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The hydrolysis of ethyl acetate needs two reactants, ester and water, yet its kinetics look exactly like a first-order decay. That is not a contradiction. It is a pseudo first-order reaction, and examiners like it because it tests whether you understand what a rate law measures. This post explains the idea, works through numbers and shows how to get the true rate constant back.
Where does this appear in the syllabus?
The rationalised NCERT Class 12 chapter on Chemical Kinetics (Reprint 2026-27) introduces pseudo first-order reactions straight after integrated rate equations, with two examples: hydrolysis of ethyl acetate and inversion of cane sugar. The JEE (Main) 2026 syllabus lists rate law, rate constant and its units, and the integrated forms of zero and first-order reactions, which is all you need here. Check the syllabus for your exam year.
Why does a second-order reaction look first order?
NCERT writes the rate law for the hydrolysis as rate = k[CH₃COOC₂H₅][H₂O]. Both concentrations matter. But water is taken in large excess. NCERT's own example uses 0.01 mol of ester with 10 mol of water. At the end of the reaction the ester has gone from 0.01 mol to 0, while water has gone from 10 mol to 9.99 mol, a change of 0.1%.
If [H₂O] barely changes, you can treat it as a constant and fold it into the rate constant:
rate = k[H₂O][ester] = k′[ester], where k′ = k[H₂O]
The reaction now follows a first-order rate law in the ester alone. It is "pseudo" because the true order is two; the first-order behaviour comes from the conditions, not the mechanism.
How does cane sugar inversion fit the same pattern?
In acid, sucrose reacts with water to give glucose and fructose. NCERT writes the rate as rate = k[C₁₂H₂₂O₁₁]. Water is the solvent, so its concentration stays essentially constant. The H⁺ ion is written over the arrow as a catalyst, so it is not used up either. The only concentration that falls noticeably is that of sucrose, and the reaction follows first-order kinetics in sucrose.
Worked example: finding k′ and k
These are sample numbers for practice. A sucrose solution in dilute acid has [sucrose] = 0.400 mol L⁻¹ at t = 0, 0.300 mol L⁻¹ at t = 60 min and 0.225 mol L⁻¹ at t = 120 min.
Test for first order. Use k′ = (1/t) ln([A]₀/[A]). At 60 min: ln(0.400/0.300) / 60 = ln(1.333) / 60 = 0.2877 / 60 = 4.79 × 10⁻³ min⁻¹.
At 120 min: ln(0.400/0.225) / 120 = ln(1.778) / 120 = 0.5754 / 120 = 4.79 × 10⁻³ min⁻¹. The constant value confirms first-order behaviour.
Half-life: t½ = 0.693 / k′ = 0.693 / (4.79 × 10⁻³) ≈ 145 min. Notice that 0.400 → 0.300 → 0.225 is a fall by the same factor (0.75) every 60 minutes, the signature of first order.
True second-order constant. Pure water is about 1000/18 ≈ 55.5 mol L⁻¹. Taking that as [H₂O], k = k′/[H₂O] = (4.79 × 10⁻³) / 55.5 ≈ 8.6 × 10⁻⁵ L mol⁻¹ min⁻¹.
Look at the units. k′ is in min⁻¹, a first-order unit. k is in L mol⁻¹ min⁻¹, a second-order unit. Our post on units of the rate constant shows how to read order from units like these.
How large must the excess be?
The approximation is only as good as the excess. With a tenfold excess, even complete reaction of the other reactant lowers the excess reagent's concentration by just 10%. With water as solvent, at about 55.5 mol L⁻¹, a 0.5 mol L⁻¹ ester solution changes [H₂O] by less than 1%. Kineticists use the same idea on purpose: Corbett (1972) describes running second-order reactions under first-order conditions to obtain accurate second-order rate constants, with an allowance for the slow depletion of the reactant in excess.
Which mistakes cost marks?
Calling the reaction first order. Its true order is two. Say "pseudo first order" and explain the excess.
Reporting k′ as k. If a question asks for the second-order constant, divide by the excess concentration.
Using pseudo first order without an excess. If both reactants are at similar concentrations, the first-order equations do not apply.
Mixing log bases. Use ln, or 2.303 log, consistently.
Forgetting that the catalyst sits inside k′. Change the acid concentration and k′ changes too, because H⁺ was treated as constant.
A practice routine
Day 1: Write the rate law for ester hydrolysis, then rewrite it as a pseudo first-order law, naming what goes into k′.
Day 2: Redo the worked example with readings at 30, 60 and 90 minutes that you generate yourself from k′ = 4.79 × 10⁻³ min⁻¹.
Day 3: Three questions on converting between k′ and k, with the units at each step.
Day 4: Revise the first-order derivation in our post on rate laws, order and half-life.
If the difference between k and k′ will not settle, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit rate laws before returning to this one.
Tonight, take NCERT's 0.01 mol ester and 10 mol water and calculate what percentage of the water is used up; then repeat with only 0.1 mol of water and see why the approximation fails.
