# For the reaction $aA+bB\rightarrow cC+dD$ the rate of formation of [C] in term of [A] is given by

$\begin{array}{1 1}(a)\;\large\frac{d[C]}{dt}=\frac{c}{a}\frac{d[A]}{dt}&(b)\;\large\frac{d[C]}{dt}=-\frac{a}{c}\frac{d[A]}{dt}\\(c)\;\large\frac{d[C]}{dt}=-\frac{d[A]}{dt}&(d)\;\large\frac{d[C]}{dt}=-\frac{c}{a}\frac{d[A]}{dt}\end{array}$

Answer: $-\large\frac{c}{a}\frac{d[A]}{dt}$
In general for the reaction: $aA+bB\rightarrow cC+dD$,
$\quad \large - \frac{1}{a}$$\large\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}=\frac{1}{d}\frac{d[D]}{dt}$
$\Rightarrow \large\frac{d[C]}{dt}=-\frac{c}{a}\frac{d[A]}{dt}$
edited Jul 24, 2014