Two moles of an equimolar mixture of two alcohols R1-OH and R2-OH are esterified with 1 mole of acetic acid - Chemistry - Chemical Kinetics

Question

Two moles of an equimolar mixture of two alcohols R1-OH and
R2-OH are esterified with 1 mole of acetic acid If 80% of the acid
is comsumed and the quantities of ester formed under eqilibrium are
in the ratio of3:2,what is the value of the equlibrium constant for
the esterification of R1-OH ?

Sumeet Malik · Accepted Answer

R1-OH + R2-OH + 2CH3COOH
â‡”CH3COOR1 + CH3COOR2 +
2H2O
2 moles 2 moles 1 mole 0 moles 0 moles
At equilibrium
0 8 moles of CH3COOH consumed
So it produced
0 8*3/5 = 0 48 moles of CH3COOR1
0 8*2/5 = 0 32 moles of CH3COOR2
Kc = [0 48] [0 32] [1 6] / [1 6] [1 6] [0 2] = 0 48

Two moles of an equimolar mixture of two alcohols R1-OH and R2-OH are esterified with 1 mole of acetic acid .If 80% of the acid is comsumed and the quantities of ester formed under eqilibrium are in the ratio of3:2,what is the value of the equlibrium constant for the esterification of R1-OH ?