Which of the following expressions is correct for the rate of reaction given below? 5Br^-(aq) + BrO₃^-(aq) + 6

NCERT Exemplar · Q12

5Br^-(aq) + BrO_3^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)

(i)

\frac{\Delta[Br^-]}{\Delta t} = 5\,\frac{\Delta[H^+]}{\Delta t}

(ii)

\frac{\Delta[Br^-]}{\Delta t} = \frac{6}{5}\,\frac{\Delta[H^+]}{\Delta t}

(iii)

\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6}\,\frac{\Delta[H^+]}{\Delta t}

(iv)

\frac{\Delta[Br^-]}{\Delta t} = 6\,\frac{\Delta[H^+]}{\Delta t}

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Concept understanding — Reaction Rate Stoichiometry

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:

$\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}$ …

Why this formula?

Reaction Rate Stoichiometry: Why the Formula Holds

Let’s start with the core idea: In a chemical reaction, the rate at which reactants disappear and products appear is not arbitrary — it is tied directly to the stoichiometric coefficients in the balanced equation.

The Key Formula

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction ( $R$ ) 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}

Where:

$[A], [B], [C], [D]$ are concentrations (in mol/L)
$t$ is time
$a, b, c, d$ are stoichiometric coefficients

Why This Formula Holds: The Reasoning

1. The Physical Meaning of Stoichiometric Coefficients

The coefficients tell us the mole ratio in which substances react or are produced. For example:

2H_2 + O_2 \rightarrow 2H_2O

2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$ .
This means: for every 2 molecules of $H_2$ that disappear, only 1 molecule of $O_2$ disappears, and 2 molecules of $H_2O$ appear.

Key insight: The number of moles changing per unit time is different for each substance, but the reaction event is the same.

2. The Problem with Raw Rates

If we simply wrote:

\text{Rate} = -\frac{d[H_2]}{dt}

This would be twice the rate of disappearance of $O_2$ (since $H_2$ disappears twice as fast). That’s inconsistent — the same reaction shouldn’t have two different numerical rates.

We need a single, unique rate that describes the reaction itself, not just one substance.

3. The Solution: Normalize by Stoichiometric Coefficients

To get a reaction rate that is the same regardless of which substance we track, we divide each substance’s rate of change by its stoichiometric coefficient.

Why division works:

If $A$ disappears at rate $-\frac{d[A]}{dt}$ , and $a$ moles of $A$ are consumed per reaction event, then the number of reaction events per unit time is:

\text{Reaction events per second} = \frac{-\frac{d[A]}{dt}}{a}

Similarly, for product $C$ appearing at rate $+\frac{d[C]}{dt}$ , with $c$ moles produced per event:

\text{Reaction events per second} = \frac{+\frac{d[C]}{dt}}{c}

Since the same reaction is happening, these must be equal. Hence:

-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

4. The Sign Convention

Reactants decrease over time → $\frac{d[\text{reactant}]}{dt} < 0$ → we add a negative sign to make the rate positive.
Products increase over time → $\frac{d[\text{product}]}{dt} > 0$ → we use a positive sign. …

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The rate of a reaction is defined per stoichiometric coefficient, so the rate of disappearance of $Br^-$ divided by 5 equals the rate of disappearance of $H^+$ divided by 6. This gives $\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6} \frac{\Delta[H^+]}{\Delta t}$ , which is option (iii).

The key idea here is that the rate of a reaction is a single, unified quantity — it doesn't depend on which reactant or product you measure, as long as you account for the stoichiometric coefficients. For the reaction

$5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$ ,

the rate can be written as:

\text{Rate} = -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{6}\frac{\Delta[H^+]}{\Delta t}

The negative signs indicate that concentrations of reactants decrease over time. Since both expressions equal the same rate, we can set them equal to each other (ignoring the negative signs, as they cancel):

\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = \frac{1}{6}\frac{\Delta[H^+]}{\Delta t}

Now multiply both sides by 5:

\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[H^+]}{\Delta t}

That matches option (iii). …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing the sign convention

The error: Students often forget that reactants have a negative sign in the rate expression. They write:

\frac{\Delta[Br^-]}{\Delta t} = +\frac{5}{6}\frac{\Delta[H^+]}{\Delta t}

But since both $Br^-$ and $H^+$ are reactants, their concentrations decrease with time — both $\Delta[Br^-]$ and $\Delta[H^+]$ are negative. The correct relationship must account for this.

How to avoid: Always write the definition of rate first:

-\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{6}\frac{\Delta[H^+]}{\Delta t}

Cancel the negative signs on both sides, then solve:

\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[H^+]}{\Delta t}

Answer: Option (iii) is correct.

Mistake 2: Inverting the stoichiometric ratio

The error: Students often write the ratio backwards — putting the coefficient of the substance they are solving for in the denominator instead of the numerator.

For example, they might write:

\frac{\Delta[Br^-]}{\Delta t} = \frac{6}{5}\frac{\Delta[H^+]}{\Delta t}

This is wrong because the rate definition gives:

-\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{6}\frac{\Delta[H^+]}{\Delta t}

Multiplying both sides by 5 gives:

\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[H^+]}{\Delta t}

How to avoid: Use the formula method:

\frac{\Delta[\text{A}]}{\Delta t} = \frac{\text{coefficient of A}}{\text{coefficient of B}} \times \frac{\Delta[\text{B}]}{\Delta t}

Here, coefficient of $Br^-$ is 5, coefficient of $H^+$ is 6, so:

\frac{\Delta[Br^-]}{\Delta t} = \frac{5}{6}\frac{\Delta[H^+]}{\Delta t}

Mistake 3: Forgetting to use the rate definition as the starting point …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2026Set eng-2026-04184 marksMCQ
$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$
›Reveal solutionSolution
Track partial pressures at 75% decomposition of $N_2O_5$ ; total $\approx 2.1$ times the initial pressure.

Reaction: $N_2O_5 \rightarrow 2NO_2 + \tfrac{1}{2}O_2$ , initial pressure $P$ (only $N_2O_5$ ).

At 75% completion, $0.75P$ of $N_2O_5$ decomposes:

$N_2O_5$ left $= 0.25P$

$NO_2$ formed $= 2\times0.75P = 1.5P$

$O_2$ formed $= 0.5\times0.75P = 0.375P$ …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$\rightarrow$
›Reveal solutionSolution

For $3A \rightarrow$ products, rate $= -\frac{1}{3}\frac{\Delta[A]}{\Delta t}$ . With $\Delta[A] = -0.3$ mol L⁻¹ over 20 min, rate $= 0.005$ mol L⁻¹ min⁻¹.

Reasoning

The stoichiometric coefficient of A is 3, so the reaction rate is

$\text{Rate} = -\frac{1}{3}\frac{\Delta[A]}{\Delta t}$ …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$2P + Q \rightleftharpoons P_2Q$
›Reveal solutionSolution
P disappears at $-0.48$ and Q at $-0.24$ mol dm $^{-3}$ s $^{-1}$ .

Concept and Intuition

Rates for reactants and products are linked by stoichiometric coefficients: rate $= -\frac{1}{2}\frac{d[P]}{dt} = -\frac{d[Q]}{dt} = +\frac{d[P_2Q]}{dt}$ .

Step-by-Step Solution

Given $\frac{d[P_2Q]}{dt} = 0.24$ mol dm $^{-3}$ s $^{-1}$ .

$-\frac{d[P]}{dt} = 2 \times 0.24 = 0.48$ (P has coefficient 2).

$-\frac{d[Q]}{dt} = 0.24$ (Q has coefficient 1). …
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Q.Which of the following expressions is correct for the rate of reaction given below?
$5Br^-(aq) + BrO_3^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)$

Concept understanding — Reaction Rate Stoichiometry

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Why this formula?

Reaction Rate Stoichiometry: Why the Formula Holds

The Key Formula

Why This Formula Holds: The Reasoning

1. The Physical Meaning of Stoichiometric Coefficients

2. The Problem with Raw Rates

3. The Solution: Normalize by Stoichiometric Coefficients

4. The Sign Convention

Method: Stoichiometric Rate Relation

Steps for this problem

Common Mistakes & How to Avoid Them

Mistake 1: Confusing the sign convention

Mistake 2: Inverting the stoichiometric ratio

Mistake 3: Forgetting to use the rate definition as the starting point …

Concept and Intuition

Step-by-Step Solution

Q.Which of the following expressions is correct for the rate of reaction given below? 5Br−(aq)+BrO3−(aq)+6H+(aq)→3Br2(aq)+3H2O(l)5Br^-(aq) + BrO_3^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)5Br−(aq)+BrO3−​(aq)+6H+(aq)→3Br2​(aq)+3H2​O(l)

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Q.Which of the following expressions is correct for the rate of reaction given below?
$5Br^-(aq) + BrO_3^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)$

Unlock everything free for 14 days