With the help of an example explain what is meant by pseudo first order reaction.

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Q. With the help of an example explain what is meant by pseudo 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.

--- …

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 pseudo first order reaction is a reaction that is actually of higher order but behaves like a first order reaction because one reactant is present in such large excess that its concentration remains effectively constant throughout the reaction. The classic example is the acid‑catalysed hydrolysis of ethyl acetate: $\ce{CH3COOC2H5 + H2O ->[H+] CH3COOH + C2H5OH}$ , where water is in vast excess.

The core idea — why a reaction can “pretend” to be first order

In chemical kinetics, the order of a reaction is determined experimentally from the rate law. A true first order reaction has a rate that depends linearly on the concentration of a single reactant: $\text{rate} = k[A]$ . But what if the actual rate law is more complicated — say, $\text{rate} = k[A][B]$ — yet under certain conditions it looks like $\text{rate} = k_{\text{obs}}[A]$ ? That is exactly what a pseudo first order reaction is.

The trick is to make one reactant so abundant that its concentration barely changes as the reaction proceeds. Because it is effectively constant, it can be absorbed into the rate constant, and the rate appears to depend only on the other reactant.

A + B \rightarrow \text{products}

Step‑by‑step: the hydrolysis of ethyl acetate

Let’s walk through the classic example that every Indian syllabus (CBSE, ISC, JEE, NEET) uses.

1. Write the balanced reaction

\ce{CH3COOC2H5 + H2O ->[H+] CH3COOH + C2H5OH}

The acid ( $\ce{H+}$ ) is a catalyst — it speeds up the reaction but does not appear in the stoichiometric equation as a consumed reactant. The actual reactants are ethyl acetate and water.

2. Write the expected rate law

For a bimolecular reaction between ethyl acetate and water, the rate law should be second order overall:

\text{rate} = k[\ce{CH3COOC2H5}][\ce{H2O}]

This is the true, underlying rate law.

3. Identify the condition that creates the “pseudo” behaviour

In a typical experiment, we take a small amount of ethyl acetate (say 0.1 M) and a huge amount of water. Water is the solvent — its concentration in pure water is about 55.5 M. Even if some water is consumed, the change is negligible relative to 55.5 M. So throughout the reaction:

[\ce{H2O}] \approx \text{constant}

4. Absorb the constant into a new rate constant

Because $[\ce{H2O}]$ does not change, we can combine it with the true rate constant $k$ :

k_{\text{obs}} = k[\ce{H2O}]

Now the rate law becomes:

\text{rate} = k_{\text{obs}}[\ce{CH3COOC2H5}]

This is mathematically identical to a first order rate law. The reaction is pseudo first order — it is not truly first order, but it behaves as one under these conditions.

Watch out

A common mistake is to think that pseudo first order means the reaction is first order. It is not — the true order is higher. The “pseudo” label reminds us that the apparent order is an artefact of the experimental conditions. If you change the concentration of the excess reactant significantly, the rate law will revert to its true form.

5. Confirm by experiment …

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Common Mistakes: Pseudo First Order Reactions

Students often lose marks on this concept due to a few recurring errors. Here's what to watch out for — and how to avoid each.

Mistake 1: Confusing "Pseudo" with "True" First Order

The error:

Students think a pseudo first order reaction is a genuine first order reaction. They forget that the reaction is actually bimolecular (second order overall), but one reactant is in such large excess that its concentration remains nearly constant.

How to avoid:

Always remember the definition:

A pseudo first order reaction is a reaction that is bimolecular (second order) but behaves like a first order reaction because one reactant is present in large excess.

Example:

Hydrolysis of ethyl acetate in acidic medium:

\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}

Here, water is the solvent — present in huge excess. So its concentration $[\text{H}_2\text{O}]$ is effectively constant. The rate law becomes:

\text{Rate} = k'[\text{ester}]

where $k' = k[\text{H}_2\text{O}]$ — a pseudo first order rate constant.

Mistake 2: Forgetting to Mention the "Large Excess" Condition

The error:

Students give the example but don't explicitly state why the reaction becomes pseudo first order. They skip the key condition: one reactant's concentration is so large that it doesn't change measurably.

How to avoid:

Always write:

"Since $[\text{H}_2\text{O}] \gg [\text{ester}]$ , the concentration of water remains almost constant throughout the reaction. Hence, it is absorbed into the rate constant."

Mistake 3: Writing the Wrong Rate Law Expression

The error:

Students write the rate law as:

\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]

and then claim it's first order — which is contradictory.

How to avoid:

Write the true rate law first, then simplify:

\text{Rate} = k[\text{ester}][\text{H}_2\text{O}]

Since $[\text{H}_2\text{O}]$ is constant, let $k' = k[\text{H}_2\text{O}]$ , giving:

\text{Rate} = k'[\text{ester}]

This is the pseudo first order form.

Mistake 4: Not Giving a Clear, Exam-Ready Example

The error:

Students give vague examples (e.g., "any reaction with excess water") without naming the reactants or writing the equation.

How to avoid:

Memorise one standard example with the full balanced equation:

Example: Acid-catalysed hydrolysis of methyl acetate …

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Q.With the help of an example explain what is meant by pseudo 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 a reaction can “pretend” to be first order

Step‑by‑step: the hydrolysis of ethyl acetate

Method: Identifying and Deriving a Pseudo First Order Rate Law

Steps

Pseudo First Order Reaction — Explained with Example

What it means

Example: Hydrolysis of Ethyl Acetate

Common Mistakes: Pseudo First Order Reactions

Mistake 1: Confusing "Pseudo" with "True" First Order

Mistake 2: Forgetting to Mention the "Large Excess" Condition

Mistake 3: Writing the Wrong Rate Law Expression

Mistake 4: Not Giving a Clear, Exam-Ready Example

Q.With the help of an example explain what is meant by pseudo first order reaction.

The core idea — why a reaction can “pretend” to be first order

Step‑by‑step: the hydrolysis of ethyl acetate

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