Factors Influencing Rate of a Reaction

Strike a match and it burns out in seconds; leave iron in damp air and it takes weeks to rust. Both are chemical reactions, yet their speeds differ by orders of magnitude. What decides this — and can it be written as an equation rather than a hunch? This is the second major section of the Chemical Kinetics chapter in CBSE (NCERT) Class 12 Chemistry, building directly on the definition of reaction rate. The syllabus asks you to state qualitatively how concentration, temperature, and catalysts affect rate; to write and use the rate law, always determined by experiment and never guessed from stoichiometry; to define order and distinguish it sharply from molecularity; and to derive the units of the rate constant $k$ for reactions of any order. These ideas set up the next topic, integrated rate equations, so getting the definitions exact here pays off later.

The core insight

The rate of a reaction depends on reactant concentrations through an experimentally determined rate law, $\text{Rate}=k[A]^x[B]^y$ , whose exponents define the order (never assumed equal to stoichiometric coefficients), while molecularity is a separate, always-whole-number count of colliding species in one elementary mechanistic step.

The core idea

1. The rate law connects rate to concentration. For a reaction $aA + bB \rightarrow \text{products}$ , experiments show

\text{Rate} = k[A]^x[B]^y

where: $k$ is the rate constant (the rate when every concentration is unity), and $x$ , $y$ are exponents found only by experiment — they need not equal the stoichiometric coefficients $a$ , $b$ . This expression is the rate law or rate equation, and it is the single most-tested idea in this section: a rate law is measured, never read off a balanced equation, except in the special case of a genuinely single-step (elementary) reaction.

2. Order is defined from the rate law. The order of a reaction is the sum of the concentration exponents: order $= x + y$ . Order can be zero, a positive integer, or even a fraction — it is purely an experimental number with no requirement of being "nice."

3. Molecularity is a different, theoretical idea. Molecularity is the number of reacting species — atoms, ions, or molecules — that must collide simultaneously in one elementary step of a proposed mechanism. It is always a small whole number (1, 2, rarely 3), applies only to a single mechanistic step, and is not defined for a multi-step overall reaction, because such a reaction is not one collision event.

	Order	Molecularity
Determined by	Experiment (rate measurements)	Proposed mechanism (theory)
Possible values	0, fraction, or integer	Always a positive whole number
Applies to	Overall reaction or any single step	Only one elementary step

4. Temperature and catalysts act on the same underlying picture. Raising the temperature increases the fraction of molecules whose energy meets or exceeds the activation energy, so the rate rises sharply — for many reactions the rate roughly doubles for every $10^\circ\text{C}$ rise, a ratio called the temperature coefficient. A catalyst opens an alternative reaction pathway of lower activation energy, speeding up the forward and reverse reactions equally without being consumed and without shifting the equilibrium position.

Figure

A bar chart comparing typical time scales for a fast explosive reaction a moderate rusting reaction and a slow geological weathering reaction on a logarithmic axis

Derivation AUnits of the Rate Constant k

The goal of this derivation is to obtain a single general formula for the units of the rate constant $k$ that works for a reaction of any order $n$ , so that units never need to be memorized case by case but can instead be reconstructed on the spot from the rate law itself using dimensional consistency.

Step 1. Start from the general rate law for a reaction whose overall order is $n$ : $\text{Rate} = k[\text{concentration}]^{n}$ , where "concentration" is collectively the product of all reactant concentration terms raised to their individual exponents, combined into one overall power $n$ for this purpose.

Step 2. Rate itself always carries the units $\text{mol L}^{-1}\text{time}^{-1}$ , because it is defined as a change in concentration per unit time, regardless of the reaction's order — this is fixed and does not depend on $n$ .

Step 3. Rearrange the rate law to isolate $k$ : $k = \dfrac{\text{Rate}}{[\text{concentration}]^{n}}$ . Substituting the units from Step 2 for Rate and writing concentration units as $\text{mol L}^{-1}$ raised to the power $n$ gives

k = \frac{\text{mol L}^{-1}\text{time}^{-1}}{(\text{mol L}^{-1})^{n}}

Step 4. Simplify the exponent on $\text{mol L}^{-1}$ by subtracting the powers in numerator and denominator ( $1 - n$ ), which yields the general result

k \; \text{has units} \; (\text{mol L}^{-1})^{1-n}\,\text{time}^{-1}

Step 5. Check the formula against familiar cases. For $n = 0$ : $(\text{mol L}^{-1})^{1}\text{time}^{-1} = \text{mol L}^{-1}\text{s}^{-1}$ . For $n = 1$ : $(\text{mol L}^{-1})^{0}\text{time}^{-1} = \text{s}^{-1}$ . For $n = 2$ : $(\text{mol L}^{-1})^{-1}\text{time}^{-1} = \text{L mol}^{-1}\text{s}^{-1}$ . All three match the standard textbook units, confirming the general formula is correct and can replace memorized lists.

Worked example

The reaction $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$ is found experimentally to obey the rate law $\text{Rate} = k[NO]^2[O_2]$ . Find the order with respect to each reactant, the overall order, the units of $k$ (concentration in $\text{mol L}^{-1}$ , time in seconds), and how the initial rate changes if $[NO]$ is doubled while $[O_2]$ is kept constant.

Step 1. Read the exponents directly off the rate law: the power on $[NO]$ is 2, so the reaction is second order in $NO$ ; the power on $[O_2]$ is 1, so it is first order in $O_2$ .

Step 2. Overall order is the sum of the exponents: $2 + 1 = 3$ , so the reaction is third order overall. It happens to match the sum of the stoichiometric coefficients here, but that agreement is coincidental — the exponents were still obtained from experimental data, never assumed from the balanced equation.

Step 3. Find the units of $k$ using the general formula from the derivation, with $n = 3$ : $(\text{mol L}^{-1})^{1-3}\text{time}^{-1} = \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$ . Direct substitution confirms this: $k = \dfrac{\text{Rate}}{[NO]^2[O_2]} = \dfrac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{3}} = \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$ .

Step 4. Determine the effect of doubling $[NO]$ at constant $[O_2]$ . Since rate $\propto [NO]^2$ , doubling $[NO]$ multiplies the rate by $2^2 = 4$ , while $[O_2]$ contributes no change since it is unaltered. The new rate is 4 times the original rate, and if $[O_2]$ were also doubled at the same time, the rate would rise by a further factor of $2^1$ , giving an overall factor of $4 \times 2 = 8$ .

Competitive stretch

If instead only $[O_2]$ is tripled while $[NO]$ is halved, the rate changes by a factor of $(1/2)^2 \times 3^1 = 0.25 \times 3 = 0.75$ , so the rate would fall to three-quarters of its original value — a reminder that each concentration change must be raised to its own exponent before combining.

Common traps

Writing the rate law straight from the balanced equation's coefficients — valid only for a genuinely single-step elementary reaction, never assumed for a multi-step overall reaction.
Treating order and molecularity as always equal — they coincide only for simple elementary reactions; molecularity has no meaning for a multi-step overall process.
Forgetting order can be zero or fractional while molecularity must always be a positive whole number.
Combining concentration-change factors additively instead of multiplying each factor raised to its own exponent, as in the worked example's Step 4.

Ask the tutor: If you are stuck distinguishing order from molecularity in a specific mechanism, or need the units of $k$ worked out for an unusual rate law, ask the AI tutor to walk through it with your own reaction.

Why can we not just look at the balanced chemical equation and write the rate law from its coefficients

Because a balanced equation only tells you the overall stoichiometry — the net atoms consumed and produced — not the actual step-by-step mechanism by which the reaction happens. Most reactions proceed through several elementary steps, and the rate is controlled by the slowest of these, whose stoichiometry can look nothing like the overall equation. Only when a reaction genuinely occurs in a single step can its rate law be written from the coefficients, and even then this must be confirmed experimentally rather than assumed.

If order and molecularity are different ideas why do they often turn out to be the same number

For a reaction that occurs in a single elementary step, the rate law is determined by exactly the species that collide in that step, so the exponents in the rate law naturally match the number of colliding species — order and molecularity coincide by construction. The mismatch only appears once a reaction has multiple steps, because then the measured order reflects the slow (rate-determining) step's kinetics combined with any fast pre-equilibria, while "molecularity of the overall reaction" simply is not a defined quantity.

Does a catalyst change the order of a reaction

A catalyst can change the numerical value of the order if it opens up a different mechanism with a different rate-determining step, because order is an experimentally measured property of whichever pathway the reaction actually follows. What a catalyst does not do is change the equilibrium position or get consumed overall — it simply provides a lower activation energy route, and if that new route has a different slow step, the observed rate law and hence order can genuinely differ from the uncatalyzed reaction.

Factors Influencing Rate of a Reaction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads