calculating k c values

Calculating Kc values

CH3COOH (aq) + CH3CH2OH(aq)

CH3COOCH2CH3 (ag) + H2O(l)

CH3COOH CH3CH2OH CH3COOCH2CH3 H2O

Initial conc

a b 0 0

From the balanced equation 1 mol of acid reacts with 1 mol of alcohol to form 1 mol of ester and 1 mol of water. This means that…..

• The amount of the two products formed is the same (x)

• The amount by which the two reactants are reduced equals the amount of each product formed (x)

Equilib conc

(a-x) (b-x) x x





Initial conc

a b 0 0

Kc = [CH3COOCH2CH3] [H2O] [CH3COOH ] [CH3CH2OH]

Kc = (x/V) x (x/V) (a-x)/V x (b-x)/V

Kc = ___x2___ (a-x) x (b-x)

Equilib conc

(a-x) (b-x) x x

Where V = volume in dm3

If the number of moles is the same on both sides of the equilibrium then the volume cancels, so we can use moles rather than concentrations





Initial moles

1.0 0.18 0 0

Moles CH3CH2OH at equilibrium = 0.18 – 0.171

= 0.009

Moles CH3COOH at equilibrium = 1.0 – 0.171

Equilib moles

0.171 0.171

= 0.829





Initial moles

1.0 0.33 0 0

Kc = 0.171 x 0.171 0.829 x 0.009

Equilib moles

0.829 0.009 0.171 0.171


= 0.0292 0.829 x 0.009

= 3.92




Initial moles

1.0 1.0 0 0

Moles CH3COOCH2CH3 at equilibrium = 1.0 – 0.333

= 0.667

Moles CH3CH2OH at equilibrium = 0.333

Equilib moles

0.333

When 1 mol each of ethanoic acid and ethanol are mixed together at a fixed temperature 0.333mol of acid remain at equilibrium. Calculate Kc

Moles H2O at equilibrium = 1.0 – 0.333

= 0.667





Initial moles

1.0 1.0 0 0

Kc = 0.667 x 0.667 0.333 x 0.333

Equilib moles

0.333 0.333 0.667 0.667


= 4.01

H2 (g) + I2g)

2HI(g)

H2 I2 2HI

Initial moles

0.206 0.144 0

Moles I2 at equilibrium = 0.144 – 0.258/2

= 0.077

Moles H2 at equilibrium = 0.206 – 0.258/2

Equilib moles

0.258

0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate Kc

From the balanced equation 1mol hydrogen reacts with 1 mol iodine to form 2 moles of hydrogen iodide

= 0.015

H2 (g) + I2g)

2HI(g)

H2 I2 2HI

Initial moles

0.206 0.144 0

Equilib moles

0.077 0.015 0.258

0.206 mol of hydrogen and 0.144 mol if iodine were heated at 683K. At equilibrium 0.258 mol of HI were present. Calculate Kc

Kc = [HI]2 [H2 ] [I2]

Kc = (0.258)2 0.077 x 0.015

Kc = 57.63

For reactions that involve a change in the number of moles, the volume must be known in order to calculate the concentrations before Kc can be calculated as the volume terms will not cancel.

N2O4(g)

2NO2(g)

N2O4 2NO2

Initial moles

1.0 0

Equilib moles

At 300K 1.0 mol of N2O4 is 20% dissociated in 2.0dm3 flask. Calculate Kc

If 20% is dissociated, then 80% remains at equilibrium

Mols of N2O4 at equilibriuum = 80/100 x 1.0 = 0.8

Moles of NO2 at equilibrium = 2 x (1.0 – 0.8) = 0.4

N2O4(g)

2NO2(g)

N2O4 2NO2

Initial moles

1.0 0

Equilib moles

0.8 0.4

At 300K 1.0 mol of N2O4 is 20% dissociated in 2.0dm3 flask. Calculate Kc

Kc = [NO2]2 [N2O4]

Kc = (0.4/V)2 0.8/V

Kc = 0.1 moldm-3

Kc = (moldm-3)2 moldm-3

= moldm-3

Kc = (0.4/2)2 0.8/2

Finding Kc Experimentally

• Known amounts of reagents are used

(Initial concentrations known)

• The system is closed and left until equilibrium is reached

• The concentration of the products is then analysed, often by titration

• Kc can then be calculated

• Different initial concentrations of alcohol were used in different experiments

• Calculate Kc for the following experiments

• All the experiments were performed at 373 k

Initial Moles Equilibrium molesKcCH3COOH CH3CH2OH CH3COOCH2CH3 H2O

1 1.0 0.33 0.171 0.171 3.92

2 1.0 0.5 0.425 0.425

3 1.0 1.0 0.667 0.667

4 1.0 2.0 0.850 0.850

5 1.0 8.0 0.967 0.967

Kc is a constant at constant temperature

Initial Moles Equilibrium molesKcCH3COOH CH3CH2OH CH3COOCH2CH3 H2O

1 1.0 0.33 0.171 0.171 3.92

2 1.0 0.5 0.425 0.425 4.12

3 1.0 1.0 0.667 0.667 4.01

4 1.0 2.0 0.850 0.850 4.12

5 1.0 8.0 0.967 0.967 4.02

Altering the concentration of the reactants will shift the equilibrium position but the value of Kc remains constant

Factors that Affect Equilibrium

• Changing the concentration of a reactant or product will shift the position of an equilibrium




Kc is the ratio of concentrations so…

• Changing the concentration of a reactant or product will not affect the value of Kc

• Changing concentration has no effect on Kc

• Changing pressure has no effect on Kc

• Catalysts have no effect on Kc

• Changing temperature will change Kc

Factors that Affect Equilibrium

calculating k c values

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