NEET (UG)

Chemical Kinetics

Rate of reaction, order & molecularity, integrated rate laws and the Arrhenius equation for NEET

1
Module 1

Rate, Rate Law and Order

Rate of Reaction and the Rate LawTopic 1

Chemical kinetics is the study of how fast reactions go and what controls their speed. The rate of a reaction is the change in concentration of a reactant or product per unit time. For a reaction $\text{A} \rightarrow \text{B}$, the rate can be expressed as the rate of disappearance of A ($-\dfrac{d[A]}{dt}$) or the rate of appearance of B ($+\dfrac{d[B]}{dt}$). For a general reaction the rate is divided by each stoichiometric coefficient so that one value describes the whole reaction — a point NEET tests with the 'rate in terms of each species' question.

We distinguish the average rate (change over a measurable time interval) from the instantaneous rate (the rate at a particular instant, the slope of the concentration–time curve at that point). Rates are usually expressed in $\text{mol L}^{-1}\,\text{s}^{-1}$.

The rate depends on several factors: the concentration of reactants (more concentrated usually means faster), temperature (higher temperature speeds reactions sharply), the presence of a catalyst, the surface area of solids, and (for photochemical reactions) light. The quantitative dependence on concentration is captured by the rate law — a relationship determined experimentally, not from the balanced equation.

For a reaction $a\text{A} + b\text{B} \rightarrow$ products, the rate law takes the form $\text{rate} = k[A]^x[B]^y$, where $k$ is the rate constant and $x, y$ are the orders with respect to A and B. Crucially, $x$ and $y$ are found by experiment and need not equal the stoichiometric coefficients $a$ and $b$. The rate constant $k$ is independent of concentration but depends strongly on temperature; its units change with the overall order, which is another reliable NEET question.

Figure — Rate of Reaction and the Rate Law
ConceptMeaning
Average ratechange over a time interval
Instantaneous rateslope of conc.–time curve at an instant
Rate law$\text{rate} = k[A]^x[B]^y$ (experimental)
Rate constant $k$independent of conc., depends on T
Worked Examples
1

For $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$, write the rate in terms of each species.

Show solution

Rate $= -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = +\dfrac{1}{2}\dfrac{d[NH_3]}{dt}$ (each divided by its coefficient).

2

The rate law of a reaction is rate $= k[A]^2[B]$. What is the overall order?

Show solution

Overall order $= 2 + 1 = 3$ (third order). The reaction is second order in A and first order in B.

✎ Self-Check — 5 questions0 / 5
Q1.

The rate of a reaction is the change in concentration per unit:

Explanation: Rate = change in concentration per unit time.
Q2.

The rate law is determined:

Explanation: Rate law must be found by experiment.
Q3.

The rate constant $k$ depends on:

Explanation: $k$ is independent of concentration but depends on temperature.
Q4.

The instantaneous rate is the slope of the:

Explanation: Slope of the concentration–time curve at an instant.
Q5.

For rate $= k[A][B]^2$, the overall order is:

Explanation: $1 + 2 = 3$.

NEET tip: the rate law is experimental — orders need not match stoichiometric coefficients. Rate constant $k$ is concentration-independent but temperature-dependent; its units depend on overall order.

Order, Molecularity and the Rate ConstantTopic 2

Two terms describe how reactions depend on the species involved, and NEET frequently tests the distinction. The order of a reaction is the sum of the powers of the concentration terms in the experimentally determined rate law. It tells us how the rate responds to concentration and can be zero, a whole number, or even fractional. The molecularity, by contrast, is the number of reacting species that come together in a single elementary step; it is a theoretical quantity, always a positive whole number (1, 2, or rarely 3), and is defined only for elementary reactions.

The key differences: order is experimental and applies to the overall reaction, while molecularity is theoretical and applies only to a single step; order can be zero or fractional, but molecularity cannot. For a multi-step reaction, the overall order is governed by the slowest step, the rate-determining step — its molecularity often equals the overall order. Confusing these two is one of the most common NEET errors.

The units of the rate constant $k$ depend on the overall order, and reading them off is a guaranteed exam point. For a zero-order reaction, $k$ has units of $\text{mol L}^{-1}\,\text{s}^{-1}$; for first order, $\text{s}^{-1}$; for second order, $\text{L mol}^{-1}\,\text{s}^{-1}$. The general rule is units of $k = (\text{mol L}^{-1})^{1-\text{order}}\,\text{s}^{-1}$. You can therefore identify the order of a reaction simply from the units of its rate constant — and vice versa.

Order is found experimentally by methods such as the initial-rates method (measuring how the initial rate changes when one reactant's concentration is varied) or by fitting concentration–time data to integrated rate equations (the next topic). Once the order is known, the rate constant can be calculated, and the reaction's behaviour fully predicted. Being clear on order versus molecularity, and matching $k$'s units to the order, equips you for most conceptual NEET questions in kinetics.

Figure — Order, Molecularity and the Rate Constant
FeatureOrder vs Molecularity
Orderexperimental; can be 0 or fractional
Molecularitytheoretical; whole number, elementary steps
Units of $k$ (1st order)$\text{s}^{-1}$
Units of $k$ (zero order)$\text{mol L}^{-1}\,\text{s}^{-1}$
Worked Examples
1

State two differences between order and molecularity.

Show solution

(1) Order is experimental and can be zero or fractional; molecularity is theoretical and always a whole number. (2) Order applies to the overall reaction; molecularity applies only to an elementary step.

2

The rate constant of a reaction has units $\text{s}^{-1}$. What is its order?

Show solution

Units of $\text{s}^{-1}$ correspond to a first-order reaction (using units of $k = (\text{mol L}^{-1})^{1-\text{order}}\,\text{s}^{-1}$).

✎ Self-Check — 5 questions0 / 5
Q1.

Molecularity of a reaction is always a:

Explanation: Molecularity is a positive whole number.
Q2.

Order of a reaction can be:

Explanation: Order can be zero, fractional or whole.
Q3.

The units of the rate constant for a first-order reaction are:

Explanation: First order: $\text{s}^{-1}$.
Q4.

The slowest step of a reaction is called the:

Explanation: The slowest step determines the overall rate.
Q5.

The units of $k$ for a zero-order reaction are:

Explanation: Zero order: $\text{mol L}^{-1}\text{s}^{-1}$.

NEET trap: order = experimental (can be 0/fractional); molecularity = theoretical (whole number, elementary step only). Identify order from $k$'s units: $\text{s}^{-1}$ = 1st, $\text{mol L}^{-1}\text{s}^{-1}$ = zero.

2
Module 2

Integrated Rate Laws and Temperature Dependence

Integrated Rate Equations and Half-LifeTopic 3

The rate law tells us the rate at an instant; integrated rate equations tell us how concentration changes with time, which is what we actually measure. Each order has its own integrated form, and NEET problems usually plug values into the first-order equation.

For a zero-order reaction, the rate is independent of concentration, so $[A] = [A]_0 - kt$. A plot of $[A]$ against time is a straight line with slope $-k$. The half-life is $t_{1/2} = \dfrac{[A]_0}{2k}$, which depends on the initial concentration. Zero-order behaviour is seen in some surface-catalysed and photochemical reactions.

For a first-order reaction, the integrated equation is $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$, and a plot of $\log[A]$ against time is a straight line with slope $-\dfrac{k}{2.303}$. The most important result is the half-life: $t_{1/2} = \dfrac{0.693}{k}$, which is independent of the initial concentration. This is why radioactive decay — a first-order process — has a constant half-life regardless of how much sample remains, a fact NEET often connects to nuclear chemistry and carbon dating.

The difference in half-life behaviour is a favourite NEET discriminator: for first order the half-life is fixed, so the amount remaining after $n$ half-lives is $[A]_0/2^n$; for zero order the half-life shrinks as the reaction proceeds. Being fluent with $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ and $t_{1/2} = \dfrac{0.693}{k}$, and knowing which graph is linear for each order, handles essentially all the quantitative kinetics questions in this chapter.

Figure — Integrated Rate Equations and Half-Life
OrderIntegrated law / half-life
Zero$[A]=[A]_0-kt$; $t_{1/2}=\dfrac{[A]_0}{2k}$
First$k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$
First-order half-life$t_{1/2}=\dfrac{0.693}{k}$ (conc.-independent)
Linear plotzero: $[A]$ vs $t$; first: $\log[A]$ vs $t$
Worked Examples
1

A first-order reaction has $k = 0.0693\ \text{s}^{-1}$. Find its half-life.

Show solution

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0693} = 10\ \text{s}$ (independent of initial concentration).

2

After how many half-lives does a first-order reactant fall to $1/8$ of its initial amount?

Show solution

$[A]_0/2^n = [A]_0/8 \Rightarrow 2^n = 8 \Rightarrow n = 3$. It takes 3 half-lives.

✎ Self-Check — 5 questions0 / 5
Q1.

The half-life of a first-order reaction is:

Explanation: $t_{1/2} = 0.693/k$ for first order.
Q2.

The first-order half-life is:

Explanation: Independent of initial concentration.
Q3.

For a first-order reaction, a straight line is obtained by plotting:

Explanation: $\log[A]$ vs $t$ is linear for first order.
Q4.

For a zero-order reaction, $[A]$ versus $t$ is:

Explanation: $[A]=[A]_0-kt$ is linear.
Q5.

After 3 half-lives, the fraction of first-order reactant remaining is:

Explanation: $1/2^3 = 1/8$.

NEET tip: first order $t_{1/2}=0.693/k$ (concentration-independent; like radioactive decay); after $n$ half-lives, $[A]_0/2^n$ remains. Linear plots: zero order $[A]$ vs $t$, first order $\log[A]$ vs $t$.

Temperature, the Arrhenius Equation and CatalystsTopic 4

Reaction rates rise sharply with temperature — as a rough rule of thumb, the rate roughly doubles for every $10\ \text{K}$ increase. This strong dependence is explained quantitatively by the Arrhenius equation: $k = A\, e^{-E_a/RT}$, where $k$ is the rate constant, $A$ the frequency (pre-exponential) factor, $E_a$ the activation energy, $R$ the gas constant and $T$ the absolute temperature. The equation is central to NEET kinetics.

The activation energy $E_a$ is the minimum energy that colliding molecules must possess for a reaction to occur — the energy barrier between reactants and products. A higher $E_a$ means a slower reaction (fewer molecules can surmount the barrier), while a lower $E_a$ means a faster one. Taking logarithms gives the linear form $\ln k = \ln A - \dfrac{E_a}{RT}$, so a plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$, from which $E_a$ can be found — a standard NEET graph question.

Comparing rate constants at two temperatures gives the useful two-point form $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$, which lets you calculate $E_a$ from $k$ at two temperatures, or predict $k$ at a new temperature. The deeper picture comes from collision theory: molecules must collide with energy at least equal to $E_a$ and with the correct orientation; only such 'effective collisions' lead to product. The fraction of molecules exceeding $E_a$ rises rapidly with temperature, explaining the Arrhenius behaviour.

A catalyst increases the rate by providing an alternative pathway with a lower activation energy, so a larger fraction of collisions are effective. Importantly, a catalyst speeds up the forward and reverse reactions equally — it does not change the enthalpy change $\Delta H$ of the reaction or shift the position of equilibrium; it only helps the system reach equilibrium faster. This ties kinetics back to thermodynamics and equilibrium, and is a perennial NEET assertion-reason theme.

Figure — Temperature, the Arrhenius Equation and Catalysts
IdeaKey point
Arrhenius equation$k = A\, e^{-E_a/RT}$
Activation energyenergy barrier; higher $E_a$ to slower
Arrhenius plot$\ln k$ vs $1/T$, slope $-E_a/R$
Catalystlowers $E_a$; no change to $\Delta H$ or equilibrium
Worked Examples
1

How does increasing the temperature affect the rate constant, according to the Arrhenius equation?

Show solution

In $k = A e^{-E_a/RT}$, raising $T$ makes the exponent $-E_a/RT$ less negative, so $e^{-E_a/RT}$ increases and $k$ increases — the reaction speeds up.

2

Does a catalyst change the activation energy and the enthalpy of a reaction?

Show solution

A catalyst lowers the activation energy by offering a new pathway, but it does not change the enthalpy change $\Delta H$ — the energies of reactants and products are unchanged, only the barrier between them is lowered.

✎ Self-Check — 5 questions0 / 5
Q1.

The Arrhenius equation is:

Explanation: $k = A e^{-E_a/RT}$.
Q2.

A higher activation energy makes a reaction:

Explanation: Higher $E_a$ means fewer effective collisions, so slower.
Q3.

An Arrhenius plot of $\ln k$ vs $1/T$ has slope:

Explanation: Slope $= -E_a/R$.
Q4.

A catalyst increases the rate by:

Explanation: It provides a lower-$E_a$ pathway.
Q5.

An effective collision requires sufficient energy and:

Explanation: Energy ≥ $E_a$ AND proper orientation.

NEET tip: $k=Ae^{-E_a/RT}$; $\ln k$ vs $1/T$ has slope $-E_a/R$. Two-point: $\log\frac{k_2}{k_1}=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$. Catalyst lowers $E_a$ only — never changes $\Delta H$ or equilibrium.

Quick Revision — Chemical Kinetics

  • Rate = change in concentration per unit time; average vs instantaneous rate.
  • Rate law: $\text{rate} = k[A]^x[B]^y$; order $= x+y$ (experimental), molecularity = number of colliding species (theoretical, for elementary steps).
  • Zero order: rate independent of concentration; $[A] = [A]_0 - kt$; units of $k$ = mol L⁻¹ s⁻¹.
  • First order: $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$; units of $k$ = s⁻¹; half-life $t_{1/2} = \dfrac{0.693}{k}$ (independent of initial concentration).
  • Arrhenius equation: $k = A e^{-E_a/RT}$ — rate constant rises with temperature and falls with activation energy.
  • Collision theory: reaction needs collisions with sufficient energy ($E_a$) and correct orientation.
  • A catalyst speeds a reaction by providing a lower-$E_a$ path; it does not change $\Delta H$ or the equilibrium.

Frequently Asked Questions

What is the difference between order and molecularity?
Order is the experimentally determined sum of the powers of concentration terms in the rate law and can be zero or fractional, whereas molecularity is the number of reacting species in an elementary step and is always a whole number. Order applies to overall reactions; molecularity only to elementary steps.
Why is the half-life of a first-order reaction independent of concentration?
For a first-order reaction the half-life is t½ = 0.693/k, which contains only the rate constant and no concentration term. So however much reactant you start with, it always takes the same time for half of it to react — which is why radioactive decay (first order) has a fixed half-life.
What does the Arrhenius equation tell us?
The Arrhenius equation k = A·e^(−Ea/RT) shows that the rate constant increases with temperature and decreases with a higher activation energy. It explains why even a small temperature rise can noticeably speed up a reaction, since the exponential term is very sensitive to T.
How does a catalyst increase the rate of a reaction?
A catalyst provides an alternative reaction pathway with a lower activation energy, so a larger fraction of molecular collisions have enough energy to react. It speeds up both forward and reverse reactions equally and is not consumed, so it does not change the enthalpy change or the position of equilibrium.
What are the two requirements for an effective collision in collision theory?
According to collision theory, a collision leads to reaction only if the colliding molecules have energy equal to or greater than the activation energy and if they collide with the correct orientation. Collisions that lack either condition do not produce products.

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