From the rate expression for the following reactions, determine their order of reaction and the dimensions of

Exercises · 3.1

Q. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i)

3NO(g) \rightarrow N_2O(g)

; Rate

= k[NO]^2

(ii)

H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-

; Rate

= k[H_2O_2][I^-]

(iii)

CH_3CHO(g) \rightarrow CH_4(g) + CO(g)

; Rate

= k[CH_3CHO]^{3/2}

(iv)

C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)

; Rate

= k[C_2H_5Cl]

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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.

Common Pitfall to Avoid

Watch out

Do not write $\frac{d[\text{reactant}]}{dt}$ as a positive number and then forget the minus sign. The rate of change of a reactant is negative (concentration falls). The minus sign in the definition flips it to a positive $r$ . If you skip the sign, you will get the wrong magnitude for other species.

Why This Matters

In exams (JEE, NEET, etc.), you will often be given the rate for one species and asked to find the rate for another. The stoichiometric relation is the only tool you need — no extra formulas. It also appears in more advanced topics like the rate law (where the exponents are not the coefficients) — but that is a separate concept. Reaction rate stoichiometry is purely about the definition of the reaction rate itself.

Final takeaway: The coefficients in the balanced equation are the conversion factors between the rates of different species. Always normalise by dividing by the coefficient to get the universal reaction rate $r$ .

Average rate of reaction is one of the first ideas introduced in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘average rate of reaction formula’ or ‘average vs instantaneous rate’ are common important-question topics in board exams and JEE Main chemistry. A solid grasp of this basic definition is also assumed in nearly every subsequent kinetics numerical asked in NEET and state CETs.

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}

$\nu_X$ is negative for reactants → the minus sign is already built in.
$\nu_X$ is positive for products.

6. Why this is the average rate (not instantaneous)

Average rate uses a finite $\Delta t$ — it’s the slope of the chord between two points.
Instantaneous rate uses $\Delta t \rightarrow 0$ — it’s the slope of the tangent at a single point.

The average rate formula is just the discrete version of the derivative:

\text{Instantaneous rate} = \frac{1}{\nu_X} \frac{d[X]}{dt}

Summary — the "why" in one sentence

The average rate formula holds because it measures change in concentration per unit time, uses a minus sign to keep rates positive for reactants, and divides by stoichiometric coefficients to give a single, comparable value for the whole reaction.

Always remember:

$\Delta [\text{reactant}]$ is negative → minus sign makes it positive.
$\Delta [\text{product}]$ is positive → no minus sign.
Divide by coefficient → normalise to "per mole of reaction".

The order of a reaction is the sum of the exponents in its rate law. The dimensions of the rate constant $k$ depend on the overall order $n$ as $[k] = (\text{concentration})^{1-n} \cdot \text{time}^{-1}$ . For each given reaction, we find:

order = 2, $[k] = \text{L mol}^{-1} \text{s}^{-1}$ .
order = 2, $[k] = \text{L mol}^{-1} \text{s}^{-1}$ .
order = 1.5, $[k] = \text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$ .
order = 1, $[k] = \text{s}^{-1}$ .

The core idea: what “order” really means

The order of a reaction is not the same as the stoichiometric coefficients in the balanced equation. It is an experimentally determined number that tells you how the rate depends on the concentration of each reactant. In a rate law like

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

the overall order is $m + n$ . The rate constant $k$ is the proportionality constant that makes the equation dimensionally consistent. Since rate always has dimensions of concentration per time (e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ), the units of $k$ must adjust to match the total exponent.

n

Let’s apply this to each case.

(i) $3NO(g) \rightarrow N_2O(g)$ ; Rate $= k[NO]^2$

Find the order. The rate law has only one reactant, $NO$ , raised to the power 2. So the overall order is simply $2$ .
Find the dimensions of $k$ .

Rate has units of $\text{mol L}^{-1} \text{s}^{-1}$ (concentration per time).

$[NO]^2$ has units of $(\text{mol L}^{-1})^2 = \text{mol}^2 \text{L}^{-2}$ .

So:

[k] = \frac{\text{rate}}{[NO]^2} = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^2 \text{L}^{-2}} = \text{mol}^{-1} \text{L} \text{s}^{-1}

Equivalently, $\text{L mol}^{-1} \text{s}^{-1}$ .

Tip

For a second-order reaction, $k$ always has units of $\text{(concentration)}^{-1} \text{time}^{-1}$ . In gas-phase problems, you might see $\text{atm}^{-1} \text{s}^{-1}$ , but here we stick with molarity.

(ii) $H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-$ ; Rate $= k[H_2O_2][I^-]$

Find the order. The exponents are 1 on $H_2O_2$ and 1 on $I^-$ . The $H^+$ ion does not appear in the rate law (its concentration may be constant or it does not affect the rate). So overall order = $1 + 1 = 2$ .
Find the dimensions of $k$ .

$[H_2O_2][I^-]$ has units of $(\text{mol L}^{-1}) \times (\text{mol L}^{-1}) = \text{mol}^2 \text{L}^{-2}$ .

Hence:

[k] = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^2 \text{L}^{-2}} = \text{mol}^{-1} \text{L} \text{s}^{-1}

Same as case (i): $\text{L mol}^{-1} \text{s}^{-1}$ .

Watch out

A common mistake is to add the stoichiometric coefficients (3 for $I^-$ , 2 for $H^+$ ) and claim the order is 6. That is wrong — order comes from the rate law, not the balanced equation. The given rate law explicitly shows only $[H_2O_2]$ and $[I^-]$ to the first power.

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ ; Rate $= k[CH_3CHO]^{3/2}$

Find the order. The exponent is $3/2 = 1.5$ . So overall order = $1.5$ .
Find the dimensions of $k$ .

$[CH_3CHO]^{3/2}$ has units of $(\text{mol L}^{-1})^{3/2} = \text{mol}^{3/2} \text{L}^{-3/2}$ .

Therefore:

[k] = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^{3/2} \text{L}^{-3/2}} = \text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}

Which is usually written as $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$ .

Note

Fractional orders are common in complex reactions (e.g., chain reactions or reactions with a pre-equilibrium step). The units of $k$ will always involve fractional powers of concentration when $n$ is not an integer.

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$ ; Rate $= k[C_2H_5Cl]$

Find the order. The exponent is 1 (implied). So overall order = $1$ .
Find the dimensions of $k$ .

$[C_2H_5Cl]$ has units of $\text{mol L}^{-1}$ .

So:

[k] = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}

For a first-order reaction, $k$ always has units of $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$ , $\text{min}^{-1}$ ).

Tip

First-order reactions are the only ones where $k$ is independent of concentration units — it’s always just per time. That’s why half-life for a first-order reaction ( $t_{1/2} = \ln 2 / k$ ) is constant.

✓Final answer

[k] = \text{L mol}^{-1} \text{s}^{-1}

Method: Order of Reaction from Rate Law Expression

This method uses the experimentally determined rate law to directly read the order with respect to each reactant and the overall order. The dimensions of the rate constant $k$ are then derived using dimensional analysis.

Steps

Identify the exponents on each concentration term in the given rate expression.
- The exponent on a reactant is its individual order.
- The overall order is the sum of all individual orders.
Write the general form of the rate law:

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

where $m$ and $n$ are the orders.
Find dimensions of $k$ using:

$\text{Dimensions of Rate} = \text{concentration} \cdot \text{time}^{-1}$

So, $[k] = \dfrac{[\text{Rate}]}{[\text{concentration}]^{\text{overall order}}}$

Applying to each reaction

(i) $3NO(g) \rightarrow N_2O(g)$ ; Rate $= k[NO]^2$

Order w.r.t. NO = 2
Overall order = 2
Dimensions of $k$ : $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L} \text{s}^{-1}$ → $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$

(ii) $H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-$ ; Rate $= k[H_2O_2][I^-]$

Order w.r.t. $H_2O_2$ = 1
Order w.r.t. $I^-$ = 1
Overall order = $1 + 1 = 2$
Dimensions of $k$ : $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L} \text{s}^{-1}$ → $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ ; Rate $= k[CH_3CHO]^{3/2}$

Order w.r.t. $CH_3CHO$ = $3/2$
Overall order = $1.5$
Dimensions of $k$ : $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^{3/2}} = \text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$ → $k$ has units of $\text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$ ; Rate $= k[C_2H_5Cl]$

Order w.r.t. $C_2H_5Cl$ = 1
Overall order = 1
Dimensions of $k$ : $[k] = \dfrac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1}$ → $k$ has units of $\text{s}^{-1}$

Final Answer Summary

Reaction	Overall Order	Units of $k$
(i)	2	$\text{L mol}^{-1} \text{s}^{-1}$
(ii)	2	$\text{L mol}^{-1} \text{s}^{-1}$
(iii)	1.5	$\text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$
(iv)	1	$\text{s}^{-1}$

Common Mistakes Students Make on "Order of Reaction & Dimensions of Rate Constants"

Mistake 1: Confusing Order with Molecularity

The error: Students often think the stoichiometric coefficient (like 3 in $3NO$ ) determines the order. Here, the order comes only from the rate law exponent, not the balanced equation.

How to avoid:

Order = sum of exponents in the rate law.
For (i): Rate $= k[NO]^2$ → order = 2 (not 3).
For (ii): Rate $= k[H_2O_2][I^-]$ → order = $1 + 1 = 2$ (ignore $H^+$ and $I_3^-$ ).

Mistake 2: Forgetting Units of $k$ Depend on Overall Order

The error: Using the same unit for all reactions (e.g., always $\text{s}^{-1}$ or always $\text{L mol}^{-1} \text{s}^{-1}$ ).

How to avoid:

General formula:

[k] = \frac{\text{concentration}^{1-n}}{\text{time}}

where $n$ = overall order.

For zero order: $k$ has units of $\text{mol L}^{-1} \text{s}^{-1}$
For first order: $k$ has units of $\text{s}^{-1}$
For second order: $k$ has units of $\text{L mol}^{-1} \text{s}^{-1}$
For order $3/2$ : $k$ has units of $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$

Mistake 3: Mishandling Fractional Orders

The error: Treating $3/2$ order as either 1 or 2, or writing wrong units.

How to avoid:

For (iii): Rate $= k[CH_3CHO]^{3/2}$ → $n = 1.5$
Units:

[k] = \frac{(\text{mol L}^{-1})^{1 - 1.5}}{\text{s}} = \frac{(\text{mol L}^{-1})^{-0.5}}{\text{s}} = \text{L}^{0.5} \text{mol}^{-0.5} \text{s}^{-1}

(Often written as $\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$ )

Mistake 4: Ignoring Concentration Units in Dimensional Analysis

The error: Using $\text{mol}$ without volume, or forgetting that concentration is $\text{mol L}^{-1}$ .

How to avoid:

Always write concentration as $[\text{mol L}^{-1}]$
For (iv): first order → $k$ has units $\text{s}^{-1}$ (time $^{-1}$ only, no concentration term)

Mistake 5: Misreading the Rate Law for Multi-reactant Reactions

The error: Including all reactants in the order calculation even if they don't appear in the rate law.

How to avoid:

In (ii), the rate law is $k[H_2O_2][I^-]$ — $H^+$ does not appear, so its exponent is 0.
Order = $1 + 1 = 2$ , not 3.

Quick Summary Table (Exam-Ready)

Reaction	Rate Law	Overall Order	Units of $k$
(i) $3NO \rightarrow N_2O$	$k[NO]^2$	2	$\text{L mol}^{-1} \text{s}^{-1}$
(ii) $H_2O_2 + 3I^- + 2H^+ \rightarrow ...$	$k[H_2O_2][I^-]$	2	$\text{L mol}^{-1} \text{s}^{-1}$
(iii) $CH_3CHO \rightarrow CH_4 + CO$	$k[CH_3CHO]^{3/2}$	1.5	$\text{L}^{1/2} \text{mol}^{-1/2} \text{s}^{-1}$
(iv) $C_2H_5Cl \rightarrow C_2H_4 + HCl$	$k[C_2H_5Cl]$	1	$\text{s}^{-1}$

Final tip: Always write the general formula for $k$ first, then substitute $n$ . This prevents unit errors even for fractional orders.

←Previous section · summarySummary →Next question3.2 — For the reaction: 2A + B arrow A₂B the rate = k[A][B]² with k = 2.0×10…

Q.From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

The core idea: what “order” really means

(i) 3NO(g)→N2O(g)3NO(g) \rightarrow N_2O(g)3NO(g)→N2​O(g); Rate =k[NO]2= k[NO]^2=k[NO]2

(ii) H2O2(aq)+3I−(aq)+2H+→2H2O(l)+I3−H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-H2​O2​(aq)+3I−(aq)+2H+→2H2​O(l)+I3−​; Rate =k[H2O2][I−]= k[H_2O_2][I^-]=k[H2​O2​][I−]

(iii) CH3CHO(g)→CH4(g)+CO(g)CH_3CHO(g) \rightarrow CH_4(g) + CO(g)CH3​CHO(g)→CH4​(g)+CO(g); Rate =k[CH3CHO]3/2= k[CH_3CHO]^{3/2}=k[CH3​CHO]3/2

(iv) C2H5Cl(g)→C2H4(g)+HCl(g)C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)C2​H5​Cl(g)→C2​H4​(g)+HCl(g); Rate =k[C2H5Cl]= k[C_2H_5Cl]=k[C2​H5​Cl]

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(i) $3NO(g) \rightarrow N_2O(g)$ ; Rate $= k[NO]^2$

(ii) $H_2O_2(aq) + 3I^-(aq) + 2H^+ \rightarrow 2H_2O(l) + I_3^-$ ; Rate $= k[H_2O_2][I^-]$

(iii) $CH_3CHO(g) \rightarrow CH_4(g) + CO(g)$ ; Rate $= k[CH_3CHO]^{3/2}$

(iv) $C_2H_5Cl(g) \rightarrow C_2H_4(g) + HCl(g)$ ; Rate $= k[C_2H_5Cl]$

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