Chemistry · Chemical Kinetics · NEET
It is a reaction whose true order is higher (usually second order), but it acts like a first order reaction. One reactant is taken in such a large amount that its concentration barely changes during the reaction. So the rate seems to depend on only one reactant. The word 'pseudo' means fake or false, so the first order is not real, it just looks that way.
The reaction is CH3COOC2H5 + H2O -> CH3COOH + C2H5OH (in acid). It needs both ethyl acetate and water, so its true order is 2. But water is taken in huge excess (for example 10 mol water with only 0.01 mol ester). The water used up is tiny, so [H2O] stays almost constant. Its constant value merges into the rate constant, and Rate = k'[ester]. This is first order behaviour, so it is called pseudo first order.
If a reactant is in large excess, only a small fraction of it reacts. So its concentration stays nearly constant from start to finish. A constant concentration acts like a fixed number, not a variable. That number gets absorbed into a new rate constant. Then the rate formula only shows the reactant that actually changes, making it look first order.
A true first order reaction really depends on only one reactant, and its order is genuinely 1. A pseudo first order reaction actually depends on two reactants (true order higher), but one is in excess, so it only appears to be first order. If you removed the excess and used equal amounts, the pseudo one would show its real higher order. For NEET, remember: true first order is real, pseudo first order is disguised.
Its true order is 2 because it needs both cane sugar and water: C12H22O11 + H2O -> glucose + fructose (in acid). But water is in large excess, so the rate depends only on cane sugar: Rate = k[C12H22O11]. So it behaves as first order and is a pseudo first order reaction. NEET can ask this as a direct example.
Start with Rate = k[A][B]. If B (like water) is in large excess, [B] is nearly constant. So Rate = k[B] x [A] = k'[A], where k' = k[B] is the new observed rate constant. Because [B] is fixed, k' is also fixed. The formula now has only [A], which is the mark of a first order reaction.
The correct option for the rate law that corresponds to an overall first order reaction is:
Which one of the following statements correctly describes the difference between a first-order and a second-order reaction?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
'Pseudo' means false or fake. The reaction is not genuinely first order; it only appears first order because one reactant is in large excess. Its true order is higher, usually second order.
1) Acid hydrolysis of ethyl acetate: CH3COOC2H5 + H2O -> CH3COOH + C2H5OH. 2) Inversion of cane sugar: C12H22O11 + H2O -> glucose + fructose. In both, water is in large excess, so the rate depends only on the organic reactant.
It is written as Rate = k'[A], where A is the reactant that changes and k' = k[B] is a new observed rate constant that already contains the constant, large-excess concentration [B].
Yes. Because it behaves as first order, its half-life t(1/2) = 0.693/k' is constant and does not depend on the starting concentration of the changing reactant.
They make hard second order rate problems simple, and NEET often asks their examples (ester hydrolysis, cane sugar inversion) and the reason one reactant is in excess. Knowing this helps you spot when order 'looks' first but is really higher.
