(a) (i) The initial concentration of N₂O₅ in the first order reaction : N₂O₅ (g) arrow 2NO₂ (g) + (1)/(2) O₂ (

Question

Q. (a)

(i) The initial concentration of

N_2O_5

in the first order reaction : $N_2O_5

(g) \rightarrow 2NO_2

(g) + \frac{1}{2} O_2 (g)

was

1\cdot2 \times 10^{-2}

mol L⁻¹. The concentration of

N_2O_5

after 60 minutes was

0\cdot2 \times 10^{-2}$ mol L⁻¹. Calculate the rate constant of the reaction at 318 K. [log 6 = 0·778]

(ii) Account for the following : (I) We cannot determine the order of a reaction by taking into consideration the balanced chemical equation. (II) A bimolecular reaction may become kinetically of first order under a specified condition.

(OR)

(b)

(i) The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate activation energy (Ea). [2·303 R = 19·15 JK⁻¹ mol⁻¹, log 2 = 0·3]

(ii) For a reaction :

2H_2O_2 \xrightarrow{I^-} 2H_2O + O_2

the proposed mechanism is as given below : (I)

H_2O_2 + I^- \rightarrow H_2O + IO^-

(slow) (II)

H_2O_2 + IO^- \rightarrow H_2O + I^- + O_2

(fast)

(1) Write rate law for the reaction.

(2) Write the overall order and molecularity of the reaction.

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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 — Arrhenius Equation

Why reactions speed up when you heat them

You already know that most reactions go faster when you raise the temperature. The intuitive reason is simple: molecules move faster, collide more often, and collide harder. But that alone doesn't explain the dramatic jump in rate — a 10 °C rise can double or triple the rate, even though the collision frequency only increases by a few percent. Something else is at work.

The missing piece is that not every collision leads to a reaction. Only collisions with enough energy — above a certain threshold — actually break bonds and form products. That threshold is the activation energy $E_a$ . Think of it as a hill the reactants must climb before they can roll down into products. At room temperature, only a tiny fraction of molecules have enough energy to get over that hill. Raise the temperature, and that fraction grows exponentially.

The Arrhenius equation

The Swedish chemist Svante Arrhenius captured this relationship in a single compact formula:

k = A e^{-E_a / (RT)}

Here:

$k$ is the rate constant (how fast the reaction proceeds at a given temperature)
$A$ is the pre-exponential factor (roughly, how often collisions happen with the right orientation)
$E_a$ is the activation energy (the energy barrier, in J/mol or kJ/mol)
$R$ is the gas constant (8.314 J/mol·K)
$T$ is the absolute temperature (in Kelvin)

Important

The exponential term $e^{-E_a/(RT)}$ is the fraction of molecules that have energy at least $E_a$ . This fraction is tiny when $E_a$ is large or $T$ is low, and it grows sharply as $T$ increases.

Why the rate jumps so sharply with temperature

The exponential is the key. Suppose $E_a = 50$ kJ/mol. At 300 K, the fraction is $e^{-50000/(8.314 \times 300)} \approx e^{-20.0} \approx 2 \times 10^{-9}$ . At 310 K, it becomes $e^{-50000/(8.314 \times 310)} \approx e^{-19.4} \approx 3.8 \times 10^{-9}$ . That's nearly double — even though the temperature rose only 3%. The collision frequency $A$ barely changed, but the exponential term nearly doubled. That's why a small temperature rise can cause a large rate increase.

Tip

A useful rule of thumb: for many reactions near room temperature, a 10 °C rise roughly doubles the rate constant. This is a consequence of the exponential, not a law — it depends on $E_a$ .

What $A$ and $E_a$ really mean

$A$ (the pre-exponential factor) accounts for how often molecules collide and whether they're oriented correctly. It depends on the size and shape of the molecules. $E_a$ is the minimum energy needed for a successful collision. A reaction with a high $E_a$ is very sensitive to temperature; one with a low $E_a$ is less sensitive.

Watch out

Do not confuse $E_a$ with the overall energy change of the reaction ( $\Delta H$ ). $E_a$ is the barrier height; $\Delta H$ is the net energy difference between reactants and products. A reaction can be highly exothermic ( $\Delta H$ large and negative) but still have a high $E_a$ — that's why some exothermic reactions (like burning wood) need a spark to start.

The logarithmic form …

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Part (a): $k = 2.99\times10^{-2}\text{ min}^{-1}$ ; order is experimental; a reaction becomes pseudo-first-order when one reactant is in large excess.

Part (b): $E_a \approx 52.7\text{ kJ mol}^{-1}$ ; Rate $=k[H_2O_2][I^-]$ , overall order $=2$ , molecularity $=2$ .

Part (a)

(i) Rate constant of a first-order reaction

k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t}

Given $[A]_0 = 1.2\times10^{-2}\ \text{mol L}^{-1}$ , $[A]_t = 0.2\times10^{-2}\ \text{mol L}^{-1}$ , $t=60$ min.

\frac{[A]_0}{[A]_t}=\frac{1.2\times10^{-2}}{0.2\times10^{-2}}=6,\qquad \log 6 = 0.778

k=\frac{2.303}{60}\times0.778 = 0.03838\times0.778 = 0.0299\ \text{min}^{-1}=2.99\times10^{-2}\ \text{min}^{-1}

The temperature (318 K) is extra data — it is not needed to compute $k$ from the concentration data.

(ii)(I) The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law. These powers are determined only by measuring how rate varies with concentration; they need not equal the stoichiometric coefficients of the balanced equation, because most reactions occur through a sequence of elementary steps and the slowest (rate-determining) step controls the kinetics. Hence order cannot be predicted from the balanced equation. …

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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 49 on this concept.

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
$Q. Which of the following represents the effect of temperature on reaction rate ?$
(a) Nernst's equation
(b) Gibbs-Helmholtz equation
(c) Arrhenius equation
(d) Van't Hoff equation
›Reveal solutionSolution

The Arrhenius equation, k = A e^(-Ea/RT), describes how the rate constant (and hence reaction rate) depends on temperature.

The Arrhenius equation k = A e^(-Ea/RT) links the rate constant k to temperature T and activation energy Ea. It quantifies the effect of temperature on reaction rate. The Nernst equation relates electrod …

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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
$Q. Activation energy of a chemical reaction can be determined by$
(a) rate constant at standard temperature
(b) rate constants at two different temperatures
(c) orientation of reactant molecules during the collision
(d) using catalyst
›Reveal solutionSolution

Activation energy is found from the Arrhenius equation by measuring the rate constant of a reaction at two (or more) different temperatures.

The Arrhenius equation is: k = A e^(-Ea/RT)

Taking the natural log at two temperatures T1 and T2 with rate constants k1 and k2, and subtracting, gives:

ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The correct form of Arrhenius equation is:$
(a) k = e^(-Ea/RT)
(b) k = Ea/RT
(c) k = log_e (Ea/R)
(d) k = A e^(-Ea/RT)
›Reveal solutionSolution

The Arrhenius equation is $k = Ae^{-E_a/RT}$ , relating the rate constant to temperature and activation energy.

The Arrhenius equation, proposed by Svante Arrhenius, expresses how the rate constant $k$ of a reaction varies with temperature:

$k = Ae^{-E_a/RT}$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The influence of temperature on the rate of a reaction is determined by$
(a) Nernst equation
(b) Gibbs-Helmholtz equation
(c) van't Hoff equation
(d) Arrhenius equation
›Reveal solutionSolution
Of the four named equations, only the Arrhenius equation directly relates the rate constant of a reaction to temperature.

Why the others are not it:

(a) Nernst equation: relates the EMF of an electrochemical cell to the concentrations (activities) of the species involved — an electrochemistry relationship, not a kinetics one.

(b) Gibbs–Helmholtz equation: relates the temperature dependence of Gibbs free energy change to enthalpy and entropy changes — a thermodynamics relationship about spontaneity, not reaction rate.

(c) van't Hoff equation: relates the equilibrium constant ( $K_c$ or $K_p$ ) to temperature — about the position of equilibrium, not how fast equilibrium is reached.

Arrhenius equation (d): …
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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 X11 mark
$Q. Arrhenius factor is also called __________ factor.$
›Reveal solutionSolution

The Arrhenius factor $A$ is also called the frequency (pre-exponential) factor.

The Arrhenius equation is $k = A\,e^{-E_a/RT}$ , where $A$ is the Arrhenius constant. It represents the frequency of collisions with correct orientation and is therefore cal …
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CBSE 2025Set D1 markMCQ
$Q. The specific rate constant of a first order reaction depends upon$
(a) concentration of reactants
(b) concentration of products
(c) time
(d) temperature
›Reveal solutionSolution

The rate constant depends on temperature (via the Arrhenius equation), not on concentration or time.

For a first order reaction rate = k[A]. The rate constant k is a proportionality constant characteristic of the reaction at a given temperature. According to the Arrhenius equation:

k = A e^(-Ea/RT) …

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Q.(a)

Part (a)

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