State a condition under which a bimolecular reaction is kinetically first order reaction.

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Q. State a condition under which a bimolecular reaction is kinetically first order reaction.

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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.

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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):

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

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A bimolecular reaction becomes kinetically first order when one reactant is present in such large excess that its concentration remains effectively constant throughout the reaction. The observed rate then depends only on the concentration of the other reactant.

The core idea: Why excess matters

A bimolecular reaction involves two molecules colliding. The rate law is typically second order — first order in each reactant. But if one reactant is so abundant that its concentration barely changes as the reaction proceeds, we can treat it as a constant. That constant gets absorbed into the rate constant, and the reaction appears to follow first-order kinetics with respect to the other reactant.

This is not a trick — it’s a practical reality in many chemical systems. Think of a hydrolysis reaction where water is the solvent. Water is present at ~55 M, and the other reactant might be at 0.1 M. Even if the reaction consumes some water, the change is negligible relative to the total. So the rate depends only on the concentration of the dilute reactant.

Step-by-step reasoning

Write the general bimolecular reaction Consider:

A + B \rightarrow \text{products}

The true rate law (assuming an elementary step) is:

\text{Rate} = k[A][B]

This is second order overall — first order in $A$ and first order in $B$ .

Identify the condition for pseudo-first-order behaviour Suppose $[B]_0 \gg [A]_0$ . As the reaction proceeds, $[B]$ changes so little that we can approximate it as constant:

[B] \approx [B]_0

This is valid when the initial concentration of $B$ is at least 10–20 times that of $A$ , though in practice much larger excess is common.

Simplify the rate law Substitute the constant $[B]_0$ into the rate expression:

\text{Rate} = k[A][B]_0 = k'[A]

where $k' = k[B]_0$ is the observed or pseudo-first-order rate constant.

The reaction now follows:

\text{Rate} = k'[A]

which is first order in $A$ .

Confirm the kinetics The integrated form becomes: …

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Common Mistakes: Bimolecular Reaction Being Kinetically First Order

The Core Concept

A bimolecular reaction involves two molecules colliding (e.g., $A + B \rightarrow \text{products}$ ). Normally, its rate law is second order:

\text{Rate} = k[A][B]

But it becomes kinetically first order when one reactant is in large excess — its concentration remains nearly constant, so the rate depends only on the other reactant.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Confusing "bimolecular" with "first order" inherently

What students do: Assume all bimolecular reactions are second order, or that "bimolecular" automatically means first order.
Why it's wrong: Bimolecular refers to the molecularity (number of molecules in the elementary step), not the order (experimentally determined exponent). They are different concepts.
How to avoid: Remember:
- Molecularity = number of reacting species in the rate-determining step (always a whole number, ≤ 3).
- Order = sum of exponents in the rate law (can be fractional, zero, etc.).
- A bimolecular step can appear first order if one reactant's concentration is effectively constant.

✗ Mistake 2: Saying "when one reactant is in excess" without specifying "large excess"

What students do: Write "when one reactant is in excess" — this is vague and often incomplete.
Why it's wrong: "Excess" alone doesn't guarantee the concentration change is negligible. Only large excess (e.g., 10–100 times) makes the concentration nearly constant over the reaction time.
How to avoid: Always use the phrase "large excess" or "in great excess" in your answer. Example:

"When one reactant is present in large excess, its concentration remains approximately constant, so the rate depends only on the other reactant, giving pseudo-first-order kinetics."

✗ Mistake 3: Forgetting to mention "pseudo-first-order" condition

What students do: State the condition but don't use the correct terminology.
Why it's wrong: The exam expects the term "pseudo-first-order reaction" — this is the standard name for such a case.
How to avoid: Learn and use the exact phrase:

"A bimolecular reaction becomes kinetically first order under pseudo-first-order conditions, i.e., when one reactant is in large excess."

✗ Mistake 4: Writing an incorrect rate law

What students do: Write $\text{Rate} = k[A]$ without explaining why $[B]$ disappears.
Why it's wrong: The original rate law is $\text{Rate} = k[A][B]$ . If $[B]$ is in large excess, $[B] \approx \text{constant}$ , so $\text{Rate} = k'[A]$ where $k' = k[B]_0$ . Students often forget to define the new rate constant.
How to avoid: Show the derivation clearly:

\text{Rate} = k[A][B] \quad \text{and if } [B] \gg [A], \text{ then } [B] \approx [B]_0

\Rightarrow \text{Rate} = k[A][B]_0 = k'[A] \quad \text{where } k' = k[B]_0

✗ Mistake 5: Giving an example without checking if it's truly bimolecular

What students do: Use hydrolysis reactions (e.g., ester hydrolysis) as examples, but these are often not elementary bimolecular steps — they may involve multiple steps. …

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

KCET 2025Set D-41 markMCQ
$E_a$
›Reveal solutionSolution
On an energy profile the forward and backward barriers are both measured to the same peak, so $E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H = 215 - 90 = 125$ kJ.

Step 1 — Read the graph.

From the described energy profile:

The reactants sit at the lower level.

The curve rises to a peak (the activated complex / transition state).

The products sit at a higher level than the reactants — so the reaction is endothermic.

$E_a(\text{forward}) = 215\ \mathrm{kJ}$ — measured from reactants up to the peak.

$\Delta H = +90\ \mathrm{kJ}$ — measured from reactants up to products.

Step 2 — Set an energy scale.

Put the reactants at zero:
$E(\text{reactants}) = 0\ \mathrm{kJ}$ $E(\text{peak}) = 0 + E_a(f) = 215\ \mathrm{kJ}$ $E(\text{products}) = 0 + \Delta H = 90\ \mathrm{kJ}$
Step 3 — The reverse activation energy.

Running the reaction backwards, the molecules start at the products and must climb to the same peak (the transition state is common to both directions — this is the principle of microscopic reversibility):
$E_a(\text{reverse}) = E(\text{peak}) - E(\text{products}) = 215 - 90 = 125\ \mathrm{kJ}$
Step 4 — Verify with the general relation.

For any reaction, …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{A}+3 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$
›Reveal solutionSolution

Rate of reaction $=4.17\times10^{-5}$ , and rate of formation of $\text{C}$ is twice this $=8.32\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$ . Official key: (B).

Working

For $\text{A} + 3\text{B} \rightarrow 2\text{C} + \text{D}$ , the rate of reaction is
$\text{Rate} = -\frac{d[\text{A}]}{dt} = +\frac{1}{2}\frac{d[\text{C}]}{dt}$
The change in $[\text{A}]$ over $1$ minute ( $=60\ \text{s}$ ):
$-\frac{\Delta[\text{A}]}{\Delta t} = \frac{0.0150 - 0.0125}{60} = \frac{0.0025}{60} = 4.17\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$
Since $\text{C}$ has a stoichiometric coefficient of $2$ , its rate of formation is twice the rate of reaction: …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{BrO}_3^{-} \text {(aq) }+5 \mathrm{Br}^{-} \text {(aq) }+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_2(\mathrm{l})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
›Reveal solutionSolution
The rate of appearance of a product and the rate of disappearance of a reactant are linked by stoichiometric coefficients, with opposite signs. For the given reaction, the correct relation is $\frac{d[\mathrm{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\mathrm{Br}^-]}{dt}$ , which corresponds to option (D).

The key idea is that for a chemical reaction, the rate can be expressed in terms of any reactant or product, but we must account for stoichiometry and sign conventions. The rate of disappearance of a reactant is negative (its concentration decreases), while the rate of appearance of a product is positive (its concentration increases). To relate them, we divide each rate by its stoichiometric coefficient and set them equal in magnitude, then adjust signs.

Let’s work through it step by step.

Write the general rate expression. For a reaction $aA + bB \rightarrow cC + dD$ , the rate is defined as:

$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$
The negative signs for reactants ensure the rate is positive (since $\frac{d[\text{reactant}]}{dt}$ is negative).

Identify the species and coefficients. In the given reaction:

$\mathrm{BrO}_3^- + 5\mathrm{Br}^- + 6\mathrm{H}^+ \rightarrow 3\mathrm{Br}_2 + 3\mathrm{H}_2\mathrm{O}$
We care about $\mathrm{Br}^-$ (reactant, coefficient 5) and $\mathrm{Br}_2$ (product, coefficient 3).

Set up the equality from the general rate. Using the definition:

$-\frac{1}{5}\frac{d[\mathrm{Br}^-]}{dt} = \frac{1}{3}\frac{d[\mathrm{Br}_2]}{dt}$
The left side has a negative sign because $\mathrm{Br}^-$ is a reactant; the right side has no negative sign because $\mathrm{Br}_2$ is a product.

Solve for the desired relation. Multiply both sides by 3:
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COMEDK 2024Set 2024-E1 markMCQ
$5 X+Y \rightarrow 3 Z$
›Reveal solutionSolution
The key idea is that reaction rates for different species are linked by stoichiometric coefficients. For the reaction $5X + Y \rightarrow 3Z$ , the rate of disappearance of X is $\frac{5}{3}$ times the rate of formation of Z, giving $4.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$ .

The core concept here is stoichiometric rate relationships. In a chemical reaction, the rate at which reactants disappear and products appear are not independent — they are tied together by the coefficients in the balanced equation. This is because for every 5 molecules of X that react, 3 molecules of Z are formed. So the rate of change of concentration for each species must be scaled by its coefficient to give the same "overall reaction rate."

Why this works:

If we define the rate of reaction as $r = -\frac{1}{5}\frac{d[X]}{dt} = -\frac{1}{1}\frac{d[Y]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$ , then knowing any one of these rates lets us find the others by simple multiplication or division by the appropriate coefficient ratio.

Write the general rate relationship For the reaction $5X + Y \rightarrow 3Z$ , the rate of reaction $r$ is:

$r = -\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
The negative sign indicates disappearance (decrease in concentration), while positive indicates formation.

Identify what we know We are given the rate of formation of Z:

$\frac{d[Z]}{dt} = 2.4 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$

Relate the rate of disappearance of X to the rate of formation of Z From the equality above:

$-\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
Multiply both sides by $-5$ :

$\frac{d[X]}{dt} = -\frac{5}{3}\frac{d[Z]}{dt}$ …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. The equation showing the decomposition of Dinitrogen pentoxide is as follows : N2O5(g) -> N2O4(g) + 1/2 O2(g) If the initial pressure is 250 mm and after half an hour of the reaction the total pressure of the gaseous mixture is 296 mm, what is the average rate of the reaction ? (A) 4.04 * 10^-3 atm min^-1 (B) -2.02 * 10^-3 atm min^-1 (C) 2.19 * 10^-2 atm min^-1 (D) 1.09 * 10^-2 atm min^-1$
›Reveal solutionSolution

The pressure rise fixes the N $_2$ O $_5$ consumed at 92 mm; converting to atm over 30 min gives $4.04\times10^{-3}$ atm min $^{-1}$ .

Pressure bookkeeping ( $\text{N}_2\text{O}_5 \to \text{N}_2\text{O}_4 + \tfrac12\text{O}_2$ ), let $x$ = mm of N $_2$ O $_5$ reacted:
$P_{total} = (250-x) + x + \tfrac{x}{2} = 250 + \tfrac{x}{2}$
$250 + \tfrac{x}{2} = 296 \Rightarrow x = 92\ \text{mm}$ …
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KCET 2020Set A-11 markMCQ
$k = PZe^{-E_a/RT}$
›Reveal solutionSolution
The rate constant $k$ increases when the activation energy $E_a$ is decreased, because $k$ depends exponentially on $-E_a/RT$ ; the correct answer is (D).

The Arrhenius equation $k = A e^{-E_a/RT}$ (where $A = PZ$ in the given notation) tells us exactly how temperature and activation energy control reaction speed. The exponential term $e^{-E_a/RT}$ is the key: it represents the fraction of molecules that have enough energy to overcome the activation barrier. To speed up a reaction, we want $k$ to be larger — so we need that exponential term to be as large as possible.

Since the exponent is negative, making $E_a$ smaller makes the exponent less negative, which increases $e^{-E_a/RT}$ . Making $T$ larger also makes the exponent less negative (because $RT$ is in the denominator), so increasing temperature speeds up the reaction. But the question asks which factor has to be decreased to speed up the reaction.

Let’s check each option carefully.

Option (A): $T$ — Temperature appears in the denominator of the exponent. Decreasing $T$ makes $E_a/RT$ larger, so the exponent becomes more negative, and $k$ gets smaller. That slows the reaction down, not speeds it up. So this is wrong.

Option (B): $Z$ — $Z$ is the collision frequency factor (part of the pre-exponential term $PZ$ ). It multiplies the exponential directly: $k = (PZ) e^{-E_a/RT}$ . Decreasing $Z$ reduces the pre-factor, which reduces $k$ . That also slows the reaction. So this is wrong too.

Option (C): Both $Z$ and $T$ — Since decreasing either one slows the reaction, decreasing both certainly won’t speed it up. This is incorrect. …
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KCET 2018Set A-11 markMCQ
$30^\circ \mathrm{C}$
›Reveal solutionSolution

Rate ratio $= (\text{temperature coefficient})^{\Delta T/10} = 2^{60/10} = 2^{6} = 64$ .

Step 1 — Meaning of the temperature coefficient.

The temperature coefficient $\mu$ of a reaction is defined as the ratio of the rate constants for a $10^\circ$ C rise:
$\mu = \frac{k_{T+10}}{k_{T}}$
For most reactions $\mu \approx 2$ – $3$ (the familiar 'rate roughly doubles for every $10^\circ$ C rise' rule). Here $\mu = 2$ .

Step 2 — Count the $10^\circ$ intervals.
$\Delta T = 90^\circ\mathrm{C} - 30^\circ\mathrm{C} = 60^\circ\mathrm{C}$ $n = \frac{\Delta T}{10} = \frac{60}{10} = 6$
Step 3 — Compound the factor.

The effect is multiplicative, not additive — each successive $10^\circ$ step multiplies the rate by 2 again: …
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Q.State a condition under which a bimolecular reaction is kinetically first order reaction.

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)

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

The core idea: Why excess matters

Step-by-step reasoning

Concept: Pseudo-First Order Reactions

Method: Isolation Method (or Flooding Technique)

Condition

Steps

Key Result …

Common Mistakes: Bimolecular Reaction Being Kinetically First Order

The Core Concept

Common Mistakes & How to Avoid Them

✗ Mistake 1: Confusing "bimolecular" with "first order" inherently

✗ Mistake 2: Saying "when one reactant is in excess" without specifying "large excess"

✗ Mistake 3: Forgetting to mention "pseudo-first-order" condition

✗ Mistake 4: Writing an incorrect rate law

✗ Mistake 5: Giving an example without checking if it's truly bimolecular

Q.State a condition under which a bimolecular reaction is kinetically first order reaction.

The core idea: Why excess matters

Step-by-step reasoning

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