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

Concept: Average Rate of Reaction — The rate depends on the concentration of both reactants, but if one reactant is in large excess, its concentration remains nearly constant.

Reasoning:

1. For a bimolecular reaction $A + B \rightarrow \text{products}$, the rate law is typically $r = k[A][B]$.
2. If $[B] \gg [A]$, then $[B]$ changes negligibly during the reaction and can be treated as constant. …

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

1. Write the general bimolecular reaction Consider:

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

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

2. 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:

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.

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

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

The reaction now follows:

which is first order in $A$.

4. Confirm the kinetics The integrated form becomes: …

Concept: Pseudo-First Order Reactions

A bimolecular reaction involves two reactants, so its expected rate law is second order overall. However, under a specific condition, it behaves like a first-order reaction.

Method: Isolation Method (or Flooding Technique)

Condition

One reactant is taken in large excess compared to the other.

When the concentration of one reactant is so high that it remains effectively constant throughout the reaction, the rate depends only on the concentration of the other reactant.

Steps

1. Write the bimolecular reaction Example:

2. Write the expected rate law

3. Apply the condition Let $[B] \gg [A]$. Since $[B]$ changes negligibly, treat it as constant:

4. Define an effective rate constant

where $k'$ is the pseudo-first-order rate constant.

5. Write the simplified rate law

This is now kinetically first order in $A$.

Key Result …

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:

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:

✗ 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. …