MHT-CET Chemistry · Chemical Kinetics

Temperature Dependence, Arrhenius and Collision Theory

k = A e^(−Ea/RT): log k against 1/T is a line of slope −Ea/2.303R and intercept log A; between two temperatures log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂); only temperature (and a catalyst) changes k.

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Why this matters

18 PYQs at 33% HARD — the chapter's only HARD questions, all six from 2025 and 2026; the three from 2025 are all the two-temperature form: Ea from a doubled rate constant between 27 °C and 37 °C, Ea from two k values, and a half-life carried from 400 K to 300 K. The rest are the slope and intercept of the Arrhenius plot, 'k rises only with temperature', and one collision-theory statement. Two formulas, both in base-10 logs.

Concept 1 of 3: The Arrhenius Equation and Its Plot: Slope −Ea/2.303R, Intercept log A

k = A\,e^{-E_a/RT}

: the fraction of collisions with energy at least

E_a

grows with

T

. Taking base-10 logs,

\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac1T

— a straight line in

\dfrac1T

Definition

Plot of $\log_{10}k$ (y) against $\dfrac1T$ (x): slope $= -\dfrac{E_a}{2.303R}$ , intercept $= \log_{10}A$ . (With $\ln k$ the slope is $-\dfrac{E_a}{R}$ .)
$k$ from $A$ and the exponent: $A = 1.6 \times 10^{13}$ s $^{-1}$ , $\dfrac{E_a}{2.303RT} = 21$ : $\log k = 13.204 - 21 = -7.796$ , $k = 1.6 \times 10^{-8}$ s $^{-1}$ .
$k$ depends on temperature (and on a catalyst, which lowers $E_a$ ) and on NOTHING else: raising [NO] or [Cl $_2$ ] in $r = k[\text{NO}]^2[\text{Cl}_2]$ raises the rate, not $k$ .
$A$ is the frequency (pre-exponential) factor; $e^{-E_a/RT}$ is the fraction of molecules with energy at least $E_a$ .

Arrhenius

k = A\,e^{-E_a/RT},\qquad \log_{10}k = \log_{10}A - \frac{E_a}{2.303R}\cdot\frac{1}{T}

Worked example

The Arrhenius plot of

\log_{10}k

against

1/T

for a reaction has slope

-5000

K. Find

E_a

(

R = 8.314

J K

^{-1}

mol

^{-1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q54Easy

Example 1 · Chemical Kinetics · Temperature Dependence, Arrhenius and Collision Theory

What is the value of slope if

\log_{10} K

(y-axis) is plotted versus

1/T

(x-axis) for the Arrhenius equation?

Dropping the 2.303 with a base-10 plot

The slope is

-\dfrac{E_a}{R}

only for

\ln k

. With

\log_{10}k

it is

-\dfrac{E_a}{2.303R}

; option (B) inverts it and drops the sign.

Drill 3 more on the arrhenius equation and its plot: slope −ea/2.303r, intercept log a

Concept 2 of 3: Two Temperatures: log(k₂/k₁) = (Ea/2.303R)·(T₂ − T₁)/(T₁T₂)

Subtract the Arrhenius equation at

T_1

from the one at

T_2

\log A

cancels and what remains links the ratio of rate constants to

E_a

. Because

t_{1/2} \propto \dfrac1k

for first order, the same equation carries a half-life between temperatures.

Definition

$k$ doubles from $300$ K to $310$ K: $E_a = \dfrac{0.301 \times 2.303 \times 8.314 \times 300 \times 310}{10} = 53{,}600$ J $= 53.6$ kJ mol $^{-1}$ .
$k = 0.026$ s $^{-1}$ at $290$ K and $0.58$ s $^{-1}$ at $300$ K: $\log 22.3 = 1.348$ ; $E_a = \dfrac{1.348 \times 2.303 \times 8.314 \times 290 \times 300}{10} = 224.5$ kJ mol $^{-1}$ .
Half-life across temperatures: $t_{1/2} = 900$ min at $400$ K with $\dfrac{E_a}{2.303R} = 1305.6$ : $\log\dfrac{k_{400}}{k_{300}} = 1305.6\left(\dfrac{1}{300} - \dfrac{1}{400}\right) = 1.088$ , ratio $12.25$ ; at $300$ K the reaction is slower, so $t_{1/2} = 900 \times 12.25 = 11{,}025$ min.
Keep $T$ in kelvin, $R = 8.314$ J K $^{-1}$ mol $^{-1}$ , and report $E_a$ in kJ when the options are in kJ.

Two-temperature Arrhenius

\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)

Worked example

The rate constant of a reaction triples when the temperature rises from

300

K to

320

K. Find

E_a

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q91Hard

Example 2 · Chemical Kinetics · Temperature Dependence, Arrhenius and Collision Theory

The rate constant is doubled when temperature increases from

27^{\circ}C

37^{\circ}C

. What is activation energy in kJ ?

Dividing the half-life by the ratio at the LOWER temperature

Cooling from

400

K to

300

K slows the reaction, so the half-life gets LONGER:

900 \times 12.25

, not

900 / 12.25

. Decide the direction before touching the ratio.

Drill 2 more on two temperatures: log(k₂/k₁) = (ea/2.303r)·(t₂ − t₁)/(t₁t₂)

Concept 3 of 3: Collision Theory: Only Effective Collisions Count

Molecules react only when they collide with energy at least

E_a

AND in the right orientation. The rate equals the frequency of EFFECTIVE collisions, which is a small fraction of all collisions — so the observed rate of a gas reaction is far below the collision frequency.

Definition

Rate $=$ (collision frequency) $\times$ (fraction with $E \ge E_a$ ) $\times$ (orientation factor).
Not every collision reacts; proper orientation IS needed; for gases the number of collisions is far MORE than the observed rate, not less.
The activation energy is the minimum extra energy the colliding molecules need to reach the activated complex; a catalyst provides a path with a lower $E_a$ .
Raising temperature raises both the collision frequency (slightly) and the fraction of energetic collisions (greatly) — the second is why $k$ roughly doubles for a $10$ K rise.

Effective collisions

\text{rate} = Z \cdot e^{-E_a/RT} \cdot P \qquad (Z \text{ collision frequency, } P \text{ orientation factor})

Worked example

Two statements: (i) every collision between reactant molecules leads to product; (ii) the rate of a reaction equals the rate of effective collisions. Which is correct?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q60Moderate

Example 3 · Chemical Kinetics · Temperature Dependence, Arrhenius and Collision Theory

Which of the following is true for a reaction as per collision theory?

'Collisions are fewer than the observed rate'

It is the other way round: collisions vastly outnumber reactions, which is the whole point of the effective-collision idea. Option (C) states the inversion.

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)

The Arrhenius Equation and Its Plot: Slope −Ea/2.303R, Intercept log A
Arrhenius
$k = A\,e^{-E_a/RT},\qquad \log_{10}k = \log_{10}A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$
Two Temperatures: log(k₂/k₁) = (Ea/2.303R)·(T₂ − T₁)/(T₁T₂)
Two-temperature Arrhenius
$\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$
Collision Theory: Only Effective Collisions Count
Effective collisions
$\text{rate} = Z \cdot e^{-E_a/RT} \cdot P \qquad (Z \text{ collision frequency, } P \text{ orientation factor})$

Watch out for (3)

Dropping the 2.303 with a base-10 plot
→ The Arrhenius Equation and Its Plot: Slope −Ea/2.303R, Intercept log A
Dividing the half-life by the ratio at the LOWER temperature
→ Two Temperatures: log(k₂/k₁) = (Ea/2.303R)·(T₂ − T₁)/(T₁T₂)
'Collisions are fewer than the observed rate'
→ Collision Theory: Only Effective Collisions Count

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Related notes

First-Order Kinetics — the half-life the two-temperature form carries →