Equilibrium constants and equilibrium concentrations

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

K_c 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 K_c 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 K_c is less than 1, then the position of equilibrium lies to the left, on the side of the reactants.

H_2(g) + I_2(g) ⇌ 2HI_(g)

At 730K, K_c 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.

N_2(g) + O_2(g) ⇌ 2NO_(g) K_c = 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 K_c must be calculated for each system. In both the examples above, the units cancel out.

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

K_c 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)   +   Br_2(g)   ⇌   COBr_2(g)

It was determined that at equilibrium the concentrations of each component were: [CO_(g)]  =  9.23 x 10^-3 mol dm^-3, [Br_2(g)]  =  3.96 x 10^-3 mol dm^-3, [COBr_2(g)]  =  3.44 x 10^-3 mol dm^-3.

Calculate the value of K_c and determine the correct units. 

Practice question

Ammonia reacts with oxygen to form nitrogen oxide.

4NH_3(g)   +   5O_2(g)   ⇌   4NO_(g)   +   6H₂O_(g)

Write an expression for K_c and determine the correct units. 

Answer

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

Example: For the equilibrium:   PCl_{5 (g)}   ⇌   PCl_3(g)  + Cl_2(g) K_c = 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.

N₂O_4(g) ⇌  2NO_2(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 K_c   = 1.06 x 10^-5 mol dm^-3  at this temperature?

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

2H₂S_(g) ⇌  2H_2(g)  +  S_2(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:  

PCl_{5 (g)}   ⇌   PCl_3(g)     +     Cl_2(g)

At 250°C in a 2.0dm³ vessel, there are 0.15mol of PCl₃ and 0.090mol Cl₂ and K_c  = 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

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

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

Calculating the equilibrium constant, K_c, 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.

SO₂Cl_2(g)  ⇌  SO_2(g)   +  Cl_2(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 K_c from this data.

CH₃COOH_(aq)  +  C₅H_10(aq)   ⇌  CH₃COOC₅H_11(aq) 

2. 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.

H_2(g)  +  I_2(g)  ⇌  2HI_(g)

3. 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

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

K_c = 491 mol^-1 dm³

2. 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

K_c = 56 (no units)

3. 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

K_c = 4.2 (no units)

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