Q.Consider a certain reaction A→ Products with k=2.0×10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.
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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 rate depends linearly on the concentration of one reactant.
For a first-order reaction, the integrated rate law is:
ln[A][A]0=kt
Step 1: Identify the given values.
k=2.0×10−2 s−1, t=100 s, [A]0=1.0 mol L−1.
Step 2: Substitute into the equation.
ln[A]1.0=(2.0×10−2)(100)=2.0
Step 3: Solve for [A]. …
This is a first-order reaction, so we use the integrated rate law ln[A]t[A]0=kt. Substituting the given values gives [A]t=1.0×e−2.0≈0.135 mol L−1.
The problem gives us a reaction A→ Products with a rate constant k=2.0×10−2 s−1. The units of k — per second — are the first big clue. For a reaction, the units of the rate constant tell you the order. If k has units of s−1, the reaction is first order. That’s a non-negotiable fact in chemical kinetics.
Why does that matter? Because each order has its own integrated rate law — the equation that tells you how concentration changes with time. For a first-order reaction, the rate depends only on the concentration of one reactant: rate=k[A]. Integrating that differential equation gives a clean, exponential decay.
For a first-order reaction A→ Products:
ln[A]t[A]0=ktor equivalently[A]t=[A]0e−kt
We know [A]0=1.0 mol L−1, k=2.0×10−2 s−1, and t=100 s. Let’s work through it.
- Plug into the logarithmic form.
ln[A]t1.0=(2.0×10−2)(100)=2.0
So ln[A]t1.0=2.0.
- Exponentiate both sides.
[A]t1.0=e2.0
Therefore,
[A]t=e2.01.0=1.0×e−2.0
- Evaluate e−2.0. e2.0≈7.389, so e−2.0≈0.1353. Hence, [A]t≈0.135 mol L−1 …
Method: Integrated Rate Law for First-Order Kinetics
This method uses the first-order integrated rate equation to directly calculate remaining concentration at a given time.
Steps
1. Identify the reaction order and relevant equation
For a first-order reaction A→Products:
k=t2.303log[A]t[A]0
where:
- k = rate constant (s−1)
- t = time elapsed (s)
- [A]0 = initial concentration
- [A]t = concentration at time t
2. Substitute the given values
- k=2.0×10−2 s−1
- t=100 s
- [A]0=1.0 mol L−1
2.0×10−2=1002.303log[A]t1.0
3. Solve for log[A]t[A]0
2.0×10−2=0.02303×log[A]t1.0
log[A]t1.0=0.023032.0×10−2≈0.868
4. Take antilog to find the ratio
[A]t1.0=100.868≈7.38 …
Here are the common mistakes students make on this First Order Kinetics problem, along with how to avoid each.
Mistake 1: Using the wrong integrated rate law
The error:
Students often plug numbers into the zero-order or second-order equation by mistake.
For example, using [A]=[A]0−kt (zero-order) or [A]1=[A]01+kt (second-order).
Why it happens:
They don’t check the units of k.
Here, k=2.0×10−2 s−1 — the unit s−1 is a dead giveaway for first-order kinetics.
How to avoid:
Always check the unit of k first:
- s−1 → first order
- mol L−1s−1 → zero order
- L mol−1s−1 → second order
Then write the correct integrated law:
ln[A][A]0=kt
Mistake 2: Forgetting to take the natural log
The error:
Students write [A]=[A]0e−kt but then plug numbers directly without using ln, or they use log10 without converting.
Example of wrong calculation:
They compute kt=2.0×10−2×100=2.0, then say [A]=1.0×e−2.0 — which is actually correct if they evaluate e−2.0 properly. But many forget the exponential step entirely and just do [A]=1.0−2.0 (wrong).
How to avoid:
Use the logarithmic form step-by-step:
ln[A][A]0=kt
Then:
- Compute kt=2.0×10−2×100=2.0
- So ln[A]1.0=2.0
- Exponentiate: [A]1.0=e2.0
- Therefore [A]=e2.01.0
Final answer:
[A]=1.0×e−2.0≈0.135 mol L−1
Mistake 3: Mis-handling the exponential calculation
The error:
Students approximate e2.0≈7.4 (correct) but then do 1.0/7.4≈0.135 — that’s fine.
But some use e2.0≈7.389 and round poorly, or they forget the negative sign and compute e2.0 instead of e−2.0.
How to avoid:
- Use the exponential function on your calculator correctly.
- Remember: e−2.0=e2.01
- Keep at least 3 significant figures: e−2.0≈0.1353
Mistake 4: Confusing concentration with amount or pressure
The error: …
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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