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ASChemistryCambridge (CIE)Revision Notes7. EquilibriaChemical Equilibria: Reversible Reactions & Dynamic EquilibriumEquilibrium Constants, Kc & Kp

Equilibrium Constants, Kc & Kp (Cambridge (CIE) AS Chemistry)

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Caroline Carroll

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Equilibrium Constant: Concentrations

Equilibrium expression & 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)

Answer

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

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

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

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

Mole Fraction & Partial Pressure

Partial pressure

  • For reactions involving mixtures of gases, the equilibrium constant Kp is used as it is easier to measure the pressure than the concentration for gases

  • The partial pressure of a gas is the pressure that the gas would have if it was in the container all by itself

  • The total pressure is the sum of the partial pressure:

Ptotal = PA + PB + PC + .......

  • Ptotal = total pressure

    • PA, PB, PC = partial pressures

 How partial pressures contribute to total pressure

Equilibria Partial Pressures, downloadable AS & A Level Chemistry revision notes

Partial pressures can be added together to calculate the total pressure

Mole fraction

  • The mole fraction of a gas is the ratio of moles of a particular gas to the total number of moles of gas present

bold Mole bold space bold fraction bold equals fraction numerator bold number bold space bold of bold space bold moles bold space bold of bold space bold a bold space bold particular bold space bold gas over denominator bold total bold space bold number bold space bold of bold space bold moles bold space bold of bold space bold all bold space bold the bold space bold gases bold space bold in bold space bold the bold space bold mixture end fraction

  • To calculate the partial pressures of each gas the following relationship can be used:

bold Partial bold space bold pressure bold equals bold mole bold space bold fraction bold cross times bold total bold space bold pressure bold space

  • The sum of the mole fractions should add up to 1.00, while the sum of the partial pressures should add up to the total pressure

Equilibrium Constant: Partial Pressures

Equilibrium expressions involving partial pressures

  • Equilibrium expressions in terms of partial pressures are written similarly to those involving concentrations with a few differences:

Comparing Kp and Kc expressions 

Equilibria Comparison Equilibrium Expressions, downloadable AS & A Level Chemistry revision notes

The process of writing the expressions is similar, but there is a different presentation and different information required

Worked Example

Deducing equilibrium expressions of gaseous reactions

Deduce the equilibrium expression for the following reactions:

  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)

Answer

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

Kp = fraction 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

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

Kp = fraction numerator p squared space N O subscript 2 over denominator p space straight N subscript 2 straight O subscript 4 end fraction

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

Kp = fraction 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

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    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    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 creating high-quality resources to help students achieve their full potential.