# How to identify a pseudo-first-order reaction?

We can easily identify a pseudo first-order reaction with the integrated rate equation, but what if only the reaction is given? For example, what if I were given a question like the following on the hydrolysis of acetic anhydride?

Identify the type of this reaction: $$\ce{(CH3CO)2O + H2O -> 2CH3COOH}$$

• A. Pseudo-first-order reaction
• B. Pseudo-second-order reaction
• C. Zero-order reaction
• D. Third-order reaction

I would like to know how I could identify the correct answer if such a question were asked.

• Look at the mechanism of the reaction Jul 3, 2016 at 12:11

There are four steps:

1. Identify the mechanism.
2. Find the rate-determining step.
3. Write the corresponding rate equation according to the stoichiometric coefficients of that step.
4. Remove abundant reactants.

# 1. Identify the mechanism Assuming that there is no acid to catalyze the reaction, it proceeds via a nucleophilic addition/elimination reaction.

An acetic anhydride molecule $(1)$ is attacked by the lone pairs on the water to form the complex $(2)$. This is the addition step.

The complex tautomerizes to give $(3)$, which is converted to our products, namely 2 acetic acid molecules $(4)$. This is the elimination step.

# 2. Find the rate-determining step.

Presumably, it is the first step.

# 3. Write the rate equation.

In the first step, one molecule of acetic anhydride reacts with one molecule of water. Therefore, it is a first-order reaction with respect to both of them:

• $\text{Rate} = k [\ce{Ac2O}] [\ce{H2O}]$

# 4. Remove abundant reactants.

Water is abundant, so it is removed from the equation:

• $\text{Rate} = k [\ce{Ac2O}]$

# Conclusion

It is a pseudo-first-order reaction [A].

The stoichiometric equation does not indicate what the reaction mechanism is unless by accident. Thus if you don't know all the individual equilibrium, unimolecular and bimolecular reactions you cant work out what is going on without doing experiments. For example by performing experiments your reaction you may turn out to be acid catalysed, but this does appear in the equation you give.