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.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{NH}_3$
›Reveal solutionSolution
The key is to use the stoichiometric ratio from the balanced reaction $\mathrm{N_2 + 3H_2 \to 2NH_3}$ and the given rate of hydrogen disappearance to find the rate of ammonia formation, then scale to the requested time. The answer is $1.2 \times 10^{-3}$ moles, option (C).

The problem gives a rate of disappearance of hydrogen and asks for the amount of ammonia formed in a different time interval. The central concept is stoichiometric proportionality in chemical kinetics: for a balanced reaction, the rates of consumption and production of species are related by their coefficients. Once we know how fast hydrogen is used up, we can directly find how fast ammonia appears, then multiply by the new time.

Write the balanced equation The synthesis of ammonia is:

$\mathrm{N_2 + 3H_2 \to 2NH_3}$
This tells us: for every 3 moles of $\mathrm{H_2}$ consumed, 2 moles of $\mathrm{NH_3}$ are produced.

Find the rate of hydrogen disappearance Given: $6 \times 10^{-2}$ moles of $\mathrm{H_2}$ disappear in 10 minutes. So the rate of disappearance of $\mathrm{H_2}$ is:

$\text{Rate}_{\mathrm{H_2}} = \frac{6 \times 10^{-2} \text{ mol}}{10 \text{ min}} = 6 \times 10^{-3} \text{ mol/min}$

Use stoichiometry to find the rate of ammonia formation From the balanced equation:

$\frac{\text{Rate of formation of } \mathrm{NH_3}}{\text{Rate of disappearance of } \mathrm{H_2}} = \frac{2}{3}$
Therefore:
$\text{Rate}_{\mathrm{NH_3}} = \frac{2}{3} \times (6 \times 10^{-3}) = 4 \times 10^{-3} \text{ mol/min}$

Calculate moles of ammonia formed in 0.3 minutes
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KCET 2025Set D-41 markMCQ
$0.2$
›Reveal solutionSolution

Count how many half-lives fit into the elapsed time and halve the concentration that many times: $80\,\mathrm{s} = 4 \times 20\,\mathrm{s}$ , so $[A] = 0.2/2^4$ .

Step 1 — Why the half-life shortcut is legitimate here.

For a first-order reaction the integrated rate law is
$[A]_t = [A]_0\, e^{-kt}$
and the half-life
$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$
is independent of the initial concentration. That is the special feature of first order: every successive $20\,\mathrm{s}$ window halves whatever is present at the start of that window.

Step 2 — Count the half-lives.
$n = \frac{t}{t_{1/2}} = \frac{80\ \mathrm{s}}{20\ \mathrm{s}} = 4$
Step 3 — Halve four times.
$[A]_t = \frac{[A]_0}{2^n} = \frac{0.2}{2^4} = \frac{0.2}{16} = 0.0125\ \mathrm{M}$
Tracking it explicitly:

Time (s) Half-lives elapsed $[A]$ (M)
0 0 0.2
20 1 0.1
40 2 0.05
60 3 0.025
80 4 0.0125
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COMEDK 2025Set 2025-M1 markMCQ
$\frac{3}{5} X(a q) \rightarrow \frac{1}{2} Y(a q)+Z(g)$
›Reveal solutionSolution
Using the standard rate definition with fractional stoichiometric coefficients gives $-\dfrac{d[X]}{dt} = +\dfrac{6}{5}\dfrac{d[Y]}{dt}$ , matching option (A).

Concept & Intuition

For a reaction $aA \to bB$ , the rate is defined consistently for every species:
$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = +\frac{1}{b}\frac{d[B]}{dt}$
This holds even when the coefficients are fractions, as here: $\dfrac{3}{5}X \to \dfrac{1}{2}Y + Z$ .

Step-by-step

Define the common rate $r$ .

For X (coefficient $3/5$ ): $r = -\dfrac{1}{3/5}\dfrac{d[X]}{dt} = -\dfrac{5}{3}\dfrac{d[X]}{dt}$ .

For Y (coefficient $1/2$ ): $r = +\dfrac{1}{1/2}\dfrac{d[Y]}{dt} = +2\dfrac{d[Y]}{dt}$ .

Equate.

$-\frac{5}{3}\frac{d[X]}{dt} = 2\frac{d[Y]}{dt}$

Solve for $-d[X]/dt$ . …
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COMEDK 2024Set 2024-M1 markMCQ
$3 X+2 Y+\frac{5}{2} Z \rightarrow P_1+P_2+P_3$
›Reveal solutionSolution
The rate of a reaction is defined as the common value obtained by dividing the rate of change of each reactant’s concentration by its stoichiometric coefficient (with a negative sign for reactants). Matching this definition to the given options shows that option (C) is correct.

The key idea is that for a reaction
$aA + bB \rightarrow \text{products},$
the rate of reaction $r$ is defined as
$r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt}.$
The negative sign is used because reactant concentrations decrease over time, making their derivatives negative, so the rate comes out positive. For products, the sign is positive.

Here, the reactants are $X$ , $Y$ , and $Z$ with coefficients $3$ , $2$ , and $\frac{5}{2}$ respectively. So the rate must satisfy:
$r = -\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt} = -\frac{1}{5/2}\frac{d[Z]}{dt}.$
We then rewrite these equalities in the form given in the options.

Write the rate equalities from definition

$-\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt} = -\frac{2}{5}\frac{d[Z]}{dt}$
(since $1/(5/2) = 2/5$ ). The negatives cancel, so we have:
$\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt} = \frac{2}{5}\frac{d[Z]}{dt}.$

Convert to the form in the options Options express relationships like $k_1 \frac{d[X]}{dt} = k_2 \frac{d[Y]}{dt} = k_3 \frac{d[Z]}{dt}$ . From $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt}$ , multiply both sides by 6:

$2\frac{d[X]}{dt} = 3\frac{d[Y]}{dt}.$
From $\frac{1}{3}\frac{d[X]}{dt} = \frac{2}{5}\frac{d[Z]}{dt}$ , multiply both sides by 15:
$5\frac{d[X]}{dt} = 6\frac{d[Z]}{dt}.$
So one valid set is:
$2\frac{d[X]}{dt} = 3\frac{d[Y]}{dt} = \frac{6}{5}\frac{d[Z]}{dt} \quad \text{(not directly in options)}.$

Match to the given options We need to find which option’s equalities are equivalent to $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt} = \frac{2}{5}\frac{d[Z]}{dt}$ . Check option (C):

$\frac{4}{15}\frac{d[X]}{dt} = \frac{4}{5}\frac{d[Y]}{dt} = \frac{8}{15}\frac{d[Z]}{dt}.$
Divide each term by its coefficient to see the common rate:

From first equality: $\frac{4}{15}\frac{d[X]}{dt} = \frac{4}{5}\frac{d[Y]}{dt}$ → multiply by 15/4: $\frac{d[X]}{dt} = 3\frac{d[Y]}{dt}$ ? Wait, let’s do carefully:

$\frac{4}{15}\frac{d[X]}{dt} = \frac{4}{5}\frac{d[Y]}{dt} \implies \frac{d[X]}{dt} = 3\frac{d[Y]}{dt}.$
But from our definition, $\frac{1}{3}\frac{d[X]}{dt} = \frac{1}{2}\frac{d[Y]}{dt}$ gives $\frac{d[X]}{dt} = \frac{3}{2}\frac{d[Y]}{dt}$, not $3$. So this seems off? Let’s re-check systematically.

Actually, the correct method: For option (C), set the common value as $k$ . Then:
$\frac{4}{15}\frac{d[X]}{dt} = k \implies \frac{d[X]}{dt} = \frac{15}{4}k,$ $\frac{4}{5}\frac{d[Y]}{dt} = k \implies \frac{d[Y]}{dt} = \frac{5}{4}k,$ $\frac{8}{15}\frac{d[Z]}{dt} = k \implies \frac{d[Z]}{dt} = \frac{15}{8}k.$
Now compute $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{3}\cdot\frac{15}{4}k = -\frac{5}{4}k$ .

$-\frac{1}{2}\frac{d[Y]}{dt} = -\frac{1}{2}\cdot\frac{5}{4}k = -\frac{5}{8}k$ . These are not equal — so (C) fails? That suggests I made a sign error: Actually, the rate definition uses negative of reactant derivatives. Let’s re-derive carefully.

Watch out

A common mistake is forgetting the negative sign when relating derivatives of reactants. The rate $r$ is positive, so for reactants we must have $r = -\frac{1}{\text{coeff}}\frac{d[\text{reactant}]}{dt}$ . If you drop the minus, you get the wrong proportionality.

Correct derivation with signs

$r = -\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt} = -\frac{2}{5}\frac{d[Z]}{dt}.$
Multiply each equality by the denominators to eliminate fractions:

From $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt}$ , multiply by $-6$ :
$2\frac{d[X]}{dt} = 3\frac{d[Y]}{dt}.$
From $-\frac{1}{3}\frac{d[X]}{dt} = -\frac{2}{5}\frac{d[Z]}{dt}$ , multiply by $-15$ :
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KCET 2023Set D-21 markMCQ
$a\mathrm{MnO}_4^- + b\mathrm{S}_2\mathrm{O}_3^{2-} + 2\mathrm{H}_2\mathrm{O} \rightarrow x\mathrm{MnO}_2 + y\mathrm{SO}_4^{2-} + z\mathrm{OH}^-$
›Reveal solutionSolution

Balance the electrons: each Mn gains 3 e $^-$ and each thiosulphate loses 8 e $^-$ , giving $a=8$ , $b=3$ , $x=8$ , $y=6$ .

Step 1 — Oxidation half-reaction (reduction of permanganate in basic medium).

In $\mathrm{MnO_4^-}$ , Mn is $+7$ ; in $\mathrm{MnO_2}$ , Mn is $+4$ .
$\mathrm{Mn}^{+7} + 3e^- \rightarrow \mathrm{Mn}^{+4} \quad (\text{3 electrons gained per Mn})$
Step 2 — Oxidation of thiosulphate.

In $\mathrm{S_2O_3^{2-}}$ the average oxidation state of S is $+2$ (since $2x + 3(-2) = -2$ ). In $\mathrm{SO_4^{2-}}$ , S is $+6$ . Each $\mathrm{S_2O_3^{2-}}$ produces two sulphate ions:
$\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{SO_4^{2-}} + 8e^- \quad (2 \times (6-2) = 8 \text{ electrons lost})$
Step 3 — Equalise the electrons.

LCM of 3 and 8 is 24, so multiply the reduction by 8 and the oxidation by 3:
$8\,\mathrm{MnO_4^-} + 3\,\mathrm{S_2O_3^{2-}} \rightarrow 8\,\mathrm{MnO_2} + 6\,\mathrm{SO_4^{2-}}$
Step 4 — Complete with water/hydroxide (charge and atom balance). …
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KCET 2021Set B-21 markMCQ
$Q. The number of angular and radial nodes in 3p orbital respectively are (A) 3, 1 (B) 1, 1 (C) 2, 1 (D) 2, 3$
›Reveal solutionSolution
Angular nodes $= l$ ; radial nodes $= n - l - 1$ . For $3p$ ( $n=3$ , $l=1$ ) both come out as 1.

Step 1 — Identify the quantum numbers.

For a $3p$ orbital:

principal quantum number $n = 3$ (from the digit),

azimuthal quantum number $l = 1$ (since $p \Rightarrow l = 1$ ; $s=0$ , $p=1$ , $d=2$ , $f=3$ ).

Step 2 — Angular (nodal-plane) nodes.

These come from the angular part $Y_{l,m}(\theta,\phi)$ of the wavefunction, and their count is exactly
$\text{angular nodes} = l = \boxed{1}$
Physically: a $p$ orbital is dumbbell-shaped with one nodal plane through the nucleus (e.g. the $xy$ -plane for $p_z$ ). Note this is independent of $n$ — $2p$ , $3p$ , $4p$ all have exactly 1 angular node.

Step 3 — Radial (spherical) nodes.

These come from the radial part $R_{n,l}(r)$ , and

$\text{radial nodes} = n - l - 1 = 3 - 1 - 1 = \boxed{1}$ …
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KCET 2021Set B-21 markMCQ
$k$
›Reveal solutionSolution

Substitute "99% reacted ⇒ 1% left" into the integrated first-order rate law $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$ .

Step 1 — The integrated first-order rate law.

Starting from $-\dfrac{d[A]}{dt} = k[A]$ and integrating,
$\ln\frac{[A]_0}{[A]} = kt \quad\Longrightarrow\quad t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}.$
(The factor $2.303$ converts $\ln$ to $\log_{10}$ .)

Step 2 — Translate "99% completion".

If 99% of A has reacted, the fraction remaining is $1\%$ :
$[A] = 0.01\,[A]_0 \quad\Longrightarrow\quad \frac{[A]_0}{[A]} = \frac{1}{0.01} = 100.$
This is the step candidates most often get wrong — the ratio uses what is left, not what has gone.

Step 3 — Substitute.
$t = \frac{2.303}{k}\log(100) = \frac{2.303}{k}\times 2 = \frac{4.606}{k}.$
Step 4 — Cross-check with half-lives. …
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KCET 2020Set A-11 markMCQ
$60\%$
›Reveal solutionSolution

First-order kinetics: equal fractional decays take equal times, and $0.064=(0.4)^3$ — so 93.6% completion takes exactly three times as long as 60% completion.

Step 1 — The first-order integrated rate law.
$k=\frac{2.303}{t}\log\frac{[A]_0}{[A]}=\frac{2.303}{t}\log\frac{a}{a-x}$
where $a$ is the initial amount and $x$ the amount reacted. The key property of first order: $k$ depends only on the ratio $a/(a-x)$ , not on the absolute concentration.

Step 2 — Use the given data to get $k$ .

60% complete ⇒ 40% remains ⇒ $\dfrac{a}{a-x}=\dfrac{100}{40}=2.5$ , with $t=50$ min:
$k=\frac{2.303}{50}\log 2.5=\frac{2.303\times0.3979}{50}=\frac{0.9163}{50}=0.01833\ \text{min}^{-1}.$
Step 3 — Set up the second condition.

93.6% complete ⇒ remaining $=100-93.6=6.4\%$ ⇒
$\frac{a}{a-x}=\frac{100}{6.4}=15.625 .$
Step 4 — Spot the elegant relation (no calculator needed).
$15.625=\left(2.5\right)^3\qquad\text{since } 2.5^3=15.625 .$
Equivalently the surviving fraction $0.064=(0.4)^3$ . Taking logs:
$\log 15.625=3\log 2.5 .$
Step 5 — Get the time. …
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Time (s)	Half-lives elapsed	$[A]$ (M)
0	0	0.2
20	1	0.1
40	2	0.05
60	3	0.025
80	4	0.0125

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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 …

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

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