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

NCERT Exemplar · Q3

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.

Sikkim CbseMCQ· 1mImportance★★★★★

44% · 52/117 Questions

✓ Free question

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.

←Previous questionQ2 — In the presence of a catalyst, the heat evolved or absorbed during the…→Next questionQ4 — A potential-energy profile (energy on the vertical axis, reaction coor…

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

Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Common Pitfall to Avoid

Why This Matters

Why this formula?

Average Rate of Reaction — Why the Formula Holds

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

2. The core idea: change per unit time

3. Applying this to a reaction

For reactant A (disappearing):

For product B (appearing):

4. The general formula for any reaction

5. Key exam point: the formula in one line

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

Summary — the "why" in one sentence

Method: Arrhenius Equation (Two-Point Method)

Why this method works

Steps

Correct answer for the given question

Common Mistakes Students Make on This Question

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

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

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

Mistake 4: Forgetting the Arrhenius equation in logarithmic form

✓ Correct Answer

Quick Summary Table

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

🎓Unlock everything free for 14 days

Unlock everything free for 14 days