Question

Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.

$\begin{array}{ccccc}\text{t/min}& 0& 30& 60& 90\\ C/mol{L}^{-1}& 0.8500& 0.8004& 0.7538& 0.7096\end{array}$

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant $55mol{L}^{-1}$, during the course of the reaction. What is the value of $k^{\prime }$ in this equation?

$Rate=k^{\prime }\left[C{H}_{3}COOC{H}_3\right]\left[H_{2}O\right]$

Answer 1

For pseudo first order reaction, the
reaction should be first order with respect to ester when $\left[H_{2}O\right]$ is constant

Values of k at different $t$

We know that, rate constant k for pseudo
first order reaction is

$k=\frac{2.303}{t}\log \frac{C_0}{C}$where $k=k^{\prime }\left[H_{2}O\right]$
Now ,
$k(att=30min)=\frac{2.303}{30}\log \frac{0.8500}{0.8004}$

$=2.004\times {10}^{-3}mi{n}^{-1}$

$k(att=60min)=\frac{2.303}{60}\log \frac{0.8500}{0.7538}$

$=2.002\times {10}^{-3}mi{n}^{-1}$

$k(att=90min)=\frac{2.303}{90}\log \frac{0.8500}{0.7096}$

$=2.005\times {10}^{-3}mi{n}^{-1}$

From the above data we note

$\begin{array}{ccc}\text{t/min}& C/mol{L}^{-1}& k/mi{n}_{-1}\\ 0& 0.8500& -\\ 30& 0.8004& 2.004\times {10}^{-3}\\ 60& 0.7538& 2.002{\times }^{-3}\\ 90& 0.7096& 2.005\times {10}^{-3}\end{array}$

Value of $k^{\prime }$
It can be seen that $k^{\prime }\left[H_{2}O\right]$ is constant and
equal to $2.004\times {10}^{-3}mi{n}^{-1}$ and hence, it
is pseudo first order reaction. We can now
determine k as:

$k^{\prime }\left[H^{2}O\right]=2.004\times {10}^{-3}mi{n}^{-1}$

$k^{\prime }\left[55mol{L}^{-1}\right]=2.004\times {10}^{-3}mi{n}^{-1}$

$k^{\prime }=3.64\times {10}^{-5}mo{l}^{-1}Lmi{n}^{-1}$

Final answer: value of $k^{\prime }$ in this equation
is $3.64\times {10}^{-5}mo{l}^{-1}Lmi{n}^{-1}$