Arrhenius Theory

8 MCQs9-step worked example

Source: NCERT Organic Chemistry — Basic PrinciplesPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap first: the Arrhenius two-temperature formula contains the term (1/T₁ − 1/T₂). Swapping T₁ and T₂ flips the sign of Eₐ to negative — a physically meaningless result that costs you the full mark plus the negative-marking penalty.

The Arrhenius equation relates the rate constant k to temperature: k = Ae^(−Eₐ/RT). Here A is the frequency factor (pre-exponential factor), Eₐ is the activation energy in J/mol, R = 8.314 J mol⁻¹ K⁻¹, and T is absolute temperature in kelvin (NCERT Class 12 Chemistry Chapter 3, page 20). The exponential term represents the fraction of molecular collisions possessing sufficient energy to cross the activation barrier. A larger Eₐ means the rate constant is more sensitive to temperature changes.

The NEET-critical working form compares two temperatures: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). This is derived by writing the Arrhenius equation at T₁ and T₂ and subtracting the logarithmic forms.

Sign discipline: assign labels so T₂ > T₁. Then k₂ > k₁ (reactions speed up at higher temperature), making ln(k₂/k₁) > 0. On the right side, 1/T₁ > 1/T₂ when T₁ < T₂, so the bracket is also positive. Both sides positive → Eₐ positive. If your computed Eₐ is negative, you swapped the temperatures.

An NCERT-stated empirical observation: for many reactions, the rate roughly doubles per 10 K rise (Class 12 Chemistry Chapter 3, page 22). This is a consequence of the Arrhenius equation, not a separate law, and holds only for typical Eₐ values (~50–100 kJ/mol).

Watch-out: NEET problems sometimes state temperatures in °C. Always convert: T(K) = T(°C) + 273.

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

In the Arrhenius equation k = Ae^(−Eₐ/RT), the quantity Eₐ represents:

MCQ 2Easy RecallPractice

The pre-exponential factor A in the Arrhenius equation has:

MCQ 3Easy RecallPractice

For many reactions, the rate constant approximately doubles for every 10 K rise in temperature. This empirical observation is a direct consequence of:

MCQ 4Direct ApplicationPractice

The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Using ln 2 = 0.693 and R = 8.314 J mol⁻¹ K⁻¹, the activation energy Eₐ is closest to:

MCQ 5Direct ApplicationPractice

For a reaction with Eₐ = 100 kJ/mol, a student calculates ln(k₂/k₁) using the two-temperature Arrhenius equation and obtains a negative value. The most likely error is:

MCQ 6Direct ApplicationPractice

For a reaction, k = 2.0 × 10⁻² s⁻¹ at 350 K and k = 4.0 × 10⁻² s⁻¹ at 400 K. Using R = 8.314 J mol⁻¹ K⁻¹ and ln 2 = 0.693, the activation energy Eₐ is:

MCQ 7CalculationPractice

The rate constant of a reaction is 1.0 × 10⁻³ s⁻¹ at 300 K. The activation energy is 83.14 kJ/mol. Using R = 8.314 J mol⁻¹ K⁻¹, the rate constant at 310 K is closest to:

MCQ 8CalculationPractice

For a reaction, Eₐ = 60.0 kJ/mol. At temperature T₁, the rate constant is k₁. The temperature must be raised to T₂ such that the rate constant becomes 10k₁. If T₁ = 300 K, using R = 8.314 J mol⁻¹ K⁻¹ and ln 10 = 2.303, T₂ is closest to:

Worked Example

Pattern: Arrhenius temperature-dependence calculation (NEET pattern: arrhenius temp dependence — observed in NEET 2021 and 2025).

1
Given
A reaction has rate constant k₁ = 2.0 × 10⁻² s⁻¹ at T₁ = 298 K and k₂ = 8.0 × 10⁻² s⁻¹ at T₂ = 318 K. R = 8.314 J mol⁻¹ K⁻¹. Find Eₐ.
2
Required
Activation energy Eₐ in kJ/mol.
3
Concept
The Arrhenius two-temperature equation relates rate constants measured at two different temperatures to the activation energy.
4
Formula
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) Rearranged: Eₐ = R × ln(k₂/k₁) / (1/T₁ − 1/T₂)
5
Substitution
ln(k₂/k₁) = ln(8.0 × 10⁻²/2.0 × 10⁻²) = ln 4 = 2 × ln 2 = 2 × 0.693 = 1.386 1/T₁ − 1/T₂ = 1/298 − 1/318 = (318 − 298)/(298 × 318) = 20/94764 = 2.1104 × 10⁻⁴ K⁻¹
6
Calculation
Eₐ = 8.314 × 1.386 / 2.1104 × 10⁻⁴ Numerator: 8.314 × 1.386 = 11.523 J/mol Eₐ = 11.523 / 2.1104 × 10⁻⁴ = 54,602 J/mol **Note on exact values:** ln 2 = 0.693 is a given constant for this calculation. The integer 20 in the numerator of the temperature bracket and the integer 2 in the rate-constant ratio are exact counting values and do not limit significant figures.
7
Final answer
Eₐ ≈ 54.6 kJ/mol (Reported to 3 significant figures, consistent with the 2-sig-fig precision of the given rate constants. The rate constants 2.0 × 10⁻² and 8.0 × 10⁻² each have 2 sig figs, but their ratio is an exact integer 4, so the precision is limited by the temperature values at 3 sig figs.)
8
Common trap
Swapping T₁ and T₂ in the bracket: writing (1/318 − 1/298) gives −2.11 × 10⁻⁴, which yields Eₐ = −54.6 kJ/mol. A negative activation energy is physically meaningless. **Quick check:** if T₂ > T₁ and k₂ > k₁, both sides of the equation must be positive.
9
Similar NEET-style question
A first-order reaction has k = 3.0 × 10⁻⁴ s⁻¹ at 350 K and k = 1.2 × 10⁻³ s⁻¹ at 400 K. Find Eₐ. (Answer: use ln 4 = 1.386, (1/350 − 1/400) = 50/140000 = 3.571 × 10⁻⁴. Eₐ = 8.314 × 1.386/3.571 × 10⁻⁴ ≈ 32.3 kJ/mol.) ---

Before solving, remember these

Law

Arrhenius equation

k = A·e^(-Ea/RT), where Ea = activation energy, A = frequency factor. ln k = ln A - Ea/(RT). For two T: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂).

-- NCERT Class 12 Chemistry, Ch. 3, p. 20

Key Fact

Effect of temperature

Rate roughly doubles for every 10°C rise (rule of thumb). True dependence: exponential via Arrhenius. Higher Ea → more T-sensitive rate.

-- NCERT Class 12 Chemistry, Ch. 3, p. 22

Formulas

Arrhenius equation

k = A e^{- E_{a} / (R T)}

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

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

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

ln \frac{[ A ] _{0}}{[ A ]} = k t; t_{1/2} = \frac{0.693}{k}

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

[A] = [A]_{0} - k t

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.

Trap

Zero-order vs first-order half-life dependence

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).

Mistake

Mixes up T1 and T2 in Arrhenius two-temperature formula.

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.

Mistake

Uses first-order half-life for zero-order reaction (or vice versa).

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).

Mistake

Mixes up zero-order vs first-order half-life formulas.

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).

Mistake

Reads rate law exponents from stoichiometric coefficients.

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

Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?

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

If the half-life (t ) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction 1/2 is closest to :

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

The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K–1 mol–1, log4 = 0.6021

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

For a certain reaction, the rate = k[A]2[B], when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would

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

Given below are two statements Statement I: Primary aliphatic amines react with HNO to give unstable diazonium salts. 2 Statement II: Primary aromatic amines react with HNO to form diazonium salts which are stable even above 300 K. In 2 the light of the above statements, choose the most appropriate answer from the options given below

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

Which one of the following methods can be used to o b ta i n h ig h ly pure metal which is liquid at room te m p e r a tu re ?

1Zone refining

2Electrolysis

3Chromatography

4Distillation

NTA Answer: Option 4(final)

NEET 2021

The slope of Arrhenius plot of first T  103K. order reaction is –5 × The value of E of the a reaction is. Choose the correct option for your answer. JK–1mol–1] [Given R = 8.314 mol–1

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.20

Test yourself on this topic with real past-paper questions:

Practice this topic →

Free NEET study resources

Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.