A pseudo first order reaction is a reaction that is really second order but
behaves as first order, because one of its reactants is present in such huge excess that its
concentration barely changes.
Story track
Imagine a small boat taking water from a lake. Every minute, the boat's pump removes a
bucketful. Does the lake level drop?
Technically, yes. In practice, no — you could never measure it. The lake is so vast compared with
the bucket that its level is, for all purposes, constant.
Now, the pumping rate genuinely depends on two things: how hard the pump works, and how much water
is in the lake. It really is a two-factor process. But since the lake never changes, the only thing
you ever see varying is the pump. So if you measured the rate over an afternoon you would
conclude, quite reasonably, that it depends on one factor alone.
That is a pseudo first order reaction. Truly two-factor, observably one-factor, because one
factor sits still.
The mathematics of the disguise
Maths track
Take a genuinely second order reaction:
Rate = k[A][B]
Now suppose B is in vast excess, so [B] stays essentially constant at [B]₀ throughout. Then:
Rate = k[A][B]₀ = (k[B]₀)[A] = k'[A]
k' = k[B]₀ — the pseudo first order rate constant
The reaction now obeys first order kinetics exactly, with an apparent rate constant k'
that quietly contains the constant concentration of B inside it. Everything from the first order unit
applies: t½ = 0.693/k', ln[A] vs t is linear, and so on.
Why chemists do this deliberately. Second order kinetics is
awkward to analyse. By flooding the system with one reactant, you force it into first order behaviour
that is far easier to measure — and you can then extract the true k afterwards from k' = k[B]₀. This
is a standard laboratory technique, not just a textbook curiosity.
NCERT gives the actual numbers, and they are worth looking at closely:
CH₃COOC₂H₅
H₂O
CH₃COOH
C₂H₅OH
At t = 0
0.01 mol
10 mol
0 mol
0 mol
At completion
0 mol
9.99 mol
0.01 mol
0.01 mol
The ester is completely consumed — a 100% change. Water goes from 10 to 9.99 mol — a change
of 0.1%. NCERT's conclusion: the concentration of water does not get altered much during
the course of the reaction, so the reaction behaves as a first order reaction.
Sucrose is split into glucose and fructose. Again water is the solvent and in vast excess, so its
concentration is effectively fixed and only sucrose appears in the rate law.
The question that catches people. "What is the order of
ester hydrolysis?" has two defensible answers, and the question always specifies which it wants.
True order: 2 (first order in ester, first order in water). Observed / pseudo order: 1.
Read whether the question says "actually", "truly", "in reality" — or "observed", "apparent",
"behaves as".
How to recognise one in an exam
A solvent appears as a reactant — water, usually. Solvents are always in vast excess.
The stated rate law has a zero exponent on one reactant that clearly does participate.
The question mentions one reactant being in large excess.
An acid catalyst (H⁺) over the arrow alongside water is a strong hint — both NCERT
examples have exactly this.
Beyond the textbook
Recovering the true rate constant. Since k' = k[B]₀,
measuring k' at several different fixed values of [B]₀ and plotting k' against [B]₀ gives a straight
line whose slope is the true second order k. This is how the technique is actually used in a
laboratory.
Units give the game away. The pseudo constant k' has units
of s⁻¹, because it is behaving as first order. The true constant k has units of L mol⁻¹ s⁻¹. If a
question gives you k' in s⁻¹ for an ester hydrolysis, it is handing you the pseudo constant, not the
real one.
Why NCERT places this here. It appears immediately after
the sentence "the order of a reaction is sometimes altered by conditions". That is the real lesson:
order is a property of the experiment, not only of the reaction.
See it move — 3 animations
The second animation is the one worth sitting with — it shows that 'order' is partly a statement about your experiment, not only about the reaction.
ANIM 1
0.01 mol of ester in 10 mol of water
Drag the reaction to completion and compare the two bars, each scaled to its own starting amount. The ester empties entirely. Water moves by a tenth of one percent — barely visible. That contrast is the whole justification for treating water's concentration as constant, and NCERT gives these exact numbers.
ANIM 2
Watch the disguise become perfect
The red curve is the true second order behaviour of A, plotted as ln[A] against t. At small excess it is visibly curved — the second order nature shows. Increase the excess and the curvature shrinks until the curve is experimentally indistinguishable from the straight first order fit. The reaction did not change; only your ability to detect its true order did.
ANIM 3
Where the hidden reactant went
The excess reactant has not left the chemistry — it has moved inside the rate constant. Slide [B]₀ and watch k′ move with it in exact proportion. Note the units too: the true k is in L mol⁻¹ s⁻¹ while k′ is in s⁻¹, so the units alone tell you which constant a question has handed you.
Formula sheet
The starred rows carry the unit. The single most common lost mark here is answering 'first' when the question asked for the true order.
Quantity / situation
Formula
When you use it
Definition
a reaction that is truly higher order but behaves as first order
t½ = 0.693/k′ ; ln[A] vs t linear ; halving ladder
Once disguised, it is first order in every way
Example 1
CH₃COOC₂H₅ + H₂O →(H⁺) CH₃COOH + C₂H₅OH
Rate = k[ester]¹[H₂O]⁰
Example 1 numbers
0.01 mol ester with 10 mol water : water falls only to 9.99 mol
A 0.1% change versus 100%
Example 2
C₁₂H₂₂O₁₁ + H₂O →(H⁺) glucose + fructose
Inversion of cane sugar
True vs observed order ★
ester hydrolysis: true order 2, observed order 1
Read which the question wants
Recognition cues
solvent as reactant ; zero exponent on a real reactant ; 'large excess' ; H⁺ catalyst
Four reliable signals
Why order changes
order describes the experiment, not only the reaction
NCERT: order is sometimes altered by conditions
Recovering true k — gap
plot k′ against [B]₀ : slope = true k
Standard laboratory method
25 NEET-type questions with worked solutions
Four graph questions and three assertion–reason questions are included, marked by their coloured left borders. Questions tagged PYQ pattern follow forms that have appeared in NEET/AIPMT papers or come directly from NCERT exercises — exact year attributions are deliberately omitted rather than guessed.
Q01PYQ pattern
The hydrolysis of ethyl acetate in the presence of excess water is an example of:
(a) a pseudo first order reaction
(b) a true first order reaction
(c) a zero order reaction
(d) a third order reaction
Given
CH₃COOC₂H₅ + H₂O →(H⁺) products, water in excess
Asked
Type of reaction
Concept
Truly second order but behaving as first order because water is in vast excess.
Formula
Rate = k[CH₃COOC₂H₅]¹[H₂O]⁰
Baby steps
The reaction genuinely involves both ester and water, so it is second order in reality.
Water is taken in large excess, so its concentration is not altered much.
The rate then depends only on the ester concentration, giving observed first order behaviour.
NCERT names such reactions pseudo first order reactions.
Answer · (a) a pseudo first order reaction
Q02PYQ pattern
For the hydrolysis of ethyl acetate, the TRUE order of the reaction is:
(a) 2
(b) 1
(c) 0
(d) 1.5
Given
Ester hydrolysis, asked for the true order
Asked
True order
Concept
Both reactants genuinely affect the rate; only the excess conceals one of them.
Formula
Rate = k[ester][water] — genuinely second order
Baby steps
NCERT states that in reality it is a second order reaction and the concentration of both ethyl acetate and water affect the rate.
The observed first order behaviour arises only because water is in vast excess.
So the true order is 2, while the observed order is 1.
Read the question wording carefully — 'true', 'actually' and 'in reality' all ask for 2.
Answer · (a) 2
Shortcut · This question exists in two forms with two different answers. The wording tells you which is wanted.
Q03PYQ pattern
In the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, the amount of water at completion is:
(a) 9.99 mol
(b) 10 mol
(c) 9.00 mol
(d) 0.01 mol
Given
0.01 mol ester, 10 mol water, reaction goes to completion
Asked
Water remaining
Concept
Stoichiometry consumes one water per ester, which is negligible against 10 mol.
Formula
water consumed = ester consumed = 0.01 mol
Baby steps
The equation consumes one mole of water per mole of ester.
All 0.01 mol of ester reacts, so 0.01 mol of water is consumed.
Water remaining = 10 − 0.01 = 9.99 mol.
That is a change of only 0.1%, which is why water's concentration is treated as constant.
Answer · (a) 9.99 mol
Q04PYQ pattern
Inversion of cane sugar, C₁₂H₂₂O₁₁ + H₂O →(H⁺) C₆H₁₂O₆ + C₆H₁₂O₆, has rate law:
(a) Rate = k[C₁₂H₂₂O₁₁]
(b) Rate = k[C₁₂H₂₂O₁₁][H₂O]
(c) Rate = k[H₂O]
(d) Rate = k[C₁₂H₂₂O₁₁]²
Given
Inversion of cane sugar in aqueous acid
Asked
The rate law
Concept
Water is the solvent and in vast excess, so it does not appear in the observed rate law.
Formula
Rate = k[C₁₂H₂₂O₁₁]
Baby steps
The reaction converts sucrose into glucose and fructose.
Water is present as the solvent, in enormous excess.
Its concentration is effectively constant, so it folds into the rate constant.
NCERT gives the rate law as Rate = k[C₁₂H₂₂O₁₁], making it pseudo first order.
Answer · (a) Rate = k[C₁₂H₂₂O₁₁]
Q05
For a pseudo first order reaction Rate = k[A][B] with B in large excess, the observed rate constant k′ equals:
(a) k[B]₀
(b) k/[B]₀
(c) k[A]₀
(d) k
Given
Rate = k[A][B] with [B] ≈ [B]₀ constant
Asked
Expression for k′
Concept
The constant concentration is absorbed into the rate constant.
Formula
Rate = k[A][B]₀ = (k[B]₀)[A] = k′[A]
Baby steps
Since [B] barely changes, replace it by the constant [B]₀.
Rate = k[B]₀ × [A], and the bracket is a constant.
Defining k′ = k[B]₀ gives Rate = k′[A], which is first order in form.
Note k′ has units of s⁻¹ while the true k has units of L mol⁻¹ s⁻¹.
Answer · (a) k[B]₀
Shortcut · k′ contains the hidden reactant. Doubling [B]₀ doubles k′ — proof B is still involved.
Q06
Which of the following indicates that a reaction is pseudo first order rather than genuinely first order?
(a) A reactant appears with a zero exponent in the rate law despite taking part
(b) The half-life is constant
(c) A plot of ln[A] against t is linear
(d) The rate constant has units of s⁻¹
Given
Distinguishing features
Asked
The indicator of pseudo first order
Concept
Once disguised, a pseudo first order reaction is indistinguishable from a true one in its kinetics.
Formula
Rate = k[A]¹[B]⁰
Baby steps
Options (b), (c) and (d) are all shared with genuine first order reactions — the disguise is complete in those respects.
The giveaway is that a species genuinely taking part in the reaction carries a zero exponent.
That can only happen because its concentration is held effectively constant.
So option (a) is the distinguishing signal.
Answer · (a) A reactant appears with a zero exponent in the rate law despite taking part
Q07
The units of the pseudo first order rate constant k′ for ester hydrolysis are:
(a) s⁻¹
(b) L mol⁻¹ s⁻¹
(c) mol L⁻¹ s⁻¹
(d) mol⁻² L² s⁻¹
Given
Pseudo first order behaviour
Asked
Units of k′
Concept
Observed behaviour is first order, so the observed constant has first order units.
Formula
Rate = k′[A] ⟹ k′ in s⁻¹
Baby steps
The observed rate law is Rate = k′[A], which is first order in form.
First order rate constants have units of time⁻¹.
So k′ is in s⁻¹.
The true constant k, from Rate = k[A][B], has units L mol⁻¹ s⁻¹ instead.
Answer · (a) s⁻¹
Shortcut · If a question hands you a constant in s⁻¹ for an ester hydrolysis, it is the pseudo constant, not the true one.
Q08
Why is water's concentration treated as constant in the hydrolysis of ethyl acetate?
(a) It is present in such large excess that its concentration changes by only about 0.1%
(b) Water does not react at all
(c) Water is a catalyst
(d) Water is regenerated during the reaction
Given
0.01 mol ester in 10 mol water
Asked
Reason for treating [H₂O] as constant
Concept
A negligible fractional change justifies the constant approximation.
Formula
water: 10 → 9.99 mol, a 0.1% change
Baby steps
Water is genuinely consumed — one mole per mole of ester.
But only 0.01 mol of the 10 mol present is used up.
That is a change of 0.1%, far too small to affect the measured rate.
Water is neither a catalyst nor regenerated, so options (c) and (d) are wrong.
Answer · (a) It is present in such large excess that its concentration changes by only about 0.1%
Q09
A pseudo first order reaction has k′ = 2 × 10⁻³ s⁻¹. Its half-life is:
(a) 346.5 s
(b) 693 s
(c) 200 s
(d) 1386 s
Given
k′ = 2 × 10⁻³ s⁻¹
Asked
t½
Concept
Pseudo first order reactions obey every first order result.
Formula
t½ = 0.693/k′
Baby steps
Since the observed kinetics are first order, the first order half-life formula applies.
t½ = 0.693/(2 × 10⁻³).
= 346.5 s.
Note the half-life is constant in time, exactly as for a true first order reaction.
Answer · (a) 346.5 s
Q10
If the true rate constant for a second order reaction is 5 × 10⁻⁴ L mol⁻¹ s⁻¹ and the excess reactant is at 2.0 M, the pseudo first order rate constant is:
(a) 1.0 × 10⁻³ s⁻¹
(b) 2.5 × 10⁻⁴ s⁻¹
(c) 5.0 × 10⁻⁴ s⁻¹
(d) 2.0 s⁻¹
Given
k = 5 × 10⁻⁴ L mol⁻¹ s⁻¹, [B]₀ = 2.0 M
Asked
k′
Concept
Multiply the true constant by the fixed concentration.
Formula
k′ = k[B]₀
Baby steps
k′ = 5 × 10⁻⁴ × 2.0.
= 1.0 × 10⁻³ s⁻¹.
Unit check: (L mol⁻¹ s⁻¹) × (mol L⁻¹) = s⁻¹. Correct for a first order constant.
Answer · (a) 1.0 × 10⁻³ s⁻¹
Shortcut · The unit cancellation is a free check: if you do not land on s⁻¹, you have multiplied the wrong things.
Q11Graph
For a pseudo first order reaction, the plot of ln[A] against t is:
(a) a straight line of slope −k′
(b) a straight line of slope −k
(c) a curve
(d) horizontal
Given
Pseudo first order reaction, ln[A] vs t
Asked
Shape and slope
Concept
The observed kinetics are first order, so the observed constant appears in the slope.
Formula
ln[A] = ln[A]₀ − k′t
Baby steps
The observed rate law is Rate = k′[A], identical in form to first order.
Integrating gives ln[A] = ln[A]₀ − k′t, a straight line.
The slope is −k′, the pseudo constant, not the true second order k.
This is why option (b) is wrong — k has different units and would not fit the axes.
Answer · (a) a straight line of slope −k′
Q12Graph
The plot shows the observed pseudo first order rate constant k′ against the concentration of the excess reactant [B]₀. The slope gives:
(a) the true second order rate constant k
(b) the half-life
(c) the activation energy
(d) the order of the reaction
Given
k′ plotted against [B]₀, straight through the origin
Asked
Meaning of the slope
Concept
Gap content — this is the laboratory method for recovering the true rate constant.
Formula
k′ = k[B]₀ — compare with y = mx
Baby steps
The relation k′ = k[B]₀ has the form y = mx with no intercept.
So the plot is a straight line through the origin, exactly as shown.
The slope m equals the true second order rate constant k.
This is how chemists deliberately use the pseudo first order technique — flood the system, measure the easy first order kinetics, then recover the true k.
Answer · (a) the true second order rate constant k
Q13Graph
Two experiments on the same reaction are run with different large excesses of B. The ln[A] versus t plots are both straight but have different slopes. This is because:
(a) k′ = k[B]₀ differs between the runs, even though the true k is the same
(b) the true rate constant changed
(c) the order changed between runs
(d) the temperature must have differed
Given
Two straight ln[A] vs t plots of different slope, different [B]₀
Asked
Explanation
Concept
The pseudo constant contains [B]₀, so it changes when the excess changes.
Formula
slope = −k′ = −k[B]₀
Baby steps
Both plots are straight, so both runs show first order behaviour.
The slope is −k′ = −k[B]₀.
With more B present, k′ is larger and the line is steeper.
The true k is a property of the reaction and does not change; only the observed constant does. This is direct evidence that B is still chemically involved.
Answer · (a) k′ = k[B]₀ differs between the runs, even though the true k is the same
Shortcut · If changing an 'absent' reactant changes the observed constant, the reaction is pseudo order, not genuinely first order.
Q14Graph
The bars compare the percentage change in concentration of ester and water during hydrolysis. What does this justify?
(a) Treating [H₂O] as constant, so it drops out of the rate law
(b) Treating water as a catalyst
(c) That water does not react
(d) That the reaction is zero order in the ester
Given
Ester changes 100%, water changes 0.1%
Asked
What this justifies
Concept
A negligible change licenses the constant-concentration approximation.
Formula
Rate = k[ester][H₂O] ≈ (k[H₂O]₀)[ester]
Baby steps
Water is genuinely consumed, so options (b) and (c) are wrong.
But its fractional change is a thousand times smaller than the ester's.
So [H₂O] may be treated as constant and absorbed into the rate constant.
The ester's concentration changes fully and remains first order in the rate law, so (d) is wrong.
Answer · (a) Treating [H₂O] as constant, so it drops out of the rate law
Q15Assertion–Reason
Assertion (A): The hydrolysis of ethyl acetate is called a pseudo first order reaction. Reason (R): Although it is truly second order, water is present in such large excess that its concentration does not alter appreciably during the reaction.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about ester hydrolysis
Asked
Truth values and explanation
Concept
The excess is precisely what causes the observed order to drop.
Formula
Rate = k[ester][H₂O] ≈ k′[ester]
Baby steps
Check A: NCERT names such reactions pseudo first order reactions. A is true.
Check R: NCERT states it is in reality a second order reaction, and that the concentration of water does not get altered much. R is true.
Does R explain A? Yes — the unchanging water concentration is exactly why the reaction behaves as first order.
Without the excess, the reaction would show ordinary second order kinetics.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q16Assertion–Reason
Assertion (A): The order of a reaction can be altered by the experimental conditions. Reason (R): Pseudo first order reactions show first order kinetics only because one reactant is held effectively constant.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about order and conditions
Asked
Truth values and explanation
Concept
Pseudo first order is the standard demonstration that order depends on conditions.
Formula
—
Baby steps
Check A: NCERT opens this section with 'the order of a reaction is sometimes altered by conditions'. A is true.
Check R: holding one reactant effectively constant is precisely what produces the apparent first order behaviour. R is true.
Does R explain A? Yes — pseudo first order reactions are the worked example NCERT uses to establish the claim in A.
The deeper lesson: order describes an experiment, not only a reaction.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q17Assertion–Reason
Assertion (A): In a pseudo first order reaction, the reactant present in excess plays no chemical role. Reason (R): Its concentration does not appear in the observed rate law.
(a) A is false but R is true
(b) Both A and R are true and R explains A
(c) Both A and R are true but R does not explain A
(d) Both A and R are false
Given
Statements about the excess reactant
Asked
Truth values
Concept
Absence from the rate law is not absence from the chemistry.
Formula
Rate = k[A][B]₀ — B is inside k′, not absent
Baby steps
Check R: the excess reactant genuinely does not appear in the observed rate law. R is true.
Check A: it is chemically essential — the reaction cannot proceed without it, and it is consumed stoichiometrically. A is false.
Its concentration is hidden inside k′ = k[B]₀, so doubling it doubles the observed rate constant.
So A is false while R is true. Vanishing from the rate law is a matter of measurement, not of chemistry.
Answer · (a) A is false but R is true
Shortcut · An excess reactant is invisible in the rate law but indispensable in the reaction. Do not confuse the two.
Q18
Which of the following would convert a pseudo first order reaction back into an observably second order one?
(a) Using comparable amounts of both reactants
(b) Raising the temperature
(c) Adding a catalyst
(d) Diluting both reactants equally
Given
A pseudo first order reaction
Asked
How to reveal the true order
Concept
The disguise depends entirely on one reactant being in vast excess.
Formula
Rate = k[A][B]
Baby steps
The first order appearance arises only because [B] stays constant.
If both reactants are present in comparable amounts, both concentrations change appreciably.
The rate then visibly depends on both, revealing the true second order behaviour.
Temperature and catalysts change k but not the order, and equal dilution preserves the excess ratio.
Answer · (a) Using comparable amounts of both reactants
Q19
A reaction A + B → products is second order overall. It will show pseudo first order kinetics if:
(a) [B] ≫ [A]
(b) [A] = [B]
(c) [A] ≫ [B] and [B] is monitored
(d) both are dilute
Given
Second order A + B → products
Asked
Condition for pseudo first order behaviour in A
Concept
The monitored species must be the scarce one; the other must be in excess.
Formula
Rate = k[A][B] ≈ (k[B]₀)[A]
Baby steps
For the rate to depend only on [A], the concentration of B must stay effectively constant.
That requires B to be present in large excess over A.
Then Rate = k′[A] with k′ = k[B]₀.
Option (c) would instead give pseudo first order behaviour in B, which is not what is being monitored here.
Answer · (a) [B] ≫ [A]
Q20
The acid H⁺ shown above the arrow in CH₃COOC₂H₅ + H₂O →(H⁺) products acts as:
(a) a catalyst
(b) a reactant
(c) an intermediate
(d) a product
Given
H⁺ above the reaction arrow
Asked
Role of H⁺
Concept
A species written above the arrow speeds the reaction without being consumed.
Formula
—
Baby steps
H⁺ appears above the arrow, not among reactants or products.
It accelerates the hydrolysis and is recovered unchanged.
That fits the definition of a catalyst.
Both of NCERT's pseudo first order examples use acid catalysis, which is a useful recognition cue.
Answer · (a) a catalyst
Q21
Which reaction is NOT pseudo first order?
(a) 2N₂O₅ → 4NO₂ + O₂
(b) Hydrolysis of ethyl acetate in excess water
(c) Inversion of cane sugar
(d) Hydrolysis of methyl acetate in excess water
Given
Four reactions
Asked
The one that is not pseudo first order
Concept
A genuinely first order reaction is not pseudo anything.
Formula
—
Baby steps
Options (b), (c) and (d) all involve water in vast excess and are pseudo first order.
The decomposition of N₂O₅ involves a single reactant and is genuinely first order.
There is no hidden second reactant to conceal.
So it is a true first order reaction, not a pseudo one.
Answer · (a) 2N₂O₅ → 4NO₂ + O₂
Q22
For a pseudo first order reaction with k′ = 1.386 × 10⁻² s⁻¹, the time for 75% completion is:
(a) 100 s
(b) 50 s
(c) 200 s
(d) 150 s
Given
k′ = 1.386 × 10⁻² s⁻¹, 75% completion
Asked
t
Concept
All first order shortcuts apply to pseudo first order reactions.
Formula
t₇₅ = 2 t½ = 2 × 0.693/k′
Baby steps
t½ = 0.693/(1.386 × 10⁻²) = 50 s.
75% completion leaves 25% = 1/4, which is two half-lives.
t = 2 × 50 = 100 s.
Answer · (a) 100 s
Q23
The observed rate constant of a pseudo first order reaction doubles when the concentration of the excess reactant is doubled. This shows that:
(a) the excess reactant is still chemically involved, hidden inside k′
(b) the reaction is zero order in the excess reactant in reality
(c) the true rate constant changed
(d) the reaction became second order
Given
k′ doubles when [B]₀ doubles
Asked
Interpretation
Concept
k′ = k[B]₀, so k′ is proportional to the excess concentration.
Formula
k′ = k[B]₀
Baby steps
Since k′ = k[B]₀, doubling [B]₀ doubles k′.
So the excess reactant does influence the observed kinetics after all.
It simply does so through the constant rather than through a visible concentration term.
The true k, being a property of the reaction, is unchanged.
Answer · (a) the excess reactant is still chemically involved, hidden inside k′
Q24
Pseudo first order reactions are useful in the laboratory because:
(a) first order kinetics is far simpler to analyse than second order
(b) they proceed faster
(c) they need no catalyst
(d) they have zero activation energy
Given
Practical value of the technique
Asked
The reason
Concept
Flooding one reactant reduces a two-variable problem to a one-variable one.
Formula
Rate = k[A][B] → k′[A]
Baby steps
Second order kinetics with two varying concentrations is awkward to analyse.
By putting one reactant in vast excess, its concentration is held constant.
The kinetics then reduce to simple first order form, with straight-line plots and constant half-life.
The true k can be recovered afterwards from k′ = k[B]₀.
Answer · (a) first order kinetics is far simpler to analyse than second order
Q25
If a question states that ester hydrolysis 'behaves as a first order reaction', the order being referred to is:
(a) the observed or apparent order
(b) the true order
(c) the molecularity
(d) the order with respect to water
Given
The phrase 'behaves as a first order reaction'
Asked
Which order is meant
Concept
Wording distinguishes true from observed order.
Formula
—
Baby steps
The word 'behaves' signals observed experimental behaviour rather than underlying reality.
The observed order is 1, while the true order is 2.
NCERT's own sentence reads: the reaction behaves as first order reaction.
Words such as 'actually', 'truly' and 'in reality' would instead point to the true order of 2.
Answer · (a) the observed or apparent order
Shortcut · Underline the qualifier in the question before answering. It alone decides between 1 and 2.