Q.The rate constant for a first order reaction is 60 s−1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
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First Order Kinetics
Imagine you have a bucket of water with a small hole at the bottom. The water drains out. At the start, the bucket is full, so the pressure at the hole is high — water gushes out fast. As the water level drops, the pressure decreases, and the water trickles out more slowly. The rate at which water leaves is directly proportional to how much water is still in the bucket.
That's the core intuition behind first order kinetics: the rate of a process depends linearly on how much of the substance is left.
The Precise Statement
In chemistry, first order kinetics describes a reaction where the rate of the reaction is directly proportional to the concentration of one reactant.
If we have a reaction: A→products, then:
Rate=−dtd[A]=k[A]
Here:
- [A] is the concentration of reactant A at any time t
- k is the rate constant (units: time−1, e.g., s−1)
- The negative sign indicates that [A] decreases over time
The key point: double the concentration, double the rate. Halve the concentration, halve the rate.
The Integrated Form — What Actually Happens Over Time
The differential equation above tells us the instantaneous rate. But what we usually want is: how does concentration change with time?
Integrating gives:
ln[A]t=ln[A]0−kt
or equivalently:
[A]t=[A]0e−kt
Where [A]0 is the initial concentration and [A]t is the concentration at time t.
This exponential decay is the hallmark of first order kinetics. The concentration drops rapidly at first, then more slowly, approaching zero asymptotically.
The Half-Life — A Beautiful Constant
For first order kinetics, the half-life (t1/2) — the time taken for half the reactant to be consumed — is independent of the starting concentration.
t1/2=kln2≈k0.693
This is a powerful result. Whether you start with 100 g or 1 g, it always takes the same time to go from that amount to half of it. This is unique to first order kinetics — no other order has this property.
For a first order process, after n half-lives, the fraction remaining is (21)n. After 1 half-life: 50% remains. After 2: 25%. After 3: 12.5%. And so on.
How to Identify First Order Kinetics Experimentally
If you plot ln[A] versus time and get a straight line with slope −k, the reaction is first order. This is the gold standard test.
Alternatively, if the half-life remains constant as you change the initial concentration, that's a strong indicator.
Real-World Examples
- Radioactive decay: Every radioactive isotope decays by first order kinetics. Carbon-14 dating works because t1/2=5730 years, regardless of how much carbon-14 is present. …
Why this formula?
First Order Kinetics: Why the Formula Holds
Let's build this from the core idea — not just memorise the equation.
The Fundamental Assumption
In a first order reaction, the rate of reaction depends linearly on the concentration of only one reactant.
If we have:
A→products
The rate law is:
Rate=−dtd[A]=k[A]
Here:
- −dtd[A] = rate of disappearance of A (negative because [A] decreases)
- k = rate constant (units: time−1, e.g., s−1)
- [A] = concentration of A at any time
Why linear? Because the probability of a single molecule reacting in a given time is constant — it doesn't depend on other molecules. This is the molecular logic behind first order.
Deriving the Integrated Rate Law
We start from the differential form:
−dtd[A]=k[A]
Step 1: Separate variables
Bring all [A] terms to one side, dt to the other:
[A]d[A]=−kdt
Step 2: Integrate both sides
Integrate from initial time t=0 (concentration [A]0) to any time t (concentration [A]t):
∫[A]0[A]t[A]d[A]=−k∫0tdt
The left side integrates to ln[A]:
ln[A]t−ln[A]0=−kt
Step 3: Rearrange
ln[A]0[A]t=−kt
Or equivalently:
ln[A]t=ln[A]0−kt
This is the integrated rate law for first order kinetics.
Why This Form Makes Sense
- Exponential decay: Taking antilog:
[A]t=[A]0e−kt
The concentration decays exponentially — a hallmark of first order processes.
- Constant half-life: The time for [A]t to become half of [A]0 is:
2[A]0=[A]0e−kt1/2
21=e−kt1/2
ln(21)=−kt1/2
t1/2=kln2
Key insight: t1/2 is independent of initial concentration — unique to first order. This is why radioactive decay (a first order process) has a fixed half-life regardless of how much you start with.
Graphical Interpretation (Exam-Ready) …
The key idea is First Order Kinetics, where the integrated rate law relates concentration to time:
t=k2.303log[A][A]0.
Step 1: The concentration is reduced to 161 of its initial value, so
[A][A]0=16.
Step 2: Substitute k=60 s−1 into the formula:
t=602.303log16.
Step 3: log16=log24=4log2≈4×0.3010=1.204. …
For a first-order reaction, the time to reduce concentration to 1/16th is four half-lives. Since t1/2=kln2=60 s−10.693≈0.01155 s, the required time is 4×0.01155=0.0462 s.
First-order kinetics is one of the cleanest models in chemical kinetics because the rate depends only on the concentration of one reactant. The key insight: the time to go from any concentration to a fraction of it is constant — that’s the half-life property. For a first-order reaction, each half-life reduces the concentration by half. So if you want to go from [A]0 to [A]0/16, you’re asking: how many half-lives does it take to drop to 1/16th?
1/16=(1/2)4, so it takes exactly 4 half-lives. That’s the conceptual shortcut. But let’s verify it formally using the integrated rate law, because exams often test both the formula and the reasoning.
- Write the integrated rate law for a first-order reaction. The standard form is:
ln[A][A]0=kt
where [A]0 is the initial concentration, [A] is the concentration at time t, and k is the rate constant.
- Plug in the given fraction. We want [A]=16[A]0. So:
ln[A]0/16[A]0=ln16=kt
- Simplify ln16. 16=24, so ln16=4ln2. Thus:
4ln2=kt
- Solve for t.
t=k4ln2
- Substitute k=60 s−1. Using ln2≈0.693: …
Method: Integrated Rate Law for First-Order Reactions
This is the standard approach for solving time–concentration problems in first-order kinetics.
Steps
- Recall the integrated rate law For a first-order reaction:
k=t2.303log[A]t[A]0
where
- k = rate constant (60 s−1)
- [A]0 = initial concentration
- [A]t = concentration at time t
- Identify the given ratio The concentration is reduced to 161 of its initial value:
[A]0[A]t=161
Therefore:
[A]t[A]0=16
- Substitute into the equation
60=t2.303log16
- Simplify the logarithm
log16=log(24)=4log2
Using log2≈0.3010:
log16=4×0.3010=1.204
- Solve for t
t=602.303×1.204
t=602.773
t≈0.0462 s
Quick Check (Alternative Method) …
Here are the most common mistakes students make when solving this first-order kinetics problem, along with how to avoid each.
Mistake 1: Using the wrong formula for the number of half-lives
Students often try to use the half-life formula directly without checking if the fraction 1/16 is a simple power of 1/2.
- The error: They might calculate t1/2=k0.693 and then multiply by 4 (since 1/16=(1/2)4). This is actually correct for this specific fraction, but the mistake is doing this blindly without verifying the fraction is a power of 1/2.
- How to avoid: Always check: 1/16=(1/2)4 → 4 half-lives. Then t=4×t1/2=4×60 s−10.693.
- Key insight: This shortcut only works when the fraction is exactly (1/2)n. For fractions like 1/10 or 1/3, you must use the integrated rate law.
Mistake 2: Forgetting to convert units or misplacing the rate constant
The rate constant is given as 60 s−1. Some students treat it as if it were in minutes or hours, or they incorrectly invert it.
- The error: Writing t=k2.303log[A][A]0 but then plugging k=60 without checking units, or writing t=600.693 and forgetting the unit is seconds.
- How to avoid:
- Always write the unit alongside the number: k=60 s−1.
- The answer will be in seconds (since k is in s−1).
- If the question expects an answer in minutes, convert at the end: divide by 60.
Mistake 3: Using the wrong logarithm base in the integrated rate law
The first-order integrated rate law is:
t=k2.303log10[A][A]0
Some students use natural log (ln) but forget the factor 2.303, or they use log10 but omit the 2.303.
- The error: Writing t=k1ln[A][A]0 (correct) but then using log10 tables without converting, or writing t=k1log10[A][A]0 (wrong).
- How to avoid:
- Stick to one form and be consistent.
- If using ln: t=k1ln[A][A]0
- If using log10: t=k2.303log10[A][A]0
- For this problem, since [A]0/[A]=16, log1016=1.2041 and ln16=2.7726. Both give the same t if you use the correct factor.
Mistake 4: Misidentifying the ratio [A]0/[A]
Students sometimes invert the fraction.
- The error: They set [A][A]0=161 instead of 16.
- How to avoid: The question says "reduce to its 1/16th value". So [A]=161[A]0. Therefore [A][A]0=161[A]0[A]0=16. Always write it out: final concentration = initial concentration divided by 16.
Mistake 5: Arithmetic errors in the final calculation …
Showing the 12 most recent of 26 on this concept.
- CBSE 2026Set 56/1/11 markMCQQ.Which of the following curve represents the first order reaction ? (A) A graph of t1/2 (y-axis) against initial concentration [R]0 (x-axis): a straight line rising from the origin (B) A graph of t1/2 against [R]0: a horizontal straight line (t1/2 independent of [R]0) (C) A graph of Rate against Concentration: a horizontal straight line (D) A graph of Rate against Concentration: a curve that falls as concentration increases
›Reveal solutionSolution
For a first-order reaction, the half-life t1/2 is independent of the initial concentration [R]0, so the correct plot is a horizontal straight line on a t1/2 vs. [R]0 graph — option (B).
The key to this question is knowing how the half-life of a reaction depends on the initial concentration — and that dependence is different for different orders. Let’s build the intuition from the Arrhenius equation and the integrated rate laws.
Why this approach works
For a first-order reaction, the rate law is:
Rate=k[R]
where k is the rate constant. The integrated form gives:
ln[R][R]0=kt
The half-life t1/2 is the time when [R]=2[R]0. Substituting:
ln[R]0/2[R]0=kt1/2⇒ln2=kt1/2
So:
t1/2=kln2
Notice: no [R]0 appears in this expression. That’s the defining feature — for a first-order reaction, the half-life is a constant, determined only by the rate constant k.
Now let’s examine each option.
-
Option (A): A straight line rising from the origin on a t1/2 vs. [R]0 graph. This would mean t1/2∝[R]0, which is true for a zero-order reaction (where t1/2=[R]0/2k). Not first-order.
-
Option (B): A horizontal straight line — t1/2 does not change as [R]0 changes. This matches t1/2=ln2/k, a constant. This is the correct plot for a first-order reaction.
-
Option (C): A graph of Rate vs. Concentration that is a horizontal straight line. That would mean Rate is independent of concentration — which is true for a zero-order reaction (Rate = k). For first-order, Rate = k[R], so the plot is a straight line through the origin, not horizontal. …
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- CBSE 2026Set A1 markMCQQ.Which of the following is not a first order reaction ?(a) CH3COOC2H5 + H2O --(H+)--> CH3COOH + C2H5OH(b) CH3COOC2H5 + NaOH --> CH3COONa + C2H5OH(c) 2H2O2 --> 2H2O + O2(d) 2N2O5 --> 4NO2 + O2
›Reveal solutionSolution
Ester hydrolysis by NaOH (saponification) is second order (first order in ester and first order in OH-), so it is NOT a first-order reaction.
- (a) Acid hydrolysis of ester with excess water is pseudo-first order. …
- CBSE 2026Set ANNUAL1 markMCQQ.Acid hydrolysis of ethyl acetate is:(a) Zero order reaction(b) First order reaction(c) Second order reaction(d) Third order reaction
›Reveal solutionSolution
Acid hydrolysis of ethyl acetate is a classic example of a pseudo first order reaction.
The reaction is: CH3COOC2H5+H2OH+CH3COOH+C2H5OH. Strictly, this reaction depends on the concentrations of BOTH the ester and water, and should be second order overall (first order in each). However, water is used as the solvent and is present in vast molar excess compared to the ester, so as the reaction proceeds its concentration barely changes and can be treated as effectively constant.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The value of rate constant of a pseudo-first-order reaction(a) depends on the concentration of reactants present in small amount(b) depends on the concentration of reactants present in excess(c) is independent of the concentration of the reaction(d) depends only on temperature
›Reveal solutionSolution
A pseudo-first-order rate constant is not a true elementary-step constant — it already has the (essentially fixed) concentration of the reactant present in excess multiplied into it, so its numerical value depends on how much of that excess reactant was used.
Example — acid-catalysed hydrolysis of ethyl acetate:
CH3COOC2H5+H2OH+CH3COOH+C2H5OH
The true rate law is rate=k[ester][H2O]. Since water is the solvent and is present in huge excess, [H2O] stays essentially constant throughout the reaction, so rate=k′[ester] where k′=k[H2O]. Experimentally the reaction looks first order (only [ester] appears), but the measured k′ is really the true rate constant k multiplied by whatever fixed [H2O] happened to be present.
Why the other options are wrong:
- (a) It is the concentration of the reactant present in excess — not the one present in a small amount — that gets folded into kobs. …
- CBSE 2025Set 56/6/11 markMCQQ.For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Hydrolysis of an ester follows first order kinetics. Reason (R) : The concentration of water does not get altered much during the reaction.
›Reveal solutionSolution
The assertion is true because ester hydrolysis is pseudo-first order; the reason correctly explains why — water is in large excess so its concentration stays nearly constant, making the observed kinetics first order.
The Concept: Why First Order Kinetics Appears
When you study reaction kinetics, the order of a reaction tells you how the rate depends on the concentrations of reactants. For a true bimolecular reaction like ester hydrolysis:
CH3COOC2H5+H2OH+CH3COOH+C2H5OH
The rate law should be:
Rate=k[ester][H2O]
That would make it second order overall — first order in ester and first order in water. But here’s the twist: in practice, the reaction is carried out in aqueous solution where water is the solvent. Its concentration is about 55.5 M, while the ester concentration is typically 0.1 M or less. So water is in huge excess.
TipWhen one reactant is present in such large excess that its concentration changes negligibly during the reaction, we can treat it as constant. The rate law then appears to depend only on the other reactant — this is called pseudo-first order kinetics.
Since [H2O] remains essentially constant, we absorb it into the rate constant:
Rate=k′[ester],where k′=k[H2O]
This is exactly the form of a first order reaction. So the assertion is correct — hydrolysis of an ester follows first order kinetics (under typical conditions).
Step-by-Step Reasoning
-
Identify the true order of the reaction.
The balanced equation shows one molecule of ester reacts with one molecule of water. The fundamental rate law is second order: Rate=k[ester][H2O].
-
Examine the reaction conditions.
In a typical lab or exam context, ester hydrolysis is done in dilute aqueous solution. Water is the solvent — its initial concentration is ~55.5 M and it barely changes because only a tiny fraction is consumed.
-
Apply the concept of excess reactant. …
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- CBSE 2025Set X11 markMCQQ.An example for pseudo first-order reaction is,(a) The decomposition of gaseous ammonia on a hot platinum surface(b) Photochemical reaction between hydrogen and chlorine(c) Inversion of cane sugar(d) Hydrogenation of ethene
›Reveal solutionSolution
Inversion (hydrolysis) of cane sugar is the standard example of a pseudo first-order reaction — water is in large excess so its concentration is effectively constant.
Hydrolysis of sucrose (cane sugar) into glucose and fructose:
sucroseC12H22O11+H2OH+glucoseC6H12O6+fructoseC6H12O6 …
- CBSE 2025Set A1 markQ.Fill in the blank: The unit of a first order rate constant is ______.
›Reveal solutionSolution
A first-order rate constant always has the unit of (time)⁻¹, e.g. s⁻¹.
For a reaction of order n, the rate constant k has general units of (mol L−1)1−ntime−1. For a first-order reaction (n=1), the concentration term's exponent becomes zero, so the concentration unit cancels out completely, leaving only:
…
- CBSE 2025Set ANNUAL1 markMCQQ.What is the concentration of the reactant in a first order reaction, when the rate of the reaction is 0.6 Ms^-1 and the rate constant is 0.035 s^-1 ?(a) 26.667 M(b) 17.143 M(c) 26.183 M(d) 17.667 M
›Reveal solutionSolution
For a first order reaction, Rate = k[R], so the reactant concentration is simply Rate divided by the rate constant.
For a first order reaction:
Rate=k[R] …
- CBSE 2025Set ANNUAL1 markMCQQ.The units of first order reaction:(a) s⁻¹(b) s(c) mol L⁻¹(d) L⁻¹s
›Reveal solutionSolution
The rate constant of a first order reaction has units of (time)⁻¹, i.e. s⁻¹.
For a general reaction of order n, rate =k[A]n, so
k=[A]nrate=(molL−1)nmolL−1s−1
For a first order reaction (n=1):
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which one of the following is a pseudo first order reaction?(a) Hydrogenation of ethene(b) Hydrolysis of ethyl acetate in the presence of dilute acid(c) Combination of H2 and Br2(d) Decomposition of NH2 on a platinum surface
›Reveal solutionSolution
Water, present in huge excess as solvent, has an essentially constant concentration during the ester's hydrolysis, so the true second-order rate law collapses to an apparent first-order one.
The acid-catalyzed hydrolysis CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH is genuinely second order overall (first order in ester, first order in water), but since water is present in vast excess (it is essentially the solvent), its concentration barely changes during the reaction and gets absorbed into the rate constant, so the reaction experimentally behaves as first order — a pseudo …
- CBSE 2024Set ANNUAL1 markMCQQ.The unit of rate constant for a first order reaction is -(a) L2Sec−1(b) Sec−1(c) MolL−1Sec−1(d) Mol−1LSec−1
›Reveal solutionSolution
A first order rate constant has units of (time)−1.
…
- CBSE 2024Set D1 markMCQQ.Which of the following is not a first order reaction?(a) CH3COOCH3 + H2O --H+--> CH3COOH + CH3OH(b) CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH(c) 2H2O2 -> 2H2O + O2(d) 2N2O5 -> 4NO2 + O2
›Reveal solutionSolution
Saponification of an ester by NaOH is second order; the other three are (pseudo/) first order.
- (a) Acid hydrolysis of an ester (with H+ and large excess water) is a pseudo-FIRST-order reaction.
- (b) Ester + NaOH (saponification) depends on both [ester] and [OH-], so rate = k[ester][OH-]: SECOND order.
- (c) Decomposition of H2O2 (2H2O2 -> 2H2O + O2) is first order. …
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