In the graph showing Maxwell Boltzmann distribution of energy, ______. (Two or more than two options may be co

NCERT Exemplar · Q53

Q. In the graph showing Maxwell Boltzmann distribution of energy, ___________. (Two or more than two options may be correct.)

(i) area under the curve must not change with increase in temperature.

(ii) area under the curve increases with increase in temperature.

(iii) area under the curve decreases with increase in temperature.

(iv) with increase in temperature curve broadens and shifts to the right hand side.

Karnataka PUCMCQ· 1mImportance★★★★★

87% · 102/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

The Maxwell-Boltzmann distribution graph shows the spread of molecular energies at a given temperature. The total area under the curve represents the total number of molecules, which is fixed, so it does not change with temperature. As temperature rises, the curve broadens and shifts to the right (higher energies). Therefore, options (i) and (iv) are correct.

The Maxwell-Boltzmann distribution is a probability density function for the kinetic energies (or speeds) of molecules in an ideal gas. The key idea is that the area under the curve is always equal to the total number of molecules in the sample. Since the number of molecules is constant (assuming no chemical reaction or leakage), the area must remain unchanged regardless of temperature.

Now, what happens when you heat the gas? The average kinetic energy increases, so more molecules acquire higher energies. This causes the peak of the curve to shift to the right (toward higher energy values) and the curve becomes broader and flatter. The broadening happens because the spread of energies increases — some molecules still have low energy, but many now have much higher energy.

Let’s examine each option carefully.

Option (i): "Area under the curve must not change with increase in temperature."

The area under the curve is proportional to the total number of molecules. Since the number of molecules is fixed, the area is constant. This is a fundamental property of any probability distribution: the total probability (or total count) is always 1 (or N). So this statement is correct.
Option (ii): "Area under the curve increases with increase in temperature."

This would imply that the number of molecules increases when you heat the gas, which is false. The area is invariant. So this is incorrect.
Option (iii): "Area under the curve decreases with increase in temperature."

Same reasoning — the area cannot decrease because the number of molecules does not change. This is also incorrect. …

Start your 14-day free trial to unlock the full solution →

Let’s break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid them.

The Concept: Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution shows the spread of molecular speeds (or kinetic energies) in a gas at a given temperature.

Key properties:

The area under the curve represents the total number of molecules in the sample.
Since the number of molecules is fixed (unless gas is added or removed), the area is constant regardless of temperature.
As temperature increases:
- The curve broadens (wider spread of speeds).
- The peak shifts to the right (higher most probable speed).
- The peak lowers (to keep area constant).

Correct Answer

From the options given:

(i) area under the curve must not change with increase in temperature. ✓
(iv) with increase in temperature curve broadens and shifts to the right hand side. ✓

So the correct choices are A and D.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking area increases with temperature

Why it happens: Students see the curve “spreading out” and assume there is more “stuff” under it.

How to avoid: Remember — the area = total number of molecules. Unless you add or remove gas, the number of molecules is fixed. The curve changes shape, but the area stays the same.

Tip: Visualise a fixed amount of dough being rolled out — it spreads but the amount doesn’t increase.

✗ Mistake 2: Thinking area decreases with temperature

Why it happens: Students notice the peak gets lower and think the area shrinks.

How to avoid: A lower peak does not mean less area — the curve also gets wider. The two effects balance exactly.

Tip: Always check both height and width. A shorter, wider shape can have the same area as a tall, narrow one.

✗ Mistake 3: Confusing “broadening” with “shifting left”

Why it happens: Some students misremember the direction of the shift.

How to avoid: At higher temperature, molecules move faster on average — so the whole distribution shifts to the right (higher energy/speed). …

Start your 14-day free trial to unlock the full solution →

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 $\frac{1}{T}$ …
Start your 14-day free trial to see the full solution →
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 $\Delta H = -135\ \text{kJ mol}^{-1}$ : …
Start your 14-day free trial to see the full solution →
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: $\log A = 8.0 \Rightarrow A = 1\times10^{8}\ \text{s}^{-1}$ . …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
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:

$n - 1 = 5 \;\Longrightarrow\; n = 6$ …
Start your 14-day free trial to see the full solution →
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.
$\text{rate} = k\,[\mathrm{A}]^{0}[\mathrm{B}]^{1} = k[\mathrm{B}]$
$\text{overall order} = 0 + 1 = 1.$ …
Start your 14-day free trial to see the full solution →
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,

$\text{rate} = Z \cdot p \cdot e^{-E_a/RT}$ …
Start your 14-day free trial to see the full solution →
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:

$\text{Rate of disappearance of } B = \frac{3}{4} \times (2.8 \times 10^{-3}) \text{ mol L}^{-1} \text{s}^{-1}$ …
Start your 14-day free trial to see the full solution →
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): …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

←Previous questionQ52 — According to Maxwell Boltzmann distribution of energy, _____. (Two or …→Next questionQ54 — Which of the following statements are in accordance with the Arrhenius…

Q.In the graph showing Maxwell Boltzmann distribution of energy, ___________. (Two or more than two options may be correct.)

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

Concept: Maxwell-Boltzmann Distribution of Molecular Energies

Method: Physical Interpretation of the Distribution Curve

Steps:

The Concept: Maxwell-Boltzmann Distribution

Correct Answer

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking area increases with temperature

✗ Mistake 2: Thinking area decreases with temperature

✗ Mistake 3: Confusing “broadening” with “shifting left”

Step 1 — Recall the relation between ΔH and activation energies

Step 2 — Substitute known values

Concept

Solution

Q.In the graph showing Maxwell Boltzmann distribution of energy, ___________. (Two or more than two options may be correct.)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days