How to find the equilibrium constant and equilibrium concentrations for a reaction

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Now it is time to introduce the concept of the equilibrium constant, Kc.

Kc is basically a ratio – the concentrations of the products at equilibrium divided by the concentration of the reactants. It tells us exactly where the position of equilibrium lies.

If Kc is greater than 1, the position of equilibrium lies to the right as the concentration of products at equilibrium must be greater than the concentration of reactants. If Kc is less than 1, then the position of equilibrium lies to the left, on the side of the reactants.

H2(g) + I2(g) ⇌ 2HI(g)

At 730K, Kc for this reaction is 46.7 which tells us that to all intents the reaction has gone to completion and there is virtually no hydrogen and iodine present in the equilibrium mixture.

N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.6 x 10-31 at 298K

By contrast, this reaction hardly gets going – virtually all the nitrogen and oxygen remain unreacted at equilibrium.

The units of Kc must be calculated for each system. In both the examples above, the units cancel out.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Kc calculations at A level range from the exceptionally straightforward to the fiendishly difficult 😈.

Example: Carbon monoxide and bromine were mixed at 308K and the system allowed to achieve equilibrium. 

CO(g)   +   Br2(g)   ⇌   COBr2(g)

It was determined that at equilibrium the concentrations of each component were: [CO(g)]  =  9.23 x 10-3 mol dm-3, [Br2(g)]  =  3.96 x 10-3 mol dm-3, [COBr2(g)]  =  3.44 x 10-3 mol dm-3.

Calculate the value of Kc and determine the correct units. 

How does changing the conditions of a system at equilibrium affect Kc?

The important point to remember is that Kc is a ratio. Changing the concentration or pressure of a substance once a system has been allowed to reach equilibrium (equilibrium position 1) will cause the rate of the forward and back reactions to change until a new equilibrium position has been reached (equilibrium position 2).

In equilibrium position 2, the concentrations or partial pressures of reactants and products will be different to those in equilibrium position 1, but the ratio stays the same. Kc remains constant.

Along as the temperature remains constant …

Changes in temperature will affect the value of Kc. If we change the temperature of a system at equilibrium, the new equilibrium position will reflect a change in the value of Kc.

e.g. N2(g)  +  3H2(g)  ⇌  2NH3(g) ΔrH = -92 kJ mol-1

The forward reaction is exothermic so increasing the temperature would cause the position of equilibrium to shift to the left in the endothermic direction, the concentration of reactants increases so Kc decreases (Kc is 4.39 x 104 mol-2 dm6 at 400K and 3.00 x 10-2 mol-2 dm6 at 800K).

Practice questions

  1. Ammonia reacts with oxygen to form nitrogen oxide.

4NH3(g)   +   5O2(g)   ⇌   4NO(g)   +   6H2O(g)

Write an expression for Kc and determine the correct units. 

2. The equilibrium system established when carbon monoxide and bromine are mixed can be represented by two different equations:

CO(g)  +  Br2(g)  ⇌  COBr2(g)

COBr2(g)   ⇌   CO(g)  +  Br2(g)

(a) What is the mathematical relationship between Kc and Kc1?

(b) Write an expression for Kc for each equation and give the units.

3. Dinitrogen tetroxide exists in an equilibrium with nitrogen dioxide.

N2O4(g)   ⇌   2NO2(g)

Given that the value of Kc is 0.490 mol dm-3 at 100°C and 18.6 mol dm-3 at 200°C, deduce with reason whether the reaction is exothermic or endothermic.

Answers

1.

2.

3. Kc increases with increasing temperature which means that as the temperature increases so does the concentration of the product, nitrogen dioxide. Since increasing temperature causes the position of equilibrium to shift to the right, this must be in the endothermic direction according to le Chatelier’s principle. The forward reaction is endothermic. 

You also need to be able to calculate the equilibrium concentration of a reactant or product given Kc.

Example: For the equilibrium:   PCl5 (g)   ⇌   PCl3(g)  + Cl2(g) Kc = 0.19 mol dm-3 at 250°C

One equilibrium mixture at this temperature contains PCl5 at a concentration of 0.20 mol dm-3  and PCl3 at a concentration of 0.010. Determine the concentration of Cl2 in the equilibrium mixture.

Practice questions

  1. Some N2O4 dissolved in chloroform was allowed to reach equilibrium at a known temperature.

N2O4(g) ⇌  2NO2(g)

At this point the concentration of NO2 was 1.85 x 10-3 mol dm-3.  What was the equilibrium     concentration of N2O4  given that Kc   = 1.06 x 10-5 mol dm-3  at this temperature?

2. At 1400K, Kc = 2.25  x 10-4 mol dm-3 for the equilibrium:

2H2S(g) ⇌  2H2(g)  +  S2(g)

In an equilibrium mixture, [H2S(g)] = 4.84 x 10-3 mol dm-3 and [S2] = 2.33 x 10-3 mol dm-3. Calculate the equilibrium concentration of H2.

3. In another equilibrium mixture of the reaction:  

PCl5 (g)   ⇌   PCl3(g)     +     Cl2(g)

At 250°C in a 2.0dm3 vessel, there are 0.15mol of PCl3 and 0.090mol Cl2 and Kc  = 0.19 mol dm-3 at 250°C.

(a)  Calculate the amount of PCl5 present at equilibrium

(b)  Calculate the mass of PCl5 present at equilibrium

4. Sulfur dioxide reacts with oxygen to form sulfur trioxide, a reaction which comes to equilibrium.

2SO2(g)    +    O2(g)   ⇌   2SO3(g)

At 400K, Kc = 0.160 dm3 mol-1 and the concentrations of [SO2(g)] and [O2(g)] are 2.40 mol dm-3 and 1.35 mol dm-3 respectively. 

(a) Explain whether the position of equilibrium lies to the left or the right at 400K.

    (b) Calculate the concentration of SO3(g) in the equilibrium mixture.

    Answers

    1.  [N2O4] = 0.323 mol dm-3

    2. [H2] = 1.5 x 10-3 mol dm-3

    3. (a)  amount of PCl5 = 0.036 mol  (b)  mass = 7.40g

    4. (a) Kc is less than 1 so the concentrations of the reactants must be greater than the concentration of the products at equilibrium. Position of equilibrium lies to the left, the reactant side.

    Moving on to the fiendishly difficult types of equilibrium constant question …

    Calculating the equilibrium constant, Kc, given the balanced equation, the initial concentration (or amount) and the equilibrium concentration of at least one substance …

    Example: 6.75g of SO2Cl2 was put into a 2.00dm3 vessel, the vessel sealed and its temperature raised to 375°C.  At equilibrium, the vessel contained 0.0345mol of Cl2.  Calculate the equilibrium constant for the reaction.

    SO2Cl2(g)  ⇌  SO2(g)   +  Cl2(g)

    Practice questions

    1. Ethanoic acid and pentene react together to form pentyl ethanoate in an inert solvent.   A solution was prepared containing 0.020 mol of pentene and 0.010 mol of ethanoic acid in 600 cm3 of solution.  At equilibrium there was 9.0 x 10-3 mol of pentyl ethanoate.  Calculate Kc from this data.

    CH3COOH(aq)  +  C5H10(aq)   ⇌  CH3COOC5H11(aq) 

    1. A mixture of 1.90 mol of hydrogen and 1.90 mol of iodine was allowed to reach equilibrium at 710K.  The equilibrium mixture was found to contain 3.00 mol of hydrogen iodide.  Calculate the equilibrium constant at 710K for the reaction.

    H2(g)  +  I2(g)  ⇌  2HI(g)

    1. If a mixture of 6.0g of ethanoic acid and 6.9g of ethanol is allowed to reach equilibrium, 7.0g of ethyl ethanoate is formed.  Calculate Kc.

    CH3COOH  +  C2H5OH   ⇌   CH3COOC2H5   +  H2O

    4.  0.4 mol of nitrogen dioxide was placed in a 2 dm3 sealed tube at 600K and the following reaction was allowed to come to equilibrium:

    2NO2(g)   ⇌   2NO(g)   +   O2(g)   ΔrH⦵  =  +180 kJ mol-1

    The amounts of O2(g) and NO2(g) over time are shown on the graph below.

    (a) Deduce the time at which the mixture reached equilibrium.

    (b) Sketch on the graph how the amount of NO(g) changed over the course of 6 minutes at 600K.

    (c) Calculate a value for Kc at 600K.

    (d) The temperature of the sealed tube was changed and the mixture allowed to establish a new equilibrium.  At this new temperature, the concentration of  [NO2(g)] was found to be 0.03 mol dm-3.  Deduce with reason whether the new temperature was higher or lower than 600K

    5.  1.0 mol of substance A and 1.0 mol of substance B reacted together in a sealed 1 dm3 container to form substance C according to the stoichiometric equation:

    A(g)   +   B(g)   ⇌   2C(g)

    At equilibrium, x mol of substance A had reacted. 

    (a) Write an expression for Kc.

      (b) Deduce in terms of x the number of moles of substances A, B and C at equilibrium.

        (c) Given that at 300K the value of Kc was found to be 4.0, calculate the equilibrium concentrations of each substance.

        Answers

        1. amount of pentene at equilibrium = (initial – reacted) = 0.020 – (9.0 x 10-3) = 0.011mol

        amount of ethanoic acid = 0.010 – (9.0 x 10-3) = 1.0 x 10-3 mol

        Kc = 491 mol-1 dm3

        1. amount of H2 at equilibrium = initial – reacted

        from balanced equation the ratio of H2:HI is 1:2, so to have made 3.00 mol of HI, 1.5 mol of H2 must have reacted and at equilibrium, the amount of H2 remaining = 1.90 – 1.5 mol = 0.40 mol and similarly for I2

        there are equal numbers of moles on either side so we can substitute in moles rather than concnetrations

        Kc = 56 (no units)

        1. initial amounts of ethanoic acid = 0.10mol, ethanol = 0.15mol

        equilibrium amount of ester = 0.079 mol

        balanced equation shows water in a 1:1 ratio with the ester, so equilibrium amount of water must be 0.079mol

        equilibrium amount of acid = initial – reacted = 0.10 – 0.079 = 0.021 mol (1:1 ratio again)

        equilibrium amount of ethanol = 0.15 – 0.079 = 0.071mol

        there are equal amounts (mol) on either side, so we can substitute mol into expression

        Kc = 4.2 (no units)

        4.

        4. (d) The forward reaction is endothermic. Increasing the temperature will shift the equilibrium position to the right, in the endothermic direction, leading to higher concentrations of products and lower concentrations of reactants.

        Concentration of NO2 has decreased from 0.1 mol dm-3 to 0.03 mol dm-3, so the new temperature must be higher than 600K.