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The Equilibrium Constant (DP IB Chemistry)

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Caroline Carroll

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DP Chemistry: SLDP Chemistry: HL

The Equilibrium Constant

The equilibirium constant, K

  • The size of the equilibrium constant, K, tells us how the equilibrium mixture is made up with respect to reactants and products

bold italic K bold space bold equals bold space stretchy left square bracket products stretchy right square bracket subscript bold eqm over stretchy left square bracket reactants stretchy right square bracket subscript bold eqm

  • If K > 1, the concentration of products is greater than the concentration of reactants and we say that the equilibrium lies to the right hand side

    • When K >> 1, equilibrium lies far over to the right hand side and the reaction almost goes to completion

  • If K < 1, then the concentration of reactants is greater than the concentration of products and we say that the equilibrium lies to the left hand side

    • When K << 1, equilibrium lies far over to the left hand side and the reaction hardly proceeds

  • When K = 1, at equilibrium, there are significant amounts of both reactants and products and equilibrium does not lie in favour of either the reactants or products

  • K is a constant at a specified temperature

  • Since temperature can affect the position of equilibrium, it follows that K is dependent on temperature

Worked Example

When the following reactions reach equilibrium, state whether the equilibrium mixture contains mostly reactants or products. Assume the value of K corresponds to the temperature of the reaction mixture

 

1.

Ag+ (aq) + Fe2+ (aq) ⇌ Ag (s) + Fe3+ (aq)

K = 7.3 x 10-26

 

2.

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

K = 2.6 x 10-18

 

3.

2SO2 (g) + O2 (g) ⇌ 2SO3 (g)   

K = 5.0 x 1013

 

Answer:

  • Reactions 1 and 2:

    • K is very much smaller than 1

    • So, the denominator in the equilibrium constant expression must be much larger than the numerator

    • This means that the concentration of the reactants is much larger than the concentration of products

    • Therefore, the equilibrium lies far to the left and the equilibrium mixture contains mostly reactants

  • Reaction 3:

    • K is very much larger than 1

    • So, the numerator in the equilibrium constant expression must be much larger than the denominator

    • This means that the concentration of the products is much larger than the concentration of reactants

    • Therefore, the equilibrium lies to the right-hand side and the reaction mixture contains mostly products

Examiner Tips and Tricks

  • Stronger acids dissociate more than weaker acids in solution, meaning that equilibrium lies towards the products

  • So, stronger acids will have a higher value of K than weaker acids.

The relationship between K values for reactions that are the reverse of each other

  • The equilibrium constant expression is dependent on a specific reaction

  • For example, take the reaction between nitrogen and hydrogen to make ammonia:

N2(g)    +    3H2(g)  ⇌     2NH3(g)

  • The equilibrium constant expression for this reaction is:

K space equals fraction numerator open square brackets NH subscript 3 close square brackets squared over denominator open square brackets straight N subscript 2 close square brackets open square brackets straight H subscript 2 close square brackets cubed end fraction

  • If we reverse the equation:

2NH3(g) ⇌ N2(g)    +    3H2(g)

  • The equilibrium constant expression for the reverse of this reaction, K', is:

K apostrophe space equals fraction numerator stretchy left square bracket straight N subscript 2 stretchy right square bracket open square brackets straight H subscript 2 close square brackets cubed over denominator stretchy left square bracket NH subscript 3 stretchy right square bracket squared end fraction

  • What is the relationship between these two K values? At the same temperature, K'  becomes the reciprocal of the original K value:

K apostrophe space equals space 1 over K   or    K apostrophe space equals space K to the power of negative 1 end exponent

Worked Example

The equilibrium constant for the following reaction is 7.1 × 1032.

2NO2 (g) + F2 (g) ⇌ 2NO2F (g)

What is the equilibrium constant for the reverse at the same temperature?

Answer:

  • K subscript open parentheses reverse close parentheses end subscript space equals space 1 over K subscript open parentheses forward close parentheses end subscript space equals space fraction numerator 1 over denominator 7.1 space cross times space 10 to the power of 32 end fraction space equals bold space bold 1 bold. bold 41 bold space bold cross times bold space bold 10 to the power of bold minus bold 33 end exponent 

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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.