A quantity of 60 g CH 3 COOH and 46 g CH 3 CH 2 OH reacts in a 4 L flask to form 66 g CH 3 COOC 2 H 5 at equilibrium. On taking 60 g CH 3 COOH and 82.8 g CH 3 CH 2 OH, CH 3 COOC 2 H 5 formed at equili

Question

A quantity of 60 g CH 3 COOH and 46 g CH 3 CH 2 OH reacts in a 4 L flask to form 66 g CH 3 COOC 2 H 5 at equilibrium. On taking 60 g CH 3 COOH and 82.8 g CH 3 CH 2 OH, CH 3 COOC 2 H 5 formed at equilibrium is

TestHub · Accepted Answer

First case: acid = 1 mol, alcohol = 1 mol, ester = 66/88 = 0.75 mol At equilibrium acid = 0.25, alcohol = 0.25, ester = 0.75, water = 0.75 mol; volume cancels since Δ n = 0

K_c = ((0.75)(0.75))/((0.25)(0.25)) = 9

Second case: acid = 1 mol, alcohol = 82.8/46 = 1.8 mol; ester formed = x mol

x²/((1-x)(1.8-x)) = 9, so 8x² - 25.2x + 16.2 = 0 and x = 0.9

Mass of ester = 0.9×88 = 79.2 g

This is option (d).

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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