Chemical Kinetics

The rate of a reaction is the change in concentration of a reactant or product per unit time. For a reaction R → P, if the concentration of R falls from [R]₁ to [R]₂ in the time t₁ to t₂, the average rate is r_av = −Δ[R]/Δt = +Δ[P]/Δt. The minus sign is there because [R] decreases; rate is always reported as a positive quantity. The unit is mol L⁻¹ s⁻¹, or, for gases, bar s⁻¹ or atm s⁻¹ if concentration is expressed as partial pressure.

The average rate depends on which interval you pick, so the chapter moves to the instantaneous rate: the rate at one particular moment, obtained by shrinking Δt to zero, r_inst = −d[R]/dt = d[P]/dt. Graphically it is the slope of the tangent to the concentration–time curve at that instant. When the examiner draws [R] against t, the tangent slopes downward and the rate is the negative of that slope; when [P] is plotted, the slope itself is the rate.

When the stoichiometric coefficients are not all one, divide each rate by its coefficient so that everyone quotes the same number. For 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O, rate = −(1/5) d[Br⁻]/dt = −d[BrO₃⁻]/dt = −(1/6) d[H⁺]/dt = (1/3) d[Br₂]/dt. For 2HI → H₂ + I₂, rate = −½ d[HI]/dt = d[H₂]/dt = d[I₂]/dt. Notice that hydrogen iodide disappears twice as fast as hydrogen appears.

The reaction rate itself falls as the reaction proceeds because reactant concentration keeps decreasing; this is the whole reason a single number cannot describe a reaction and why we need the rate law and the rate constant of the next section.

The rate of a reaction depends on the concentration of the reactants (or pressure for gases), the temperature, the presence of a catalyst, and for heterogeneous reactions the surface area of the solid. The chapter treats concentration first, then temperature and the catalyst in Section 3.4.

Experiment shows that the rate of most reactions falls as the concentration of reactants decreases and rises as it increases. The quantitative statement of this is the rate law: an expression in which the rate is written in terms of the molar concentrations of the reactants, each raised to some power. For aA + bB → cC + dD the general form is rate = k[A]^x[B]^y. The key point, and the one the examiner tests, is that x and y need not be equal to a and b. They are found by experiment, not by reading the balanced equation.

NCERT gives examples that show this. For 2NO(g) + O₂(g) → 2NO₂(g), the rate law is rate = k[NO]²[O₂], which happens to match the coefficients. But for CHCl₃ + Cl₂ → CCl₄ + HCl the rate law is rate = k[CHCl₃][Cl₂]^1/2, and for CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH it is rate = k[CH₃COOC₂H₅][H₂O]⁰. Neither can be predicted from the equation.

The way the rate law is found in the exam is the method of initial rates: keep every concentration fixed except one, see how the initial rate changes, and read off the power. If doubling [A] doubles the rate, x = 1; if it quadruples the rate, x = 2; if the rate does not change, x = 0. Once the powers are known, substitute one full row of data to get k.

The differential rate equation writes the instantaneous rate in terms of concentrations: −d[R]/dt = k[A]^x[B]^y. The proportionality constant k is the rate constant or specific reaction rate. It is the rate of the reaction when the concentration of every reactant is unity (1 mol L⁻¹). Two facts about k are tested again and again: it does not change with the concentration of reactants, and it does change with temperature (and with a catalyst). At a given temperature it is a characteristic of the reaction.

The units of k depend on the overall order because rate always has the unit mol L⁻¹ s⁻¹. Rearranging, k = rate/[concentration]ⁿ, so the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹. For n = 0 that is mol L⁻¹ s⁻¹; for n = 1 it is s⁻¹; for n = 2 it is L mol⁻¹ s⁻¹; for n = 3 it is L² mol⁻² s⁻¹. A fractional order of 3/2 gives L^1/2 mol^−1/2 s⁻¹. If a question gives the unit of k and asks the order, work backwards: s⁻¹ means first order, L mol⁻¹ s⁻¹ means second order, mol L⁻¹ s⁻¹ means zero order.

For gas-phase reactions followed by pressure, the same logic holds with bar or atm in place of mol L⁻¹. In NCERT Exercise 3.4 the decomposition of dimethyl ether has rate = k(p_CH3OCH3)^3/2; with pressure in bar and time in minutes the unit of rate is bar min⁻¹ and the unit of k is bar^−1/2 min⁻¹.

Do not confuse rate and rate constant. Rate depends on concentration and falls during the reaction; the rate constant does not. Rate has one fixed unit; the unit of the rate constant tells you the order.

In the rate law rate = k[A]^x[B]^y, x and y are the orders with respect to A and B, and the sum x + y is the overall order of the reaction. So order is the sum of the powers of the concentration terms in the experimentally determined rate law. It can be 0, 1, 2, 3, or even a fraction; it applies to both elementary and complex reactions; and a zero order reaction means the rate is independent of concentration.

NCERT's examples for units and orders: 3NO → N₂O with rate = k[NO]² is second order; H₂O₂ + 3I⁻ + 2H⁺ → 2H₂O + I₃⁻ with rate = k[H₂O₂][I⁻] is second order overall (first in each); CH₃CHO → CH₄ + CO with rate = k[CH₃CHO]^3/2 is of order 1.5; C₂H₅Cl → C₂H₄ + HCl with rate = k[C₂H₅Cl] is first order.

Reactions are also classified as elementary or complex. An elementary reaction takes place in a single step. A complex reaction is a sequence of elementary steps; the overall reaction is their sum, and the slowest step, the rate-determining step, decides the rate law. NCERT's example is the oxidation of HBr by oxygen, 4HBr + O₂ → 2H₂O + 2Br₂. The rate law is rate = k[HBr][O₂] because the first, slow step is HBr + O₂ → HOOBr; the fast steps HOOBr + HBr → 2HOBr and HOBr + HBr → H₂O + Br₂ do not appear in the rate law. Species like HOOBr and HOBr that are formed and consumed in between are intermediates.

In a numerical you may be asked to write the rate law from a proposed mechanism: use the slow step, with the concentration of each species in it raised to its coefficient in that step. If the slow step contains an intermediate, substitute its concentration from the preceding fast equilibrium.

The molecularity of an elementary reaction is the number of reacting species (atoms, ions or molecules) that must collide simultaneously to bring about the reaction. A unimolecular reaction has one species: NH₄NO₂ → N₂ + 2H₂O. A bimolecular reaction has two: 2HI → H₂ + I₂. A trimolecular (termolecular) reaction has three: 2NO + O₂ → 2NO₂. The chance of more than three molecules colliding at the same instant with the right energy and orientation is negligible, so reactions of molecularity greater than three are not observed. This is the explanation for why a reaction such as KClO₃ + 6FeSO₄ + 3H₂SO₄ → KCl + 3Fe₂(SO₄)₃ + 3H₂O, which looks like it involves ten species, actually proceeds in steps and is a second order reaction.

The comparison between order and molecularity is one of the most frequently asked 2- or 3-mark questions. Write it as clear pairs. Order is an experimental quantity; molecularity is a theoretical concept. Order can be zero or fractional; molecularity is always a positive whole number, never zero or fractional. Order applies to both elementary and complex reactions; molecularity is defined only for elementary reactions and has no meaning for the overall complex reaction. For an elementary reaction the two are usually equal; for a complex reaction the order is given by the slowest step and the molecularity of that slowest step is what matters.

The decomposition of hydrogen peroxide catalysed by iodide ion, 2H₂O₂ → 2H₂O + O₂, is NCERT's illustration. It looks bimolecular in H₂O₂, but the rate law is rate = k[H₂O₂][I⁻], first order in each: the reaction goes through the slow step H₂O₂ + I⁻ → H₂O + IO⁻ followed by the fast step H₂O₂ + IO⁻ → H₂O + O₂ + I⁻, and both steps are bimolecular.

A differential rate law tells us the rate at a given concentration, but to compare theory with experiment we need concentration as a function of time. Integrating the rate law gives the integrated rate equation, and each order gives a different one.

In a zero order reaction the rate is independent of the concentration of the reactant. For R → P, rate = −d[R]/dt = k[R]⁰ = k. Separating the variables, d[R] = −k dt, and integrating, [R] = −kt + I, where I is the constant of integration. At t = 0, [R] = [R]₀, so I = [R]₀. Hence [R] = −kt + [R]₀, or k = ([R]₀ − [R])/t. This is the derivation the board asks for; write each of those lines.

The equation has the form of a straight line y = mx + c. A plot of [R] against t is a straight line with slope −k and intercept [R]₀. The unit of k is mol L⁻¹ s⁻¹, the same as the unit of rate.

Zero order reactions are rare and usually involve a saturated surface or a saturated catalyst. NCERT's examples are the decomposition of ammonia on a hot platinum surface at high pressure, 2NH₃(g) → N₂(g) + 3H₂(g) with rate = k[NH₃]⁰ = k (the metal surface is saturated, so raising the gas pressure changes nothing), the thermal decomposition of HI on a gold surface, and some enzyme-catalysed reactions.

In a numerical, the typical form is: k = 0.0030 mol L⁻¹ s⁻¹, find the time for [R] to fall from 0.10 to 0.075 mol L⁻¹. Then t = ([R]₀ − [R])/k = 0.025/0.0030 = 8.33 s.

In a first order reaction the rate is proportional to the first power of the concentration of the reactant: for R → P, rate = −d[R]/dt = k[R]. This is the derivation that appears most often in the 5-mark question, so learn every line.

Rearranging, d[R]/[R] = −k dt. Integrating, ln[R] = −kt + I. At t = 0, [R] = [R]₀, so I = ln[R]₀. Therefore ln[R] = −kt + ln[R]₀, which rearranges to ln([R]₀/[R]) = kt, or k = (1/t) ln([R]₀/[R]). Converting to base-10 logarithms (ln x = 2.303 log x): k = (2.303/t) log([R]₀/[R]). Equivalently [R] = [R]₀ e^−kt, which shows the exponential decay of the reactant. Writing the equation for two times t₁ and t₂ and subtracting gives ln([R]₁/[R]₂) = k(t₂ − t₁).

Graphs: ln[R] versus t is a straight line with slope −k and intercept ln[R]₀; log[R] versus t has slope −k/2.303; and log([R]₀/[R]) versus t is a straight line through the origin with slope k/2.303. The last one is what NCERT uses to test first order kinetics from data.

Examples: the hydrogenation of ethene, C₂H₄ + H₂ → C₂H₆ with rate = k[C₂H₄]; all natural and artificial radioactive decay, such as ²²⁶Ra → ⁴He + ²²²Rn; and the decomposition of N₂O₅ and of N₂O.

For gas-phase reactions followed by total pressure, say A(g) → B(g) + C(g) with initial pressure p_i: if the pressure of A that has decomposed at time t is x, then p_A = p_i − x and p_total = p_i + x, so p_A = 2p_i − p_t. Then k = (2.303/t) log[p_i/(2p_i − p_t)]. This is exactly the working for the azoisopropane and SO₂Cl₂ exercises.

The half-life t_1/2 of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. Its dependence on concentration is different for each order, and that difference is what the board tests.

Zero order. From k = ([R]₀ − [R])/t, put t = t_1/2 and [R] = [R]₀/2: k = ([R]₀ − ½[R]₀)/t_1/2, so t_1/2 = [R]₀/2k. The half-life is directly proportional to the initial concentration and inversely proportional to k.

First order. From k = (2.303/t) log([R]₀/[R]), put t = t_1/2 and [R] = [R]₀/2: k = (2.303/t_1/2) log 2, so t_1/2 = 2.303 × 0.301/k = 0.693/k. The half-life of a first order reaction is independent of the initial concentration; it is a constant for the reaction at a given temperature. This is why 'the half-life does not change when the initial concentration is doubled' is the exam's signal for first order. Because each half-life is the same, after n half-lives the fraction remaining is (½)ⁿ: 1/16 remains after four half-lives, 1/8 after three.

General pattern, worth quoting: for an nth order reaction t_1/2 ∝ [R]₀¹⁻ⁿ. So for second order the half-life is inversely proportional to the initial concentration.

Worked pair: k = 200 s⁻¹ gives t_1/2 = 0.693/200 = 3.47 × 10⁻³ s; a first order reaction with t_1/2 = 693 s has k = 10⁻³ s⁻¹, and the time for 90% completion is (2.303/10⁻³) log 10 = 2303 s. Also learn that the time for 99% completion is twice that for 90%: t_99% = (2.303/k) log 100 = 4.606/k and t_90% = (2.303/k) log 10 = 2.303/k, so the ratio is 2.

The order of a reaction can be altered by the conditions. When one reactant is present in such large excess that its concentration hardly changes during the reaction, that concentration behaves like a constant, gets absorbed into the rate constant, and the reaction shows first order kinetics even though it is really of higher order. Such a reaction is called a pseudo first order reaction.

The standard examples, and the ones to quote, are the acid-catalysed hydrolysis of ethyl acetate and the inversion of cane sugar. In CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, the reaction is bimolecular and would be expected to be second order, but water is the solvent and is present in enormous excess (NCERT's illustration: 0.01 mol of ester in 10 mol of water; at the end there are still 9.99 mol of water). So rate = k′[CH₃COOC₂H₅][H₂O] becomes rate = k[CH₃COOC₂H₅] with k = k′[H₂O]. Similarly for C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose) in acid, rate = k[C₁₂H₂₂O₁₁].

In the 2-mark form the examiner wants the definition, an example and the reason (one reactant in large excess so its concentration is effectively constant). In NCERT Exercise 3.8 you are given [A] at 0, 30, 60 and 90 s for a pseudo first order reaction in water and asked for the average rate between 30 and 60 s and the rate constant; use k = (2.303/t) log([A]₀/[A]) at each time and average the values.

The rate of most reactions rises sharply with temperature. A useful rule of thumb from NCERT: for many reactions a 10° rise in temperature nearly doubles the rate constant. The quantitative statement is the Arrhenius equation, k = A e^−E_a/RT, where A is the Arrhenius factor or frequency factor (pre-exponential factor), R is the gas constant, T the absolute temperature and E_a the activation energy in J mol⁻¹. The factor e^−E_a/RT is the fraction of molecules with kinetic energy equal to or greater than E_a.

Why does the rate rise? Reactants must first form an unstable, high-energy activated complex (the transition state) before they turn into products. The energy needed to lift the reactants to this state is the activation energy. In the hydrogen–iodine reaction the H–H and I–I bonds are partly broken and the H–I bonds partly formed in the activated complex, which then breaks down into 2HI. On a plot of potential energy against the reaction coordinate, E_a is the height of the peak above the reactants, and the difference between reactant and product levels is ΔH.

According to the Maxwell–Boltzmann distribution of kinetic energies, only the molecules in the tail of the curve, beyond the threshold energy, can react. Raising the temperature broadens the curve, shifts its peak to higher energy, and roughly doubles the area under the curve beyond E_a. So even a small rise in temperature greatly increases the number of effective collisions and hence the rate.

Taking the natural logarithm of the Arrhenius equation, ln k = −E_a/RT + ln A. A plot of ln k against 1/T is a straight line with slope −E_a/R and intercept ln A; with base-10 logs the slope of log k versus 1/T is −E_a/2.303R. For rate constants k₁ at T₁ and k₂ at T₂, subtracting gives ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂), or log(k₂/k₁) = (E_a/2.303R) [(T₂ − T₁)/(T₁T₂)]. This two-temperature form is what every Arrhenius numerical uses. Example: the rate doubles from 298 K to 308 K, so log 2 = E_a/(2.303 × 8.314) × (10/(298 × 308)); E_a = 0.301 × 19.15 × 91,784/10 = 52,900 J ≈ 52.9 kJ mol⁻¹.

A catalyst is a substance that increases the rate of a reaction without itself undergoing any permanent chemical change. NCERT's example is the decomposition of KClO₃, which is slow on heating alone but rapid in the presence of a small amount of MnO₂.

The intermediate complex theory explains how it works: the catalyst forms a temporary bond with the reactants, giving an intermediate complex of lower energy that decomposes to the products and regenerates the catalyst. The result is a new reaction path with a lower activation energy. On the potential energy diagram the catalysed path has a lower peak than the uncatalysed path, while the energies of the reactants and products, and hence ΔH, are unchanged. Because E_a is smaller, the fraction of molecules with energy above E_a (e^−E_a/RT) is larger and the rate rises at the same temperature.

Three facts about a catalyst are marking points in any question. A small amount of catalyst can catalyse a large amount of reactants. A catalyst does not change the Gibbs energy ΔG of a reaction: it catalyses spontaneous reactions but not non-spontaneous ones. A catalyst does not change the equilibrium constant or the position of equilibrium; it increases the rates of the forward and backward reactions equally and so helps the equilibrium to be reached faster.

Do not say that a catalyst 'lowers the activation energy of the reaction'. It provides an alternative pathway with lower activation energy; the activation energy of the original path is what it was. Also, a catalyst does not initiate a reaction that would not otherwise take place.

Collision theory, developed by Max Trautz and William Lewis in 1916–18, treats reacting molecules as hard spheres and says that a reaction happens when molecules collide. The number of collisions per second per unit volume of the reaction mixture is the collision frequency Z. For a bimolecular elementary reaction A + B → products, rate = Z_AB e^−E_a/RT, where Z_AB is the collision frequency of A and B and e^−E_a/RT is the fraction of molecules with energy equal to or greater than E_a. Comparing with the Arrhenius equation shows that A, the frequency factor, corresponds to the collision frequency.

Not every collision leads to product. Two conditions must be met. First, the colliding molecules must have kinetic energy at least equal to the threshold energy, the minimum energy required for an effective collision; the activation energy is the difference between the threshold energy and the average energy of the reactants. Second, they must be correctly oriented. NCERT's example is the formation of ethanol from bromoethane: C₂H₅Br + OH⁻ → C₂H₅OH + Br⁻. Only when OH⁻ approaches the carbon from the side opposite the bromine does the collision give products; other orientations give no reaction. A collision that satisfies both the energy and the orientation conditions is an effective collision.

To account for orientation, a probability or steric factor P is introduced: rate = P Z_AB e^−E_a/RT. Thus the activation energy and the proper orientation together decide the rate.

Limitations: the theory treats molecules as hard spheres and ignores their structure, so it works reasonably for simple atomic or small-molecule reactions but gives significant deviations for complex molecules. This elementary picture is all the syllabus asks; no further mathematics is needed.