For a reaction $A + B \rightarrow$ Products, the rate law is — Rate $= k[A][B]^{3/2}$ Can the reaction be an elementary reaction? Explain.

> [!TLDR] > For an elementary reaction, the rate law must match the stoichiometric coefficients exactly. Here the rate law has fractional order ($3/2$), which is impossible for a single-step elementary process. Therefore, the reaction **cannot** be elementary. The question asks whether a reaction with the given rate law can be an elementary reaction. This is a classic test of the definition of an elementary reaction. An elementary reaction is one that occurs in a single step, exactly as written in the chemical equation. For such a reaction, the rate law is directly determined by the molecularity — the number of molecules that must collide simultaneously. The exponents in the rate law must be the stoichiometric coefficients of the reactants in that step. > [!FORMULA] > For an elementary reaction $aA + bB \rightarrow$ products, the rate law is: > $$ \text{Rate} = k[A]^a[B]^b $$ > where $a$ and $b$ are the stoichiometric coefficients (small whole numbers: 1, 2, or rarely 3). Now look at the given reaction: $A + B \rightarrow$ Products. The stoichiometric coefficients are 1 for A and 1 for B. If this were an elementary reaction, the rate law would have to be: $$ \text{Rate} = k[A]^1[B]^1 = k[A][B] $$ But the actual rate law given is: $$ \text{Rate} = k[A][B]^{3/2} $$ The exponent on $[B]$ is $3/2$, which is a fraction. This is the critical point. 1. **Fractional orders are impossible in elementary reactions.** Why? Because an elementary step represents a physical collision event. You cannot have half a molecule colliding. The number of molecules involved must be an integer — 1, 2, or 3. So the exponent must be a whole number. 2. **The exponent 3/2 is not a whole number.** This alone is sufficient to rule out an elementary reaction. Even if the exponent were a whole number but didn't match the stoichiometric coefficient (e.g., Rate $= k[A]^2[B]$), it would still not be elementary. Here, the mismatch is even more fundamental. …

For a reaction A + B arrow Products, the rate law – Class 12 Chemistry

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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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For an elementary reaction, the rate law must match the stoichiometric coefficients exactly. Here the rate law has fractional order ( $3/2$ ), which is impossible for a single-step elementary process. Therefore, the reaction cannot be elementary.

The question asks whether a reaction with the given rate law can be an elementary reaction. This is a classic test of the definition of an elementary reaction.

An elementary reaction is one that occurs in a single step, exactly as written in the chemical equation. For such a reaction, the rate law is directly determined by the molecularity — the number of molecules that must collide simultaneously. The exponents in the rate law must be the stoichiometric coefficients of the reactants in that step.

aA + bB \rightarrow

Now look at the given reaction: $A + B \rightarrow$ Products. The stoichiometric coefficients are 1 for A and 1 for B. If this were an elementary reaction, the rate law would have to be:

\text{Rate} = k[A]^1[B]^1 = k[A][B]

But the actual rate law given is:

\text{Rate} = k[A][B]^{3/2}

The exponent on $[B]$ is $3/2$ , which is a fraction. This is the critical point.

Fractional orders are impossible in elementary reactions. Why? Because an elementary step represents a physical collision event. You cannot have half a molecule colliding. The number of molecules involved must be an integer — 1, 2, or 3. So the exponent must be a whole number.
The exponent 3/2 is not a whole number. This alone is sufficient to rule out an elementary reaction. Even if the exponent were a whole number but didn't match the stoichiometric coefficient (e.g., Rate $= k[A]^2[B]$ ), it would still not be elementary. Here, the mismatch is even more fundamental. …

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Here are the common mistakes students make with this question, along with how to avoid each.

Mistake 1: Confusing "Molecularity" with "Order"

The Mistake: Students see the rate law $\text{Rate} = k[A][B]^{3/2}$ and immediately say "The order is $1 + 1.5 = 2.5$ , so it cannot be elementary because the molecularity must be a whole number."

While the conclusion is correct, the reasoning is often incomplete or muddled. They forget that molecularity (the number of molecules colliding) must be a positive integer (1, 2, or rarely 3), while order is an experimental value that can be fractional.

How to Avoid:

Define both terms clearly before answering.
- Molecularity: The number of reactant particles taking part in an elementary step. It is always a whole number (1, 2, or 3).
- Order: The sum of the exponents in the experimentally determined rate law. It can be fractional, zero, or negative.
Connect them correctly: For an elementary reaction, the order equals the molecularity for each reactant. Therefore, the exponents in the rate law must be integers (1, 2, or 3) that sum to a small integer.

Mistake 2: Ignoring the Fractional Exponent

The Mistake: Students see the $[B]^{3/2}$ term and think "It's just a power, maybe it's possible." They fail to recognize that a fractional exponent ( $3/2$ ) is a red flag for an elementary reaction.

How to Avoid:

Memorize the rule: For an elementary reaction, the exponents in the rate law are the stoichiometric coefficients of the balanced equation. Since coefficients are whole numbers, the exponents must be whole numbers.
Spot the contradiction: The exponent $3/2$ means the reaction would require 1.5 molecules of B to collide simultaneously. This is physically impossible. A molecule is a discrete entity; you cannot have half a molecule in a collision.

Mistake 3: Forgetting to State the "Why"

The Mistake: Students simply answer "No" without explaining why the fractional order makes it impossible. In an exam, the explanation is worth more marks than the yes/no answer.

How to Avoid:

Structure your answer in two parts:
1. Conclusion: No, it cannot be an elementary reaction. …

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Q.For a reaction $A + B \rightarrow$ Products, the rate law is — Rate $= k[A][B]^{3/2}$
Can the reaction be an elementary reaction? Explain.

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

Method: Elementary Reaction Check Using Molecularity and Rate Law Consistency

Concept First — Why This Method Works

Steps

Mistake 1: Confusing "Molecularity" with "Order"

Mistake 2: Ignoring the Fractional Exponent

Mistake 3: Forgetting to State the "Why"

Q.For a reaction A+B→A + B \rightarrowA+B→ Products, the rate law is — Rate =k[A][B]3/2= k[A][B]^{3/2}=k[A][B]3/2 Can the reaction be an elementary reaction? Explain.

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Q.For a reaction $A + B \rightarrow$ Products, the rate law is — Rate $= k[A][B]^{3/2}$
Can the reaction be an elementary reaction? Explain.

Unlock everything free for 14 days