Kc & Kp (OCR A Level Chemistry)

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Kc & Kp Expressions

  • As previously discussed, the equilibrium expression links the equilibrium constantKc, 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

Kc = fraction numerator left square bracket straight C right square bracket to the power of straight c space left square bracket straight D right square bracket to the power of straight d over denominator left square bracket straight A right square bracket to the power of straight a space left square bracket straight B right square bracket to the power of straight b end fraction

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 Expressions, downloadable AS & A Level Chemistry revision notes

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

  • 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 Tip

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. N(g) + 3H(g) ⇌ 2NH(g)
  2. 2SO(g) + O(g) ⇌ 2SO(g)

Answer 1

Worked Example Answer 1, downloadable AS & A Level Chemistry revision notes

Answer 2

Worked Example Answer 2, downloadable AS & A Level Chemistry revision notes

 

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

WE1 Answer, downloadable AS & A Level Chemistry revision notes

  • 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? 

WE2 Answer 1, downloadable AS & A Level Chemistry revision notes

WE2 Answer 2, downloadable AS & A Level Chemistry revision notes

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Richard

Author: Richard

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.