Acidic Hydrolysis Pseudo First Order Reaction

1. True rate law for acidic hydrolysis

For an ester like methyl acetate ($CH_3COOCH_3$) in dilute acid, the elementary overall reaction is
$CH_3COOCH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3OH$
The experimentally observed rate law (before any approximation) is generally
$\text{Rate} = k\,[CH_3COOCH_3]^1[H_2O]^1$
so the overall order is second (first order in ester, first order in water).

2. Why the order appears to drop

• In the laboratory we usually take the ester in a very small amount and mix it with a huge excess of water (often the reaction is done in aqueous solution!).
• Because water is the solvent, its molar concentration is roughly constant at $55\,\text{mol L}^{-1}$ throughout the reaction.
• Any constant can be clubbed with the rate constant:
$k' = k[H_2O] = \text{constant}$
• The rate law becomes
$\text{Rate} = k'[CH_3COOCH_3]$
This looks like a plain first-order rate law. The reaction is therefore called pseudo (apparent) first-order.

3. Integrated form

Starting from
$\frac{d[CH_3COOCH_3]}{dt} = -k'[CH_3COOCH_3]$
Integrate between $[CH_3COOCH_3]_0$ and $[CH_3COOCH_3]$ at time $t$:

$\ln\left(\frac{[CH_3COOCH_3]_0}{[CH_3COOCH_3]}\right)=k' t$
A straight line of $\ln([\text{ester}])$ vs $t$ confirms pseudo first-order behaviour.

4. Where does the acid come in?

In acidic hydrolysis, $H^+$ acts as a catalyst. Its concentration is kept constant (buffer/strong acid). Therefore $[H^+]$ also folds into the observed rate constant if needed.

Imagine breaking cookies in lots of water

Think of a biscuit (the ester) that breaks into two pieces when dipped in lots and lots of water with a few drops of lemon juice (acid). Because there is so much water around, its amount never really changes while the biscuit breaks. So, even though both biscuit and water are needed, it looks like the speed depends only on the biscuit. Scientists call such a situation pseudo first-order – ‘pseudo’ means looks like, and ‘first-order’ means rate depends on the first power of just one reactant.

Example calculation

Problem: Initial ester concentration $=0.25\,\text{mol L}^{-1}$. After $40\,\text{min}$ it drops to $0.10\,\text{mol L}^{-1}$. Calculate the pseudo first-order rate constant $k'$.

Using

$\ln\left(\frac{[E]_0}{[E]}\right)=k' t$

$\ln\left(\frac{0.25}{0.10}\right)=k'\,(40\times60\,\text{s})$

$\ln(2.5)=k'\,(2400\,\text{s})$

$k'=\frac{\ln(2.5)}{2400}=4.20\times10^{-4}\,\text{s}^{-1}$

Hence $k'\approx4.2\times10^{-4}\,s^{-1}$.