JEE Mains Chemistry · Chemical Kinetics

Temperature and the Arrhenius Equation

k = A e^(−Ea/RT): the activation energy is read from the equation or from the slope of ln k against 1/T, found from rate constants at two temperatures, or compared between two reactions with the same A.

All Chemical Kinetics notes JEE Mains Chemistry strategy

Why this matters

Twenty-six PYQs, twenty of them numerical, and nine from 2026 — more recent questions than any other page. Fifteen read Ea or A from an equation or a plot, or test statements about the equation; seven use the two-temperature form; four compare two reactions with the same A, such as a catalysed and an uncatalysed path. Three ideas cover the page.

Concept 1 of 3: Reading Ea and A from the Arrhenius equation or its plot

Only molecules that collide with at least the activation energy

E_a

react, and the share of such molecules is

e^{-E_a/RT}

. Taking logs turns the equation into a straight line in

1/T

: the slope carries

E_a

, the intercept carries

A

. Most slips come from mixing the natural-log form with the base-10 form.

Definition

$\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac1T$ : the coefficient of $\dfrac1T$ is $-\dfrac{E_a}{R}$ .
$\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac1T$ : the coefficient is $-\dfrac{E_a}{2.303R}$ . With $R = 8.314$ , $2.303R = 19.15$ J K $^{-1}$ mol $^{-1}$ .
$\ln k$ against $\dfrac{10^3}{T}$ : slope $= -\dfrac{E_a}{10^3R}$ , so multiply the slope by $10^3R$ .
$e^{-E_a/RT}$ is the fraction of molecules with energy AT LEAST $E_a$ .
$A$ has the unit of $k$ . $A$ and $E_a$ are taken as independent of temperature.
At a given $T$ , lower $E_a$ means larger $k$ . Higher $E_a$ means $k$ is MORE sensitive to temperature, and a given rise in $T$ changes $k$ more at low temperature than at high.
$k$ rises with $T$ for every reaction, endothermic or exothermic.

Arrhenius equation

k = A\,e^{-E_a/RT}\qquad \ln k = \ln A - \frac{E_a}{RT}\qquad \log k = \log A - \frac{E_a}{2.303RT}

Worked example

For a first-order reaction,

\log k = 12.5 - \dfrac{9000\ \text{K}}{T}

, with

k

in s

^{-1}

. Find

E_a

and

A

. (

R = 8.314

J K

^{-1}

mol

^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q50Moderate

Example 1 · Chemical Kinetics · Temperature and the Arrhenius Equation

Consider

A\overset{k_{1}\ }{\rightarrow}\text{ }B

and

C\overset{k_{2}\ }{\rightarrow}D

are two reactions. If the rate constant

\left( k_{1} \right)

of the

A \rightarrow B

reaction can be expressed by the following equation

\log_{10}k = 14.34 - \frac{1.5 \times 10^{4}}{\text{ }T/K}

and activation energy of

C \rightarrow D

reaction

\left( {Ea}_{2} \right)

\frac{1}{5}

th of the

A \rightarrow B

reaction

\left( {Ea}_{1} \right)

, then the value of

\left( {Ea}_{2} \right)

\_\_\_\_

kJ{mol}^{- 1}

. (Nearest Integer)

The ln form read as the log form

The slope of

\ln k

against

1/T

-\dfrac{E_a}{R}

; the slope of

\log k

against

1/T

-\dfrac{E_a}{2.303R}

. Using the wrong one puts

E_a

out by a factor of 2.303.

Fraction BELOW the activation energy

e^{-E_a/RT}

is the fraction of molecules with energy equal to or MORE than

E_a

— the ones that can react. A statement calling it the fraction with less than

E_a

is false.

Endothermic reactions slowing on heating

The sign of

\Delta H

does not enter

k = Ae^{-E_a/RT}

. For every reaction

k

rises with temperature, along a curve that bends upward at first.

Drill 14 more on reading ea and a from the arrhenius equation or its plot

Concept 2 of 3: Two-temperature form of the Arrhenius equation

Write the log form at two temperatures and subtract:

A

cancels. What is left links the ratio of the two rate constants to

E_a

and the two temperatures. Any one of

E_a

k_2

T_2

can then be found from the rest.

Definition

$\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ , with $k_2$ at $T_2$ .
$\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{T_2 - T_1}{T_1T_2}$ . Temperatures in kelvin.
Near room temperature a 10 °C rise roughly doubles $k$ when $E_a \approx 53$ kJ mol $^{-1}$ .
kcal to kJ: multiply by 4.184 (or by the 4.2 the question gives).
A half-life gives $k$ first: $k = \dfrac{0.693}{t_{1/2}}$ .

Two-temperature form

\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Worked example

The rate constant of a reaction doubles between 300 K and 320 K. Find

E_a

. (

R = 8.314

J K

^{-1}

mol

^{-1}

\log 2 = 0.301

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q49Moderate

Example 2 · Chemical Kinetics · Temperature and the Arrhenius Equation

For reaction

A \rightarrow P

, rate constant

k = 1.5 \times 10^{3}{\text{ }s}^{- 1}

27^{\circ}C

. If activation energy for the above reaction is

60\text{ }kJ{\text{ }mol}^{- 1}

, then the temperature (in

\ ^{\circ}C

) at which rate constant,

k = 4.5 \times 10^{3}{\text{ }s}^{- 1}

\_\_\_\_

. (Nearest integer) Given :

\log2 = 0.30,\log3 = 0.48,R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1}

\ln10 = 2.3

Celsius in the formula

\dfrac{1}{T}

must be in kelvin. Convert 27 °C to 300 K before subtracting reciprocals, and convert an answer in kelvin back to °C only if the question asks.

The reciprocals subtracted the wrong way round

With

k_2

at the higher temperature

T_2

\dfrac{1}{T_1} - \dfrac{1}{T_2}

is positive, and so is

\log\dfrac{k_2}{k_1}

. A negative

E_a

means one of the two was flipped.

Drill 6 more on two-temperature form of the arrhenius equation

Concept 3 of 3: Ratio of rate constants for two reactions with the same A

When two reactions share the same

A

and temperature, only their activation energies differ. Dividing the two Arrhenius equations leaves

e^{\Delta E_a/RT}

. A catalyst that lowers

E_a

is the commonest case: the catalysed and uncatalysed paths share

A

Definition

$\dfrac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/RT}$ , so $\ln\dfrac{k_2}{k_1} = \dfrac{E_{a1} - E_{a2}}{RT}$ and $\log\dfrac{k_2}{k_1} = \dfrac{E_{a1} - E_{a2}}{2.303RT}$ .
Catalyst lowering $E_a$ by $\Delta E_a$ : $\dfrac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{\Delta E_a/RT}$ .
Catalysed at $T_1$ as fast as uncatalysed at $T_2$ (same $A$ ): $\dfrac{E_{a,\text{cat}}}{T_1} = \dfrac{E_a}{T_2}$ .
Temperature at which two rate constants are equal: $A_1e^{-E_1/RT} = A_2e^{-E_2/RT}$ , so $T = \dfrac{E_1 - E_2}{R\ln(A_1/A_2)}$ .

Same A, different Ea

\ln\frac{k_2}{k_1} = \frac{E_{a1} - E_{a2}}{RT}

Worked example

At 400 K a catalyst lowers the activation energy of a reaction by 15 kJ mol

^{-1}

, with no change in

A

. Find

\ln\dfrac{k_{\text{cat}}}{k_{\text{uncat}}}

and the ratio itself. (

R = 8.3

J K

^{-1}

mol

^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q46Moderate

Example 3 · Chemical Kinetics · Temperature and the Arrhenius Equation

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by

20\text{ }kJ{\text{ }mol}^{- 1}

. If

k_{1}

and

k_{2}

are the rate constants of first and second reaction respectively at 300 K , then In

\frac{k_{2}}{k_{1}}

will be (nearest integer)

\left\lbrack R = 8.3\text{ }J{\text{ }K}^{- 1}{\text{ }mol}^{- 1} \right\rbrack

ln of the ratio given when log was asked

\log\dfrac{k_2}{k_1} = \dfrac{\Delta E_a}{2.303RT}

and

\ln\dfrac{k_2}{k_1} = \dfrac{\Delta E_a}{RT}

differ by a factor of 2.303. Also keep

\Delta E_a

in joules when

R

is in J K

^{-1}

mol

^{-1}

; kJ puts the answer out by 1000.

Drill 3 more on ratio of rate constants for two reactions with the same a

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Reading Ea and A from the Arrhenius equation or its plot
Arrhenius equation
$k = A\,e^{-E_a/RT}\qquad \ln k = \ln A - \frac{E_a}{RT}\qquad \log k = \log A - \frac{E_a}{2.303RT}$
Two-temperature form of the Arrhenius equation
Two-temperature form
$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
Ratio of rate constants for two reactions with the same A
Same A, different Ea
$\ln\frac{k_2}{k_1} = \frac{E_{a1} - E_{a2}}{RT}$

Watch out for (6)

The ln form read as the log form
→ Reading Ea and A from the Arrhenius equation or its plot
Fraction BELOW the activation energy
→ Reading Ea and A from the Arrhenius equation or its plot
Endothermic reactions slowing on heating
→ Reading Ea and A from the Arrhenius equation or its plot
Celsius in the formula
→ Two-temperature form of the Arrhenius equation
The reciprocals subtracted the wrong way round
→ Two-temperature form of the Arrhenius equation
ln of the ratio given when log was asked
→ Ratio of rate constants for two reactions with the same A

Test yourself on Chemical Kinetics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Kinetics chapter test Drill all 26 questions