JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Arrhenius Collision
TOPIC 12 OF 23
Arrhenius Collision
🎓 Class 12
Chemistry
CBSE
Theory
Ch 3 – Chemical Kinetics
⏱ ~14 min
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Arrhenius Collision
3.9 Temperature Dependence of Reaction Rate
Most chemical reactions go faster as the temperature rises. A useful empirical observation, called the temperature coefficient, says that for many reactions the rate roughly doubles for every 10 K rise in temperature:
\[ \mu \;=\; \frac{k_{(T+10)}}{k_T} \;\approx\; 2 \text{ to } 3 \]
The quantitative relationship between rate constant and temperature was given by Arrhenius in 1889.
3.10 The Arrhenius Equation
Arrhenius Equation:
\[ \boxed{\; k \;=\; A\,e^{-E_a / RT} \;} \]
where:
- k = rate constant at temperature T (in Kelvin)
- A = Arrhenius factor (frequency factor or pre-exponential factor) — units same as k
- Eₐ = activation energy of the reaction (J mol⁻¹)
- R = universal gas constant = 8.314 J K⁻¹ mol⁻¹
Activation Energy
The activation energy \(E_a\) is the minimum energy that reactants must acquire so that their collisions can be productive. The activated complex (or transition state) is the high-energy species at the top of the energy barrier between reactants and products.
Logarithmic Forms of the Arrhenius Equation
Taking natural logs of \(k = A e^{-E_a/RT}\):
\[ \ln k \;=\; \ln A \;-\; \frac{E_a}{RT} \]
or in base-10 form:
\[ \log k \;=\; \log A \;-\; \frac{E_a}{2.303\,RT} \]
Hence a plot of log k vs 1/T is a straight line with
- slope = \(-E_a/(2.303\,R)\) → gives \(E_a\)
- intercept = log A
Two-Temperature Form
If \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\), subtracting the Arrhenius equations:
\[ \boxed{\; \log\frac{k_2}{k_1} \;=\; \frac{E_a}{2.303\,R}\,\left[\frac{T_2 - T_1}{T_1 T_2}\right] \;} \]
This is the most useful form for problems that give two rate-constant–temperature pairs.
Worked Example 3.14 L3 Apply
The rate constants of a reaction at 500 K and 700 K are \(0.02\) s⁻¹ and \(0.07\) s⁻¹ respectively. Calculate the values of \(E_a\) and A.
\(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left[\dfrac{T_2 - T_1}{T_1 T_2}\right]\)
\(\log\dfrac{0.07}{0.02} = \dfrac{E_a}{2.303 \times 8.314}\left[\dfrac{700 - 500}{500 \times 700}\right]\)
\(\log(3.5) = \dfrac{E_a}{19.147}\left[\dfrac{200}{350000}\right]\)
\(0.544 = \dfrac{E_a}{19.147}(5.71 \times 10^{-4})\)
\(E_a = \dfrac{0.544 \times 19.147}{5.71 \times 10^{-4}} = 18230.8\) J mol⁻¹ ≈ 18.23 kJ mol⁻¹.
For A: from \(k = A e^{-E_a/RT}\) at T = 500 K:
\(0.02 = A \cdot e^{-18230.8/(8.314 \times 500)} = A \cdot e^{-4.385} = A \cdot 0.01246\)
\(A = 0.02/0.01246 \approx \mathbf{1.61}\) s⁻¹.
Worked Example 3.15 L3 Apply
The first-order rate constant for decomposition of ethyl iodide \(C_2H_5I(g) \to C_2H_4(g) + HI(g)\) at 600 K is \(1.60 \times 10^{-5}\) s⁻¹. Its activation energy is 209 kJ mol⁻¹. Calculate the rate constant at 700 K.
\(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left[\dfrac{T_2 - T_1}{T_1 T_2}\right]\)
\(= \dfrac{209000}{2.303 \times 8.314}\left[\dfrac{100}{600 \times 700}\right] = 10901.5 \times 2.381 \times 10^{-4} = 2.595\)
\(k_2/k_1 = 10^{2.595} = 393.4\)
\(k_2 = 393.4 \times 1.60 \times 10^{-5} = \mathbf{6.30 \times 10^{-3}}\) s⁻¹.
Interactive: Arrhenius Equation Explorer
Vary activation energy and temperature to see how the rate constant responds.
Rate constant k = — s⁻¹
Fraction with energy ≥ Eₐ: \(e^{-E_a/RT}\) = —
3.11 Effect of a Catalyst
A catalyst increases the rate of a reaction without being consumed in the overall process. According to the intermediate complex theory the catalyst forms a temporary complex of lower energy with reactants, providing an alternative pathway with a smaller activation energy.
Important Notes about Catalysts:
- A small amount of catalyst can speed up large amounts of reaction.
- A catalyst does not change the equilibrium constant — it only helps equilibrium be reached sooner.
- A catalyst lowers Eₐ for both forward and reverse directions equally.
- It does not change ΔG, ΔH or ΔS of the overall reaction.
3.12 Collision Theory of Chemical Reactions
Although Arrhenius's equation is empirical, a microscopic theory was developed by Max Trautz and William Lewis (1916–18) called the collision theory. It treats reactant molecules as hard spheres and assumes that reaction occurs when they collide.
For a bimolecular elementary reaction \(A + B \to \text{products}\), the rate is given by:
\[ \text{Rate} \;=\; Z_{AB}\,e^{-E_a/RT} \]
where \(Z_{AB}\) is the collision frequency per unit volume per unit time, and the exponential factor is the fraction of collisions with energy ≥ \(E_a\) (Boltzmann factor).
Steric Factor
For complex molecules, even an energetic collision may not lead to reaction unless the molecules are correctly oriented. We introduce a steric (or probability) factor \(P\):
\[ \text{Rate} \;=\; P\,Z_{AB}\,e^{-E_a/RT} \]
Two requirements emerge as essential:
- Energy criterion: the colliding molecules must have at least the threshold energy \(E_a\) — only the high-energy tail of the Maxwell–Boltzmann distribution contributes.
- Orientation criterion: the molecules must collide in a productive geometry.
Interactive: Maxwell–Boltzmann Distribution and Effect of T
Increasing temperature shifts the Maxwell–Boltzmann curve so that a much larger fraction of molecules has energy exceeding \(E_a\). This is why rate increases sharply with T.
Fraction with energy ≥ Eₐ ≈ —
Activity 3.4 — Estimate the Activation Energy of an Aspirin Tablet's Decomposition
Setup: Suppose two batches of aspirin tablets are stored — one at 25 °C (298 K) and one at 35 °C (308 K). Tests show that aspirin decomposes about 2.5 times faster at the higher temperature.
Predict: Calculate Eₐ from the temperature coefficient using the Arrhenius equation.
\(\log(k_2/k_1) = \log 2.5 = 0.398\)
\(\log(k_2/k_1) = \dfrac{E_a}{2.303 R}\dfrac{T_2 - T_1}{T_1 T_2} = \dfrac{E_a}{2.303 \times 8.314} \cdot \dfrac{10}{298 \times 308}\)
\(0.398 = \dfrac{E_a}{19.147} \times 1.089 \times 10^{-4}\)
\(E_a = \dfrac{0.398 \times 19.147}{1.089 \times 10^{-4}} = 6.99 \times 10^{4}\) J mol⁻¹ ≈ 70 kJ mol⁻¹.
This is why pharmaceuticals are stored 'in a cool, dry place' — every 10 K rise more than doubles the rate of degradation, drastically shortening shelf life.
Worked Example 3.16 L3 Apply
The activation energy for the reaction \(2HI(g) \to H_2 + I_2\) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than the activation energy.
Fraction = \(e^{-E_a/RT}\)
\(= e^{-209500 / (8.314 \times 581)} = e^{-43.36}\)
\(\log(\text{fraction}) = -43.36 / 2.303 = -18.83\)
\(\text{fraction} = \mathbf{1.47 \times 10^{-19}}\). Only about 1 molecule in \(10^{19}\) has enough energy — yet the reaction still proceeds because there are about \(10^{23}\) molecules per mole.
Competency-Based Questions
Q1. The Arrhenius factor A in the equation k = A e^(−Eₐ/RT) represents: L2
(b) A includes both collision frequency and the steric (orientation) factor.
Q2. A catalyst affects: L2
(b) A catalyst lowers Eₐ. ΔH and Keq remain unchanged.
Q3. (Short answer) Why does a slight increase in temperature produce a large increase in the rate of a chemical reaction even though the increase in average kinetic energy is small? L4
A small rise in T shifts the high-energy tail of the Maxwell-Boltzmann distribution by a large amount. The fraction of molecules with energy ≥ Eₐ rises dramatically (because the exponential factor \(e^{-E_a/RT}\) is very sensitive to T when Eₐ ≫ RT). So the rate constant — and hence rate — increases sharply.
Q4. (True/False) The slope of an Arrhenius plot of log k vs 1/T equals \(-E_a/R\). L2
False. Since we use base-10 log, slope = \(-E_a/(2.303\,R)\). For ln k vs 1/T the slope is \(-E_a/R\).
Q5. (Long answer) Two reactions A and B have the same A but \(E_a(A) = 30\) kJ/mol while \(E_a(B) = 60\) kJ/mol. At 298 K, by what factor is reaction A faster than B? L5
\(k_A/k_B = e^{-(E_{a,A} - E_{a,B})/RT} = e^{(60000-30000)/(8.314 \times 298)} = e^{12.10} \approx 1.8 \times 10^{5}\). Reaction A is about 180,000 times faster at room temperature. This explains why catalysts that lower Eₐ by even modest amounts can be enormously effective.
Assertion–Reason Questions
(A) Both true & R explains A. (B) Both true but R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion: A catalyst increases the rate of the forward and the backward reactions to the same extent.
Reason: A catalyst lowers the activation energy of the forward and reverse reactions by the same amount.
(A) Both true; the second statement explains why a catalyst does not change Keq.
Assertion: The rate of reaction increases sharply with temperature.
Reason: The activation energy of a reaction decreases with temperature.
(C) Assertion is true. Reason is false — Eₐ is essentially independent of T over moderate ranges; what increases is the fraction of molecules with energy ≥ Eₐ.
Assertion: For a bimolecular reaction collisions of molecules with sufficient energy is necessary but not sufficient for product formation.
Reason: Molecules must also collide in a favourable orientation (steric factor).
(A) Both true; R explains A. Energy + orientation are both required as per collision theory.
Frequently Asked Questions - Arrhenius Collision
What is the main concept covered in Arrhenius Collision?
In NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics), "Arrhenius Collision" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Arrhenius Collision useful in real-life or applied chemistry?
Real-life applications of "Arrhenius Collision" from NCERT Class 12 Chemistry Chapter 3 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Arrhenius Collision?
Key reactions in "Arrhenius Collision" (NCERT Class 12 Chemistry Chapter 3 Chemical Kinetics) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics) is structured so each part builds chemical understanding sequentially. "Arrhenius Collision" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Arrhenius Collision?
CBSE board questions from "Arrhenius Collision" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Arrhenius Collision" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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