Pseudo-First-Order Reactions and the Isolation Method

By Aniket Bhardwaj · 14 October 2026 · Updated 14 October 2026 · IIT-JAM Chemistry

Many reactions that obey a simple first-order law are not really "one-reactant" reactions. Acid-catalysed ester hydrolysis, the inversion of sucrose and many solvolysis reactions all involve a second species, yet they show clean first-order kinetics. The reason is the pseudo-first-order approximation. It appears in kinetics sections of IIT-JAM, GATE, CSIR-NET and CUET-PG preparation, and it is also the standard way to measure the order of a reaction in the lab.

A three-row isolation-method table shows [B]₀ values 0.50, 1.0, and 2.0 M, corresponding k′ values 1.2 × 10⁻³, 2.4 × 10⁻³, and 4.8 × 10⁻³ s⁻¹, and a constant ratio of 2.4 × 10⁻³. — Doubling [B]₀ doubles k′; the constant ratio confirms first order in B.

The idea

Take a reaction A + B → products with rate law rate = k[A][B]. This is second order overall. Now keep B in a very large excess, for example 100 times the amount of A. Even when all of A has reacted, B has changed by only about 1%, so [B] stays practically equal to its initial value [B]₀. The rate law collapses:

rate = k[A][B] ≈ k[B]₀[A] = k′[A]
k′ = k[B]₀ (pseudo-first-order rate constant, unit s⁻¹)
ln[A] = ln[A]₀ − k′t t_½ = ln 2 / k′

The true second-order constant k has units L mol⁻¹ s⁻¹. It is recovered by k = k′ / [B]₀.

What each term means

[A]₀, [B]₀ are the starting concentrations. The condition for the method is [B]₀ ≫ [A]₀.
k′ is the observed first-order constant. It changes if you change [B]₀, so it is not a true constant of the reaction.
k is the real rate constant. It depends on temperature (Arrhenius) but not on [B]₀.
Solvent as reagent. In hydrolysis in water, [H₂O] is about 55 M and does not change measurably, so it is folded into k′. The same applies to a catalyst whose concentration stays constant (such as H⁺ in acid-catalysed ester hydrolysis).

Worked examples

Example 1: getting k′ and the half-life from data. In acid, the ester concentration falls from 0.100 M to 0.080 M in 30 minutes. Water is in large excess. Find k′, the half-life, and the ester left after 90 minutes.
Step 1: k′ = (1/t) ln([A]₀/[A]) = (1/30) ln(0.100/0.080) = (1/30)(0.22314) = 7.44 × 10⁻³ min⁻¹.
Step 2: t_½ = 0.6931 / 7.438 × 10⁻³ = 93.2 min.
Step 3: [A] at 90 min = 0.100 × e^{−(7.438×10⁻³)(90)} = 0.100 × e^−0.6694 = 0.0512 M.
Check: 90 min is slightly less than one half-life (93.2 min), so a little more than half should remain. 0.0512 M is just above 0.050 M ✓
Answer: k′ = 7.44 × 10⁻³ min⁻¹, t_½ ≈ 93 min, [A] ≈ 0.051 M.

Example 2: from k′ to the true k. For A + B → P, [A]₀ = 0.010 M and [B]₀ = 1.0 M. The measured first-order constant is k′ = 2.4 × 10⁻³ s⁻¹. Find k, the half-life, and check the approximation.
Step 1: k = k′ / [B]₀ = 2.4 × 10⁻³ / 1.0 = 2.4 × 10⁻³ L mol⁻¹ s⁻¹.
Step 2: t_½ = 0.6931 / 2.4 × 10⁻³ = 289 s (about 4.8 min).
Step 3: Excess ratio = 1.0 / 0.010 = 100. B can change by at most 0.010 M, which is 1% of [B]₀.
Step 4: At t = 300 s, the pseudo-first-order result is [A] = 0.010 × e^−0.72 = 4.868 × 10⁻³ M. The exact second-order solution, which allows [B] to fall, gives 4.878 × 10⁻³ M.
Answer: k = 2.4 × 10⁻³ L mol⁻¹ s⁻¹. The two [A] values differ by about 0.2%, so the approximation is excellent here.

Example 3: finding the order in B by the isolation method. With A as the limiting reagent, k′ is measured at three values of [B]₀ (data constructed for this exercise).

[B]₀ (M)	k′ (s⁻¹)	k′ / [B]₀
0.50	1.2 × 10⁻³	2.4 × 10⁻³
1.0	2.4 × 10⁻³	2.4 × 10⁻³
2.0	4.8 × 10⁻³	2.4 × 10⁻³

Step 1: When [B]₀ doubles from 0.50 to 1.0, k′ doubles from 1.2 to 2.4 (× 10⁻³). Doubling again doubles k′ again.
Step 2: So k′ ∝ [B]₀¹, and the order in B is 1. The ratio k′/[B]₀ is constant, which confirms it.
Step 3: Together with first order in A (a straight ln[A] vs t line), the overall order is 2 and k = 2.4 × 10⁻³ L mol⁻¹ s⁻¹.
Answer: first order in B. In general, a plot of log k′ against log [B]₀ has slope equal to the order in B.

Checking whether a reaction is pseudo-first-order

Observation	Meaning
ln[A] vs t is a straight line	First order in A under these conditions
Slope (−k′) changes when [B]₀ changes	B takes part in the rate law
Slope does not change with [B]₀	Zero order in B
k′ ∝ [B]₀	First order in B, true k = k′/[B]₀
k′ ∝ [B]₀²	Second order in B

Calling the reaction "first order" without the word pseudo. The molecularity is still two. Order and molecularity are different ideas, and examiners test that difference.
Reporting k′ as the rate constant. k′ depends on [B]₀. Always divide by it to get k.
Using the shortcut when the excess is small. If [B]₀ is only 2 or 3 times [A]₀, B falls noticeably and ln[A] vs t curves. Use the full second-order integrated equation.
Wrong units. k′ is in s⁻¹, k is in L mol⁻¹ s⁻¹.
Forgetting the catalyst. In acid-catalysed hydrolysis, k′ also contains [H⁺], which stays constant. Changing the acid concentration changes k′.

Exam relevance

Typical question	Key step
Why does ester hydrolysis follow first-order kinetics?	Water is in large excess, so [H₂O] is constant
Find k from a pseudo-first-order k′	k = k′ / [B]₀
Design an experiment to find the order in B	Excess of A or B, vary the excess reagent, plot log k′ against log [B]₀
Difference between order and molecularity	Order is experimental, molecularity is the number of species in the elementary step

Check the current official syllabus of your target exam for the depth expected in chemical kinetics.

The integrated first-order law is what the half-life tool applies. Use it to check your t_½ and decay values once you have k′.

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