# 7.18     Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:                $CH_{3}COOH _{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$(ii)  At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

M manish

Let the volume of the mixture will be V.
$CH_{3}COOH _{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$
initial conc.                                     1/V                             0.18/V                          0                                   0

At equilibrium                          $\frac{1-.171}{V}$  (= 0.829/V)  $\frac{1-.171}{V}$               $0.171$                         $0.171$

So the equilibrium constant for the reaction can be calculated as;

$K_c = \frac{(0.171)^2}{(0.829)(0.009)}$

$=3.92$ (approx)

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