Compounds 'A' and 'B' react according to the following chemical equation. $A(g) + 2B(g) \rightarrow 2C(g)$ Concentration of either 'A' or 'B' were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction. Experiment 1: initial [A] $= 0.30$ mol L$^{-1}$, initial [B] $= 0.30$ mol L$^{-1}$, initial rate of formation of [C] $= 0.10$ mol L$^{-1}$ s$^{-1}$ Experiment 2: initial [A] $= 0.30$ mol L$^{-1}$, initial [B] $= 0.60$ mol L$^{-1}$, initial rate of formation of [C] $= 0.40$ mol L$^{-1}$ s$^{-1}$ Experiment 3: initial [A] $= 0.60$ mol L$^{-1}$, initial [B] $= 0.30$ mol L$^{-1}$, initial rate of formation of [C] $= 0.20$ mol L$^{-1}$ s$^{-1}$ (i) Rate $= k[A]^2[B]$ (ii) Rate $= k[A][B]^2$ (iii) Rate $= k[A][B]$ (iv) Rate $= k[A]^2[B]^0$

Compounds 'A' and 'B' react according to the following chemical equation. A(g) + 2B(g) arrow 2C(g) Concentrati

NCERT Exemplar · Q17

A(g) + 2B(g) \rightarrow 2C(g)

(i) Rate

= k[A]^2[B]

(ii) Rate

= k[A][B]^2

(iii) Rate

= k[A][B]

(iv) Rate

= k[A]^2[B]^0

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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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The rate law is found by comparing experiments where one reactant’s concentration is held constant. Doubling [B] quadruples the rate (order 2 in B), while doubling [A] doubles the rate (order 1 in A). The correct rate equation is Rate = k[A][B]², which is option (ii).

The average rate of a reaction tells us how fast reactants disappear or products appear. But to write a rate law — an equation linking rate to concentrations — we need to find the order with respect to each reactant. The order is not given by the stoichiometric coefficients; it must be determined experimentally. Here, the data is set up perfectly: in each pair of experiments, only one concentration changes while the other stays constant. That lets us isolate the effect of each reactant.

Let’s work through it step by step.

Write the general rate law. The rate of formation of C depends on [A] and [B] raised to some unknown powers:

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

We need to find $m$ and $n$ .

Find the order with respect to B.

Compare Experiment 1 and Experiment 2. Here, [A] is constant at 0.30 mol L⁻¹, while [B] doubles from 0.30 to 0.60 mol L⁻¹.
- Rate in Exp 1: $0.10$ mol L⁻¹ s⁻¹
- Rate in Exp 2: $0.40$ mol L⁻¹ s⁻¹ The rate increases by a factor of $0.40 / 0.10 = 4$ . Since [B] doubled and the rate quadrupled, we have $2^n = 4$ , so $n = 2$ . The reaction is second order in B.
Find the order with respect to A.

Compare Experiment 1 and Experiment 3. Here, [B] is constant at 0.30 mol L⁻¹, while [A] doubles from 0.30 to 0.60 mol L⁻¹.
- Rate in Exp 1: $0.10$ mol L⁻¹ s⁻¹
- Rate in Exp 3: $0.20$ mol L⁻¹ s⁻¹ The rate increases by a factor of $0.20 / 0.10 = 2$ . …

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Common Mistakes in Average Rate of Reaction Problems

Mistake 1: Confusing "Rate of Formation of C" with "Rate of Reaction"

The error: Students directly use the given rate ( $0.10$ , $0.40$ , $0.20$ mol L $^{-1}$ s $^{-1}$ ) as the rate of reaction without adjusting for stoichiometric coefficients.

Why it's wrong: The rate of formation of C is not the same as the rate of disappearance of A or B. From the equation:

A(g) + 2B(g) \rightarrow 2C(g)

The relationship is:

-\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} = \frac{1}{2}\frac{d[C]}{dt}

So the rate of reaction $= \frac{1}{2} \times \text{(rate of formation of C)}$

How to avoid: Always write the stoichiometric relationship first. For every problem, ask: "Is the given rate the rate of reaction or the rate of formation/disappearance of a specific species?"

Mistake 2: Forgetting to convert the given rate before comparing experiments

The error: Students compare raw values like $0.10$ and $0.40$ directly to find order with respect to B.

Why it's wrong: The rate law uses rate of reaction, not rate of formation of C. You must divide each given rate by 2 first.

Correct approach:

Experiment 1: Rate of reaction $= 0.10/2 = 0.05$ mol L $^{-1}$ s $^{-1}$
Experiment 2: Rate of reaction $= 0.40/2 = 0.20$ mol L $^{-1}$ s $^{-1}$
Experiment 3: Rate of reaction $= 0.20/2 = 0.10$ mol L $^{-1}$ s $^{-1}$

How to avoid: Make a habit of writing "Rate of reaction = ?" as your first step, then calculate it explicitly.

Mistake 3: Incorrectly identifying which experiments to compare

The error: Students compare experiments where both concentrations change, making it impossible to isolate the effect of one reactant.

Why it's wrong: To find order with respect to B, you need experiments where [A] is constant and [B] changes. Comparing Exp 1 and Exp 3 (where both change) gives no useful information.

Correct pairs:

Order w.r.t. B: Compare Exp 1 and Exp 2 ([A] constant at $0.30$ , [B] doubles from $0.30$ to $0.60$ )
Order w.r.t. A: Compare Exp 1 and Exp 3 ([B] constant at $0.30$ , [A] doubles from $0.30$ to $0.60$ )

How to avoid: Before calculating, circle the constant concentration in each pair. If no concentration is constant, that pair is useless for finding order.

Mistake 4: Misinterpreting the rate change when concentration doubles

The error: When [B] doubles and rate quadruples (from $0.05$ to $0.20$ ), students incorrectly conclude order = 1 (because $2^1 = 2$ ).

Why it's wrong: The rate increased by a factor of $4$ , not $2$ . Since $2^{\text{order}} = \text{factor change in rate}$ :

2^n = \frac{0.20}{0.05} = 4 \implies n = 2

So order with respect to B is 2, not 1.

Correct analysis:

For B: $[B] \times 2$ , rate $\times 4 \implies$ order = 2
For A: $[A] \times 2$ , rate $\times 2 \implies$ order = 1 …

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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: Initial Rates Method

Steps

Common Mistakes in Average Rate of Reaction Problems

Mistake 1: Confusing "Rate of Formation of C" with "Rate of Reaction"

Mistake 2: Forgetting to convert the given rate before comparing experiments

Mistake 3: Incorrectly identifying which experiments to compare

Mistake 4: Misinterpreting the rate change when concentration doubles

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