Chemistry - Chemical Equilibrium Question with Solution | TestHub
If in this esterification, initially equal moles of alcohol and acid are taken. At equilibrium, unreacted alcohol and acid, on treatment with sodium, produced H2 which occupied 33.6 L at 0°C and 1 atm. The percentage esterification of acid and alcohol is
Options:
Answer:
Solution:
Let a mol each of acid and alcohol be taken initially and y mol of ester form At equilibrium: acid = alcohol = a - y, ester = water = y, total = 2a
Mole fraction of ester = y/(2a) = 0.375, so y = 0.75a
Percentage esterification = (y/a) × 100 = 75%
Check with the gas data: 33.6 L at 0°C and 1 atm is 1.5 mol of H2, so 3 mol of molecules with an active H are present
2(a - y) + y = 2a - y = 1.25a = 3, so a = 2.4 mol, which gives unreacted 0.6 mol each and 1.8 mol of water, consistent
K = (0.75/0.25)2 = 9, matching the passage
This is option (b).
