Q. $CH_{3}COOH(l) + C_{2}H_{5}OH(l) \rightleftharpoons CH_{3}COOC_{2}H_{5}(l) + H_{2}O(l)$ In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc.$H_{2}SO_{4}$. On equilibrium being attained
A) 1 mole of ethyl acetate is formed
B) 2 mole of ethyl acetate is formed
C) $\frac{2}{3}$mole of ethyl acetate is formed
D) $\frac{1}{2}$ mole of ethyl acetate is formed
Explanation
(c)
In the presence of little $H_{2}SO_{4}$ (as catalyst) about 2/3 mole of each of $CH_{3}COOH$ and $C_{2}H_{5}OH$ to form $\frac{2}{3}$ mole of the product at equilibrium
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.