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}

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

52% · 61/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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

AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$2A \to$
›Reveal solutionSolution
The rate of reaction must be defined per mole of reactant consumed, using the stoichiometric coefficient of A; the answer is $5\times10^{-3}$ mol L⁻¹ min⁻¹.

Concept and Intuition

For a reaction $2A \to$ products, the rate of reaction is not simply the rate of disappearance of A — it must be divided by A's stoichiometric coefficient so that the rate is the same regardless of which species you track: $\text{Rate} = -\dfrac{1}{2}\dfrac{d[A]}{dt}$ .

Step-by-Step Solution

Change in concentration of A: $\Delta[A] = 0.4-0.5 = -0.1$ M over $\Delta t = 10$ min.

Rate of disappearance of A $= \dfrac{0.1}{10} = 0.01$ mol L⁻¹ min⁻¹. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$N_2O_5$
›Reveal solutionSolution
From the O $_2$ evolved, back-calculate the N $_2$ O $_5$ consumed via stoichiometry, giving the unreacted concentration $X=1.5$ M and the average rate of N $_2$ O $_5$ disappearance $Y=1.66\times10^{-2}$ mol L $^{-1}$ min $^{-1}$ .

Concept and Intuition

For a reaction with known stoichiometry, measuring how much of ONE product forms lets you find how much reactant disappeared, using mole ratios from the balanced equation. Average rate over a finite time interval is simply $\Delta(\text{concentration})/\Delta t$ , referenced to whichever species' stoichiometric coefficient is being tracked (here, N $_2$ O $_5$ , coefficient 1).

Step-by-Step Solution

Initial moles of $N_2O_5$ : $50\,\text{mL} \times 2\,M = 0.1$ mol; initial concentration $=2$ M.

Moles of $O_2$ formed at STP: $\dfrac{0.28\,L}{22.4\,L/mol} = 0.0125$ mol.

From $N_2O_5 \rightarrow N_2O_4 + \tfrac12 O_2$ : every 1 mol $O_2$ formed corresponds to 2 mol $N_2O_5$ consumed. So $N_2O_5$ consumed $= 2\times0.0125 = 0.025$ mol.

Unreacted $N_2O_5$ $= 0.1-0.025 = 0.075$ mol; in 50 mL this is $X = 0.075/0.050 = 1.5$ M. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{A + 2B \longrightarrow 3C + 2D}$
›Reveal solutionSolution
Using the stoichiometric definition of "rate of reaction" for A + 2B → 3C + 2D, the ratio of the overall rate to the rate of appearance of C is 1:3. Answer: (A).

Concept and Intuition

For a general reaction, the single unambiguous "rate of reaction" is defined by dividing each species' rate of change by its own stoichiometric coefficient: $\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt}$ . This ensures the reported rate is the same number regardless of which species you measure, since a species with a larger coefficient appears/disappears proportionally faster.

Step-by-Step Solution

For $A + 2B \rightarrow 3C + 2D$ , the rate of reaction is $\text{Rate} = -\dfrac{d[A]}{dt} = -\dfrac{1}{2}\dfrac{d[B]}{dt} = \dfrac{1}{3}\dfrac{d[C]}{dt} = \dfrac{1}{2}\dfrac{d[D]}{dt}$ .

So Rate of appearance of C $= \dfrac{d[C]}{dt} = 3 \times \text{Rate of reaction}$ .

Therefore, $\dfrac{\text{Rate of reaction}}{\text{Rate of appearance of C}} = \dfrac{\text{Rate}}{3\,\text{Rate}} = \dfrac{1}{3}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$_3$
›Reveal solutionSolution
Heating MHCO3 releases CO2 (half a mole per mole of MHCO3); that CO2 reacts 1:1 with MOH to regenerate MHCO3, so working backward from 0.2 mol MOH gives $x=33.6$ g.

Concept and Intuition

Metal bicarbonates decompose on heating to the carbonate, releasing water and carbon dioxide: $2MHCO_3 \rightarrow M_2CO_3 + H_2O + CO_2$ . That released CO2, when passed into a metal hydroxide solution in a 1:1 ratio (as opposed to a 1:2 ratio which would give the carbonate), reforms the bicarbonate: $CO_2+MOH\rightarrow MHCO_3$ . This is analogous to the well-known $CO_2+NaOH\rightarrow NaHCO_3$ reaction (in excess CO2).

Step-by-Step Solution

From decomposition, moles $CO_2$ produced $=\tfrac12\times$ (moles $MHCO_3$ initially).

From the CO2 + MOH reaction (1:1 stoichiometry), moles $CO_2$ = moles $MOH$ reacted completely $= 0.2$ mol.

So moles of original $MHCO_3 = 2\times0.2=0.4$ mol. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$-\dfrac{\Delta[Br^-]}{\Delta t}$
›Reveal solutionSolution
Dividing the rate of change of each species by its stoichiometric coefficient must give the same overall rate; since Br⁻ has coefficient 5, the reaction rate equals x/5.

Concept and Intuition

For a reaction aA + bB → cC + dD, the rate of reaction is defined so that it's independent of which species you track: Rate = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = ... Each species' own rate of change must be divided by its stoichiometric coefficient.

Step-by-Step Solution

Reaction: 5Br⁻ + BrO3⁻ + 6H⁺ → 3Br2 + 3H2O, so Br⁻ has stoichiometric coefficient 5.

Given: −Δ[Br⁻]/Δt = x mol L⁻¹ min⁻¹.

Overall rate = −(1/5)Δ[Br⁻]/Δt = (1/5)·x = x/5.

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$2A \rightleftharpoons 4B + C$
›Reveal solutionSolution
This tests relating rates of different species in a reaction using their stoichiometric coefficients. Answer: rate of disappearance of A = $2.5\times10^{-4}$ mol L⁻¹ s⁻¹.

Concept and Intuition

For a reaction $2A \rightleftharpoons 4B + C$ , the overall rate of reaction is the same whichever species you track, once you divide by its stoichiometric coefficient: $\text{Rate} = -\dfrac{1}{2}\dfrac{d[A]}{dt} = \dfrac{1}{4}\dfrac{d[B]}{dt} = \dfrac{d[C]}{dt}$ .

Step-by-Step Solution

Rate of increase of B: $\dfrac{\Delta[B]}{\Delta t} = \dfrac{5\times10^{-3}}{10} = 5\times10^{-4}$ mol L⁻¹ s⁻¹.

Overall rate of reaction $= \dfrac{1}{4}\times \dfrac{d[B]}{dt} = \dfrac{1}{4}\times5\times10^{-4} = 1.25\times10^{-4}$ mol L⁻¹ s⁻¹. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$3X(g) \rightarrow 2Y(g) + Z(g)$
›Reveal solutionSolution
Using the stoichiometric rate relation for $3X\to2Y+Z$ , the rate of formation of Y is $4.8\times10^{-3}$ mol L⁻¹s⁻¹.

Concept and Intuition

For a balanced reaction, the rates of consumption/formation of different species are not numerically equal — they're tied together through the stoichiometric coefficients, because for every 3 molecules of X consumed, exactly 2 molecules of Y are formed (and 1 of Z). Dividing each species' rate of change by its own coefficient gives one common "rate of reaction" that all species share.

Step-by-Step Solution

Write the rate of reaction using stoichiometric coefficients: $r=-\dfrac{1}{3}\dfrac{d[X]}{dt}=\dfrac{1}{2}\dfrac{d[Y]}{dt}=\dfrac{d[Z]}{dt}$ .

Given $-\dfrac{d[X]}{dt}=7.2\times10^{-3}$ mol L⁻¹s⁻¹, compute $r=\dfrac{7.2\times10^{-3}}{3}=2.4\times10^{-3}$ mol L⁻¹s⁻¹. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$\mathrm{H_2}$
›Reveal solutionSolution
This tests converting a mass-based rate of disappearance of one reactant into the mass-based rate of formation of a product, using the reaction stoichiometry after first converting to moles.

Concept and Intuition

Stoichiometric rate relations ( $-\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{b}\frac{d[B]}{dt}$ ) are strictly in terms of moles, not mass. Since the rates here are given in g L $^{-1}$ s $^{-1}$ , we must first convert to mol L $^{-1}$ s $^{-1}$ using molar masses, apply the mole ratio from the balanced equation, then convert the answer back to mass units using the product's molar mass.

Step-by-Step Solution

Reaction: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ .

Rate of disappearance of H $_2$ (mass) $= 6\,\text{g L}^{-1}\text{s}^{-1}$ . Molar mass of H $_2 = 2\,\text{g/mol}$ , so in moles: $6/2 = 3\,\text{mol L}^{-1}\text{s}^{-1}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$\mathrm{N_2O_5}$
›Reveal solutionSolution
Using the stoichiometric coefficients to relate species rates, $\mathrm{NO_2}$ (coefficient 4) forms twice as fast as $\mathrm{N_2O_5}$ (coefficient 2) disappears.

Concept and Intuition

For a reaction $a A \to b B$ , the rates of consumption/production of each species are tied together through the coefficients: $-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$ (the single, unique "rate of reaction"). A species with a larger coefficient changes concentration proportionally faster.

Step-by-Step Solution

Reaction: $2\mathrm{N_2O_5(g)} \to 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}$ .

Rate of reaction $= -\dfrac{1}{2}\dfrac{d[\mathrm{N_2O_5}]}{dt} = \dfrac{1}{4}\dfrac{d[\mathrm{NO_2}]}{dt}$ .

Given $-\dfrac{d[\mathrm{N_2O_5}]}{dt} = 1.2\times10^{-5}\ \mathrm{mol\,L^{-1}s^{-1}}$ , so rate of reaction $= \dfrac{1.2\times10^{-5}}{2} = 0.6\times10^{-5}$ .

$\dfrac{d[\mathrm{NO_2}]}{dt} = 4 \times (\text{rate of reaction}) = 4 \times 0.6\times10^{-5} = 2.4\times10^{-5}\ \mathrm{mol\,L^{-1}s^{-1}}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$NO_2 + CO \rightarrow CO_2 + NO$
›Reveal solutionSolution
The rate law of a multi-step mechanism is written from the slowest (rate-determining) elementary step; here that is $NO_2+NO_2\to NO+NO_3$ , giving $rate=k[NO_2]^2$ .

Concept and Intuition

In a mechanism made of several elementary steps, the overall reaction rate is controlled by the slowest step because it acts as a bottleneck — no matter how fast the later steps are, the overall reaction cannot proceed faster than the slow step allows. For an elementary step, the rate law can be written directly from its molecularity (unlike for the overall multi-step reaction, where the rate law must be determined experimentally or derived from the mechanism).

Step-by-Step Solution

Identify the rate-determining step: since Step 1 is explicitly labelled "slow" and Step 2 is "fast," Step 1 controls the overall rate.

Step 1 is $NO_2 + NO_2 \xrightarrow{slow} NO + NO_3$ — this is a bimolecular elementary step involving two molecules of $NO_2$ .

For an elementary step, the rate law follows directly from the stoichiometry of that step: $rate = k[NO_2][NO_2] = k[NO_2]^2$ .

The fast second step, $NO_3+CO \to CO_2+NO_2$ , consumes the intermediate $NO_3$ as fast as it forms, so it does not appear in (and does not limit) the overall rate expression. …
Start your 14-day free trial to see the full solution →

←Previous questionQ11 — Which of the following statements is correct? (i) The rate of a reacti…→Next questionQ13 — Three reaction energy profiles (energy on the vertical axis versus rea…

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)

🎓Unlock everything free for 14 days

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