ChemistryCBSE Class 12 · Chemical Kinetics

Arrhenius Equation and Activation Energy: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

The Arrhenius equation, $k = Ae^{-E_a/RT}$, shows how the rate constant rises with temperature. $E_a$, the activation energy, is the minimum extra energy colliding molecules need to react. From rate constants at two temperatures, $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1T_2}\right)$. A catalyst lowers $E_a$ and so speeds up the reaction.

Notice the hotter curve is flatter and shifted right, so the shaded tail beyond Ea, the molecules able to react, roughly doubles.

Illustration drawn for QuestionBuddy.

Understand Arrhenius Equation and Activation Energy

Most reactions speed up when heated. For many, a 10 K rise roughly doubles the rate constant; this ratio $\frac{k _{T + 10}}{k _{T}}$ is called the temperature coefficient. The Arrhenius equation explains why.

Activation energy. Arrhenius proposed that reactant molecules must first reach an unstable, high-energy arrangement, the activated complex, before they turn into products. The energy needed to climb from the reactants to this peak is the activation energy, $E_{a}$ . On an energy diagram it is the height of the hump between reactants and products. A reaction with a small $E_{a}$ is fast; a large $E_{a}$ makes it slow, regardless of whether the overall reaction releases or absorbs energy.

Why temperature matters so much. Molecules in a sample do not all have the same kinetic energy; their energies are spread according to the Maxwell–Boltzmann distribution. Only the molecules in the tail of the curve, with energy at least $E_{a}$ , can react on collision. The fraction of such molecules is about $e^{- E_{a} / R T}$ . When the temperature rises by just 10 K, the curve flattens and shifts right, and the area beyond $E_{a}$ roughly doubles. That is why a small rise in temperature can double the rate.

The equation. $k = A e^{- E_{a} / R T}$ . Here $A$ is the Arrhenius (frequency) factor, related to how often molecules collide in a suitable way, and has the same units as $k$ . Taking natural logs: $ln k = ln A - \frac{E _{a}}{R T}$ . A plot of $ln k$ against $1/ T$ is a straight line with slope $- \frac{E _{a}}{R}$ and intercept $ln A$ . With base-10 logs the slope is $- \frac{E _{a}}{2.303 R}$ . Writing the equation at two temperatures and subtracting removes $A$ : $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}]$ . This is the working formula for nearly every board numerical.

Effect of a catalyst. A catalyst provides a different pathway with a lower activation energy. More molecules can then cross the barrier at the same temperature, so the rate increases. It is not used up, does not change $Δ G$ of the reaction, and speeds up the forward and backward reactions equally, so the equilibrium constant is unchanged; equilibrium is just reached sooner. A catalyst can only speed up reactions that are already spontaneous.

Collision theory. Developed by Max Trautz and William Lewis, this theory treats molecules as hard spheres. The number of collisions per second per unit volume is the collision frequency, $Z$ . For $A + B \to$ products, rate $= Z_{A B} e^{- E_{a} / R T}$ . But even energetic collisions fail if the molecules hit in the wrong orientation, so a probability (steric) factor $P$ is added: rate $= P Z_{A B} e^{- E_{a} / R T}$ . An effective collision needs both enough energy (at least the threshold energy) and proper orientation. Comparing with Arrhenius, $A$ corresponds to $P Z_{A B}$ .

Arrhenius Equation and Activation Energy: formulas and key facts

Arrhenius equation: $k = A e^{- E_{a} / R T}$; $E_{a}$ in J mol⁻¹ with $R = 8.314$ J K⁻¹ mol⁻¹; $T$ in kelvin.
Logarithmic form: $ln k = ln A - \frac{E _{a}}{R T}, lo g k = lo g A - \frac{E _{a}}{2.303 R T}$; Plot of $ln k$ vs $1/ T$ : slope $= - E_{a} / R$ .
Two-temperature form: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$; Works equally with rates or with times taken for the same change (inverted).
Fraction of molecules with E ≥ Ea: $f = e^{- E_{a} / R T}$; The exponential part of the Arrhenius equation.
Collision theory: $rate = P Z_{A B} e^{- E_{a} / R T}$; $P$ = probability (steric) factor; $Z_{A B}$ = collision frequency.
Temperature coefficient: $\frac{k _{T + 10}}{k _{T}} \approx 2 to 3$; Typical, not universal; use the Arrhenius equation for exact work.

How to solve Arrhenius Equation and Activation Energy questions

Board numericals on this topic come in three forms: find $E_{a}$ from two rate constants, find a new rate constant from $E_{a}$ , or interpret the Arrhenius equation and graph. Units decide most marks.

1Convert all temperatures to kelvin ( $T = t ° C + 273$ ).
2Identify $k_{1}$ at $T_{1}$ and $k_{2}$ at $T_{2}$ (or the ratio $k_{2} / k_{1}$ ; 'rate doubles' means the ratio is 2).
3Write $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ with $R = 8.314$ J K⁻¹ mol⁻¹.
4To find $E_{a}$ : $E_{a} = \frac{2.303 R T _{1} T _{2}}{T _{2} - T _{1}} lo g \frac{k _{2}}{k _{1}}$ . Give it in J mol⁻¹ and kJ mol⁻¹.
5To find $k_{2}$ : work out the right-hand side, take the antilog to get the ratio, multiply by $k_{1}$ .
6If the equation is given as $k = A e^{- x / T}$ , then $E_{a} = x R$ directly.
7For graph questions, slope of $lo g k$ vs $1/ T$ is $- \frac{E _{a}}{2.303 R}$ ; the intercept gives $lo g A$ .

Arrhenius Equation and Activation Energy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding activation energy from a doubling of rate

Suppose your family sets curd at home. In summer, at about 35 °C, it sets in 4 hours, while in a cooler room at 25 °C it takes 8 hours. Assume the setting time is inversely proportional to the rate constant.

Estimate the activation energy of the curd-setting process. ( $R = 8.314$ J K⁻¹ mol⁻¹, log 2 = 0.3010)

Try it first, then show the solutionSolution

$T_{1} = 298$ K, $T_{2} = 308$ K. Half the time means double the rate: $\frac{k _{2}}{k _{1}} = 2$ .
$lo g 2 = \frac{E _{a}}{2.303 \times 8.314} (\frac{10}{298 \times 308})$ .
$E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$ .
$E_{a} \approx 52900$ J mol⁻¹ $= 52.9$ kJ mol⁻¹.

Answer: About 53 kJ mol⁻¹.

Memory hook: Rate doubles for 10 degrees near room temperature: that is an activation energy of about 53 kJ mol⁻¹.

Tricks to remember Arrhenius Equation and Activation Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hill, height, heat
Activation energy is the Hill's Height; Heat (temperature) gives more molecules the energy to climb it; a catalyst digs a tunnel through it.
Use it when: Explaining temperature and catalyst effects together.
2.303 R T₁T₂ over ΔT
$E_{a} = 2.303 R \times \frac{T _{1} T _{2}}{Δ T} \times lo g (ratio)$ : big product on top, small difference below.
Use it when: Rearranging without errors.
Catalyst changes the road, not the destination
Lower $E_{a}$ , faster both ways; same $Δ H$ , $Δ G$ and $K$ .
Use it when: Assertion-reason and MCQ traps.

Arrhenius Equation and Activation Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope of the Arrhenius plot · 1 mark · MCQBoard level

A graph of $ln k$ against $1/ T$ for a reaction is a straight line. Its slope is:
(a) $E_{a} / R$ (b) $- E_{a} / R$ (c) $- E_{a} /2.303 R$ (d) $ln A$

Try it first, then show the solutionSolution

$ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ , which is $y = c + m x$ with $m = - \frac{E _{a}}{R}$ .

Answer: (b) $- E_{a} / R$ .

Example 2: Ea when the rate triples · 2 marks · Very short answerBoard level

The rate constant of a reaction becomes three times when the temperature is raised from 300 K to 320 K. Calculate the activation energy. ( $R = 8.314$ J K⁻¹ mol⁻¹, log 3 = 0.4771)

Try it first, then show the solutionSolution

$lo g 3 = \frac{E _{a}}{2.303 \times 8.314} (\frac{20}{300 \times 320})$ (1 mark).
$E_{a} = \frac{0.4771 \times 2.303 \times 8.314 \times 96000}{20} = 43850$ J mol⁻¹ $\approx 43.9$ kJ mol⁻¹ (1 mark).

Answer: About 43.9 kJ mol⁻¹.

Example 3: New rate constant from Ea · 3 marks · Short answerHarder board level

A first-order reaction has $k = 0.020$ s⁻¹ at 500 K and an activation energy of 100 kJ mol⁻¹. Calculate the rate constant at 600 K. ( $R = 8.314$ J K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

$lo g \frac{k _{2}}{0.020} = \frac{100 000}{2.303 \times 8.314} (\frac{100}{500 \times 600})$ (1 mark).
$= 5222.6 \times 3.333 \times 1 0^{- 4} = 1.741$ (1 mark).
$\frac{k _{2}}{0.020} = antilog (1.741) = 55.1$ , so $k_{2} = 1.10$ s⁻¹ (1 mark).

Answer: $k_{600} \approx 1.1$ s⁻¹ (about 55 times faster).

Example 4: Reading Ea and k from an Arrhenius expression · 3 marks · Short answerHarder board level

The rate constant of a decomposition follows $k = (2.0 \times 1 0^{13} s^{- 1}) e^{- 12000 K / T}$ . Calculate (a) the activation energy and (b) the rate constant at 400 K. ( $R = 8.314$ J K⁻¹ mol⁻¹; $e^{- 30} = 9.36 \times 1 0^{- 14}$ )

Try it first, then show the solutionSolution

Compare with $k = A e^{- E_{a} / R T}$ : $\frac{E _{a}}{R} = 12000$ K (1 mark).
(a) $E_{a} = 12000 \times 8.314 = 99768$ J mol⁻¹ $\approx 99.8$ kJ mol⁻¹ (1 mark).
(b) At 400 K: $k = 2.0 \times 1 0^{13} \times e^{- 30} = 2.0 \times 1 0^{13} \times 9.36 \times 1 0^{- 14} = 1.87$ s⁻¹ (1 mark).

Answer: (a) About 99.8 kJ mol⁻¹. (b) About 1.87 s⁻¹.

Example 5: How much a catalyst helps · 3 marks · Short answerHOTS / competency

At 300 K a catalyst lowers the activation energy of a reaction from 75 kJ mol⁻¹ to 50 kJ mol⁻¹, with no change in $A$ . (a) By what factor does the rate constant increase? (b) Does the catalyst change the equilibrium constant or $Δ_{r} G$ ? Explain. (c) Sketch, in words, the energy profile with and without the catalyst. ( $R = 8.314$ J K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

(a) $\frac{k _{c a t}}{k} = e^{(E_{a} - E_{a}^{'}) / R T} = e^{25000/ (8.314 \times 300)} = e^{10.02} \approx 2.3 \times 1 0^{4}$ (1 mark).
(b) No. It lowers the barrier for the forward and the backward reactions by the same amount, so both rate constants rise by the same factor and $K = k_{f} / k_{b}$ is unchanged; $Δ_{r} G$ depends only on reactants and products (1 mark).
(c) Reactant and product energy levels stay the same; the catalysed path has a lower hump between them (1 mark).

Answer: (a) About $2.3 \times 1 0^{4}$ times faster. (b) Neither changes. (c) Same start and end levels, lower peak.

Where students lose marks in Arrhenius Equation and Activation Energy

The slips examiners see most often, and what to write instead.

Mistake: Using kJ with R in J
$E_{a}$ given in kJ mol⁻¹ is substituted directly.
Wrong working
$lo g \frac{k _{2}}{k _{1}} = \frac{100}{2.303 \times 8.314} (\dots)$
Right working
$lo g \frac{k _{2}}{k _{1}} = \frac{100 000}{2.303 \times 8.314} (\dots)$
Do this: Convert $E_{a}$ to J mol⁻¹ whenever $R = 8.314$ J K⁻¹ mol⁻¹.
Usually costs: Final answer mark.
Mistake: Using °C instead of K
Wrong working
$\frac{T _{2} - T _{1}}{T _{1} T _{2}} = \frac{10}{25 \times 35}$
Right working
$\frac{10}{298 \times 308}$
Do this: Convert every temperature to kelvin in the first line.
Usually costs: 1-2 marks.
Mistake: Saying a catalyst shifts equilibrium
Faster product formation is confused with more product.
Do this: A catalyst speeds up forward and backward reactions equally; $K$ is unchanged, equilibrium is only reached sooner.
Usually costs: 1 mark in MCQ or AR.
Mistake: Wrong sign or factor in the slope
Mixing $ln$ and $lo g$ forms.
Do this: $ln k$ vs $1/ T$ : slope $- E_{a} / R$ ; $lo g k$ vs $1/ T$ : slope $- E_{a} /2.303 R$ .
Usually costs: 1 mark.

Exam tips

Write the two-temperature formula fully before substituting; it is often worth a mark by itself.
Give $E_{a}$ in both J mol⁻¹ and kJ mol⁻¹ in the final line.
In theory answers, mention 'effective collisions', 'threshold energy' and 'proper orientation' by name.
Draw the energy profile with labelled reactants, products, activated complex and $E_{a}$ when asked about catalysts.
Use the log values supplied in the paper; do not change them.

Practice questions on Arrhenius Equation and Activation Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A catalyst increases the rate of a reaction by:
(a) increasing $Δ_{r} H$
(b) lowering the activation energy
(c) increasing the equilibrium constant
(d) making $Δ_{r} G$ more negative
Hint
It changes the path, not the end points.
Answer
(b).
Q2 · 3 marks · Short answerBoard level
Define activation energy. Explain, with reference to the distribution of molecular energies, why a 10 K rise in temperature can nearly double the rate of a reaction.
Hint
Maxwell–Boltzmann curve and the area beyond $E_{a}$ .
Answer
Activation energy is the minimum extra energy that reactant molecules must absorb to form the activated complex and change into products. At a higher temperature the energy distribution curve broadens and shifts to higher energies, so the fraction of molecules with energy equal to or above $E_{a}$ (the area under the tail) roughly doubles for a 10 K rise; more collisions are effective and the rate nearly doubles.
Q3 · 2 marks · Very short answerHarder board level
Calculate the fraction of molecules having energy equal to or greater than an activation energy of 50 kJ mol⁻¹ at 300 K. ( $R = 8.314$ J K⁻¹ mol⁻¹)
Hint
$f = e^{- E_{a} / R T}$ .
Answer
$\frac{E _{a}}{R T} = \frac{50000}{2494.2} = 20.05$ ; $f = e^{- 20.05} \approx 2.0 \times 1 0^{- 9}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For many reactions the rate constant nearly doubles for a 10 K rise in temperature.
Reason (R): The fraction of molecules with energy equal to or greater than the activation energy roughly doubles for such a rise.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Arrhenius exponential term.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
The rate constants of a reaction are $2.0 \times 1 0^{- 5}$ s⁻¹ at 300 K and $1.0 \times 1 0^{- 4}$ s⁻¹ at 325 K. Calculate the activation energy. (log 5 = 0.6990)
Hint
Ratio = 5.
Answer
$E_{a} = \frac{2.303 \times 8.314 \times 300 \times 325 \times 0.6990}{25} = 52200$ J mol⁻¹ $\approx 52.2$ kJ mol⁻¹.
Q6 · 2 marks · Very short answerHarder board level
A plot of $lo g k$ against $1/ T$ for a reaction gives a straight line of slope $- 6000$ K. Calculate the activation energy.
Hint
Slope $= - \frac{E _{a}}{2.303 R}$ .
Answer
$E_{a} = 6000 \times 2.303 \times 8.314 = 114900$ J mol⁻¹ $\approx 114.9$ kJ mol⁻¹.
Q7 · 3 marks · Short answerHOTS / competency
Suppose the spoiling of a cooked dish is controlled by a reaction with activation energy 60 kJ mol⁻¹. How many times longer will the dish keep in a refrigerator at 4 °C than on a kitchen shelf at 25 °C? State one assumption you make.
Hint
Keeping time is inversely proportional to the rate constant.
Answer
$lo g \frac{k _{298}}{k _{277}} = \frac{60000}{2.303 \times 8.314} (\frac{21}{277 \times 298}) = 0.797$ ; ratio $\approx 6.3$ . The dish keeps about 6 times longer. Assumption: the same reaction with the same $E_{a}$ and $A$ controls spoiling at both temperatures (and time to spoil ∝ 1/k).
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student models the gas reaction $A + B \to$ products with collision theory. At 300 K the collision frequency is enormous, yet the reaction is slow. Raising the temperature to 310 K increases the collision frequency by less than 2%, but the rate almost doubles. The activation energy is 50 kJ mol⁻¹.
(a) Give two conditions for a collision to be effective. (1 mark)
(b) What does the probability factor $P$ account for? (1 mark)
(c) Calculate the factor by which $e^{- E_{a} / R T}$ increases from 300 K to 310 K, and use it to explain the observation. (2 marks)
Hint
Ratio $= e^{\frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{310})}$ .
Answer
(a) The colliding molecules must have energy at least equal to the threshold energy, and they must be properly oriented. (b) The fraction of sufficiently energetic collisions that also have the correct orientation. (c) Ratio $= e^{6014 \times 1.075 \times 1 0^{- 4}} = e^{0.647} \approx 1.91$ . The rate nearly doubles because the fraction of molecules with $E \geq E_{a}$ nearly doubles, not because of more collisions.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the Arrhenius equation.Show answer
$k = A e^{- E_{a} / R T}$ .
2.What does A represent?Show answer
The frequency (pre-exponential) factor, related to collision frequency and orientation.
3.Effect of a catalyst on $E_{a}$ and $K$ ?Show answer
Lowers $E_{a}$ ; leaves $K$ unchanged.
4.Two conditions for an effective collision?Show answer
Sufficient energy (≥ threshold) and proper orientation.
5.What is $P$ in collision theory?Show answer
The probability or steric factor.
6.Ea if rate doubles from 298 K to 308 K?Show answer
About 52.9 kJ mol⁻¹.

Arrhenius Equation and Activation Energy: frequently asked questions

What is the Arrhenius equation in Class 12 Chemistry?

It is k = Ae^{-E_a/RT}, relating the rate constant to temperature through the activation energy E_a and the frequency factor A.

What is activation energy?

The minimum extra energy that reacting molecules must have, above their average energy, to form the activated complex and react.

Why does a catalyst not change the equilibrium constant?

It lowers the activation energy of the forward and backward reactions by the same amount, so both rates increase equally and their ratio, the equilibrium constant, stays the same.

How do I calculate activation energy from two temperatures?

Use \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1T_2}\right) with temperatures in kelvin and R = 8.314 J K⁻¹ mol⁻¹.

Why don't all collisions lead to a reaction?

Only collisions with enough energy and the right orientation are effective; most collisions lack one or both.