1) Starting from $1.0\;mol$ of a carboxylic acid A and $1.0\;mol$ of an alcohol B , we find that the yield of the esterification is $0.80$. Calculate the quantities of all reactants and products at equilibrium and the equilibrium constant. 2) Calculate the yield in relation to alcohol, if we start from $2.0\;mol$ of A and $1.0\;$ mol of B. 3) Compare 1) and 2) and interpret!
$K$ $=$
$\frac{0.8^2}{0.2^2}$ $=$ $16$
2)
$16$ $=$ $\frac{x^2}{(2-x)(1-x)}$
$x$ $=$ $0.95$
$r$ $=$ $\frac{0,95}{1}$ $=$ $0.95$
The yield increased because the addition of a reagent has moved the equilibrium to the right!