Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex

NCERT Exemplar · Q66

Q. Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules. Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

(i) Both assertion and reason are correct and the reason is correct explanation of assertion.

(ii) Both assertion and reason are correct but reason does not explain assertion.

(iii) Assertion is correct but reason is incorrect.

(iv) Both assertion and reason are incorrect.

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Concept understanding — Arrhenius Equation Plot

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

You already know that heating things up makes reactions go faster. A cold chai takes forever to dissolve sugar; hot chai does it in seconds. But how much faster? And is there a pattern that holds for every reaction?

That pattern is the Arrhenius equation, and plotting it in a clever way reveals something fundamental about how molecules need to collide to react.

The core idea: an energy barrier

Imagine a ball sitting in a valley. To get to the next valley, it must first be pushed up over a hill. That hill is the activation energy ( $E_a$ ) — the minimum energy two molecules need to have when they collide, for the reaction to happen.

At a low temperature, most molecules move slowly. Only a tiny fraction have enough energy to climb that hill. Raise the temperature, and suddenly many more molecules have the required energy. The fraction of molecules with energy $\ge E_a$ is given by the Boltzmann distribution:

\text{fraction} = e^{-E_a / RT}

where $R$ is the gas constant and $T$ is the absolute temperature (in Kelvin). This exponential is the heart of the story.

The Arrhenius equation (precise statement)

The rate constant $k$ of a reaction depends on temperature as:

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

$k$ = rate constant (how fast the reaction proceeds)
$A$ = pre-exponential factor (frequency of collisions, times a steric factor — how often molecules hit in the right orientation)
$E_a$ = activation energy (J/mol or kJ/mol)
$R$ = 8.314 J/(mol·K)
$T$ = temperature in Kelvin

Important

$k$ is not the reaction rate itself — it's the proportionality constant in the rate law. But for a fixed concentration, a larger $k$ means a faster reaction.

Why plot it? The linear trick

The equation $k = A e^{-E_a / RT}$ is exponential in $1/T$ . That's hard to eyeball. But take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is the equation of a straight line:

y = c + m x

where:

$y = \ln k$
$x = 1/T$
slope $m = -E_a / R$
intercept $c = \ln A$

So if you measure $k$ at several temperatures and plot $\ln k$ versus $1/T$ , you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).

Tip

Always use Kelvin for $T$ . Celsius will give you a curved mess because $1/T$ is not linear in Celsius.

What the plot tells you

From the slope, you get $E_a$ :

E_a = - (\text{slope}) \times R

A steep negative slope means a large $E_a$ — the reaction is very sensitive to temperature. A shallow slope means a small $E_a$ — temperature doesn't affect it much.

From the intercept, you get $A$ :

A = e^{\text{intercept}}

This tells you about the collision frequency and orientation factor. A high $A$ means molecules are colliding often and in the right geometry.

A typical Arrhenius plot looks like this

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
300	0.0012	0.00333	-6.72
310	0.0028	0.00323	-5.88
320	0.0061	0.00313	-5.10
330	0.0125	0.00303	-4.38

Plot $\ln k$ (y-axis) vs $1/T$ (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute $E_a$ . …

Why this formula?

Arrhenius Equation Plot: Why It Holds

The Arrhenius equation is not a guess — it emerges from a deep physical picture of how molecules react. Let's build that understanding step by step.

The Core Idea: Molecules Need Energy to React

For a reaction to occur, molecules must collide with enough energy to break existing bonds and form new ones. This minimum energy is called the activation energy ( $E_a$ ).

But not all collisions succeed — only those with kinetic energy $\geq E_a$ lead to a reaction.

The Key Formula

The Arrhenius equation is:

k = A e^{-E_a / (RT)}

Where:

$k$ = rate constant
$A$ = pre-exponential factor (frequency of collisions with correct orientation)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol·K)
$T$ = absolute temperature (K)

Why the Exponential Term Appears

Step 1: The Boltzmann Distribution

Molecules in a gas or liquid have a distribution of kinetic energies. The fraction of molecules with energy $\geq E$ is given by the Boltzmann factor:

\text{Fraction} = e^{-E / (k_B T)}

For molar quantities, replace $k_B$ with $R$ :

\text{Fraction} = e^{-E_a / (RT)}

This is not arbitrary — it comes from statistical mechanics. The exponential arises because the probability of a molecule having energy $E$ decreases exponentially as $E$ increases.

Step 2: Rate Depends on This Fraction

The rate constant $k$ is proportional to:

The collision frequency (how often molecules meet)
The fraction of collisions with energy $\geq E_a$

Thus:

k \propto \text{(collision frequency)} \times e^{-E_a / (RT)}

The collision frequency is captured by $A$ , giving:

k = A e^{-E_a / (RT)}

Why the Plot is Linear

Take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is of the form $y = mx + c$ , where:

$y = \ln k$
$x = 1/T$
Slope $m = -E_a / R$
Intercept $c = \ln A$

Thus, plotting $\ln k$ vs $1/T$ gives a straight line — this is the Arrhenius plot.

What the Slope Tells Us

From the slope: …

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The Arrhenius equation gives accurate rate constants for simple reactions, but for complex molecules, steric factors and orientation matter — so the assertion is false. The reason is also false because molecules must have proper orientation for a reaction to occur. Both statements are incorrect.

The key here is to understand what the Arrhenius equation actually models and where it falls short. The equation $k = A e^{-E_a/RT}$ assumes that every collision with sufficient energy leads to a reaction — but that's only true for simple, small molecules. For complex molecules, the orientation during collision becomes critical, and the Arrhenius equation overestimates the rate unless corrected by a steric factor.

Let's break down each statement.

Assertion: "Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules."

This is incorrect. For simple molecules (like two atoms colliding), the Arrhenius equation works well because almost every energetic collision leads to reaction. But for complex molecules (large organic compounds, for instance), the molecule must hit the reactive site in the correct orientation. The Arrhenius equation ignores this — it assumes all collisions with enough energy are effective. That's why we introduce the steric factor $P$ in collision theory: $k = P Z e^{-E_a/RT}$ , where $P$ is often much less than 1 for complex molecules. So the assertion is false.
Reason: "Reactant molecules undergo chemical change irrespective of their orientation during collision." …

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Here’s a breakdown of the common mistakes students make on this specific assertion-reason question about the Arrhenius equation, along with how to avoid each.

Mistake 1: Confusing “Accuracy” with “Universality”

The Error: Students think the Arrhenius equation works perfectly for all reactions, including complex ones. They then mark the assertion as correct without checking the nuance.
Why It’s Wrong: The Arrhenius equation is empirical and works well for simple, elementary reactions. For complex molecules or multi-step reactions, the rate constant often deviates because the equation assumes a single activation energy barrier, which isn’t true for complex pathways.
How to Avoid: Remember: Arrhenius is accurate for simple, single-step reactions. For complex reactions, the plot of $\ln k$ vs. $1/T$ may be curved, not linear. The assertion says “fairly accurate for simple as well as complex molecules” — this is incorrect because it overstates the equation’s range.

Mistake 2: Misinterpreting the Reason (Orientation Factor)

The Error: Students think the reason sounds scientific and matches the topic of collision theory, so they assume it must be correct.
Why It’s Wrong: The reason states: “Reactant molecules undergo chemical change irrespective of their orientation during collision.” This is false. In reality, proper orientation is critical for effective collisions (the steric factor in collision theory). If orientation is wrong, no reaction occurs even if energy is sufficient.
How to Avoid: Link the reason to collision theory:
- Effective collision = sufficient energy (Arrhenius) + proper orientation.
- The reason denies the orientation requirement, which is a classic mistake. Always check if a statement contradicts basic collision theory.

Mistake 3: Assuming “Both Correct” Without Checking the Link

The Error: Students see two statements that both sound plausible and pick option (i) or (ii) without verifying if the reason actually explains the assertion.
Why It’s Wrong: Even if both were correct (they aren’t here), the reason (orientation is irrelevant) does not explain why the Arrhenius equation is accurate. The accuracy of Arrhenius depends on the reaction being elementary, not on orientation.
How to Avoid: For assertion-reason questions, always ask: “Does the reason directly cause or justify the assertion?” Here, the reason is about orientation during collision, while the assertion is about the equation’s accuracy — they are unrelated concepts.

Mistake 4: Forgetting the “Complex Molecules” Trap …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$\ln k$
›Reveal solutionSolution
The Arrhenius equation gives a linear relationship between $\ln k$ and $1/T$ , with slope $-E_a/R$ , so plotting $\ln k$ vs $1/T$ yields the activation energy. The correct option is (A).

The key idea is the Arrhenius equation, which describes how the rate constant $k$ depends on temperature $T$ and activation energy $E_a$ :
$k = A e^{-E_a/(RT)}$
Taking the natural logarithm gives:
$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$
This is in the form $y = mx + c$ , where $y = \ln k$ , $x = 1/T$ , the slope $m = -E_a/R$ , and the intercept $c = \ln A$ . So a plot of $\ln k$ vs $1/T$ gives a straight line whose slope directly yields $E_a$ .

Now let’s examine each option:

Option (A): $\ln k$ vs $1/T$

This matches the linear form above. The slope is $-E_a/R$ , so $E_a = -(\text{slope}) \times R$ . This is the standard method.

Option (B): $\ln k$ vs $T$

The Arrhenius equation is exponential in $T$ , not linear. Plotting $\ln k$ vs $T$ gives a curve, not a straight line, so the slope is not constant and cannot be used to find $E_a$ directly.

Option (C): $\frac{\ln k}{T}$ vs $T$

Substituting $\ln k = \ln A - E_a/(RT)$ gives $\frac{\ln k}{T} = \frac{\ln A}{T} - \frac{E_a}{R T^2}$ . This is not linear in $T$ ; it’s a more complicated function. No simple slope gives $E_a$ .

Option (D): $\frac{T}{\ln k}$ vs $\frac{1}{T}$ …
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KCET 2026Set D31 markMCQ
$\rightarrow$
›Reveal solutionSolution

Use the energy-diagram relation between enthalpy change and the forward/reverse activation energies: $\Delta H = E_{a(\text{forward})} - E_{a(\text{reverse})}$ .

Step 1 — Recall the relation between ΔH and activation energies

On a reaction energy profile, the enthalpy change of the reaction equals the difference between the forward and reverse activation energies:
$\Delta H = E_{a(\text{forward})} - E_{a(\text{reverse})}$
Step 2 — Substitute known values

Here, $E_{a(\text{forward}, X \to Y)} = 150\ \text{kJ mol}^{-1}$ and $\Delta H = -135\ \text{kJ mol}^{-1}$ : …
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COMEDK 2024Set 2024-A1 markMCQ
$(\mathrm{T})$
›Reveal solutionSolution

Matching the given line to the logarithmic Arrhenius equation: the intercept 8.0 gives $A = 10^8\ \text{s}^{-1}$ , and the slope $-2000$ gives $E_a = 2000 \times 2.303 \times R = 38.3\ \text{kJ mol}^{-1}$ .

The Arrhenius equation in logarithmic form is
$\log k = \log A - \frac{E_a}{2.303\,R\,T}$
Comparing with the experimental fit $\log k = -(2000)\dfrac{1}{T} + 8.0$ :

Intercept: $\log A = 8.0 \Rightarrow A = 1\times10^{8}\ \text{s}^{-1}$ . …
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COMEDK 2023Set 2023-E1 markMCQ
$Q. Given below are graphs showing the variation in velocity constant with temperature on Kelvin scale. Identify the graph which represents Arrhenius equation. (A) D (B) C (C) A (D) B$
›Reveal solutionSolution
[!TLDR]

Taking logs of the Arrhenius equation gives a straight line of negative slope on $\ln k$ vs $1/T$ axes, which is graph [A]; the option letter for graph A is (C).

Concept

This is the standard CBSE/NCERT Class-12 Chemical Kinetics result. The Arrhenius equation relates the rate (velocity) constant to temperature as $k = A\,e^{-E_a/RT}$ , where $A$ is the frequency factor and $E_a$ the activation energy. Its usefulness comes from its linearised (logarithmic) form.

Solution

Start from
$k = A\,e^{-E_a/RT}.$
Take the natural logarithm of both sides:
$\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right).$
Comparing with $y = c + m x$ (with $y=\ln k$ , $x = 1/T$ ), this is a straight line whose slope is $m = -E_a/R$ (negative, since $E_a>0$ ) and whose intercept is $\ln A$ .

Examining the four graphs:

[A] $\ln k$ vs $1/T$ , straight line of negative slope — matches the linearised Arrhenius form exactly.

[B] $k$ vs $1/T$ , rising exponential curve — wrong (not a $\ln k$ axis, wrong trend).

[C] $k$ vs $1/T$ , decaying curve — this is $k$ itself vs $1/T$ , not the linear Arrhenius plot. …
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KCET 2022Set B-31 markMCQ
$Q. A first order reaction is half completed in 45 min. How long does it need 99.9% of the reaction to be completed? (A) 10 Hours (B) 20 Hours (C) 5 Hours (D) 7.5 Hours$
›Reveal solutionSolution

For a first-order reaction, 99.9% completion is a fall to $1/1000$ of the initial concentration — almost exactly 10 half-lives, i.e. $10 \times 45\ \text{min} = 7.5$ hours.

Step 1 — The first-order rate law.

For a first-order reaction,
$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$
and the half-life is independent of concentration:
$t_{1/2} = \frac{0.693}{k}$
Step 2 — Find $k$ from the given half-life.
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{45\ \text{min}} = 0.0154\ \text{min}^{-1}$
Step 3 — Translate "99.9% complete" into a concentration ratio.

If 99.9% of $A$ has reacted, only 0.1% remains:
$[A] = 0.001\,[A]_0 \quad \Longrightarrow \quad \frac{[A]_0}{[A]} = 1000$
Step 4 — Substitute.
$t_{99.9\%} = \frac{2.303}{0.0154}\log(1000) = \frac{2.303}{0.0154} \times 3$ $t_{99.9\%} = 149.5 \times 3 = 448.6\ \text{min} \approx 450\ \text{min}$
Step 5 — Convert to hours.
$t = \frac{450}{60} = 7.5\ \text{hours}$
Step 6 — The elegant shortcut (worth knowing). …
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KCET 2022Set B-31 markMCQ
$Q. Half-life of a reaction is found to be inversely proportional to the fifth power its initial concentration, the order of reaction is (A) 5 (B) 6 (C) 3 (D) 4$
›Reveal solutionSolution

Use the general half-life–concentration relation for an $n^{th}$ order reaction, $t_{1/2} \propto [A]_0^{\,1-n}$ , and match the given exponent.

Step 1 — Why half-life depends on the initial concentration at all.

For a reaction of order $n$ in a single reactant $A$ , the rate law is
$-\frac{d[A]}{dt} = k\,[A]^{n}$
Integrating this from $[A]_0$ at $t=0$ to $[A]$ at time $t$ (for $n \neq 1$ ) gives
$\frac{1}{n-1}\left(\frac{1}{[A]^{\,n-1}} - \frac{1}{[A]_0^{\,n-1}}\right) = k\,t$
Step 2 — Put $[A] = [A]_0/2$ to get the half-life.

By definition $t = t_{1/2}$ when half the reactant is consumed:
$k\,t_{1/2} = \frac{1}{n-1}\left(\frac{2^{\,n-1}}{[A]_0^{\,n-1}} - \frac{1}{[A]_0^{\,n-1}}\right) = \frac{2^{\,n-1}-1}{(n-1)\,[A]_0^{\,n-1}}$ $\boxed{\;t_{1/2} = \frac{2^{\,n-1}-1}{k\,(n-1)}\cdot \frac{1}{[A]_0^{\,n-1}}\;}$
Everything in front of the concentration term is a constant, so the whole concentration dependence is
$t_{1/2} \propto \frac{1}{[A]_0^{\,n-1}} \quad\text{i.e.}\quad t_{1/2} \propto [A]_0^{\,1-n}$
Step 3 — Match the exponent given in the question.

The problem states the half-life is inversely proportional to the fifth power of the initial concentration:
$t_{1/2} \propto \frac{1}{[A]_0^{\,5}}$
Comparing exponents of $[A]_0$ in the denominator:

$n - 1 = 5 \;\Longrightarrow\; n = 6$ …
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KCET 2021Set B-21 markMCQ
$\ce{A + 2B \rightarrow Products}$
›Reveal solutionSolution

Deduce the order in each reactant from the two clues, add them to get the overall order, then read off the units from $\text{rate} = k[\,]^{n}$ .

Step 1 — Order with respect to B.

The half-life of a reaction of order $n$ scales as
$t_{1/2} \propto \frac{1}{[\,]^{\,n-1}}.$
It is independent of concentration only when $n - 1 = 0$ , i.e. $n = 1$ . Since "when concentration of B alone is increased, half-life remains the same",
$\text{order in B} = 1.$
(For a first-order reaction $t_{1/2} = 0.693/k$ — no concentration in it, which is exactly the observation.)

Step 2 — Order with respect to A.

"If concentration of A alone is doubled, rate remains the same." If the rate depended on $[\mathrm{A}]^{a}$ , doubling would multiply the rate by $2^{a}$ . Rate unchanged ⇒ $2^{a} = 1$ ⇒
$a = 0 \quad (\text{zero order in A}).$
Step 3 — Write the rate law and overall order.
$\text{rate} = k\,[\mathrm{A}]^{0}[\mathrm{B}]^{1} = k[\mathrm{B}]$
$\text{overall order} = 0 + 1 = 1.$ …
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KCET 2021Set B-21 markMCQ
$>3$
›Reveal solutionSolution
Higher-order steps demand a simultaneous many-body collision, whose probability falls away steeply as the number of colliding molecules rises.

Step 1 — Link order to molecularity for an elementary step.

For an elementary reaction, the order equals the molecularity — the number of species that must collide simultaneously in one act for the reaction to occur. So an order-4 elementary step requires a four-body simultaneous collision.

Step 2 — Apply collision theory.

Collision theory says a collision is fruitful only if the molecules:

meet at the same point at the same instant,

carry energy $\geq E_a$ (the threshold energy), and

arrive with the correct mutual orientation.

A binary (two-body) collision is common. A termolecular (three-body) collision is already rare — three particles must converge on one point at one instant. Requiring four or more to do so drives the probability down catastrophically: each extra body multiplies in another small factor for the coincidence of position, timing and orientation.

Step 3 — The consequence.

Because the collision frequency $Z$ for such an encounter is effectively negligible, and the rate is proportional to it,

$\text{rate} = Z \cdot p \cdot e^{-E_a/RT}$ …
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KCET 2020Set A-11 markMCQ
$2A + 3B \rightarrow 4C + D$
›Reveal solutionSolution

For an elementary reaction, the rate of disappearance of a reactant is related to the rate of appearance of a product by the stoichiometric coefficients. The rate of disappearance of $B$ is $\frac{3}{4} \times 2.8 \times 10^{-3}$ mol L $^{-1}$ s $^{-1}$ , which matches option (C).

The key idea here is that for an elementary reaction, the rate can be written in terms of any reactant or product using the stoichiometric coefficients. The rate of the reaction itself is a single, unique value — but the rate of disappearance of a reactant or appearance of a product is scaled by its coefficient.

For the reaction $2A + 3B \rightarrow 4C + D$ , the rate of reaction $r$ is defined as:
$r = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{1}{3}\frac{d[B]}{dt} = \frac{1}{4}\frac{d[C]}{dt} = \frac{d[D]}{dt}$
The negative signs indicate disappearance (decrease in concentration), and the positive signs indicate appearance (increase in concentration). The fractions ensure that no matter which species you track, you get the same numerical value for $r$ .

We are given the rate of appearance of $C$ : $\frac{d[C]}{dt} = 2.8 \times 10^{-3}$ mol L $^{-1}$ s $^{-1}$ .

From the definition above:
$\frac{1}{4}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[B]}{dt}$
So:
$-\frac{d[B]}{dt} = \frac{3}{4} \cdot \frac{d[C]}{dt}$
The left side $-\frac{d[B]}{dt}$ is the rate of disappearance of $B$ (a positive quantity). Substituting the given value:

$\text{Rate of disappearance of } B = \frac{3}{4} \times (2.8 \times 10^{-3}) \text{ mol L}^{-1} \text{s}^{-1}$ …
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KCET 2019Set A-11 markMCQ
$\mathrm{CH_3COOH}$
›Reveal solutionSolution

The rate law is first order in each reactant; halving both concentrations by dilution multiplies the rate by $\left(\tfrac12\right)^2=\tfrac14$ .

Step 1 — Write the rate law. The acid-catalysed esterification
$\mathrm{CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O}$
is first order in each reactant (overall second order):
$\text{Rate}=k\,[\mathrm{CH_3COOH}]\,[\mathrm{C_2H_5OH}] .$
Step 2 — Initial concentrations (after the original mixing).

Moles: acid $=2\ \mathrm{M}\times 1\ \mathrm L = 2$ mol; alcohol $=3\ \mathrm{M}\times 1\ \mathrm L=3$ mol. Total volume $=2$ L.
$[\mathrm{acid}]_0=\frac{2}{2}=1\ \mathrm M,\qquad [\mathrm{alcohol}]_0=\frac{3}{2}=1.5\ \mathrm M$ $r_0 = k(1)(1.5)=1.5k .$
Step 3 — After each solution is diluted with an equal volume of water.

Each 1 L solution becomes 2 L, so the total volume is now $2+2=4$ L, while the moles are unchanged (water only dilutes): …
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KCET 2018Set A-11 markMCQ
$Q. The value of rate constant of a pseudo first order reaction (A) Depends only on temperature (B) Depends on the concentration of reactants present in small amounts (C) Depends on the concentration of reactants present in excess (D) Is independent of the concentration of reactants$
›Reveal solutionSolution
In $k' = k[B]_{\text{excess}}$ , the excess reactant's (effectively constant) concentration is folded INTO the observed rate constant — so $k'$ carries its value.

Step 1 — What "pseudo first order" means.

Consider the classic acid-catalysed hydrolysis of an ester:
$\mathrm{CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH}$
The true rate law is second order:
$\text{rate}=k[\mathrm{ester}][\mathrm{H_2O}]$
But water is the solvent — it is present in enormous excess, so $[\mathrm{H_2O}]$ is essentially unchanged as the reaction proceeds.

Step 2 — Fold the constant term into the rate constant.

Because $[\mathrm{H_2O}]$ is effectively constant, define
$k'=k[\mathrm{H_2O}]\qquad\Longrightarrow\qquad \text{rate}=k'[\mathrm{ester}]$
The reaction now behaves as first order (hence "pseudo"). The key point: $k'$ literally contains $[\mathrm{H_2O}]$ as a factor. Change the amount of the excess reactant and $k'$ changes proportionally.

Step 3 — Eliminate the other options.

(A) "only on temperature" — true for a genuine rate constant $k$ (Arrhenius, $k=Ae^{-E_a/RT}$ ), but $k'$ is a composite constant that also carries $[B]_{\text{excess}}$ . So "only" is wrong. …
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Q.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

Concept understanding — Arrhenius Equation Plot

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

The core idea: an energy barrier

The Arrhenius equation (precise statement)

Why plot it? The linear trick

What the plot tells you

A typical Arrhenius plot looks like this

Why this formula?

Arrhenius Equation Plot: Why It Holds

The Core Idea: Molecules Need Energy to React

The Key Formula

Why the Exponential Term Appears

Step 1: The Boltzmann Distribution

Step 2: Rate Depends on This Fraction

Why the Plot is Linear

What the Slope Tells Us

Method: Conceptual Evaluation of Assertion-Reason Statements

Step 1: Evaluate the Assertion

Step 2: Evaluate the Reason

Mistake 1: Confusing “Accuracy” with “Universality”

Mistake 2: Misinterpreting the Reason (Orientation Factor)

Mistake 3: Assuming “Both Correct” Without Checking the Link

Mistake 4: Forgetting the “Complex Molecules” Trap …

Step 1 — Recall the relation between ΔH and activation energies

Step 2 — Substitute known values

Concept

Solution

Q.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules. Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

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Q.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

Unlock everything free for 14 days