Equation to calculate concentration from number of moles and volume
Ethanoic acid and ethanol react according to the following equation:
CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)
At equilibrium, 500 cm3 of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.
Calculate a value of Kc for this reaction
Answer
Kc =
Kc
Kc =
All units cancel out
Therefore, Kc = 4.03
Calculating Kc of ethyl ethanoate
Ethyl ethanoate is hydrolysed by water:
CH3COOC2H5(I) + H2O(I) ⇌ CH3COOH(I) + C2H5OH(I)
0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1dm3. At equilibrium 0.0654 mol of water are present. Use this data to calculate a value of Kc for this reaction.
Answer
Kc = 0.28
All units cancel out
Therefore, Kc = 0.288
Calculating Kp of a gaseous reaction:
In the reaction:
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
the equilibrium partial pressures at constant temperature are
SO2 = 1.0 × 106 Pa, O2 = 7.0 × 106 Pa, SO3 = 8.0 × 106 Pa
Calculate the value for Kp for this reaction.
Answer
Kp = 9.1 x 10-6
The units of Kp are Pa-1
Therefore, Kp = 9.1 x 10-6 Pa-1
Calculating Kp of a hydrogen iodide equilibrium reaction:
The equilibrium between hydrogen, iodine and hydrogen iodide at 600 K is as follows:
H2 (g) + I2 (g) ⇌ 2HI (g)
At equilibrium the number of moles present are:
H2 = 1.71 × 10-3
I2 = 2.91 × 10-3
HI = 1.65 × 10-2
The total pressure is 100 kPa.
Calculate the value of Kp for this reaction.
Answer
Total number of moles = 1.71 x 10-3 + 2.91 x 10-3 + 1.65 x 10-2
= 2.112 x 10-2
H2 = 0.0810 x 100 = 8.10 kPa
I2 = 0.1378 x 100 = 13.78 kPa
HI = 0.7813 x 100 = 78.13 kPa
Kp = 54.7
All units cancel out
Therefore, Kp = 54.7
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