Rate ∝ collision frequency × fraction of collisions with energy ≥ Ea × steric factor. k = P·Z·e^(-Ea/RT). Steric factor accounts for orientation; usually < 1.
-- NCERT Class 12 Chemistry, Ch. 3, p. 26Collision Theory
8 MCQs2 revision cards9-step worked example
Source: NCERT Organic Chemistry — Basic PrinciplesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
Collision theory explains why reactions happen at the molecular level and why the rate constant k depends on temperature — the bridge between kinetic-molecular theory and the Arrhenius equation that NEET tests directly.
The core idea. For a bimolecular gaseous reaction, two molecules must (1) collide, (2) collide with energy ≥ the activation energy Eₐ (threshold energy), and (3) collide with the correct spatial orientation (steric factor). Only a fraction of total collisions satisfy both conditions; these are called effective collisions.
The rate of reaction is then:
Rate = Z_AB × f × p
where Z_AB is the collision frequency (total collisions per unit volume per second), f = e^(−Eₐ/RT) is the fraction of collisions with energy ≥ Eₐ (the Boltzmann factor), and p is the steric or probability factor (orientation requirement). NCERT Class 12 Chemistry Chapter 3, page 26 states this explicitly.
Connection to Arrhenius. The pre-exponential factor A in k = A·e^(−Eₐ/RT) bundles Z_AB and p together. Collision theory therefore predicts the Arrhenius form: the exponential temperature dependence comes from the Boltzmann energy distribution, and A captures how often and how favorably molecules meet.
The high-frequency confusion. When using the two-temperature Arrhenius form ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂), students swap T₁ and T₂ and get a sign error. The physical check: raising temperature increases k, so k₂ > k₁ when T₂ > T₁, and ln(k₂/k₁) must be positive. Since 1/T₁ > 1/T₂ for T₂ > T₁, the subtraction (1/T₁ − 1/T₂) is positive. Both sides positive — sign confirmed.
Watch-out. Collision theory slightly overestimates rates for complex molecules because it treats molecules as hard spheres and ignores the steric factor's magnitude. NEET may ask why collision theory overestimates — the answer is the steric factor p < 1 for non-spherical molecules.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to collision theory, which of the following is NOT a necessary condition for an effective collision in a bimolecular gaseous reaction?
MCQ 2Easy RecallPractice
In the Arrhenius equation k = A·e^(−Eₐ/RT), the pre-exponential factor A accounts for which aspects of collision theory?
MCQ 3Easy RecallPractice
Why does collision theory overestimate the rate of reaction for complex (non-spherical) molecules?
MCQ 4Direct ApplicationPractice
The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂), what is the approximate activation energy? (R = 8.314 J mol⁻¹ K⁻¹)
MCQ 5Direct ApplicationPractice
For a reaction with Eₐ = 75 kJ mol⁻¹, the fraction of molecules possessing energy ≥ Eₐ at 500 K is e^(−Eₐ/RT). If the temperature is raised to 600 K, by what factor does this fraction increase? (R = 8.314 J mol⁻¹ K⁻¹)
MCQ 6Direct ApplicationPractice
Two reactions have the same pre-exponential factor A. Reaction X has Eₐ = 40 kJ mol⁻¹ and reaction Y has Eₐ = 80 kJ mol⁻¹. At a given temperature, how does the rate constant of X compare to Y?
MCQ 7Concept TrapPractice
A student claims that increasing temperature increases the rate of reaction solely because molecules move faster and collide more frequently. According to collision theory, what is the primary flaw in this reasoning?
MCQ 8CalculationPractice
The rate constant of a reaction is 1.5 × 10⁻³ s⁻¹ at 400 K and 4.5 × 10⁻³ s⁻¹ at 450 K. Calculate Eₐ for this reaction. (R = 8.314 J mol⁻¹ K⁻¹, ln 3 = 1.099)
Quick recall before you leave
Worked Example
- 1
Given
The rate constant of a gaseous reaction is k₁ = 2.0 × 10⁻² s⁻¹ at T₁ = 300 K and k₂ = 8.0 × 10⁻² s⁻¹ at T₂ = 350 K. R = 8.314 J mol⁻¹ K⁻¹.
- 2
Required
Find the activation energy Eₐ.
- 3
Concept
Collision theory predicts that k depends exponentially on temperature through the Boltzmann factor. The two-temperature Arrhenius form lets us extract Eₐ from two (k, T) data points without knowing A.
- 4
Formula
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)
- 5
Substitution
ln(8.0 × 10⁻²/2.0 × 10⁻²) = (Eₐ/8.314)(1/300 − 1/350) ln(4) = (Eₐ/8.314)(350 − 300)/(300 × 350) 1.386 = (Eₐ/8.314)(50/105000) 1.386 = (Eₐ/8.314)(4.762 × 10⁻⁴)
- 6
Calculation
Eₐ = 1.386 × 8.314 / (4.762 × 10⁻⁴) Eₐ = 11.523 / 4.762 × 10⁻⁴ Eₐ = 24,198 J mol⁻¹ **Note on exact constants:** The integers 50, 300, 350, and 105000 are exact arithmetic results. R = 8.314 J mol⁻¹ K⁻¹ is a defined constant. ln(4) = 1.3863 is a mathematical constant. These do not limit significant figures. The given k values (2 sig figs each) control the final precision.
- 7
Final answer
Eₐ ≈ 24 kJ mol⁻¹ (2 significant figures, matching the precision of the given k values).
- 8
Common trap
Swapping T₁ and T₂ in the subtraction (1/T₁ − 1/T₂) gives a negative value, leading to a negative Eₐ — which is physically meaningless. The sign check: if T₂ > T₁ and k₂ > k₁, both ln(k₂/k₁) and (1/T₁ − 1/T₂) are positive, so Eₐ comes out positive. If your Eₐ is negative, you swapped the temperatures (mistake mistake: arrhenius t subtraction).
- 9
Similar NEET-style question
"The rate constant of a reaction increases from 0.02 s⁻¹ to 0.18 s⁻¹ when temperature is raised from 290 K to 340 K. Calculate the activation energy." (Same method: compute ln(k₂/k₁), compute 1/T₁ − 1/T₂, solve for Eₐ.)
Before solving, remember these
Formulas
Arrhenius equation
Temperature dependence of rate constant. Higher Ea → more T-sensitive rate.
| Symbol | Quantity | SI Unit |
|---|---|---|
| A | frequency factor | same as k |
| Ea | activation energy | J/mol |
| R | gas constant | J/mol/K |
| T | temp | K |
Valid when
- T in kelvins
- Most reactions in modest T range
Arrhenius for two temperatures
Compare rate constants at two temperatures to find Ea.
| Symbol | Quantity | SI Unit |
|---|---|---|
| k1, k2 | rate constants | same units |
| T1, T2 | temperatures | K |
| Ea | activation energy | J/mol |
Valid when
- A constant across temperature range
- T in kelvins
First-order kinetics
Concentration decays exponentially. Half-life independent of [A]_0.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [A] | conc at time t | mol/L |
| k | rate constant | 1/s |
| t | time | s |
Valid when
- First-order reaction (rate = k[A])
Zero-order kinetics
Concentration decays linearly. Half-life depends on initial concentration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| [A]_0 | initial conc | mol/L |
| k | rate constant | mol/L/s |
| t | time | s |
Valid when
- Zero-order reaction (rate = k, no concentration dependence)
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Zero-order t_1/2 depends on [A]_0. First-order t_1/2 INDEPENDENT of [A]_0. Student uses wrong formula.
When it triggers
Half-life question with order specified.
How to avoid
1st order: t_1/2 = 0.693/k (constant). Zero order: t_1/2 = [A]_0/(2k) (varies with initial conc). Second order: t_1/2 = 1/(k[A]_0).
Root cause: sign error
Correction
ln(k2/k1) = (Ea/R) × (1/T1 - 1/T2). At higher T2, k2 > k1, so ln(k2/k1) > 0. Need 1/T1 > 1/T2.
Root cause: formula misuse
Correction
First order: t_1/2 = 0.693/k (constant). Zero order: t_1/2 = [A]_0/(2k) (depends on [A]_0). Second order: t_1/2 = 1/(k[A]_0).
Root cause: formula misuse
Correction
Zero-order: t_1/2 = [A]_0/(2k) (depends on initial conc). First-order: t_1/2 = 0.693/k (independent of [A]_0).
Root cause: concept gap
Correction
Rate law from EXPERIMENT, not stoichiometry. Order = sum of exponents in rate = k[A]^x[B]^y. Molecularity follows mechanism.
Past Year Questions
10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1[Co(NH 3 ) 5 Cl]Cl
2[Co(NH 3 ) 3 Cl 3 ]
3[Co(NH ) Cl ]
4[Co(NH ) ]Cl 3 4 2 3 6 3 Answer (2, 3)
NTA Answer: Option multi:2,3(final)
NEET 2025
110 minutes
22 minutes
34 minutes
45 minutes
NTA Answer: Option 1(final)
NEET 2024Revised key
Activation energy of any chemical reaction can be calculated if one knows the value of
1rate constant at standard temperature
2probability of collision
3orientation of reactant molecules during collision
4rate constant at two different temperatures
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
Which reaction is NOT a redox reaction?
1Zn + CuSO → ZnSO + Cu 4 4
22KClO 3 + I 2 → 2KIO 3 + Cl 2
3H + Cl → 2HCl 2 2
4BaCl + Na SO → BaSO + 2NaCl 2 2 4 4
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
138.04 kJ/mol
2380.4 kJ/mol
33.80 kJ/mol
43804 kJ/mol
NTA Answer: Option 1(revised_final)
NEET 2023
Which one is an example of heterogenous catalysis?
1Hydrolysis of sugar catalysed by H+ ions
2Decomposition of ozone in presence of nitrogen monoxide
3Combination between dinitrogen and dihydrogen to form ammonia in the presence of finely divided iron
4Oxidation of sulphur dioxide into sulphur trioxide in the presence of oxides of nitrogen
NTA Answer: Option 3(final)
NEET 2023
1Increase by a factor of six
2Increase by a factor of nine
3Increase by a factor of three
4Decrease by a factor of nine
NTA Answer: Option 2(final)
NEET 2022
1Statement I is incorrect but Statement II is correct.
2Both Statement I and Statement II are correct.
3Both Statement I and Statement II are incorrect.
4Statement I is correct but Statement II is incorrect.
NTA Answer: Option 4(final)
NEET 2021
1Zone refining
2Electrolysis
3Chromatography
4Distillation
NTA Answer: Option 4(final)
NEET 2021
1–83 kJ mol–1
241.5 kJ mol–1
383.0 kJ mol–1
4166 kJ
NTA Answer: Option 2(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Find Ea from rate-constant ratio at two temperatures using ln(k2/k1) = Ea/R · (1/T1 − 1/T2).
Multi StepMedium
Common distractors
wrong sign of 1 over t
Mixes T1 and T2 in subtraction
First-order: t_1/2 = 0.693/k. Apply or invert to find k or remaining concentration.
Direct ApplicationEasy
Common distractors
uses zero order formula
Plugs into [A]_0/(2k) wrong-order formula
Sources
NCERT refs: Class 12 Chemistry Chapter 3, p.26
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