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.

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 $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$ . …
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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: $\dfrac{|x-y|}{100}$ . …
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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 $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ . …
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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 $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ . …
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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). …
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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 $M\,s^{-1}$ . …
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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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