In acidic medium $0.58 \;mol$ of ethanol reacts with $0.33 \;mol$ of methanoic acid. At equilibrium, one obtains $0.26 \;mol$ ester.
Calculate the equilibrium constant!
Determine the number of moles of acid, alcohol and water at equilibrium.
Follow the order shown in the following table!
a) $+0.26 \;mol$ b) $+0.26 \;mol$ c) $-0.26 \;mol$ d) $-0.26 \;mol$ e) $ 0.33 - 0.26$ $=$ $0.07 \;mol$ f) $0.58 - 0.26$ $=$ $0.32 \;mol$ g) $0+0.26$ $=$ $0.26 \;mol$
Now calculate the equilibrium constant $K$ !
$K$= $\frac{n_{ester}n_{eau}}{n_{acide}n_{alcool}}$ = $\frac{0.26\cdot 0.26}{0.07\cdot0.32}$ $\approx 3$
