Activation energy of a chemical reaction can be determined by _______. (i) determining the rate constant at st

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Q. Activation energy of a chemical reaction can be determined by _____________.

(i) determining the rate constant at standard temperature.

(ii) determining the rate constants at two temperatures.

(iii) determining probability of collision.

(iv) using catalyst.

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Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

Common Pitfall to Avoid

Watch out

Do not write $\frac{d[\text{reactant}]}{dt}$ as a positive number and then forget the minus sign. The rate of change of a reactant is negative (concentration falls). The minus sign in the definition flips it to a positive $r$ . If you skip the sign, you will get the wrong magnitude for other species.

Why This Matters

In exams (JEE, NEET, etc.), you will often be given the rate for one species and asked to find the rate for another. The stoichiometric relation is the only tool you need — no extra formulas. It also appears in more advanced topics like the rate law (where the exponents are not the coefficients) — but that is a separate concept. Reaction rate stoichiometry is purely about the definition of the reaction rate itself.

Final takeaway: The coefficients in the balanced equation are the conversion factors between the rates of different species. Always normalise by dividing by the coefficient to get the universal reaction rate $r$ .

Average rate of reaction is one of the first ideas introduced in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘average rate of reaction formula’ or ‘average vs instantaneous rate’ are common important-question topics in board exams and JEE Main chemistry. A solid grasp of this basic definition is also assumed in nearly every subsequent kinetics numerical asked in NEET and state CETs.

Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}

$\nu_X$ is negative for reactants → the minus sign is already built in.
$\nu_X$ is positive for products.

6. Why this is the average rate (not instantaneous)

Average rate uses a finite $\Delta t$ — it’s the slope of the chord between two points.
Instantaneous rate uses $\Delta t \rightarrow 0$ — it’s the slope of the tangent at a single point.

The average rate formula is just the discrete version of the derivative:

\text{Instantaneous rate} = \frac{1}{\nu_X} \frac{d[X]}{dt}

Summary — the "why" in one sentence

The average rate formula holds because it measures change in concentration per unit time, uses a minus sign to keep rates positive for reactants, and divides by stoichiometric coefficients to give a single, comparable value for the whole reaction.

Always remember:

$\Delta [\text{reactant}]$ is negative → minus sign makes it positive.
$\Delta [\text{product}]$ is positive → no minus sign.
Divide by coefficient → normalise to "per mole of reaction".

The activation energy of a reaction is found by measuring how the rate constant changes with temperature — specifically, by determining rate constants at two different temperatures and applying the Arrhenius equation.

The activation energy $E_a$ is the minimum energy that reactant molecules must possess for a collision to result in a reaction. You cannot get it from a single rate constant at one temperature — that only tells you how fast the reaction is at that specific temperature, not how sensitive it is to temperature changes. The key is that $E_a$ governs the temperature dependence of the rate constant.

The Arrhenius equation gives the relationship:

k = A e^{-E_a / RT}

where $k$ is the rate constant, $A$ is the pre-exponential factor (frequency of collisions with proper orientation), $R$ is the gas constant, and $T$ is the absolute temperature. If you take the natural logarithm:

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

This is a straight line when $\ln k$ is plotted against $1/T$ . The slope is $-E_a/R$ , so you need at least two points to determine the slope — i.e., rate constants at two different temperatures.

Let’s walk through why each option works or fails.

Option (i): Determining the rate constant at standard temperature.

One temperature gives one point on the $\ln k$ vs $1/T$ line. You cannot find the slope from a single point — infinite lines pass through it, each with a different $E_a$ . So this is insufficient.
Option (ii): Determining the rate constants at two temperatures.

With two temperatures $T_1$ and $T_2$ , you have two rate constants $k_1$ and $k_2$ . Subtract the Arrhenius equations:

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

Rearranging:

E_a = \frac{R \ln(k_2/k_1)}{\frac{1}{T_1} - \frac{1}{T_2}}

This directly gives $E_a$ . This is the standard method — simple, reliable, and exam-relevant.

Option (iii): Determining probability of collision.

Collision probability relates to the pre-exponential factor $A$ , not to $E_a$ . Even if you knew the collision frequency, you’d still need temperature variation to separate $A$ from $E_a$ in the Arrhenius equation. So this alone cannot determine $E_a$ .
Option (iv): Using catalyst.

A catalyst lowers $E_a$ by providing an alternative pathway — it changes the activation energy, it doesn’t help you measure the original $E_a$ . So this is irrelevant for determination.

Watch out

A common mistake is to think that a single rate constant at a known temperature is enough. But $E_a$ is a temperature sensitivity parameter — you must observe how $k$ changes with $T$ . One temperature gives no information about that sensitivity.

Tip

In exam problems, you’ll often be given $k_1$ at $T_1$ and $k_2$ at $T_2$ , and asked to find $E_a$ . The formula above is all you need. If they give a graph of $\ln k$ vs $1/T$ , the slope is $-E_a/R$ — read it carefully.

✓Final answer

The correct option is (ii) — determining the rate constants at two temperatures.

Common Mistakes Students Make on This Question

Mistake 1: Choosing Option (i) — "Determining the rate constant at standard temperature"

Why students pick this:

They think one rate constant at a fixed temperature is enough because the Arrhenius equation $k = A e^{-E_a/RT}$ contains $E_a$ .

Why it's wrong:

The Arrhenius equation has two unknowns: $A$ (frequency factor) and $E_a$ (activation energy). With only one $k$ at one $T$ , you cannot solve for $E_a$ uniquely — infinite pairs of $(A, E_a)$ satisfy the equation.

How to avoid:

Remember: one equation, two unknowns → need two data points. Always ask: "How many unknowns are there?"

Mistake 2: Choosing Option (iii) — "Determining probability of collision"

Why students pick this:

They confuse the collision theory (which involves collision frequency and orientation) with the experimental determination of $E_a$ .

Why it's wrong:

Collision probability is a theoretical concept, not a direct experimental method to calculate $E_a$ . It appears in the pre-exponential factor $A$ , but doesn't give $E_a$ alone.

How to avoid:

Distinguish between theory (what $E_a$ means) and experimental determination (how to measure it). The question asks for determination, not explanation.

Mistake 3: Choosing Option (iv) — "Using catalyst"

Why students pick this:

Catalysts lower activation energy — so students think they help "determine" it.

Why it's wrong:

A catalyst changes $E_a$ , it doesn't measure it. You'd need to know $E_a$ before and after to see the effect — which brings you back to needing rate constants at two temperatures.

How to avoid:

Think: "Does this tool give me a number for $E_a$ ?" A catalyst gives a new reaction path, not a measurement.

Mistake 4: Forgetting the Arrhenius equation in logarithmic form

Why students miss Option (ii):

They don't recall that the two-point form is the standard method:

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

How to avoid:

Memorise this formula. It directly shows: two temperatures + two rate constants → solve for $E_a$ .

✓ Correct Answer

(ii) determining the rate constants at two temperatures.

Quick Summary Table

Option	Why students choose it	Why it's wrong
(i) One temperature	"One equation should be enough"	Two unknowns ( $A$ and $E_a$ )
(iii) Collision probability	Confuses theory with measurement	Not a direct experimental method
(iv) Catalyst	"Catalyst affects $E_a$ "	Changes $E_a$ , doesn't measure it
(ii) Two temperatures	✓ Correct	Gives two equations → solve for $E_a$

Final tip: Whenever you see "determination of activation energy," immediately think two temperatures and the logarithmic Arrhenius equation.

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

AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$
›Reveal solutionSolution
This tests the definition of "rate of reaction" that is normalised by stoichiometric coefficients so it gives the same value regardless of which species you track. The rate works out to $2.5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Concept and Intuition

For a general reaction $aA \rightarrow bB + cC$ , different species are consumed/produced at different numerical speeds proportional to their coefficients. To get one unambiguous rate of reaction, each species' rate of change is divided by its own coefficient (with a minus sign for reactants, since their concentration falls): $\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = \dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt}$ .

Step-by-Step Solution

Reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ , so the coefficient of $N_2O_5$ is 2.

Concentration of $N_2O_5$ changes from $2$ to $1.5\ mol\,L^{-1}$ , so $\Delta[N_2O_5] = 1.5 - 2 = -0.5\ mol\,L^{-1}$ , over $\Delta t = 100$ min.

Average rate of disappearance of $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Rate of the reaction $= \dfrac{1}{2}\times(\text{rate of disappearance of } N_2O_5) = \dfrac{1}{2}\times 5\times10^{-3} = 2.5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Common Mistakes

Reporting the raw rate of disappearance of $N_2O_5$ ( $5\times10^{-3}$ ) without dividing by its coefficient 2 -- that gives option (A), a common trap.

Sign errors: forgetting the negative sign convention for a reactant, which doesn't change the magnitude here but can confuse students on which direction is positive.

✓Final answer
$2.5\times10^{-3}$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{A \rightarrow B}$
›Reveal solutionSolution
This tests the basic definition of average rate of reaction as the change in concentration of a reactant over the time interval.

Concept and Intuition

The average rate of a reaction over a time interval is simply how much the concentration of a reactant (or product) changes, divided by how long it took — with a sign convention so the rate is always reported as a positive quantity.

Step-by-Step Solution

Average rate of disappearance of A $= -\dfrac{\Delta[A]}{\Delta t} = -\dfrac{[A]_{final} - [A]_{initial}}{t}$ .

Here $[A]$ decreases from $x$ to $y$ over $t = 100$ min, so $\Delta[A] = y - x$ (negative, since $y < x$ ).

Average rate $= -\dfrac{y-x}{100} = \dfrac{x-y}{100}$ , and writing it generally (regardless of which is larger) as a positive quantity: $\dfrac{|x-y|}{100}$ .

This has correct units of concentration/time (mol L $^{-1}$ min $^{-1}$ ), unlike the other options which either invert the ratio or square a term.

Common Mistakes

Inverting the fraction (time over concentration change) — this gives the wrong units.

Squaring the concentration difference, which is dimensionally and physically meaningless here.

✓Final answer
$\dfrac{|x-y|}{100}$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$\mathrm{R \longrightarrow P}$
›Reveal solutionSolution
Average rate is simply the change in concentration of reactant divided by the time elapsed (converted to seconds); the 'first order' label is not even needed for this particular calculation. Answer: (B).

Concept and Intuition

The average rate of a reaction over a finite time interval is defined purely from the measured change in concentration and elapsed time — it does not require knowing the rate law or order of the reaction (order only matters for finding the instantaneous rate constant $k$ via the integrated rate law). Here we are only asked for the average velocity, so a direct $\Delta[\text{conc}]/\Delta t$ calculation suffices, with careful unit conversion from minutes to seconds since the answer must be in $\mathrm{mol\,L^{-1}\,s^{-1}}$ .

Step-by-Step Solution

Change in concentration of R: $\Delta[\mathrm{R}]=0.04-0.03=0.01\ \mathrm{mol\,L^{-1}}$ (a decrease, since R is being consumed).

Time elapsed: $40$ min $=40\times 60=2400$ s.

Average rate $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

$\dfrac{0.01}{2400}=4.1667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

This matches option (B).

Common Mistakes

Leaving time in minutes instead of converting to seconds, which would give $2.5\times10^{-4}$ (option A) — a common careless-unit trap built into the distractors.

Forgetting the negative sign convention or misplacing a decimal, landing on $2.5\times10^{-5}$ (option D) instead of the correctly converted value.

✓Final answer
$4.167\times10^{-6}$

ANSWER: B
AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$L^{-1}$
›Reveal solutionSolution
Rate is simply the drop in concentration divided by the elapsed time (converted to seconds) — $6.667\times10^{-6}\ \mathrm{mol\,L^{-1}s^{-1}}$ , answer (A).

Concept and Intuition

The (average) rate of a reaction over a time interval is defined as the change in concentration of a reactant divided by the time elapsed (with a negative sign convention for reactants, but the magnitude is what's asked here). Care must be taken to convert all times to consistent units (here, minutes to seconds, since the rate is required in $\mathrm{s^{-1}}$ ).

Step-by-Step Solution

Concentration drop: $\Delta[\text{reactant}] = 0.03 - 0.02 = 0.01\ \mathrm{mol\,L^{-1}}$ .

Time elapsed: $25\ \mathrm{min} = 25\times 60 = 1500\ \mathrm s$ .

Average rate $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

Common Mistakes

Forgetting to convert minutes to seconds (would give $4\times10^{-4}$ , a distractor option).

Confusing this simple average-rate calculation with computing the first-order rate constant $k$ (which needs the log form) — the question only asks for the rate, not $k$ .

✓Final answer
$6.667\times10^{-6}$

ANSWER: A
AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$\mathrm{A} \rightarrow \mathrm{P}$
›Reveal solutionSolution
On a plain concentration-vs-time graph, the instantaneous rate at any point is just the magnitude of the tangent's slope there — here that is simply $m$ , regardless of the reaction being first order.

Concept and Intuition

The instantaneous rate of a reaction $A \to P$ is defined purely graphically as $\text{Rate} = -\dfrac{d[A]}{dt}$ , which is exactly the negative of the slope of the concentration-vs-time curve at that instant. This definition holds for a reaction of ANY order — it is a direct consequence of the definition of rate, not something that depends on the rate law. The factor of $2.303$ only enters when working with $\log_{10}[A]$ vs $t$ plots (used to extract the first-order rate constant $k$ from the slope $-k/2.303$ ) — it is irrelevant here since the axes are plain concentration and time, not logarithmic concentration.

Step-by-Step Solution

The graph plots $[A]$ (concentration of A) on the y-axis against time on the x-axis — a decaying curve, as expected since A is being consumed.

At point C, a tangent line is drawn with slope magnitude $m$ (the curve is decreasing, so the true signed slope is $-m$ , but this magnitude labelled $m$ represents how fast concentration is falling at that instant).

By definition, instantaneous rate $= -\dfrac{d[A]}{dt}\Big|_C = m$ (taking the tangent's slope magnitude as the rate).

The mention of 'first order' is a distractor here — the rate-from-tangent relationship on a $[A]$ vs $t$ graph does not depend on reaction order; that information would only matter if the graph or question asked for the rate constant $k$ .

So the instantaneous rate at C is simply $m$ , with no $2.303$ factor needed.

Common Mistakes

Applying the $2.303$ conversion factor here — that factor belongs to $\log[A]$ vs $t$ plots used for finding $k$ in first-order kinetics, not to reading the rate off a plain $[A]$ vs $t$ curve.

Being distracted by 'first order' into thinking the answer must involve the rate constant relation $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ , which is unrelated to reading a tangent slope.

✓Final answer
$The correct option is (B) — m.$

ANSWER: B
AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$M\ s^{-1}$
›Reveal solutionSolution
This tests the general rule "unit of $k$ = $(mol\,L^{-1})^{1-n} s^{-1}$ " for a reaction of order $n$ . The false statement is (D): first order $k$ has units $s^{-1}$ , not $M\,s^{-1}$ .

Concept and Intuition

For a reaction of order $n$ : $\text{rate} = k[A]^n$ , and rate always has units $M\,s^{-1}$ . Rearranging for $k$ :
$k = \dfrac{\text{rate}}{[A]^n} \;\Rightarrow\; \text{units of }k = (M)^{1-n} s^{-1}$
So the units of $k$ change with order — this is exactly why chemists use the units of an experimentally measured $k$ to identify the reaction order.

Step-by-Step Solution

Rate of reaction / rate of disappearance of reactant: both are $\dfrac{\Delta[\text{conc}]}{\Delta t}$ , units $M\,s^{-1}$ — (A), (B) correct.

Units of $k$ depend on order: for order $n$ , units are $M^{1-n}s^{-1}$ — (C) correct.

For $n=1$ (first order): units of $k = M^{1-1}s^{-1} = M^0 s^{-1} = s^{-1}$ , not $M\,s^{-1}$ .

So statement (D), which claims $M\,s^{-1}$ for a first-order $k$ , is false — that unit actually belongs to a zero-order reaction ( $n=0 \Rightarrow M^{1}s^{-1}$ ).

Common Mistakes

Confusing the zero-order rate-constant unit ( $M\,s^{-1}$ ) with the first-order one ( $s^{-1}$ ).

Thinking rate constant units are always the same as rate units.

✓Final answer
$k$

ANSWER: D

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