MHT-CET Chemistry · Chemical Kinetics

Rate Law, Order, Molecularity and Rate Expression

rate = k[A]ˣ[B]ʸ with x + y the (experimental) order; k is the rate at unit concentrations, depends only on temperature, and its unit depends on the order; molecularity is the number of species in an elementary step and equals the order only for elementary reactions.

All Chemical Kinetics notes MHT-CET Chemistry strategy

Why this matters

59 PYQs, none HARD — the chapter's second-largest page. A third are k from a rate and concentrations (or the reverse), a third are the order read off a rate law or a concentration experiment, and the rest are the change in rate when concentrations are doubled or halved, the order-versus-molecularity contrast, and three named examples (H₂O₂ decomposition first order, H₂ + I₂ second, H₂ + Br₂ order 3/2). Nothing here needs more than substitution; the traps are conceptual — order is experimental, k is concentration-independent.

Concept 1 of 4: The Rate Law and the Order: Exponents Come From Experiment, Not From the Equation

The rate law

r = k[A]^x[B]^y

is measured, and

x + y

is the order. The exponents need not match the stoichiometric coefficients, can be fractional or zero, and a species with exponent zero does not affect the rate at all.

Definition

$2\text{NO} + 2\text{H}_2 \to \text{N}_2 + 2\text{H}_2\text{O}$ with $r = k[\text{NO}]^2[\text{H}_2]$ : first order in H $_2$ , overall order $3$ . $r = k[\text{H}_2][\text{I}_2]$ : overall $2$ , first in each. $r = k[A]^{1.5}[B]^{2.5}$ : order $4$ .
Rate proportional to $[\text{NO}_2]^2$ and independent of [CO]: $r = k[\text{NO}_2]^2[\text{CO}]^0$ . Second order in NO, first in Cl $_2$ : $r = k[\text{NO}]^2[\text{Cl}_2]$ . First in CHCl $_3$ , half in Cl $_2$ : $r = k[\text{CHCl}_3][\text{Cl}_2]^{1/2}$ , order $\tfrac32$ .
From an experiment: tenfold $[A]$ giving $100\times$ the rate means $10^x = 100$ , $x = 2$ .
Order is an experimental quantity; it may be an integer, a fraction or zero; it is NOT a theoretical quantity.
Named examples: $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ and CH $_3$ CHO → CH $_4$ + CO are first order; $\text{H}_2 + \text{I}_2 \to 2\text{HI}$ and $2\text{NO}_2 \to 2\text{NO} + \text{O}_2$ are second; $2\text{NO} + 2\text{H}_2$ and $2\text{NO}_2 + \text{F}_2$ are third.

Rate law

r = k[A]^x[B]^y,\qquad \text{order} = x + y \ (\text{experimental; may be } 0, \text{ fractional})

Worked example

Doubling

[A]

doubles the rate; doubling

[B]

quadruples it. Write the rate law and the overall order.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q53Easy

Example 1 · Chemical Kinetics · Rate Law, Order, Molecularity and Rate Expression

For the reaction

2\text{NO}(g) + 2\text{H}_{2}(g) \longrightarrow \text{N}_{2}(g) + 2\text{H}_{2}\text{O}(s)

rate

= k[\text{NO}]^{2}[\text{H}_{2}]

. What is the order of reaction with respect to

\text{H}_{2}

and overall order of reaction respectively?

Reading the order off the balanced equation

2\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2

is first order, and H

_2

+ Br

_2

is order

\tfrac32

. The coefficients give molecularity for an elementary step; the order is measured.

Drill 12 more on the rate law and the order: exponents come from experiment, not from the equation

Concept 2 of 4: The Rate Constant: k = rate/([A]ˣ[B]ʸ), Its Units and Its Properties

Substitute the given concentrations into the rate law and divide the rate by the concentration product. The unit of

k

is whatever makes the equation balance — (mol dm

^{-3}

)

^{1 - n}

^{-1}

for order

n

— so a unit of s

^{-1}

means first order.

Definition

$r = k[A][B]^2$ : $k = \dfrac{7.2 \times 10^{-2}}{0.4 \times 0.01} = 18$ , $\dfrac{3.6 \times 10^{-2}}{0.2 \times 0.01} = 18$ mol $^{-2}$ dm $^6$ s $^{-1}$ ; $r = k[A]^2[B]$ : $\dfrac{0.22}{1 \times 0.25} = 0.88$ , $\dfrac{1.8 \times 10^{-2}}{0.04 \times 0.1} = 4.5$ , $\dfrac{0.24}{0.25 \times 0.2} = 4.8$ .
Rate from $k$ : $6.25 \times 1^2 \times 0.2 = 1.25$ ( $r = k[A]^2[B]$ ); $6.25 \times 1 \times 0.2^2 = 0.25$ ( $r = k[A][B]^2$ ). Concentration from a rate: $[A] = \dfrac{0.25}{6.25 \times 0.25} = 0.16$ .
Units: zero order mol dm $^{-3}$ s $^{-1}$ ; first s $^{-1}$ ; second mol $^{-1}$ dm $^3$ s $^{-1}$ ; third mol $^{-2}$ dm $^6$ s $^{-1}$ . A $k$ in hour $^{-1}$ or s $^{-1}$ is first order.
$k$ is independent of concentration, varies with temperature, equals the rate at unit concentrations; its unit DEPENDS on the order. For a first-order reaction the slope of rate against concentration is $k$ .

Rate constant

k = \frac{r}{[A]^x[B]^y},\qquad [k] = (\text{mol dm}^{-3})^{1-n}\,\text{s}^{-1}

Worked example

For

r = k[A]^2[B]

, the rate is

3.2 \times 10^{-3}

mol dm

^{-3}

^{-1}

when

[A] = 0.4

M and

[B] = 0.05

M. Find

k

with its unit.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q75Easy

Example 2 · Chemical Kinetics · Rate Law, Order, Molecularity and Rate Expression

Rate of reaction,

A + B \to

product, is

7.2 \times 10^{-2}\,\text{mol dm}^{-3}\text{s}^{-1}

[A] = 0.4\,\text{mol dm}^{-3}

and

[B] = 0.1\,\text{mol dm}^{-3}

. The reaction is first order in A and second order in B. Calculate rate constant.

Squaring the wrong concentration

r = k[A][B]^2

squares

[B]

. With

[A] = 1

and

[B] = 0.2

, squaring the wrong one turns

0.25

into

1.25

— and both are on the list, because the two rate laws appear on twin stems.

Drill 12 more on the rate constant: k = rate/([a]ˣ[b]ʸ), its units and its properties

Concept 3 of 4: How the Rate Changes When Concentrations Change: Multiply the Factors

Scale each concentration by its factor raised to its order and multiply. Doubling a first-order reactant doubles the rate; doubling a second-order one quadruples it; halving one and doubling another with equal orders leaves it unchanged.

Definition

$r = k[A][B]^2$ : doubling $[A]$ → $\times 2$ ; doubling both → $\times 8$ ; doubling A, B and a zero-order C → still $\times 8$ .
$r = k[A][B]$ : doubling both → $\times 4$ ; doubling $[A]$ alone → $\times 2$ ; doubling $[B]$ and halving $[A]$ → $\times 1$ ; halving $[B]$ alone → $\times \tfrac12$ .
'Which change does NOT affect the rate' is the one whose factors multiply to $1$ .

Rate factor

\frac{r'}{r} = \left(\frac{[A]'}{[A]}\right)^x\left(\frac{[B]'}{[B]}\right)^y

Worked example

For

r = k[A]^2[B]

, by what factor does the rate change if

[A]

is tripled and

[B]

is halved?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q61Easy

Example 3 · Chemical Kinetics · Rate Law, Order, Molecularity and Rate Expression

For the reaction,

\text{CH}_3\text{Br(aq)}+\text{OH}^-\text{(aq)}\to\text{CH}_3\text{OH(aq)}+\text{Br}^-\text{(aq)}

, The rate law is rate

=k[\text{CH}_3\text{Br}][\text{OH}^-]

. What is change in rate of reaction if concentration of both reactants is doubled?

Adding the factors

Doubling A and B in

r = k[A][B]^2

gives

2 \times 4 = 8

, not

2 + 4 = 6

. Factors multiply because the rate law is a product.

Drill 5 more on how the rate changes when concentrations change: multiply the factors

Concept 4 of 4: Order Versus Molecularity

Molecularity counts the species that collide in ONE elementary step — always a positive whole number, never zero or fractional — and is a theoretical idea. Order is measured from the rate law and can be anything. For an elementary reaction the two coincide.

Definition

$\text{O}_3 + \text{O} \to 2\text{O}_2$ , elementary, $r = k[\text{O}_3][\text{O}]$ : order $2$ , molecularity $2$ . $\text{NO}_2 + \text{NO}_2 \to 2\text{NO} + \text{O}_2$ : order $2$ , molecularity $2$ .
$\text{C}_2\text{H}_5\text{I} \to \text{C}_2\text{H}_4 + \text{HI}$ , $r = k[\text{C}_2\text{H}_5\text{I}]$ : order $1$ , molecularity $1$ (unimolecular).
$\text{H}_2 + \text{Br}_2 \to 2\text{HBr}$ , $r = k[\text{H}_2][\text{Br}_2]^{1/2}$ : bimolecular, order $\tfrac32$ — the standard example of the two differing.
For a complex (multistep) reaction, molecularity is defined per step and the overall order comes from the slow step.

The contrast

\text{molecularity} \in \{1, 2, 3\} \text{ (theoretical, per step)};\qquad \text{order} \in \mathbb{R}_{\ge 0} \text{ (experimental)}

Worked example

For the elementary reaction

2\text{NO} + \text{O}_2 \to 2\text{NO}_2

, state the molecularity, the rate law and the order.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q57Easy

Example 4 · Chemical Kinetics · Rate Law, Order, Molecularity and Rate Expression

What is the order and molecularity respectively for the elementary reaction given below?

O_{3(\text{ }g)}+O_{(g)}\longrightarrow 2O_{2(\text{ }g)}

r = k\left\lbrack O_{3} \right\rbrack\lbrack O\rbrack

Calling H₂ + Br₂ 'monomolecular' because the order is fractional

Two molecules collide, so it is bimolecular whatever the order. Options pairing 'monomolecular' with

\tfrac32

are the planted confusion of the two ideas.

Drill 3 more on order versus molecularity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The Rate Law and the Order: Exponents Come From Experiment, Not From the Equation
Rate law
$r = k[A]^x[B]^y,\qquad \text{order} = x + y \ (\text{experimental; may be } 0, \text{ fractional})$
The Rate Constant: k = rate/([A]ˣ[B]ʸ), Its Units and Its Properties
Rate constant
$k = \frac{r}{[A]^x[B]^y},\qquad [k] = (\text{mol dm}^{-3})^{1-n}\,\text{s}^{-1}$
How the Rate Changes When Concentrations Change: Multiply the Factors
Rate factor
$\frac{r'}{r} = \left(\frac{[A]'}{[A]}\right)^x\left(\frac{[B]'}{[B]}\right)^y$
Order Versus Molecularity
The contrast
$\text{molecularity} \in \{1, 2, 3\} \text{ (theoretical, per step)};\qquad \text{order} \in \mathbb{R}_{\ge 0} \text{ (experimental)}$

Watch out for (4)

Reading the order off the balanced equation
→ The Rate Law and the Order: Exponents Come From Experiment, Not From the Equation
Squaring the wrong concentration
→ The Rate Constant: k = rate/([A]ˣ[B]ʸ), Its Units and Its Properties
Adding the factors
→ How the Rate Changes When Concentrations Change: Multiply the Factors
Calling H₂ + Br₂ 'monomolecular' because the order is fractional
→ Order Versus Molecularity

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