(a) Reactant 'A' underwent a decomposition reaction. The concentration of 'A' was measured periodically and re

Question

Q. (a) Reactant 'A' underwent a decomposition reaction. The concentration of 'A' was measured periodically and recorded in the table given below : Time/Hours | [A]/M 0 | 0\cdot40 1 | 0\cdot20 2 | 0\cdot10 3 | 0\cdot05 Based on the above data, predict the order of the reaction and write the expression for the rate law.

(OR)

(b) The reaction between

H_2 (g)

and

I_2 (g)

was carried out in a sealed isothermal container. The rate law for the reaction was found to be : Rate =

k[H_2][I_2]

If 1 mole of

H_2 (g)

was added to the reaction chamber and the temperature was kept constant, then predict the change in rate of the reaction and the rate constant.

CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — First Order Kinetics

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: $A \rightarrow \text{products}$ , then:

\text{Rate} = -\frac{d[A]}{dt} = 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

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 ( $t_{1/2}$ ) — the time taken for half the reactant to be consumed — is independent of the starting concentration.

t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}

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.

Note

For a first order process, after $n$ half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^n$ . 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 $t_{1/2} = 5730$ 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 \rightarrow \text{products}

The rate law is:

\text{Rate} = -\frac{d[A]}{dt} = k[A]

Here:

$-\frac{d[A]}{dt}$ = 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:

-\frac{d[A]}{dt} = k[A]

Step 1: Separate variables

Bring all $[A]$ terms to one side, $dt$ to the other:

\frac{d[A]}{[A]} = -k \, dt

Step 2: Integrate both sides

Integrate from initial time $t=0$ (concentration $[A]_0$ ) to any time $t$ (concentration $[A]_t$ ):

\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt

The left side integrates to $\ln[A]$ :

\ln[A]_t - \ln[A]_0 = -kt

Step 3: Rearrange

\ln \frac{[A]_t}{[A]_0} = -kt

Or equivalently:

\boxed{\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]_0 e^{-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:

\frac{[A]_0}{2} = [A]_0 e^{-k t_{1/2}}

\frac{1}{2} = e^{-k t_{1/2}}

\ln\left(\frac{1}{2}\right) = -k t_{1/2}

\boxed{t_{1/2} = \frac{\ln 2}{k}}

Key insight: $t_{1/2}$ 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) …

Part (b)Concept understanding — Effect of Concentration

Effect of Concentration: The Intuition

Imagine you're in a large, empty hall with just one other person. The two of you are trying to bump into each other accidentally. It'll take a while, right? Now imagine the same hall packed with a thousand people. Bumping into someone becomes almost certain within seconds.

That's the core idea behind the effect of concentration on reaction rates. Concentration simply means how much of a substance is packed into a given space. Higher concentration = more particles in the same volume.

When particles are more crowded, they collide more frequently. And since chemical reactions happen only when particles collide with enough energy and the right orientation, more collisions mean more reactions per second. The reaction speeds up.

The Precise Statement

For most chemical reactions, the rate of a reaction is directly proportional to the molar concentration of the reactants (raised to some power, which we'll get to).

\text{Rate} \propto [\text{Reactant}]^n

Here, $[ \ ]$ means "concentration in moles per litre" (mol/L or M), and $n$ is the order of reaction with respect to that reactant.

What does "order" mean?

For a simple reaction like $A \rightarrow \text{Products}$ :

First order ( $n=1$ ): Double the concentration of A → double the rate.
Second order ( $n=2$ ): Double the concentration of A → quadruple the rate ( $2^2 = 4$ ).
Zero order ( $n=0$ ): Changing concentration has no effect on the rate. This happens when the reaction is limited by something else (like a catalyst surface that's already fully covered).

Watch out

The order $n$ is not the same as the stoichiometric coefficient from the balanced equation. It must be determined experimentally. For example, the reaction $2A \rightarrow B$ could be first order in A, not second order.

Why does this happen? The collision theory

The rate depends on two things:

Collision frequency — how often particles meet.
Fraction of effective collisions — how many of those collisions have enough energy (activation energy) and the right orientation.

Doubling the concentration doubles the number of particles per unit volume. This roughly doubles the collision frequency. For a first-order reaction, that directly doubles the rate. For higher orders, the effect compounds because multiple reactant particles must meet simultaneously.

A concrete example

Consider the reaction between hydrochloric acid and sodium thiosulphate:

$\text{Na}_2\text{S}_2\text{O}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{S}(s) + \text{SO}_2(g) + \text{H}_2\text{O}(l)$ …

Why this formula?

Effect of Concentration on Reaction Rate — The Reasoning

The Effect of Concentration is rooted in collision theory. The key idea is simple:

More particles in the same volume → more frequent collisions → higher reaction rate.

Let's break down why the mathematical relationships hold.

1. The Rate Law — Why $r = k[A]^m[B]^n$ ?

This is not derived from theory alone — it is empirical (found experimentally). But the reasoning behind its form comes from collision probability.

For an elementary reaction (one step):

Consider:

A + B \rightarrow \text{products}

The rate depends on how often A and B molecules meet.
In a given volume, the number of A molecules is proportional to $[A]$ , and the number of B molecules is proportional to $[B]$ .
The number of A–B collisions per second is proportional to the product:

\text{Collision frequency} \propto [A] \times [B]

Therefore:

r \propto [A][B]

r = k[A][B]

For a reaction with coefficient $aA + bB$ :

If the reaction is elementary, the stoichiometric coefficients become the exponents:

r = k[A]^a[B]^b

Why? Because for $2A$ to react, two A molecules must collide simultaneously — the probability of that happening is proportional to $[A] \times [A] = [A]^2$ .

2. The Integrated Rate Laws — Why These Forms?

These come from solving the differential equation $r = -\frac{d[A]}{dt} = k[A]^n$ .

Zero-order ( $n=0$ ):

-\frac{d[A]}{dt} = k

Reasoning: Rate is independent of concentration. This happens when the reaction is limited by something else (e.g., a saturated catalyst surface).
Integrate:

\int_{[A]_0}^{[A]} d[A] = -k \int_0^t dt

\Rightarrow [A] = [A]_0 - kt

First-order ( $n=1$ ):

-\frac{d[A]}{dt} = k[A]

Reasoning: Rate is directly proportional to $[A]$ . Each molecule has a constant probability of reacting per unit time (like radioactive decay).
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k \int_0^t dt

\Rightarrow \ln[A] = \ln[A]_0 - kt

[A] = [A]_0 e^{-kt}

Second-order ( $n=2$ ):

-\frac{d[A]}{dt} = k[A]^2

Reasoning: Rate depends on two molecules of A colliding. Doubling $[A]$ quadruples the collision frequency.
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k \int_0^t dt

\Rightarrow \frac{1}{[A]} = \frac{1}{[A]_0} + kt

3. The Half-Life — Why It Depends on Order …

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Part (a): [A] halves each hour → first-order reaction, $\text{Rate} = k[A]$ , $k \approx 0.693\ \text{h}^{-1}$ . Part (b): adding $H_2$ increases $[H_2]$ so the rate increases, but $k$ (temperature-dependent only) is unchanged.

Part (a)

Read the data:

Time (h)	$[A]$ (M)
0	0.40
1	0.20
2	0.10
3	0.05

Each hour the concentration is cut to half — the half-life is constant at 1 h and does not depend on the starting concentration. That constant half-life is the signature of a first-order reaction.

Watch out

Do not mistake this for zero order. A zero-order reaction loses a constant amount per unit time; here it loses a constant fraction (50% per hour).

For first order, $t_{1/2} = \dfrac{0.693}{k}$ , so …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$Q. Assertion (A) : Order of reaction is applicable to elementary as well as complex reactions. Reason (R) : Order of a reaction is an experimental quantity. Options : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
The assertion is false because order is defined only for simple rate laws (elementary steps or overall reactions with a simple rate expression), not for all complex reactions. The reason is true: order is experimental. So the correct option is (D).

Concept First: What "Order" Really Means

The order of a reaction is the sum of the exponents of concentration terms in the experimentally determined rate law. For an elementary reaction (a single step), the order equals the molecularity — that’s straightforward. But for a complex reaction (a sequence of steps), the overall rate law can be messy: it might involve fractional exponents, negative exponents, or even terms that don’t look like a simple power law at all. In such cases, the concept of "order" simply doesn’t apply in the usual sense.

The reason given is a fundamental truth: order is always found by experiment, never deduced from the balanced equation (except for elementary steps). That’s correct.

Now let’s examine the assertion carefully.

Step-by-Step Reasoning

What does "applicable" mean here?

The assertion says order is "applicable" to both elementary and complex reactions. If a reaction has a rate law of the form $r = k[\text{A}]^m[\text{B}]^n$ , then we can define order $= m+n$ . For an elementary reaction, this always works. For a complex reaction, it works only if the overall rate law happens to be a simple power law — which is not guaranteed.

Counterexample: a complex reaction where order is not defined

Consider the reaction $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ . Its mechanism involves a pre-equilibrium, and the experimental rate law is $r = k[\text{NO}]^2[\text{O}_2]$ . Here order = 3, so it is applicable. But take the decomposition of $\text{N}_2\text{O}_5$ : the rate law is $r = k[\text{N}_2\text{O}_5]$ , so order = 1 — again applicable.

However, consider a reaction like $\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}$ . The experimental rate law is:

$r = \frac{k[\text{H}_2][\text{Br}_2]^{1/2}}{1 + k'[\text{HBr}]/[\text{Br}_2]}$
This is not of the form $k[\text{A}]^m[\text{B}]^n$ — it has a denominator with a concentration term. You cannot assign a single "order" to this reaction. The concept of order is simply not applicable here. …
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CBSE 2026Set 56/1/11 markMCQ
$t_{1/2}$
›Reveal solutionSolution
For a first-order reaction, the half-life $t_{1/2}$ is independent of the initial concentration $[R]_0$ , so the correct plot is a horizontal straight line on a $t_{1/2}$ 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:
$\text{Rate} = k[R]$
where $k$ is the rate constant. The integrated form gives:
$\ln \frac{[R]_0}{[R]} = kt$
The half-life $t_{1/2}$ is the time when $[R] = \frac{[R]_0}{2}$ . Substituting:
$\ln \frac{[R]_0}{[R]_0/2} = k t_{1/2} \quad \Rightarrow \quad \ln 2 = k t_{1/2}$
So:
$t_{1/2} = \frac{\ln 2}{k}$
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 $t_{1/2}$ vs. $[R]_0$ graph. This would mean $t_{1/2} \propto [R]_0$ , which is true for a zero-order reaction (where $t_{1/2} = [R]_0 / 2k$ ). Not first-order.

Option (B): A horizontal straight line — $t_{1/2}$ does not change as $[R]_0$ changes. This matches $t_{1/2} = \ln 2 / k$ , 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 markMCQ
$Q. Which of the following is not a first order reaction ?$
(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. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Acid hydrolysis of ethyl acetate is:$
(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: $CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$ . 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.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The value of rate constant of a pseudo-first-order reaction$
(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:

$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$

The true rate law is $\text{rate}=k[\text{ester}][H_2O]$ . Since water is the solvent and is present in huge excess, $[H_2O]$ stays essentially constant throughout the reaction, so $\text{rate}=k'[\text{ester}]$ where $k'=k[H_2O]$ . Experimentally the reaction looks first order (only $[\text{ester}]$ appears), but the measured $k'$ is really the true rate constant $k$ multiplied by whatever fixed $[H_2O]$ 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 $k_{obs}$ . …
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CBSE 2025Set 56/6/11 markMCQ
$Q. For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Hydrolysis of an ester follows first order kinetics. Reason (R) : The concentration of water does not get altered much during the reaction.$
›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:
$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$
The rate law should be:
$\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]$
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.

Tip

When 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 $[\text{H}_2\text{O}]$ remains essentially constant, we absorb it into the rate constant:
$\text{Rate} = k'[\text{ester}], \quad \text{where } k' = k[\text{H}_2\text{O}]$
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: $\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]$ .

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 markMCQ
$Q. An example for pseudo first-order reaction is,$
(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:

$\underset{\text{sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{fructose}}{\text{C}_6\text{H}_{12}\text{O}_6}$ …
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CBSE 2025Set A1 mark
$Q. Fill in the blank: The unit of a first order rate constant is ______.$
›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 $(\text{mol L}^{-1})^{1-n}\,\text{time}^{-1}$ . For a first-order reaction ( $n=1$ ), the concentration term's exponent becomes zero, so the concentration unit cancels out completely, leaving only:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the concentration of the reactant in a first order reaction, when the rate of the reaction is 0.6 Ms^-1 and the rate constant is 0.035 s^-1 ?$
(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:

$\text{Rate} = k[R]$ …

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CBSE 2025Set ANNUAL1 mark
$Q. The conversion of molecule X to Y follows third order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?$
›Reveal solutionSolution

For a third-order reaction, Rate ∝ [X]^3, so tripling [X] multiplies the rate by 3^3 = 27.

Rate law: Rate = k[X]^3 (third order in X).

Original rate: Rate1 = k[X]^3 …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The units of first order reaction:$
(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

$k = \frac{rate}{[A]^n} = \frac{mol\,L^{-1}\,s^{-1}}{(mol\,L^{-1})^n}$

For a first order reaction ( $n=1$ ):

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one of the following is a pseudo first order reaction?$
(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 …

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Part (a)

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