Question

The equilibrium constant of ester formation of propionic acid with ethyl alcohol is 7.36 at ${50}^{0}C$. Calculate the weight of ethyl propionate in gram existing in an equilibrium mixture when 0.5 mole of propionic acid is heated with 0.5 mole of ethyl alcohol at ${50}^{0}C$.

Answer 1

Propionic acid $+$ ethyl alcohol $\rightleftharpoons$ ethylpropionate $+$ water

water assumed to be in excess, so need not consider in $K$ equation.

$K=\frac{\left(ethylpropionate\right)}{\left(propionicacid\right)\left(ethylalcohol\right)}$

[propionic acid] $=0.5$ mol initially

[ethyl alcohol] $=0.5$ mol initially

$C_2{H}_{5}COOH+C_2{H}_{5}OH\rightleftharpoons C_2{H}_{5}COO{C}_2{H}_5+H_{2}O$

At $0.5$ $0.5$ $0$

$t=0$

At $0.5-a$ $0.5-a$ $a$

$t=t_{eq}$

$K_C=\frac{a}{[0.5-a][0.5-a]}=736$

$\frac{a}{{(0.5-a)}^2}=736=\frac{a}{0.25+a^2-a}=7.36$

$a=7.36a^2-7.36a+1.84$

${7.36a}^2-8.36a+1.84=0$

$a=0.2985$ (or) $0.837$

$0.837$ not valid because it is greater than $0.5$ which is maximum that could be formed. So $a=0.2985$

weight $=$ moles $\times$ molecular weight

$=0.2985$ $\times$ $102.13$

$=30.48g$