kpfraction numerator p squared space NH subscript 3 over denominator p cubed space straight H subscript 2 space cross times space p space straight N subscript 2 end fraction

    • Units = fraction numerator kPa squared over denominator kPa cubed cross times kPa end fraction = kPa-2

  2. N2O4 (g) rightwards harpoon over leftwards harpoon 2NO2 (g)

    • Kpfraction numerator p squared space N O subscript 2 over denominator p space straight N subscript 2 straight O subscript 4 end fraction

    • Units = kPa squared over kPa = kPa

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

    • Kpfraction numerator p squared space SO subscript 3 over denominator p squared space SO subscript 2 space cross times space p space straight O subscript 2 end fraction

    • Units = fraction numerator kPa squared over denominator kPa squared cross times kPa end fraction = kPa-1

Kp, Equilibrium & Condition Changes

  • Le Chatelier's Principle can be applied to gaseous equilibria in the same way it is applied to aqueous systems

    • It states that if a change is made to a system in dynamic equilibrium, the position of the equilibrium moves to counteract this change

  • It can be used to predict the qualitative effects of changes in temperature and pressure on:

    • The position of equilibrium

    • The value of Kp

How temperature affects Kp

  • For a reaction that is exothermic in the forward direction:

    • Increasing the temperature pushes the equilibrium to the left

    • This means that there are more reactants and less products

    • The ratio of products to reactants decreases

    • Therefore, the value of Kp decreases

  • For a reaction that is endothermic in the forward direction:

    • Increasing the temperature pushes the equilibrium to the right

    • This means that there are less reactants and more products

    • The ratio of products to reactants increases

    • Therefore, the value of Kp increases

How pressure affects Kp

  • The value of Kp is not affected by any changes in pressure.

  • Changes in pressure cause a shift in the position of equilibrium to a new position which restores the value of K

  • This is similar to Kc when you change concentration in an aqueous equilibrium; a shift restores equilibrium to a new position maintaining Kc

How a catalyst affects Kp

  • If all other conditions stay the same, Kp 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

  • Catalysts only cause a reaction to reach equilibrium faster

  • Catalysts therefore have no effect on the position of the equilibrium once this is reached

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Richard Boole

Author: Richard Boole

Expertise: Chemistry Content Creator

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

Reviewer: Stewart Hird

Expertise: Chemistry Content Creator

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.