Chemical Kinetics | ISC Class 12 Chemistry Notes

This note covers reaction rates, concentration and rate laws, order and molecularity, integrated rate equations, half-life, reaction mechanisms, factors affecting rates, collision theory, activation energy, catalysts and the Arrhenius equation.

What does chemical kinetics tell us about a reaction?

Chemical kinetics studies the rates of chemical reactions and their mechanisms. A reaction mechanism is the sequence of steps through which reactants become products. Kinetics asks how quickly a change occurs and which conditions alter its speed.

A reactant is a substance consumed during a chemical reaction; a product is a substance formed. Knowing their identities and the balanced equation does not by itself tell us how long the change will take.

How do feasibility, extent and speed differ?

Thermodynamics addresses whether a reaction is feasible under specified conditions. Chemical equilibrium, the state in which forward and backward reaction rates are equal, helps describe how far a reversible reaction proceeds. Kinetics addresses its speed.

Diamond can convert into graphite, but the conversion is so slow that the change is not perceptible. A feasible change need not be rapid. Predicting whether a reaction can occur and measuring how rapidly it occurs therefore answer different questions.

What are fast, slow and moderately paced reactions?

Mixing aqueous silver nitrate and sodium chloride produces silver chloride instantaneously. Here, aqueous means dissolved in water, and a precipitate is an insoluble solid formed from a solution. Rusting of iron in air and moisture is slow.

Inversion of cane sugar, its conversion into glucose and fructose, and hydrolysis of starch proceed at moderate speeds. Hydrolysis means a chemical breakdown involving reaction with water. These examples show why a single timescale cannot describe all chemical changes.

Reactions involving ions already present in solution can occur rapidly on contact. Reactions requiring substantial bond breaking and rearrangement may proceed more slowly. Bond changes and the energy needed to reach a reacting configuration help explain such differences; the balanced equation alone cannot supply the rate.

How are average and instantaneous rates measured?

Consider a constant-volume reaction R → P, where R denotes the reactant and P the product, with one mole of R forming one mole of P. Molar concentration is amount of substance per unit solution volume; square brackets, such as [R], denote it.

The symbol t denotes time. The subscripts 1 and 2 identify two measurements, and Δ means a finite change: Δt = t₂ − t₁ and Δ[R] = [R]₂ − [R]₁. Concentration is commonly measured in mol L⁻¹, moles per litre.

Definition: Average rate is the change in concentration over a specified time interval. Instantaneous rate is the rate at a particular moment.

rₐᵥ = −Δ[R]/Δt = Δ[P]/Δt, where rₐᵥ denotes average reaction rate. The negative sign before the reactant change makes its disappearance rate positive. Product concentration increases, so its appearance rate has a positive sign.

rᵢₙₛₜ = −d[R]/dt = d[P]/dt, where rᵢₙₛₜ denotes instantaneous rate and d[R]/dt is the limiting concentration change per time as the interval approaches zero. Rate has units of concentration divided by time, commonly mol L⁻¹ s⁻¹, where s denotes second.

In SI, the International System of Units, the SI unit of time is s, the second. An average rate represents its whole interval; it need not equal the rate at a chosen instant inside that interval. As reactants are consumed, the rate often changes.

How do the graphs distinguish the two rates?

The reactant concentration curve falls with time, while the product concentration curve rises. Lines joining two points give average slopes. Tangents touching the curves at a selected instant give instantaneous slopes; the reactant slope is negative.

See Fig. 3.1 in your NCERT textbook

A slope is vertical change divided by horizontal change. A tangent follows the curve's direction at the chosen point. Use the negative of a reactant concentration slope, or the positive product slope, for this one-to-one reaction.

Worked example 1. During hydrolysis of butyl chloride, C₄H₉Cl, its concentration falls from 0.100 to 0.0905 mol L⁻¹ between 0 and 50 s. Calculate its average disappearance rate.

Formula: rₐᵥ = −Δ[C₄H₉Cl]/Δt. Substitute: rₐᵥ = (0.100 − 0.0905)/50. Answer: 0.000190 mol L⁻¹ s⁻¹, or 1.90 × 10⁻⁴ mol L⁻¹ s⁻¹. The concentration decrease is divided by the full time interval.

Why must reaction rates include stoichiometric coefficients?

A stoichiometric coefficient is the number before a formula in a balanced equation. It states the relative amount consumed or formed. Different substances may disappear or appear at different rates even though they participate in the same reaction.

For 2HI → H₂ + I₂, HI is hydrogen iodide, H₂ is hydrogen and I₂ is iodine. Two moles of hydrogen iodide disappear for each mole of hydrogen or iodine formed. The unadjusted disappearance and appearance rates are therefore unequal.

r = −½d[HI]/dt = d[H₂]/dt = d[I₂]/dt, where r is the reaction rate adjusted for stoichiometry. Dividing each concentration change by its coefficient gives a common value. The same adjustment applies to average rates.

How is this used in calculations?

First write the balanced equation. Identify whether the question asks for the reaction rate or the rate of appearance or disappearance of a particular substance. Apply the coefficient before converting the time unit. This avoids confusing a species rate with the common reaction rate.

Worked example 2. Dinitrogen pentoxide, N₂O₅, decomposes according to 2N₂O₅ → 4NO₂ + O₂, where NO₂ is nitrogen dioxide and O₂ is oxygen. Its concentration falls from 2.33 to 2.08 mol L⁻¹ in 184 min at 318 K. Find the average reaction rate and nitrogen dioxide formation rate. Here min denotes minute and K denotes kelvin.

Formula: r = ([N₂O₅]₁ − [N₂O₅]₂)/(2Δt); v = 4r, where v denotes the nitrogen dioxide formation rate. Substitute: r = (2.33 − 2.08)/(2 × 184); v = 4 × 6.79 × 10⁻⁴. Answer: r = 0.000679 mol L⁻¹ min⁻¹ and v = 0.00272 mol L⁻¹ min⁻¹.

The SI unit of thermodynamic temperature is K. The SI unit of amount of substance is mol, the mole. One minute equals 60 seconds, so divide a rate per minute by 60 to express it per second.

For gases at constant temperature, concentration is proportional to partial pressure, the pressure contributed by an individual gas in a mixture. Rate may consequently be followed through pressure changes, provided the measured pressure is related correctly to the reacting species.

How do active mass and the rate law describe concentration effects?

Active mass is the concentration term used in the law of mass action. In the elementary concentration treatment, it is represented by molar concentration. Concentration describes how much reactant is available per unit volume, rather than its total amount alone.

The kinetic statement of the law of mass action makes the rate of an elementary reaction, at constant temperature, proportional to the product of reactant active masses raised to the numbers participating in that step. An elementary reaction takes place in one step.

For the elementary step A + B → products, A and B denote two reacting species, and rate = k[A][B]. The rate constant, k, is the proportionality factor. It is also called the specific rate constant.

Why is the experimental rate law essential?

A rate law expresses rate in terms of concentrations, each raised to an experimentally determined power. For a general reaction involving A and B, write r = k[A]ˣ[B]ʸ. The exponents x and y describe concentration dependence.

These powers may or may not match coefficients in the overall balanced equation. The equation records the net chemical change, while the measured rate reflects the reaction pathway. A rate law cannot be predicted merely from that overall equation.

What do experimental concentration changes show?

For 2NO + O₂ → 2NO₂, NO denotes nitrogen monoxide. The following measurements give the initial rate of NO₂ formation. Initial rate means the rate measured at the start, before appreciable reactant consumption.

Doubling [NO] at constant [O₂] multiplies the rate by four. Doubling [O₂] at constant [NO] doubles it. Thus the measured formation rate is proportional to [NO]²[O₂]. Define which rate is used consistently when evaluating its proportionality constant.

How do order, molecularity and rate-constant units differ?

The order of a reaction is the sum of the powers of concentration terms in its rate law. For r = k[A]ˣ[B]ʸ, the orders with respect to A and B are x and y, and the overall order n is their sum.

n = x + y. Order is determined experimentally and can be zero, an integer or a fraction. For r = k[A]¹ᐟ²[B]³ᐟ², the overall order is 2. For r = k[A]³ᐟ²[B]⁻¹, it is ½.

Molecularity counts the reacting atoms, ions or molecules participating in an elementary step. A unimolecular step involves one species, a bimolecular step involves two, and a termolecular step involves three. Molecularity cannot be zero or fractional.

The probability of more than three molecules colliding and reacting simultaneously is very small. Termolecular reactions are very rare and slow to proceed. Molecularity has no meaning for an overall complex reaction, which occurs through multiple elementary steps.

How are the units of k obtained?

Divide rate units by the concentration units raised to the overall order. Thus units of k = (concentration)¹⁻ⁿ(time)⁻¹. With concentration in mol L⁻¹ and time in seconds, this becomes (mol L⁻¹)¹⁻ⁿ s⁻¹.

The SI unit of a first-order rate constant is s⁻¹. Unlike rate itself, the units of k change with order. A first-order constant may also be expressed per minute when the time measurements use minutes.

Note: Adding coefficients in an overall equation does not determine reaction order. Establish the measured rate law first, and reserve molecularity for an elementary step.

How is the integrated equation for a zero-order reaction derived?

A zero-order reaction has a rate independent of reactant concentration under the conditions in which zero-order behaviour holds. For R → P, r = −d[R]/dt = k[R]⁰ = k. An integrated rate equation relates concentration directly to elapsed time.

Let [R]₀ denote initial concentration at t = 0 and [R]ₜ denote concentration at time t. Keep temperature and the relevant reaction conditions constant, so that the same k applies throughout the interval.

Derivation: Zero-order concentration equation

1. Start with −d[R]/dt = k, and rearrange to d[R] = −k dt.
2. Integrate over the concentration change from [R]₀ to [R]ₜ and time from 0 to t.
3. The result is [R]ₜ − [R]₀ = −kt. Rearrange to make the remaining concentration the subject.

[R]ₜ = [R]₀ − kt. Hence k = ([R]₀ − [R]ₜ)/t. The rate constant has concentration-per-time units because the concentration difference is divided by elapsed time.

What does the graph show?

Concentration of R is on the vertical axis and time on the horizontal axis. A straight line slopes down from the initial concentration [R]₀. The line is labelled k = −slope.

See Fig. 3.3 in your NCERT textbook

The vertical intercept, where the line meets the vertical axis, is [R]₀. Equal time intervals correspond to equal concentration decreases while the zero-order law applies. A straight concentration-time plot therefore differs from the logarithmic plot used for first-order kinetics.

Zero-order reactions are relatively uncommon and occur under special conditions. Decomposition of gaseous ammonia, NH₃, on hot platinum at high pressure is an example. Platinum is the solid metal surface on which the reaction occurs.

At high pressure, the surface is saturated with gas molecules. A further increase in ammonia concentration does not increase the amount on the occupied surface, so the rate becomes independent of its concentration. These conditions are essential.

How is the first-order rate equation derived and used?

In a first-order reaction, rate is proportional to the first power of reactant concentration. For R → P, r = −d[R]/dt = k[R]. As concentration decreases, rate decreases even though k remains constant at the given temperature.

The notation ln means natural logarithm, with base e; e is the base of natural logarithms. The notation log means logarithm to base 10; an antilogarithm reverses a logarithm. The conversion is ln z = 2.303 log z for a positive number z, using the rounded conversion factor.

Derivation: First-order concentration equation

1. Separate concentration and time in −d[R]/dt = k[R], giving d[R]/[R] = −k dt.
2. Integrate between [R]₀ and [R]ₜ, and between 0 and t, giving ln([R]ₜ/[R]₀) = −kt.
3. Reverse the concentration ratio to obtain ln([R]₀/[R]ₜ) = kt, then divide by elapsed time.

k = (1/t)ln([R]₀/[R]ₜ), or k = (2.303/t)log([R]₀/[R]ₜ). The equivalent exponential form is [R]ₜ = [R]₀e⁻ᵏᵗ. Use remaining concentration, not concentration consumed, in the denominator.

Which plots identify first-order behaviour?

The ln[R] against time plot slopes downward, with intercept ln[R]₀ and k = −slope. The log([R]₀/[R]) against time plot rises from the origin, with slope k/2.303.

See Figs. 3.4 and 3.5 in your NCERT textbook

Worked example 3. A first-order N₂O₅ decomposition at 318 K starts at 1.24 × 10⁻² mol L⁻¹. After 60 min the concentration is 0.20 × 10⁻² mol L⁻¹. Find k.

Formula: k = (2.303/t)log([R]₀/[R]ₜ). Substitute: k = (2.303/60)log(1.24/0.20). Answer: k = 0.0304 min⁻¹, equivalent to approximately 0.000507 s⁻¹ using 1 min = 60 s. Both concentration powers of ten cancel.

Worked example 4. A first-order reactant has k = 1.15 × 10⁻³ s⁻¹. Find the time for its mass to decrease from 5 g to 3 g at constant volume, where g denotes gram.

Formula: t = (2.303/k)log(initial mass/remaining mass). Substitute: t = [2.303/(1.15 × 10⁻³)]log(5/3). Answer: t ≈ 444 s. For the same substance at constant volume, the mass ratio equals the concentration ratio.

What does half-life reveal about zero- and first-order reactions?

The half-life, t₁/₂, is the time needed for reactant concentration to fall to half its initial value. Substitute [R]ₜ = [R]₀/2 into the appropriate integrated equation. The resulting dependence on initial concentration distinguishes zero-order and first-order behaviour.

Derivation: First-order half-life

1. Begin with kt = ln([R]₀/[R]ₜ), the integrated first-order equation.
2. At t = t₁/₂, replace [R]ₜ with [R]₀/2. The concentration ratio becomes 2.
3. Thus kt₁/₂ = ln 2 = 0.693, with the logarithm rounded, so divide by k.

t₁/₂ = 0.693/k. Initial concentration cancels, making first-order half-life independent of it. The rate constant and half-life are inversely related: a larger first-order k gives a shorter half-life.

For zero order, substitution into [R]ₜ = [R]₀ − kt gives [R]₀/2 = [R]₀ − kt₁/₂. Therefore t₁/₂ = [R]₀/(2k). At fixed k, zero-order half-life is directly proportional to initial concentration.

Worked example 5. A first-order reaction has k = 5.5 × 10⁻¹⁴ s⁻¹. Calculate its half-life.

Formula: t₁/₂ = 0.693/k. Substitute: t₁/₂ = 0.693/(5.5 × 10⁻¹⁴). Answer: t₁/₂ = 1.26 × 10¹³ s, or approximately 12 600 000 000 000 s. The time unit follows from the inverse-second unit of k.

How do reaction mechanisms and excess reactants affect the rate law?

A complex reaction proceeds through a sequence of elementary steps. Its overall equation is obtained by combining those steps. An intermediate is formed in one step and consumed in another, so it does not appear in the net equation.

The rate-determining step is the slowest step controlling the overall reaction rate. Studying the mechanism helps connect the experimentally measured concentration dependence with the species participating in this step.

What does iodide-catalysed peroxide decomposition illustrate?

Hydrogen peroxide, H₂O₂, decomposes to water, H₂O, and oxygen in an alkaline medium containing iodide ions, I⁻. An alkaline medium is a basic reaction environment. A catalyst increases reaction rate without undergoing permanent chemical change.

Evidence suggests the following two-step mechanism. The first step is slow: H₂O₂ + I⁻ → H₂O + IO⁻. The second step is H₂O₂ + IO⁻ → H₂O + I⁻ + O₂. Here IO⁻ is the hypoiodite intermediate.

Adding the steps cancels IO⁻ and I⁻, leaving 2H₂O₂ → 2H₂O + O₂. Iodide is regenerated, while IO⁻ is produced and then consumed. Both individual steps are bimolecular.

The measured peroxide disappearance rate is proportional to [H₂O₂][I⁻], consistent with the slow first step. It is first order in each of these species and second order overall. No molecularity should be assigned to the combined equation.

What is a pseudo-first-order reaction?

A pseudo-first-order reaction behaves as first order because another reactant is present in such large excess that its concentration changes very little. During ethyl acetate hydrolysis, water can be the excess reactant.

With 0.01 mol ethyl acetate and 10 mol water initially, complete hydrolysis consumes 0.01 mol water, leaving 9.99 mol. Water concentration is not altered much, so the observed rate depends on ethyl acetate concentration only. Large excess is essential.

Which experimental factors change reaction rate?

Reaction rate depends on the nature and concentrations of reacting substances and on experimental conditions. Changing one factor while holding the others fixed helps isolate its effect. Concentration, temperature, catalysts, exposed surface and radiation must be considered in the appropriate reaction context.

How do contact and concentration matter?

Nature of reactants includes their bonding and the changes required to form products. Ionic precipitation can be rapid, whereas reactions requiring extensive rearrangement may be slow. A rate comparison needs the actual reaction and its conditions.

Surface area matters when reaction occurs at a solid surface. Increasing exposed surface provides more contact between reacting substances and can increase rate. This explanation applies to surface reactions, rather than assuming every reaction depends on the size of solid particles.

Rates generally increase when reactant concentrations increase. The measured order determines the magnitude of this effect. For zero order, rate is independent of that concentration under the relevant conditions; for first order, doubling it doubles the rate at fixed k.

For gases, increasing pressure can increase concentration at constant temperature. This connects pressure effects with the rate law. Pressure is not an extra concentration exponent to insert without knowing which gaseous reactants influence the rate.

How do temperature, catalysts and radiation matter?

Most chemical reactions are accelerated by increasing temperature. A rise of 10° has been found to nearly double the rate constant for a chemical reaction. Treat this as an approximate observation, not an exact universal multiplication rule.

A catalyst offers a way to increase rate without simply increasing reactant concentration. Its role is connected to the energy barrier along the reaction pathway. An inhibitor is an added substance that reduces the reaction rate.

Radiation can initiate or accelerate a photochemical reaction, meaning a chemical change driven by absorbed light. The effect depends on suitable radiation being absorbed; illumination is not a general guarantee that every reaction will become faster.

How do energy barriers, collisions and catalysts explain reaction rates?

Activation energy, Eₐ, is the energy needed to reach the activated complex from the reactants. The activated complex is an unstable, short-lived configuration during reaction. The energy barrier is the rise in energy separating reactants from this configuration.

Threshold energy is the minimum energy required for an effective collision. In the elementary description, threshold energy equals activation energy plus the energy already possessed by the reacting species. Activation energy is therefore the additional energy required.

The SI unit of molar activation energy is J mol⁻¹, joules per mole. Potential energy here describes energy associated with the reacting configuration. A reaction coordinate tracks progress from reactants to products, rather than elapsed time.

How are exothermic and endothermic profiles read?

An exothermic reaction releases heat and has products at lower enthalpy than reactants. An endothermic reaction absorbs heat and has products at higher enthalpy. Enthalpy is the thermodynamic energy quantity whose change gives heat exchange at constant pressure.

Write ΔH = H(products) − H(reactants), where H denotes enthalpy. Thus ΔH is negative for an exothermic reaction and positive for an endothermic reaction. This overall change differs from the upward activation barrier.

Potential energy is plotted vertically against reaction coordinate. The curve rises from H₂ + I₂ to a peak labelled activated complex C, then falls to 2HI. The activation energy is marked from the reactant level to the peak, and ΔH marks the reactant-product difference.

See Fig. 3.7 in your NCERT textbook

Draw potential energy against reaction coordinate in two panels. In each, draw a peak and label the activated complex and the upward activation-energy gap. Place products below reactants for the exothermic profile and above them for the endothermic profile; label the corresponding negative or positive ΔH.

Which collisions are effective?

Collision theory treats reacting particles as hard spheres and relates reaction to their collisions. Collision frequency is the number of collisions per second per unit volume. An effective collision forms products through sufficient energy and proper orientation, the relative alignment of reacting particles.

Contact alone does not ensure reaction. Collisions need sufficient energy to cross the barrier and suitable orientation to break and form the required bonds. Treating molecules as hard spheres ignores structural detail, which is a limitation of this simple model.

What changes when a catalyst is present?

It is believed that a catalyst provides an alternative pathway with lower activation energy. It can participate through temporary intermediate formation and is regenerated when products form. A small amount can catalyse a large amount of reactants.

Two potential-energy curves share the same reactant and product levels. The pathway with catalyst has the lower peak. Vertical arrows compare the activation energies with and without catalyst.

See Fig. 3.11 in your NCERT textbook

A catalyst does not change the equilibrium constant, the concentration relationship characterising equilibrium at a given temperature. It accelerates forward and backward reactions to the same extent, so the same equilibrium state is reached sooner.

How does the Arrhenius equation connect temperature and rate constant?

The Arrhenius equation relates k to temperature and activation energy. Here A denotes the pre-exponential factor, also called the Arrhenius or frequency factor, rather than a reactant label. It is specific to the reaction.

Let T denote absolute temperature in kelvin, and R denote the gas constant, rather than the earlier reactant label. Use R = 8.314 J mol⁻¹ K⁻¹. The SI unit of the gas constant is J mol⁻¹ K⁻¹.

k = A exp(−Eₐ/(RT)), where exp denotes the exponential function. This factor represents the fraction of molecules with sufficient energy in this treatment. Use Eₐ in joules per mole with the stated value of R. The factor A has the same units as k.

Why does temperature affect the energetic fraction?

Molecules do not all have the same kinetic energy, the energy of motion. Their most probable kinetic energy is the energy possessed by the largest fraction of molecules. Raising temperature broadens the energy distribution and increases the proportion with high energies.

Fraction of molecules is plotted against kinetic energy. A distribution rises to a peak and then tails off. The higher-temperature curve is broader, with a lower peak farther right and a larger area beyond the marked activation energy.

See Figs. 3.8 and 3.9 in your NCERT textbook

The total area remains constant because total probability remains one. Increasing temperature increases the fraction able to cross the barrier. Counting collisions alone does not fully explain the rate change.

How are Eₐ and A obtained from a graph?

Taking logarithms gives ln k = ln A − Eₐ/(RT), or log k = log A − Eₐ/(2.303RT). The natural-log plot uses a different slope factor from the base-10 plot.

The graph plots ln k vertically against 1/T horizontally. A straight line slopes downward. Its slope is labelled −Eₐ/R and its vertical intercept ln A.

See Fig. 3.10 in your NCERT textbook

Let m denote the slope and b the vertical intercept of this plot. Then Eₐ = −mR and A = eᵇ. An ln k plot requires no extra factor of 2.303 in its slope calculation.

For rate constants k₁ and k₂ at temperatures T₁ and T₂, assuming A and Eₐ remain constant over the interval, subtract the two logarithmic equations: log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ − 1/T₂).

Worked example 6. First-order decomposition of ethyl iodide, C₂H₅I, has k₁ = 1.60 × 10⁻⁵ s⁻¹ at T₁ = 600 K and Eₐ = 209 kJ mol⁻¹, where kJ denotes kilojoule. Find k₂ at T₂ = 700 K, using R = 8.314 J mol⁻¹ K⁻¹ and 1 kJ = 1000 J.

Formula: log k₂ = log k₁ + (Eₐ/2.303R)(1/T₁ − 1/T₂); k₂ = antilog(log k₂). Substitute: log k₂ = −4.796 + [209000/(2.303 × 8.314)](1/600 − 1/700) = −2.197, using rounded logarithms. Answer: k₂ = 0.00636 s⁻¹, or 6.36 × 10⁻³ s⁻¹.

Glossary

• Chemical kinetics — The study of reaction rates, the factors affecting them and the steps through which reactions occur.
• Average rate — Concentration change over a stated interval divided by the time taken for that change.
• Instantaneous rate — The reaction rate at a particular moment, obtained from a concentration-time curve's tangent.
• Active mass — The concentration term used in the law of mass action, represented here by molar concentration.
• Rate constant — The proportionality constant in a rate law, whose units depend on the overall reaction order.
• Reaction order — The sum of the powers of concentration terms in an experimentally determined rate law.
• Molecularity — The number of reacting species taking part in a single elementary reaction step.
• Half-life — The time required for a reactant concentration to decrease to half its initial value.
• Rate-determining step — The slowest step in a reaction mechanism that controls the overall reaction rate.
• Activation energy — The energy required to reach the activated complex from the initial reacting state.
• Effective collision — A collision with sufficient energy and suitable orientation that leads to formation of products.
• Pre-exponential factor — The reaction-specific factor multiplying the exponential energy term in the Arrhenius equation.

Common errors and misconceptions

• Misconception: A negative reactant concentration change means a negative reaction rate. Correct: The minus sign in the disappearance-rate expression makes the rate positive.
• Misconception: All species in a balanced equation change concentration at the same rate. Correct: Divide each change by its stoichiometric coefficient to obtain the common reaction rate.
• Misconception: Reaction order is obtained by adding coefficients in the overall equation. Correct: Add powers in the experimental rate law; molecularity applies to elementary steps.
• Misconception: The rate constant and reaction rate are interchangeable. Correct: Rate may change as concentration changes while k remains constant at the specified temperature and conditions.
• Misconception: The first-order logarithm uses the amount consumed. Correct: Its ratio uses the initial amount divided by the amount remaining, under conditions allowing amounts to represent concentrations.
• Misconception: Every reaction has a concentration-independent half-life. Correct: First-order half-life is independent of initial concentration, whereas zero-order half-life is proportional to it.
• Misconception: A catalyst changes the final equilibrium composition by lowering product energy. Correct: It lowers the activation barrier and reaches the same equilibrium sooner.
• Misconception: A 10° temperature rise must exactly double every rate constant. Correct: Nearly doubling is an approximate observation; use Arrhenius data for a quantitative calculation.

Exam-style questions with model answers

Q1. Distinguish average rate from instantaneous rate using concentration-time graphs. [2 marks]

1. Average rate describes a finite interval and is obtained from the slope of the line joining its two concentration-time points.
2. Instantaneous rate describes one moment and uses the tangent slope there; for a reactant, take the negative slope.

Q2. For 2N₂O₅ → 4NO₂ + O₂ at constant volume, [N₂O₅] decreases from 2.33 to 2.08 mol L⁻¹ in 184 min. Calculate its average disappearance rate, the average reaction rate and the NO₂ formation rate in mol L⁻¹ min⁻¹. [3 marks]

1. The N₂O₅ concentration decrease is 2.33 − 2.08 = 0.25 mol L⁻¹. Its average disappearance rate is 0.25/184 = 1.36 × 10⁻³ mol L⁻¹ min⁻¹.
2. The balanced equation consumes two moles of N₂O₅ per reaction amount. Divide its disappearance rate by two: the average reaction rate is 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹.
3. Four moles of NO₂ form per reaction amount. Its formation rate is four times the reaction rate, giving 2.72 × 10⁻³ mol L⁻¹ min⁻¹.

Q3. First-order decomposition of N₂O₅ at 318 K reduces its concentration from 1.24 × 10⁻² to 0.20 × 10⁻² mol L⁻¹ in 60 min. Calculate k in min⁻¹ and state whether k depends on initial concentration at this temperature. Use ln z = 2.303 log z and log 6.2 = 0.7924. [4 marks]

1. For first-order decomposition, the rate constant is k = (2.303/t)log([R]₀/[R]ₜ), where t is elapsed time and the bracketed terms are initial and remaining concentrations.
2. The concentration ratio is (1.24 × 10⁻²)/(0.20 × 10⁻²) = 6.2; its units cancel because both concentrations have identical units.
3. Substitution gives k = (2.303/60) × 0.7924 = 0.0304 min⁻¹, using the time unit specified in the question.
4. At this temperature and under the same conditions, the first-order rate constant does not depend on the initial concentration.

Q4. For a first-order reaction R → P at constant temperature and volume, derive its integrated rate equation and half-life expression. Explain the half-life's concentration dependence. Let [R]₀ and [R]ₜ denote initial and remaining concentrations, t elapsed time and k the rate constant; use ln 2 = 0.693. [5 marks]

1. The first-order differential equation is −d[R]/dt = k[R]. Its negative sign makes the rate positive as reactant concentration falls during the reaction.
2. Separate the variables to give d[R]/[R] = −k dt. Integrate between the initial and remaining concentrations, with time limits 0 and t.
3. Integration gives ln([R]ₜ/[R]₀) = −kt. Hence k = (1/t)ln([R]₀/[R]ₜ), which is the required integrated equation.
4. Half-life t₁/₂ means [R]ₜ = [R]₀/2. Substituting this concentration into the integrated equation gives kt₁/₂ = ln 2 = 0.693.
5. Therefore t₁/₂ = 0.693/k. The initial concentration cancels from the expression, so the first-order half-life is independent of it at the specified temperature.

Q5. For 2NO + O₂ → 2NO₂, the initial NO₂ formation rate is 0.096 mol L⁻¹ s⁻¹ when [NO] = 0.30 and [O₂] = 0.30 mol L⁻¹. It is 0.384 when [NO] = 0.60 and [O₂] = 0.30, and 0.192 when [NO] = 0.30 and [O₂] = 0.60, in the same units and at the same temperature. Determine each reactant order, overall order, formation-rate law and units of its rate constant. [5 marks]

1. Compare the first and second measurements: doubling nitrogen monoxide concentration while holding oxygen constant multiplies the measured rate by four. The order with respect to NO is therefore two.
2. Compare the first and third measurements: doubling oxygen concentration while holding nitrogen monoxide constant doubles the measured rate. The order with respect to O₂ is one.
3. The overall order is the sum of these experimental powers, namely 2 + 1 = 3, making this a third-order rate law.
4. For the specified NO₂ formation rate v, the rate law is v = k[NO]²[O₂], where k is the proportionality constant for that measured rate.
5. Divide rate units by concentration cubed: (mol L⁻¹ s⁻¹)/(mol L⁻¹)³. The resulting units of this k are L² mol⁻² s⁻¹.

Q6. A first-order reaction has k = 5.5 × 10⁻¹⁴ s⁻¹. Calculate its half-life using ln 2 = 0.693, and explain what happens to that half-life if the initial concentration is doubled at unchanged temperature and conditions. [3 marks]

1. For a first-order reaction, half-life is given by t₁/₂ = 0.693/k. Substitute the supplied rate constant: t₁/₂ = 0.693/(5.5 × 10⁻¹⁴).
2. The half-life is 1.26 × 10¹³ s. The result is in seconds because division by a rate constant in inverse seconds produces a time.
3. Doubling the initial concentration leaves this half-life unchanged. Initial concentration does not appear in the first-order half-life equation, and k is unchanged under the stated conditions.

Q7. First-order ethyl iodide decomposition has k₁ = 1.60 × 10⁻⁵ s⁻¹ at 600 K and activation energy 209 kJ mol⁻¹. Calculate k₂ at 700 K, assuming constant activation energy and pre-exponential factor. Use R = 8.314 J mol⁻¹ K⁻¹, 1 kJ = 1000 J, ln z = 2.303 log z and log(1.60 × 10⁻⁵) = −4.796. [4 marks]

1. Convert the activation energy to match the gas constant: Eₐ = 209000 J mol⁻¹. Use both temperatures directly in kelvin.
2. The two-temperature relation is log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ − 1/T₂), where the numbered subscripts identify the two temperatures and their corresponding constants.
3. Substituting gives log(k₂/k₁) ≈ 2.599. Thus log k₂ = −4.796 + 2.599 = −2.197, using the rounded logarithmic values.
4. Taking the base-10 antilogarithm gives k₂ = 6.36 × 10⁻³ s⁻¹. This is larger than the original constant at 600 K.

Q8. Explain effective collisions and the effect of a catalyst on activation energy and equilibrium at fixed temperature. [4 marks]

1. An effective collision leads to products. The reacting particles must collide with enough energy to overcome the activation barrier separating reactants from the activated complex.
2. The particles also need proper orientation, meaning a suitable relative alignment for the required bonds to break and new bonds to form.
3. It is believed that a catalyst provides an alternative pathway with lower activation energy. The catalyst increases reaction rate without undergoing permanent chemical change.
4. The catalyst accelerates forward and backward reactions to the same extent. It reaches equilibrium sooner while leaving the equilibrium constant and final equilibrium state unchanged.

Key takeaways

• Average rate describes a time interval, whereas instantaneous rate comes from the tangent to a concentration-time curve at one moment.
• Divide each species' concentration change by its stoichiometric coefficient when calculating a common reaction rate.
• Reaction order comes from the measured rate law; molecularity counts reacting species in an elementary step.
• Zero-order concentration falls linearly with time, and its half-life depends directly on initial concentration.
• First-order kinetics gives a straight ln concentration-time plot and a half-life of 0.693 divided by the rate constant.
• Effective collisions require sufficient energy and suitable orientation; contact between particles alone does not guarantee product formation.
• A catalyst lowers the activation barrier and accelerates attainment of equilibrium without changing the equilibrium constant.
• Arrhenius calculations require kelvin temperatures, consistent energy units and careful distinction between natural and base-10 logarithms.

Test yourself

Why is a minus sign used for reactant disappearance rate?

Reactant concentration decreases with time. The minus sign converts this negative concentration change into a positive disappearance rate.

For 2HI → H₂ + I₂, how does HI disappearance compare with H₂ formation?

Hydrogen iodide disappears twice as fast as hydrogen forms, as shown by their stoichiometric coefficients.

What is the overall order of r = k[A]¹ᐟ²[B]³ᐟ²?

Add the concentration powers: one-half plus three-halves equals two, so the overall reaction order is two.

Which graph is linear for a zero-order reaction?

Reactant concentration against time is linear, with slope equal to negative k and intercept equal to initial concentration.

Does doubling initial concentration change a first-order half-life at fixed conditions?

No. The half-life remains 0.693/k because initial concentration cancels from the first-order half-life expression.

Why can hydrolysis with a large excess of water appear first order?

Water concentration changes very little, so its contribution is effectively constant and the measured rate depends on the other reactant only.

What are the slope and intercept of an ln k against 1/T plot?

The slope is −Eₐ/R and the vertical intercept is ln A, giving activation energy and the pre-exponential factor.

Why does a catalyst not change the final equilibrium state?

It accelerates forward and backward reactions to the same extent, helping the system reach the same equilibrium sooner.