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. 83Collision Theory
8 MCQs5 revision cards9-step worked example
Source: NCERT Chemical KineticsPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 27 Sep 2026
Collision Theory, explained for NEET
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 83 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.
Can you answer these Collision Theory 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?
Show answer and why every option is right or wrong
Answer: D. Collision theory for bimolecular gaseous reactions requires (1) collision, (2) threshold energy ≥ Eₐ, and (3) proper orientation. Being in the same phase is a precondition of the gaseous-reaction context, not a separate condition stated in the theory (NCERT Class 12 Chemistry Chapter 3, page 83).
Why A is wrong: A describes the threshold energy requirement, which is one of the core conditions for an effective collision.
Why B is wrong: B describes the orientation (steric) requirement, which is a core condition of collision theory.
Why C is wrong: C describes the basic requirement that molecules must physically collide — the foundational condition of the theory.
MCQ 2Easy RecallPractice
In the Arrhenius equation k = A·e^(−Eₐ/RT), the pre-exponential factor A accounts for which aspects of collision theory?
Show answer and why every option is right or wrong
Answer: D. A = Z_AB × p, combining collision frequency (how often molecules meet) and the steric factor (fraction with correct orientation). The exponential term handles the energy barrier separately (NCERT Class 12 Chemistry Chapter 3, page 83).
Why A is wrong: A is incorrect — the activation energy barrier is captured by the exponential term e^(−Eₐ/RT), not by the pre-exponential factor.
Why B is wrong: B is incorrect — A has no relation to the equilibrium constant. The Arrhenius equation describes kinetics (rate constant), not thermodynamic equilibrium.
Why C is wrong: C is incorrect — the Boltzmann distribution of energies is represented by the exponential factor e^(−Eₐ/RT), not by A.
MCQ 3Easy RecallPractice
Why does collision theory overestimate the rate of reaction for complex (non-spherical) molecules?
Show answer and why every option is right or wrong
Answer: B. Collision theory treats molecules as hard spheres, effectively setting p = 1. For complex molecules, only a fraction of collisions have the correct spatial orientation (p < 1), so the actual rate is lower than predicted (NCERT Class 12 Chemistry Chapter 3, page 83).
Why A is wrong: A is incorrect — collision theory does not underestimate Eₐ; the overestimation arises from ignoring orientation constraints, not from the energy barrier.
Why C is wrong: C is incorrect — collision theory does account for temperature's effect on collision frequency through the kinetic-molecular theory framework.
Why D is wrong: D is incorrect — the Boltzmann factor e^(−Eₐ/RT) is correctly derived from kinetic theory and is not the source of the overestimation.
MCQ 4Direct ApplicationPractice
For a reaction with Eₐ = 50 kJ mol⁻¹ at T = 400 K, the exponent −Eₐ/RT in the Boltzmann factor f = e^(−Eₐ/RT) is closest to (R = 8.314 J mol⁻¹ K⁻¹)
Show answer and why every option is right or wrong
Answer: A. A is correct. Eₐ/RT = 50000/(8.314 × 400) = 50000/3325.6 ≈ 15.0, so the exponent −Eₐ/RT ≈ −15.0 — a single substitution into the Boltzmann factor's exponent.
Why B is wrong: B is wrong because it drops the negative sign; the exponent of the Boltzmann factor for an activation barrier is always negative.
Why C is wrong: C is wrong because it uses Eₐ = 50 directly in the exponent instead of converting to 50000 J, understating the exponent by a factor of 1000.
Why D is wrong: D is wrong because it moves the decimal point one place, as if T were 40 K instead of 400 K, overstating the exponent tenfold.
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)
Show answer and why every option is right or wrong
Answer: A. Ratio = e^(−Eₐ/RT₂) / e^(−Eₐ/RT₁) = e^((Eₐ/R)(1/T₁ − 1/T₂)). Exponent = (75000/8.314)(1/500 − 1/600) = 9022.8 × (3.33 × 10⁻⁴) ≈ 3.01. Factor = e^(3.01) ≈ 20.3. The fraction increases because more molecules clear the energy barrier at higher T.
Why B is wrong: B has the sign reversed — swapping T₁ and T₂ gives a negative exponent, which would imply the fraction decreases with temperature. That contradicts the physical expectation (trap: mistake: arrhenius t subtraction).
Why C is wrong: C results from using Eₐ/R without dividing by the temperatures — a units/dimensional error in the exponent calculation.
Why D is wrong: D doubles the correct exponent, possibly from an error in the temperature subtraction step.
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?
Show answer and why every option is right or wrong
Answer: B. Since k = A·e^(−Eₐ/RT) and A is the same, the reaction with lower Eₐ has a larger exponential factor (less negative exponent) and hence a larger rate constant. k_X > k_Y at any temperature (NCERT Class 12 Chemistry Chapter 3, page 82).
Why A is wrong: A reverses the relationship — higher Eₐ means a more negative exponent and hence a smaller k, not a larger one.
Why C is wrong: C is incorrect — different Eₐ values with the same A will always give different k values at any temperature above 0 K.
Why D is wrong: D is incorrect — while the absolute magnitudes of k_X and k_Y depend on T, the inequality k_X > k_Y holds at every temperature because the exponent difference is always in X's favor.
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?
Show answer and why every option is right or wrong
Answer: C. While collision frequency Z does increase with T (roughly as √T), the exponential Boltzmann factor e^(−Eₐ/RT) increases far more steeply. The dominant contribution to the rate increase is the sharply larger fraction of molecules exceeding the threshold energy (NCERT Class 12 Chemistry Chapter 3, page 80).
Why A is wrong: A is incorrect — temperature does increase collision frequency (Z ∝ √T), but this is the minor effect, not the dominant one.
Why B is wrong: B is incorrect — orientation (steric factor p) is independent of temperature; faster molecules do not have systematically worse orientation.
Why D is wrong: D is incorrect — activation energy Eₐ is a property of the reaction's potential energy surface and does not change with temperature. A catalyst lowers Eₐ, not heat.
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)
Show answer and why every option is right or wrong
Answer: A. A is correct. The ratio k₂/k₁ = (4.5 × 10⁻³)/(1.5 × 10⁻³) = 3, so ln(k₂/k₁) = 1.099. The temperature bracket is 1/400 − 1/450 = 50/180000 = 1/3600 K⁻¹. From ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂): Eₐ = 1.099 × 8.314 × 3600 = 3.29 × 10⁴ J mol⁻¹ = 32.9 kJ mol⁻¹ (NCERT Class 12 Chemistry Chapter 3).
Why B is wrong: B is wrong because 14.3 kJ mol⁻¹ comes from using log₁₀ 3 = 0.477 in the natural-log equation. The log₁₀ form of the equation carries a factor of 2.303 in the denominator.
Why C is wrong: C is wrong because 75.8 kJ mol⁻¹ comes from applying a 2.303 to the natural-log form, which multiplies the right answer by 2.303. That factor converts log₁₀ to ln and belongs only with log₁₀. (trap: mistake: arrhenius t subtraction)
Why D is wrong: D is wrong because 89.8 kJ mol⁻¹ comes from using the ratio k₂/k₁ = 3 itself where its natural log, 1.099, belongs. The Arrhenius equation is linear in ln k, not in k.
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Collision Theory: quick recall before you leave
How do you solve a Collision Theory question? A 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.)
What to remember before solving Collision Theory questions
More in Chemical Kinetics: 5 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Collision Theory questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys.
NEET 2020
An increase in the concentration of the reactants of a reaction leads to change in :
1collision frequency
2activation energy
3heat of reaction
4threshold energy
NTA Answer: Option 1(final)
Sources
NCERT refs: Class 12 Chemistry Chapter 3, p.83
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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