ChemistryISC Class 12 · Chemical Kinetics

Arrhenius Equation and Activation Energy: ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

The Arrhenius equation, k = A e^(−Ea/RT), shows that the rate constant rises steeply with temperature because more molecules have energy at least equal to the activation energy Ea. Using two temperatures, log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂). A catalyst lowers Ea without changing ΔH.

Understand Arrhenius Equation and Activation Energy

Why temperature matters so much. For many reactions a rise of 10 K roughly doubles the rate constant, yet collisions become only a few per cent more frequent. The reason is that only molecules with enough energy react. Threshold energy is the minimum energy colliding molecules must have for a collision to cause reaction. Activation energy $E_{a}$ is the extra energy, above the average energy of the reactants, needed to reach the threshold: threshold energy = average energy of reactants + $E_{a}$ . Molecules with this energy form an unstable activated complex (transition state) at the top of the energy barrier, which then breaks up into products.

Energy profile. Plot potential energy against reaction coordinate: reactants on the left, a hump (activated complex) in the middle, products on the right. The height of the hump above the reactants is $E_{a}$ ; the difference between products and reactants is $Δ H$ . For an exothermic reaction the products lie lower. Labels to include: reactants, products, activated complex, $E_{a}$ (forward), $Δ H$ , and with a catalyst a lower hump.

Maxwell-Boltzmann distribution. At a higher temperature the distribution of kinetic energies flattens and shifts right; the area under the curve beyond $E_{a}$ (the fraction of molecules able to react) increases sharply. That fraction is $e^{- E_{a} / R T}$ .

The Arrhenius equation. $k = A e^{- E_{a} / R T}$ . Here $A$ is the pre-exponential (frequency) factor, related to the frequency of collisions with the right orientation, and $e^{- E_{a} / R T}$ is the fraction of molecules with energy $\geq E_{a}$ . Taking logs: $ln k = ln A - \frac{E _{a}}{R T}$ , or $lo g k = lo g A - \frac{E _{a}}{2.303 R T}$ . A plot of $lo g k$ against $\frac{1}{T}$ is a straight line of slope $- \frac{E _{a}}{2.303 R}$ and intercept $lo g A$ .

Two temperatures. Writing the equation at $T_{1}$ and $T_{2}$ and subtracting removes $A$ :

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ ,

with $R = 8.314$ J K⁻¹ mol⁻¹ and $E_{a}$ in J mol⁻¹. This is the equation behind almost every ISC activation energy numerical.

Catalysts. A catalyst provides an alternative pathway with a lower activation energy, so a larger fraction of molecules can react and the rate rises. It does not change $Δ H$ or $Δ G$ of the reaction, does not shift the position of equilibrium (it speeds up the forward and backward reactions equally), and is regenerated at the end. Enzymes are biological catalysts.

Collision theory (elementary idea). Reactant molecules are pictured as hard spheres; reaction happens only when they collide. The rate is $Z_{A B} e^{- E_{a} / R T}$ , where $Z_{A B}$ is the collision frequency. Not every collision with enough energy works: the molecules must also hit with the correct orientation. A probability (steric) factor $P$ allows for this: rate $= P Z_{A B} e^{- E_{a} / R T}$ . Collisions that lead to products are called effective collisions. For example, when CH₃Br meets OH⁻, only an approach of OH⁻ from the side opposite the Br atom forms methanol. Factors that increase effective collisions, such as higher concentration, larger surface area and higher temperature, increase the rate.

Arrhenius Equation and Activation Energy: formulas and key facts

Arrhenius equation: $k = A e^{- E_{a} / R T}$; $A$ = frequency factor (same units as k); $e^{- E_{a} / R T}$ = fraction of molecules with energy ≥ $E_{a}$ .
Logarithmic form: $lo g k = lo g A - \frac{E _{a}}{2.303 R T}$; Plot of log k against 1/T: slope $- E_{a} /2.303 R$ , intercept log A.
Two-temperature form: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$; $R = 8.314$ J K⁻¹ mol⁻¹; $E_{a}$ comes out in J mol⁻¹.
Threshold energy: $E_{t h r es h o l d} = E_{a v er a g e (r e a c t an t s)} + E_{a}$; Minimum energy for an effective collision.
Collision theory rate: $rate = P Z_{A B} e^{- E_{a} / R T}$; $Z_{A B}$ = collision frequency; $P$ = probability (steric) factor.
Effect of a catalyst at fixed T: $\frac{k _{c a t}}{k} = e^{(E_{a} - E_{a, c a t}) / R T}$; Assumes $A$ is unchanged.

How to solve Arrhenius Equation and Activation Energy questions

ISC activation energy questions are almost always two-temperature calculations, plus reasoning about catalysts and energy profiles. Keep the units consistent and the rest is arithmetic.

1List the data: $k_{1}$ at $T_{1}$ , $k_{2}$ at $T_{2}$ (temperatures in kelvin), and $E_{a}$ if given (convert kJ to J).
2Write $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ with $R = 8.314$ J K⁻¹ mol⁻¹.
3Put the larger k with the higher temperature so that both sides are positive.
4If the question says 'the rate doubles' or 'triples', use the ratio $k_{2} / k_{1} = 2$ or 3 without needing the values of k.
5To find a new k, solve for $lo g \frac{k _{2}}{k _{1}}$ , take the antilog and multiply by $k_{1}$ .
6For a catalyst at the same temperature, use $ln \frac{k _{c a t}}{k} = \frac{Δ E _{a}}{R T}$ .
7State $E_{a}$ in kJ mol⁻¹ to three significant figures.

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.

A 10 K rise can more than double a rate

Suppose a baker's bread dough takes 120 minutes to rise in a kitchen at 20 °C (293 K). Assume the rising is controlled by yeast enzyme reactions with an overall activation energy of 60 kJ mol⁻¹, and that the time needed is inversely proportional to the rate constant.

Estimate how long the dough would take to rise in a warm room at 30 °C (303 K).

Try it first, then show the solutionSolution

$lo g \frac{k _{2}}{k _{1}} = \frac{60 000}{2.303 \times 8.314} (\frac{303 - 293}{293 \times 303}) = 3133.6 \times 1.126 \times 1 0^{- 4} = 0.353$ .
$\frac{k _{2}}{k _{1}} = 1 0^{0.353} = 2.25$ .
Time is inversely proportional to k: $t_{2} = \frac{120}{2.25} = 53$ minutes.
Very high temperatures would kill the yeast, so the Arrhenius trend holds only over a moderate range.

Answer: The rate rises about 2.25 times, so the dough rises in about 53 minutes.

Memory hook: Warmth gives more molecules enough energy to cross the barrier: a small temperature rise, a big rate rise.

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.

Arrhenius = 'A times e to the minus E over RT'
Say it as a sentence: 'k equals A, e to the minus Ea by RT'. Bigger barrier (Ea) or lower T means smaller k.
Use it when: Writing the equation correctly.
Big over small, high over low
In $lo g \frac{k _{2}}{k _{1}}$ , put the bigger k on top and match it with the higher T in $T_{2} - T_{1}$ .
Use it when: Avoiding negative activation energies.
Catalyst: lower hill, same valley
A catalyst lowers the hill (Ea) but leaves the starting and finishing heights (reactants, products, ΔH) unchanged.
Use it when: Energy profile and equilibrium questions.

Arrhenius Equation and Activation Energy: solved examples

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

Example 1: Meaning of the exponential term · 1 mark · MCQBoard level

In the Arrhenius equation $k = A e^{- E_{a} / R T}$ , the factor $e^{- E_{a} / R T}$ represents:
(a) the total number of collisions per second
(b) the fraction of molecules with energy equal to or greater than $E_{a}$
(c) the enthalpy change of the reaction
(d) the fraction of collisions with correct orientation

Try it first, then show the solutionSolution

Collision frequency and orientation are contained in A.
$e^{- E_{a} / R T}$ is the Boltzmann fraction of molecules having at least the activation energy.

Answer: (b).

Example 2: Activation energy from two rate constants · 2 marks · Short answerBoard level

The rate constant of a reaction is $2.0 \times 1 0^{- 5}$ s⁻¹ at 300 K and $8.0 \times 1 0^{- 5}$ s⁻¹ at 320 K. Calculate the activation energy. ( $R = 8.314$ J K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

$lo g \frac{8.0 \times 1 0 ^{- 5}}{2.0 \times 1 0 ^{- 5}} = lo g 4 = 0.602 = \frac{E _{a}}{2.303 \times 8.314} (\frac{320 - 300}{300 \times 320})$ (1 mark).
$E_{a} = \frac{0.602 \times 19.147 \times 96 000}{20} = 5.53 \times 1 0^{4}$ J mol⁻¹ $= 55.3$ kJ mol⁻¹ (1 mark).

Answer: $E_{a} \approx 55.3$ kJ mol⁻¹.

Example 3: Rate constant at a higher temperature · 3 marks · Short answerHarder board level

A reaction has an activation energy of 75 kJ mol⁻¹ and a rate constant of $3.0 \times 1 0^{- 4}$ s⁻¹ at 310 K. Calculate the rate constant at 330 K. ( $R = 8.314$ J K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

$lo g \frac{k _{2}}{k _{1}} = \frac{75 000}{2.303 \times 8.314} (\frac{330 - 310}{310 \times 330})$ (1 mark).
$= 3917 \times 1.955 \times 1 0^{- 4} = 0.766$ , so $\frac{k _{2}}{k _{1}} = 1 0^{0.766} = 5.83$ (1 mark).
$k_{2} = 5.83 \times 3.0 \times 1 0^{- 4} = 1.75 \times 1 0^{- 3}$ s⁻¹ (1 mark).

Answer: $k_{330} \approx 1.75 \times 1 0^{- 3}$ s⁻¹.

Example 4: Reading A and Ea from an equation · 3 marks · Short answerHarder board level

For a first order reaction, the rate constant varies with temperature as $ln k = 30.5 - \frac{1.20 \times 1 0 ^{4} K}{T}$ (k in s⁻¹). Calculate (a) the activation energy, (b) the pre-exponential factor and (c) the rate constant at 600 K. ( $R = 8.314$ J K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

Compare with $ln k = ln A - \frac{E _{a}}{R T}$ (½ mark).
(a) $\frac{E _{a}}{R} = 1.20 \times 1 0^{4}$ K, so $E_{a} = 1.20 \times 1 0^{4} \times 8.314 = 9.98 \times 1 0^{4}$ J mol⁻¹ $\approx 99.8$ kJ mol⁻¹ (1 mark).
(b) $ln A = 30.5$ , so $A = e^{30.5} \approx 1.76 \times 1 0^{13}$ s⁻¹ (½ mark).
(c) $ln k = 30.5 - 20.0 = 10.5$ , so $k = e^{10.5} \approx 3.6 \times 1 0^{4}$ s⁻¹ (1 mark).

Answer: (a) About 99.8 kJ mol⁻¹. (b) $A \approx 1.8 \times 1 0^{13}$ s⁻¹. (c) $k \approx 3.6 \times 1 0^{4}$ s⁻¹.

Example 5: Temperature versus catalyst · 5 marks · Long answerHOTS / competency

A reaction has $E_{a} = 80$ kJ mol⁻¹. ( $R = 8.314$ J K⁻¹ mol⁻¹; assume A stays the same throughout.)
(a) By what factor does the rate constant increase when the temperature is raised from 300 K to 310 K? [2]
(b) A catalyst lowers $E_{a}$ by 20 kJ mol⁻¹. By what factor does it increase the rate constant at 300 K? [1]
(c) Without the catalyst, at what temperature would the reaction have the same rate constant as the catalysed reaction at 300 K? [1]
(d) Does the catalyst change ΔH of the reaction? Explain using an energy profile. [1]

Try it first, then show the solutionSolution

(a) $lo g \frac{k _{2}}{k _{1}} = \frac{80 000}{2.303 \times 8.314} (\frac{10}{300 \times 310}) = 4178 \times 1.075 \times 1 0^{- 4} = 0.449$ (1 mark); factor $= 1 0^{0.449} \approx 2.8$ (1 mark).
(b) $ln \frac{k _{c a t}}{k} = \frac{20 000}{8.314 \times 300} = 8.02$ , factor $= e^{8.02} \approx 3.0 \times 1 0^{3}$ (1 mark).
(c) Same k (same A) needs the same $E_{a} / T$ : $\frac{80 000}{T} = \frac{60 000}{300}$ , so $T = 400$ K (1 mark).
(d) No. The catalyst lowers only the height of the energy barrier (activated complex); the energies of reactants and products, and hence ΔH, are unchanged (1 mark).

Answer: (a) About 2.8 times. (b) About 3000 times. (c) 400 K. (d) No; only Ea is lowered.

Where students lose marks in Arrhenius Equation and Activation Energy

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

Mistake: Putting temperatures where rate constants belong
ISC examiners reported temperatures substituted in place of rate constants in the Arrhenius equation (2024).
Wrong working
$lo g \frac{700}{500} = \frac{E _{a}}{2.303 R} (\dots)$
Right working
$lo g \frac{k _{2}}{k _{1}} = lo g \frac{0.07}{0.02} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$
Do this: The log term holds rate constants; temperatures go only in the bracket.
Usually costs: The whole calculation.
Mistake: Using the wrong value of R
Examiners noted R = 0.0821 L atm K⁻¹ mol⁻¹ used instead of 8.314 J K⁻¹ mol⁻¹ (2024).
Do this: Activation energy is an energy: use $R = 8.314$ J K⁻¹ mol⁻¹ and give $E_{a}$ in J or kJ mol⁻¹.
Usually costs: 1 to 2 marks.
Mistake: Temperatures in °C
Students substitute 27 and 37 instead of 300 and 310 K.
Do this: Convert every temperature to kelvin first.
Usually costs: The answer mark.
Mistake: Saying a catalyst changes ΔH or the equilibrium constant
Students think a lower energy barrier means a different energy change.
Do this: A catalyst lowers Ea for both forward and backward reactions equally; ΔH, ΔG and K stay the same.
Usually costs: 1 mark.

Exam tips

Write the two-temperature equation in full before substituting; it is the method mark.
Simplify $\frac{T _{2} - T _{1}}{T _{1} T _{2}}$ separately to avoid calculator slips.
When the question gives log values (log 2 = 0.3010, log 3 = 0.4771), use them; it shows the examiner your working.
For energy profile questions, label both axes and mark Ea with and without a catalyst.
Remember that the pre-exponential factor A has the same units as k.

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
Threshold energy is equal to:
(a) activation energy only
(b) average energy of reactants + activation energy
(c) activation energy − average energy of reactants
(d) enthalpy of reaction + activation energy
Hint
Activation energy is the extra energy above the average.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Define activation energy. Describe the energy profile of an exothermic reaction, naming the points you would label.
Hint
Think of a hump between two levels.
Answer
Activation energy is the minimum extra energy, above the average energy of the reactants, that reacting molecules must acquire to form the activated complex and react. Profile: potential energy (y-axis) against reaction coordinate (x-axis); reactants at a higher level on the left, products lower on the right, a peak (activated complex) between them; label Ea as the height of the peak above the reactants and ΔH (negative) as the drop from reactants to products.
Q3 · 2 marks · Short answerHarder board level
The rate of a reaction triples when the temperature is raised from 300 K to 320 K. Calculate its activation energy. ( $R = 8.314$ J K⁻¹ mol⁻¹, log 3 = 0.4771)
Hint
$k_{2} / k_{1} = 3$ .
Answer
$0.4771 = \frac{E _{a}}{19.147} \times \frac{20}{96 000}$ , so $E_{a} = 4.38 \times 1 0^{4}$ J mol⁻¹ $\approx 43.8$ kJ mol⁻¹.
Q4 · 3 marks · Short answerHarder board level
A first order reaction has $k = 1.5 \times 1 0^{- 4}$ s⁻¹ at 350 K and $6.0 \times 1 0^{- 3}$ s⁻¹ at 400 K. Calculate (a) the activation energy and (b) the half-life at 400 K.
Hint
$\frac{k _{2}}{k _{1}} = 40$ , log 40 = 1.602.
Answer
(a) $1.602 = \frac{E _{a}}{19.147} \times \frac{50}{140 000}$ , so $E_{a} = 8.59 \times 1 0^{4}$ J mol⁻¹ $\approx 85.9$ kJ mol⁻¹. (b) $t_{1/2} = \frac{0.693}{6.0 \times 1 0 ^{- 3}} = 116$ s.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A 10 K rise in temperature may double the rate of a reaction, although the collision frequency increases by only a few per cent.
Reason (R): The fraction of molecules with energy equal to or greater than the activation energy increases sharply with temperature.
(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
Think of the area under the Maxwell-Boltzmann curve beyond Ea.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
According to collision theory, why is the rate of a reaction much smaller than the rate of collisions? Explain the two requirements for an effective collision with one example.
Hint
Energy and orientation.
Answer
Only effective collisions give products. (i) The colliding molecules must have energy at least equal to the threshold energy; at ordinary temperatures only a small fraction ( $e^{- E_{a} / R T}$ ) do. (ii) They must collide with the correct orientation; for example, OH⁻ must approach CH₃Br from the side opposite the bromine atom to form methanol. These are allowed for in rate $= P Z_{A B} e^{- E_{a} / R T}$ .
Q7 · 3 marks · Short answerHOTS / competency
For a reaction with $E_{a} = 50$ kJ mol⁻¹, calculate the fraction of molecules with energy at least equal to $E_{a}$ at 300 K and at 310 K. By what factor does this fraction change? ( $R = 8.314$ J K⁻¹ mol⁻¹)
Hint
Fraction $= e^{- E_{a} / R T}$ .
Answer
At 300 K: $e^{- 20.05} \approx 1.97 \times 1 0^{- 9}$ . At 310 K: $e^{- 19.40} \approx 3.76 \times 1 0^{- 9}$ . The fraction nearly doubles (factor about 1.9), which explains why the rate nearly doubles for a 10 K rise.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a hiker boils an egg at a hill station where water boils at 90 °C (363 K) instead of 100 °C (373 K). At home she uses a pressure cooker, in which water boils at about 120 °C (393 K). Take the activation energy for the cooking (protein denaturation) to be 100 kJ mol⁻¹ and assume cooking time is inversely proportional to the rate constant. ( $R = 8.314$ J K⁻¹ mol⁻¹)
(i) How many times longer does the egg take at the hill station than at 100 °C? [1]
(ii) How many times faster is cooking in the pressure cooker than in an open pan at 100 °C? [1]
(iii) Why does a small change in temperature have such a large effect? [1]
Hint
Use the two-temperature Arrhenius equation for each pair.
Answer
(i) $ln \frac{k _{373}}{k _{363}} = \frac{100 000}{8.314} (\frac{1}{363} - \frac{1}{373}) = 0.888$ , factor 2.4: about 2.4 times longer. (ii) $ln \frac{k _{393}}{k _{373}} = 1.64$ , factor about 5.2 times faster. (iii) The rate constant depends exponentially on $- E_{a} / R T$ ; with a large $E_{a}$ , a small rise in T greatly increases the fraction of molecules able to react.

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.Slope of log k against 1/T?Show answer
$- \frac{E _{a}}{2.303 R}$ .
3.What does a catalyst change: Ea, ΔH or K?Show answer
Only Ea (it lowers it).
4.What is an activated complex?Show answer
The unstable, high-energy arrangement of atoms at the top of the energy barrier, which breaks up to give products or reactants.
5.Two conditions for an effective collision?Show answer
Energy at least equal to the threshold energy, and correct orientation.
6.What is the rough temperature coefficient of many reactions?Show answer
The rate constant roughly doubles (or more) for a 10 K rise.

Arrhenius Equation and Activation Energy: frequently asked questions

What is the difference between threshold energy and activation energy?

Threshold energy is the minimum total energy colliding molecules need to react. Activation energy is the extra energy above the average energy of the reactants needed to reach it.

How do you calculate activation energy from two rate constants?

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 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 the position of equilibrium is unchanged.

What is the frequency factor A in the Arrhenius equation?

A constant for the reaction that reflects how often molecules collide with the correct orientation; it has the same units as the rate constant.

Why does the rate of reaction increase with temperature?

At higher temperature more molecules have energy equal to or greater than the activation energy, so a larger fraction of collisions is effective.