Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 22

Suppose that the equation $\displaystyle [c] = \frac{a^2kt}{(akt + 1)}$ means that one mole of the product $c$ is formed from one molecule of the reactant $A$ and one molecule of the reactant $B$, and the initial concentrations of $A$ and $B$ have a common value $\displaystyle [A] = [B] = a \frac{\text{moles}}{L}$. Where $k$ is constant
a.) Find the rate of reaction at time $t$.
b.) Show that if $x = [c]$, then
$\displaystyle \frac{dx}{dt} = k (a-x)^2$


a.) Using Quotient Rule,


$
\begin{equation}
\begin{aligned}
\frac{d[c]}{dt} &= \frac{(akt + 1) \frac{d}{dt}(a^2kt)-(a^2kt) \frac{d}{dt}(akt +1)}{(akt+1)^2}\\
\\
\frac{d[c]}{dt} &= \frac{(akt+1)(a^2k) - (a^2kt)(ak)}{(akt+1)^2}\\
\\
\frac{d[c]}{dt} &= \frac{\cancel{a^3k^2t} + a^2k - \cancel{a^3k^2t}}{(akt+1)^2}\\
\\
\frac{d[c]}{dt} &= \frac{a^2k}{(akt+1)^2}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{b.) } \frac{dx}{dt} = \frac{d[c]}{dt} &= k (a-x)^2\\
\\
\frac{d[c]}{dt} &= k (a-x)^2 && \text{;where } [c] = \frac{a^2kt}{(akt+1)}\\
\\
\frac{d[c]}{dt} &= k \left( \frac{\cancel{a^2kt} + a - \cancel{a^2kt}}{akt+1}\right)^2\\
\\
\frac{d[c]}{dt} &= k \left( \frac{a}{akt+1} \right)^2\\
\\
\frac{d[c]}{dt} &= \frac{a^2k}{(ak+1)^2} = \frac{dx}{dt} = k(a-x)^2
\end{aligned}
\end{equation}
$