NCERT Solutions for Class 12 Chemistry Chapter 3: Chemical Kinetics
Complete NCERT Class 12 Chemistry Chapter 3 Chemical Kinetics — all 9 intext questions and 30 exercises answered in topper-style detail, with full revision notes on rate of reaction, rate law, order and molecularity, integrated rate equations, half-life, Arrhenius equation, catalysis and collision theory.
Chapter at a glance — complete revision notes
Introduction
- Thermodynamics tells whether a reaction is feasible (Δ_rG < 0) but not how fast it occurs; chemical kinetics studies the rate of reaction, the factors affecting it (concentration, temperature, catalyst), and the mechanism (the sequence of elementary steps). Some reactions are very fast (precipitation of AgCl from AgNO₃ + NaCl — instantaneous), some very slow (rusting of iron), and some moderate (inversion of cane sugar, hydrolysis of starch).
3.1 Rate of a chemical reaction
- Rate = change in concentration of a reactant or product per unit time. For R → P: rate of disappearance of R = −Δ[R]/Δt (negative Δ[R] is multiplied by −1 to make the rate positive); rate of appearance of P = +Δ[P]/Δt. These are average rates (r_av) over a finite interval (Table 3.1, hydrolysis of butyl chloride: C₄H₉Cl + H₂O → C₄H₉OH + HCl; r_av falls from 1.90 × 10⁻⁴ to 0.4 × 10⁻⁴ mol L⁻¹ s⁻¹ as [C₄H₉Cl] falls from 0.100 to 0.017 M).
- Instantaneous rate = limit of average rate as Δt → 0: r_inst = −d[R]/dt = +d[P]/dt; found as the slope of the tangent to the concentration–time curve at time t (Fig 3.1, 3.2; e.g. r_inst at 600 s = 5.12 × 10⁻⁵ mol L⁻¹ s⁻¹). Units of rate: concentration time⁻¹, i.e. mol L⁻¹ s⁻¹ (atm s⁻¹ for gases using partial pressures).
- Stoichiometry: for Hg(l) + Cl₂(g) → HgCl₂(s) rates of loss and gain are equal; for 2HI → H₂ + I₂, Rate = −½Δ[HI]/Δt = Δ[H₂]/Δt = Δ[I₂]/Δt (divide each by its stoichiometric coefficient); for 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O: Rate = −⅕Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt = −⅙Δ[H⁺]/Δt = ⅓Δ[Br₂]/Δt = ⅓Δ[H₂O]/Δt. In general aA + bB → cC + dD: Rate = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = (1/d)Δ[D]/Δt. Example 3.2: for 2N₂O₅ → 4NO₂ + O₂ in CCl₄ at 318 K, average rate = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹ = 1.13 × 10⁻⁵ mol L⁻¹ s⁻¹ and rate of NO₂ formation = 4 × 6.79 × 10⁻⁴ = 2.72 × 10⁻³ mol L⁻¹ min⁻¹.
3.2 Factors influencing the rate of a reaction
- Rate depends on concentration of reactants (pressure for gases), temperature and catalyst (also surface area, light for photochemical reactions).
3.2.1–3.2.2 Rate law and rate constant
For aA + bB → cC + dD: Rate = k[A]ˣ[B]ʸ (rate law / rate expression); differential form −d[R]/dt = k[A]ˣ[B]ʸ; k is the rate constant (rate when all concentrations are 1 mol L⁻¹). The exponents x and y need not equal the stoichiometric coefficients — the rate law is determined experimentally and cannot be predicted from the balanced equation. 2NO + O₂ → 2NO₂: Rate = k[NO]²[O₂] (Table 3.2: doubling [NO] quadruples the rate, doubling [O₂] doubles it). Examples where they differ: CHCl₃ + Cl₂ → CCl₄ + HCl, Rate = k[CHCl₃][Cl₂]^½; CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, Rate = k[CH₃COOC₂H₅]¹[H₂O]⁰.3.2.3 Order of a reaction
The sum of the powers of the concentration terms in the rate law (x + y); x and y are orders with respect to A and B. Order can be 0, 1, 2, 3, fractional or even negative (Example 3.3: Rate = k[A]^½[B]^3/2 is of order 2; Rate = k[A]^3/2[B]⁻¹ is of order ½). Zero order — the rate is independent of concentration. Elementary reactions occur in one step; complex reactions occur through a mechanism (consecutive, reverse and side reactions).- Units of k: k = rate ÷ (concentration)ⁿ — zero order: mol L⁻¹ s⁻¹; first order: s⁻¹; second order: L mol⁻¹ s⁻¹ (mol⁻¹ L s⁻¹); n-th order: (mol L⁻¹)¹⁻ⁿ s⁻¹ (Table 3.3).
3.2.4 Molecularity
The number of reacting species (atoms, ions, molecules) that must collide simultaneously in an elementary reaction: unimolecular (NH₄NO₂ → N₂ + 2H₂O), bimolecular (2HI → H₂ + I₂), termolecular (2NO + O₂ → 2NO₂). Beyond 3 is practically impossible. Reactions with many species in the equation (e.g. KClO₃ + 6FeSO₄ + 3H₂SO₄ → KCl + 3Fe₂(SO₄)₃ + 3H₂O, apparently tenth order but actually second order) proceed in several steps; the slowest step is the rate-determining step. Example: 2H₂O₂ → 2H₂O + O₂ catalysed by I⁻ in alkaline medium, Rate = k[H₂O₂][I⁻]: step 1 (slow) H₂O₂ + I⁻ → H₂O + IO⁻; step 2 (fast) H₂O₂ + IO⁻ → H₂O + I⁻ + O₂; IO⁻ is an intermediate.- Order vs molecularity: order is experimental, can be 0 or fractional, applies to elementary and complex reactions; molecularity is theoretical, a whole number 1–3, defined only for elementary reactions (meaningless for a complex reaction); for a complex reaction the order is determined by the slowest step, whose molecularity equals the overall order.
3.3 Integrated rate equations
3.3.1 Zero order
R → P, rate = −d[R]/dt = k → [R] = −kt + [R]₀ (a plot of [R] against t is a straight line, slope −k, intercept [R]₀); k = ([R]₀ − [R])/t. Examples: decomposition of NH₃ on hot platinum at high pressure (2NH₃ → N₂ + 3H₂, rate = k, as the Pt surface is saturated), decomposition of HI on gold, some enzyme-catalysed reactions.3.3.2 First order
R → P, rate = −d[R]/dt = k[R] → ln[R] = −kt + ln[R]₀; k = (1/t) ln([R]₀/[R]) = (2.303/t) log([R]₀/[R]); [R] = [R]₀e^(−kt); between two times k = (2.303/(t₂ − t₁)) log([R]₁/[R]₂). Plots: ln[R] against t is a straight line (slope −k); log([R]₀/[R]) against t is a straight line (slope k/2.303). Examples: hydrogenation of ethene (rate = k[C₂H₄]), radioactive decay (²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He), decomposition of N₂O₅ and N₂O. Example 3.5: N₂O₅ 1.24 × 10⁻² → 0.20 × 10⁻² M in 60 min gives k = 0.0304 min⁻¹.- Gas-phase first order, A(g) → B(g) + C(g): p_t = p_i + x; p_A = 2p_i − p_t; k = (2.303/t) log[p_i/(2p_i − p_t)] (Example 3.6: N₂O₅, p_i = 0.5 atm, p_t = 0.512 atm at 100 s: k = 4.98 × 10⁻⁴ s⁻¹ using p(N₂O₅) = 1.5 − 2p_t for 2N₂O₅ → 2N₂O₄ + O₂).
3.3.3 Half-life
Time for the concentration to fall to one half. Zero order: t½ = [R]₀/2k (proportional to initial concentration). First order: t½ = 0.693/k (independent of initial concentration; k = 0.693/t½). Example 3.7: k = 5.5 × 10⁻¹⁴ s⁻¹ gives t½ = 1.26 × 10¹³ s. Example 3.8: for 99.9 % completion t = 6.909/k = 10 × t½. Table 3.4 summarises: zero order — d[R]/dt = −k, kt = [R]₀ − [R], plot [R] vs t, t½ = [R]₀/2k, k in mol L⁻¹ s⁻¹; first order — d[R]/dt = −k[R], [R] = [R]₀e^(−kt) or kt = ln([R]₀/[R]), plot ln[R] vs t, t½ = ln2/k, k in s⁻¹.- Pseudo first order reactions: reactions of higher true order that behave as first order because one reactant is in large excess so its concentration is practically constant — hydrolysis of ethyl acetate (CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH with 0.01 mol ester in 10 mol water, rate = k′[ester]) and inversion of cane sugar (C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose) in H⁺, rate = k[C₁₂H₂₂O₁₁]).
3.4 Temperature dependence of the rate of a reaction
- For most reactions, a 10 °C rise nearly doubles the rate constant (N₂O₅ decomposition t½: 12 min at 50 °C, 5 h at 25 °C, 10 days at 0 °C; KMnO₄ + oxalic acid decolourises faster when hot). Arrhenius equation: k = A e^(−Ea/RT) (A = Arrhenius (pre-exponential/frequency) factor, Ea = activation energy in J mol⁻¹, R = 8.314 J K⁻¹ mol⁻¹); proposed by van’t Hoff and justified by Arrhenius.
- Activated complex and Ea: reactants must collide to form an unstable activated complex (intermediate) (e.g. H₂ + I₂ → [H···I···I···H] → 2HI); Ea = energy of the activated complex − energy of reactants (Fig 3.7); the enthalpy change depends on the reactants and products. Molecules have a spread of kinetic energies (Maxwell–Boltzmann distribution, Fig 3.8); at a higher temperature (t + 10) the peak shifts to higher energy and broadens, so the fraction of molecules with energy ≥ Ea (the area beyond Ea) roughly doubles (Fig 3.9). The factor e^(−Ea/RT) is that fraction.
- ln k = −Ea/RT + ln A (3.19); a plot of ln k against 1/T is a straight line with slope = −Ea/R and intercept = ln A (Fig 3.10); log k = log A − Ea/(2.303RT). For two temperatures: log(k₂/k₁) = (Ea/2.303R)·(1/T₁ − 1/T₂) = (Ea/2.303R)·(T₂ − T₁)/(T₁T₂) (3.22). Examples: k = 0.02 s⁻¹ at 500 K and 0.07 s⁻¹ at 700 K ⇒ Ea = 18230.8 J mol⁻¹ and A = 1.61 (Example 3.9); C₂H₅I → C₂H₄ + HI, k = 1.60 × 10⁻⁵ s⁻¹ at 600 K, Ea = 209 kJ mol⁻¹ ⇒ k(700 K) = 6.36 × 10⁻³ s⁻¹ (Example 3.10).
3.4.1 Effect of catalyst
A catalyst increases the rate without being permanently changed (MnO₂ in 2KClO₃ → 2KCl + 3O₂); a substance that decreases the rate is an inhibitor. By the intermediate complex theory it provides an alternative pathway with a lower activation energy (Fig 3.11), so k rises. It does not change Δ_rG or the equilibrium constant; it catalyses spontaneous reactions only and speeds up forward and backward reactions equally.
3.5 Collision theory of chemical reactions
- Developed by Max Trautz and William Lewis (1916–18) from the kinetic theory of gases: reactant molecules are hard spheres and the reaction occurs on collision. The number of collisions per second per unit volume is the collision frequency Z. For a bimolecular elementary reaction A + B → products, Rate = Z_AB e^(−Ea/RT); so A is related to the collision frequency.
- Not every collision leads to reaction: only effective collisions — those in which molecules have energy ≥ threshold energy (threshold energy = activation energy + energy possessed by the reacting species) and the proper orientation (Fig 3.12; e.g. CH₃Br + OH⁻ → CH₃OH + Br⁻ depends on the orientation of the molecules). A probability or steric factor P is introduced: Rate = P Z_AB e^(−Ea/RT) (k = P Z_AB e^(−Ea/RT)). Limitation: it treats molecules as hard spheres and ignores structure.
Summary (NCERT)
- Rate = decrease in reactant (or increase in product) concentration per unit time; instantaneous vs average rate; rate law is experimental; order = sum of powers; molecularity applies only to elementary reactions (1–3); integrated equations for zero and first order, half-lives; Arrhenius equation k = Ae^(−Ea/RT); catalyst lowers Ea; collision theory with steric factor P.
Intext 3.1 — For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Concept. Average rate of disappearance of R: r_av = −Δ[R]/Δt = −([R]₂ − [R]₁)/(t₂ − t₁); the minus sign makes the rate positive because [R] decreases.
Δ[R] = 0.02 − 0.03 = −0.01 mol L⁻¹; Δt = 25 min = 25 × 60 s = 1500 s.
r_av = 0.01 mol L⁻¹ ÷ 25 min = 4.0 × 10⁻⁴ mol L⁻¹ min⁻¹
r_av = 0.01 mol L⁻¹ ÷ 1500 s = 6.67 × 10⁻⁶ mol L⁻¹ s⁻¹ (6.66 × 10⁻⁶ M s⁻¹ in NCERT)
Check: 4.0 × 10⁻⁴ ÷ 60 = 6.67 × 10⁻⁶ ✔ (units convert directly from per minute to per second).
Extra depth. Average rate is the slope of the secant joining two points on the concentration–time curve and therefore depends on the chosen interval, whereas the instantaneous rate is the slope of the tangent at one instant (Fig. 3.1). Rates are always positive, which is why the negative sign is attached to −Δ[R]/Δt for reactants; in the SI-friendly unit mol L⁻¹ s⁻¹ the rate here is 6.67 × 10⁻⁶.
Conclusion: Average rate = 4.0 × 10⁻⁴ mol L⁻¹ min⁻¹ = 6.67 × 10⁻⁶ mol L⁻¹ s⁻¹.
Intext 3.2 — In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval.
Concept. For a reaction with stoichiometric coefficient 2 for A, the rate of reaction is defined as Rate = −(1/2) Δ[A]/Δt, so that it is the same whichever species is followed.
Rate of disappearance of A = −Δ[A]/Δt = (0.5 − 0.4) ÷ 10 = 0.01 mol L⁻¹ min⁻¹.
Rate of reaction = ½ × 0.01 = 0.005 mol L⁻¹ min⁻¹ (= 8.3 × 10⁻⁵ mol L⁻¹ s⁻¹)
Extra depth. Defining the rate of reaction with 1/coefficient factors makes it independent of which species is followed: −½ d[A]/dt for 2A → products. A common error is to quote the rate of disappearance of A as the rate of reaction; the two differ by the stoichiometric factor, 0.01 versus 0.005 mol L⁻¹ min⁻¹ here.
Conclusion: Rate of reaction = 0.005 mol L⁻¹ min⁻¹.
Intext 3.3 — For a reaction A + B → Product; the rate law is given by r = k[A]^½[B]². What is the order of the reaction?
Concept. The overall order is the sum of the powers of the concentration terms in the experimentally determined rate law.
Order = ½ + 2 = 2.5
The reaction is of order ½ in A, second order in B and 2.5 (two and a half) overall; a fractional order shows that the reaction proceeds by a complex mechanism.
Extra depth. The order must be found from experiment, not from the balanced equation; fractional orders (such as ½ or 3/2) usually signal a chain or multi-step mechanism, as in the decomposition of acetaldehyde (rate = k[CH₃CHO]^3/2) or the reaction of CHCl₃ with Cl₂ (rate = k[CHCl₃][Cl₂]^½). The order with respect to each reactant may be 0, 1, 2 or fractional and the units of k depend on the overall order.
Conclusion: The overall order is 2.5.
Intext 3.4 — The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Concept. Second order in X: Rate = k[X]².
Initial rate r₁ = k[X]². When [X] is tripled, [X]′ = 3[X], so r₂ = k(3[X])² = 9k[X]² = 9 r₁.
r₂ ÷ r₁ = (3)² = 9
Extra depth. The rate–concentration relationship is the basis for controlling reactions in industry: tripling the concentration of a second-order reactant increases the rate nine-fold, whereas for a first-order reaction it would increase only three-fold, and for zero order not at all. The result is general: if the order is n, the rate changes by the factor (new concentration/old concentration)ⁿ.
Conclusion: The rate of formation of Y increases nine times.
Intext 3.5 — A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
Concept. First order: k = (2.303/t) log([R]₀/[R]); the mass of reactant is proportional to its concentration (same volume), so masses can be used.
t = (2.303 ÷ 1.15 × 10⁻³ s⁻¹) × log (5 ÷ 3)
log (5/3) = log 1.6667 = 0.2218. 2.303 ÷ 1.15 × 10⁻³ = 2002.6 s.
t = 2002.6 × 0.2218 = 444.2 s ≈ 444 s (≈ 7.4 min)
Extra depth. The integrated first-order equation can also be written [R] = [R]₀ e^(−kt); using the mass of the reactant in place of concentration works because the volume cancels in the ratio. A rate constant of 1.15 × 10⁻³ s⁻¹ corresponds to a half-life of 603 s, so a fall from 5 g to 3 g (40 % decomposition) takes a little less than one half-life, 444 s.
Conclusion: The reactant takes about 444 s to fall from 5 g to 3 g.
Intext 3.6 — Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Concept. For a first order reaction t½ = 0.693/k, so k = 0.693/t½.
k = 0.693 ÷ 60 min = 1.155 × 10⁻² min⁻¹
k = 0.693 ÷ (60 × 60 s) = 0.693 ÷ 3600 s = 1.925 × 10⁻⁴ s⁻¹
The reaction is SO₂Cl₂(g) → SO₂(g) + Cl₂(g).
Extra depth. Half-life is the time for the concentration to fall to half; for a first-order reaction it is independent of the initial concentration, which is a diagnostic test for first-order kinetics. For SO₂Cl₂ (SO₂Cl₂ → SO₂ + Cl₂) t½ = 60 min gives k = 1.155 × 10⁻² min⁻¹, so after 4 half-lives (240 min) only 1/16 of the reactant remains.
Conclusion: k = 1.925 × 10⁻⁴ s⁻¹ (1.155 × 10⁻² min⁻¹).
Intext 3.7 — What will be the effect of temperature on rate constant?
Answer. The rate constant increases with increase in temperature. For many reactions, a 10 °C (10 K) rise nearly doubles the rate constant (and hence the rate). The quantitative dependence is given by the Arrhenius equation:
k = A e^(−Ea/RT) ⇒ ln k = ln A − Ea/RT ⇒ log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)
Explanation. The factor e^(−Ea/RT) is the fraction of molecules whose energy is equal to or greater than the activation energy Ea. On raising the temperature, the Maxwell–Boltzmann energy-distribution curve shifts to higher energy and flattens, so the fraction of molecules with energy ≥ Ea increases (roughly doubles for +10 K), more collisions are effective, and k increases exponentially. The effect is larger for reactions with larger Ea. (A plot of ln k against 1/T is a straight line with slope −Ea/R.)
Extra depth. Rates of chemical reactions increase with temperature because a greater fraction of molecules has energy ≥ Ea — the area under the Maxwell–Boltzmann curve beyond Ea. The doubling per 10 K rule is approximate: the rate-constant ratio is exp[(Ea/R)(1/T₁ − 1/T₂)] and equals 2 exactly only for Ea ≈ 52.9 kJ mol⁻¹ near 298 K (intext 3.8), larger Ea giving a bigger increase.
Conclusion: k increases (almost exponentially) with temperature, doubling for about every 10 K rise; k = Ae^(−Ea/RT).
Intext 3.8 — The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Data. T₁ = 298 K, T₂ = 308 K, k₂/k₁ = 2, R = 8.314 J K⁻¹ mol⁻¹.
Formula. log (k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
log 2 = 0.3010 = Ea × 10 ÷ (2.303 × 8.314 × 298 × 308)
Ea = 0.3010 × 2.303 × 8.314 × 298 × 308 ÷ 10 = 52897 J mol⁻¹ ≈ 52.9 kJ mol⁻¹
Check of the arithmetic: 2.303 × 8.314 = 19.147; 19.147 × 0.3010 = 5.763; 5.763 × (298 × 308 ÷ 10) = 5.763 × 9178.4 ≈ 5.29 × 10⁴ J mol⁻¹ ✔.
Extra depth. The activation energy of 52.9 kJ mol⁻¹ is typical for a reaction whose rate doubles for a 10 K rise at room temperature; reactions with larger Ea are more temperature-sensitive. It is useful to remember that a catalyst lowers Ea, which has the same exponential effect on k as a rise in temperature.
Conclusion: Ea = 52.9 kJ mol⁻¹ (52.897 kJ mol⁻¹).
Intext 3.9 — The activation energy for the reaction 2HI(g) → H₂ + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Concept. From the Arrhenius equation, the fraction of molecules with energy ≥ Ea is x = e^(−Ea/RT), i.e. ln x = −Ea/RT; log x = −Ea/(2.303RT).
log x = −209500 ÷ (2.303 × 8.314 × 581) = −209500 ÷ 11124.6 = −18.833
x = antilog (−18.833) = 1.471 × 10⁻¹⁹
A fraction of only about 1.5 in 10¹⁹ molecules possesses enough energy at 581 K — which explains why the decomposition of HI is slow at this temperature despite being thermodynamically feasible.
Extra depth. The fraction e^(−Ea/RT) = 1.47 × 10⁻¹⁹ is so small that only about one molecule in 10¹⁹ is energetic enough at 581 K; yet the reaction proceeds at a measurable rate because of the enormous number of collisions occurring every second. Raising the temperature by about 10 K would double this fraction.
Conclusion: Fraction of molecules with E ≥ Ea = 1.471 × 10⁻¹⁹.
Exercise 3.1 — From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N₂O(g), Rate = k[NO]² (ii) H₂O₂(aq) + 3I⁻(aq) + 2H⁺ → 2H₂O(l) + I₃⁻, Rate = k[H₂O₂][I⁻] (iii) CH₃CHO(g) → CH₄(g) + CO(g), Rate = k[CH₃CHO]^3/2 (iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g), Rate = k[C₂H₅Cl]
Method. Order = sum of the exponents of the concentration terms in the rate law. Units of k = (concentration)^(1−n) time⁻¹, i.e. k = rate ÷ [concentration]ⁿ with rate in mol L⁻¹ s⁻¹.
| Reaction | Rate law | Order n | Dimensions (units) of k |
|---|---|---|---|
| (i) 3NO → N₂O | k[NO]² | 2 (second order) | mol⁻¹ L s⁻¹ (= L mol⁻¹ s⁻¹) — from (mol L⁻¹ s⁻¹) ÷ (mol L⁻¹)² |
| (ii) H₂O₂ + 3I⁻ + 2H⁺ → 2H₂O + I₃⁻ | k[H₂O₂][I⁻] | 1 + 1 = 2 (second order) | mol⁻¹ L s⁻¹ |
| (iii) CH₃CHO → CH₄ + CO | k[CH₃CHO]^3/2 | 3/2 (one and a half order) | mol^(−1/2) L^(1/2) s⁻¹ — from (mol L⁻¹ s⁻¹) ÷ (mol L⁻¹)^3/2 |
| (iv) C₂H₅Cl → C₂H₄ + HCl | k[C₂H₅Cl] | 1 (first order) | s⁻¹ |
Note that in (i) and (ii) the order differs from the stoichiometric sum of reactants (3 for 3NO; 6 for H₂O₂ + 3I⁻ + 2H⁺), showing that the rate law is experimental.
Extra depth. Dimensional analysis of k is a quick way to identify the order from the units: s⁻¹ means first order, L mol⁻¹ s⁻¹ second order, mol L⁻¹ s⁻¹ zero order. Note that in (i) the stoichiometric coefficient of NO is 3 (3NO → N₂O) but the order is 2, which shows that the rate law is experimental and not read from the balanced equation.
Conclusion: (i) 2, mol⁻¹ L s⁻¹; (ii) 2, mol⁻¹ L s⁻¹; (iii) 1.5, mol^−½ L^½ s⁻¹; (iv) 1, s⁻¹.
Exercise 3.2 — For the reaction 2A + B → A₂B the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.
(a) Initial rate.
Rate = k[A][B]² = 2.0 × 10⁻⁶ × 0.1 × (0.2)² = 2.0 × 10⁻⁶ × 0.1 × 0.04 = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹
(b) After [A] falls to 0.06 mol L⁻¹. [A] has decreased by 0.1 − 0.06 = 0.04 mol L⁻¹. From the stoichiometry 2A + B → A₂B, 2 mol of A react with 1 mol of B, so B decreases by 0.04 ÷ 2 = 0.02 mol L⁻¹: [B] = 0.2 − 0.02 = 0.18 mol L⁻¹.
Rate = 2.0 × 10⁻⁶ × 0.06 × (0.18)² = 2.0 × 10⁻⁶ × 0.06 × 0.0324 = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹
Extra depth. Rates of reactions fall as the reactants are consumed, so “the rate after [A] is reduced to 0.06 M” is smaller (3.89 × 10⁻⁹) than the initial rate (8.0 × 10⁻⁹). The stoichiometry 2A + B → A₂B shows that [B] decreases by half the decrease in [A], an essential step.
Conclusion: Initial rate = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹; rate when [A] = 0.06 M is 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹.
Exercise 3.3 — The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol⁻¹ L s⁻¹?
Reaction. 2NH₃(g) → N₂(g) + 3H₂(g) (on hot Pt). Zero order: Rate = k[NH₃]⁰ = k. (Printed unit of k in the book: mol⁻¹ L s⁻¹; for a zero-order reaction the correct unit is mol L⁻¹ s⁻¹ — the numerical value 2.5 × 10⁻⁴ is used.)
Rate of reaction = −½ d[NH₃]/dt = d[N₂]/dt = ⅓ d[H₂]/dt = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
Rate of production of N₂ = d[N₂]/dt = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
Rate of production of H₂ = d[H₂]/dt = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹
Rate of disappearance of NH₃ = 2 × 2.5 × 10⁻⁴ = 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹.
Extra depth. Zero-order kinetics arise when the reaction occurs on a catalyst surface that is saturated with reactant, as with NH₃ on hot platinum at high pressure; the rate is then constant until the reactant is almost exhausted. The rates of formation of the products are in the ratio of their stoichiometric coefficients (1 : 3 for N₂ : H₂).
Conclusion: N₂ is produced at 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ and H₂ at 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
Exercise 3.4 — The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by Rate = k[CH₃OCH₃]^3/2. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k(p_CH₃OCH₃)^3/2. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Units of rate. Rate = −dp/dt, i.e. pressure per unit time: bar min⁻¹.
Units of k. k = Rate ÷ p^3/2 = (bar min⁻¹) ÷ (bar)^3/2 = bar^(−1/2) min⁻¹.
Reaction: CH₃OCH₃(g) → CH₄(g) + H₂(g) + CO(g); the order is 3/2, so k has the dimensions (pressure)^(1−3/2) time⁻¹ = (pressure)^(−½) time⁻¹.
Extra depth. When gases are used the concentration may be replaced by the partial pressure, because p = (n/V)RT, and the rate constant then has different units from that in terms of concentration (the two k’s are related by factors of RT). The order 3/2 makes the units of k bar^(−½) min⁻¹.
Conclusion: Rate: bar min⁻¹; k: bar^−½ min⁻¹.
Exercise 3.5 — Mention the factors that affect the rate of a chemical reaction.
The rate of a chemical reaction depends on the following factors:
- Concentration of the reactants (pressure in the case of gases): rate = k[A]ˣ[B]ʸ; rate generally increases with concentration because the frequency of collisions rises (Table 3.1 shows the rate falling as reactants are used up).
- Temperature: the rate constant roughly doubles for every 10 K rise, as given by k = Ae^(−Ea/RT); the fraction of molecules with energy ≥ Ea increases.
- Presence of a catalyst: a catalyst provides an alternative path with lower activation energy and increases the rate (inhibitors decrease it).
- Nature of the reactants and products: ionic reactions (AgNO₃ + NaCl) are instantaneous, whereas reactions involving bond breaking in covalent molecules are slow.
- Surface area of solid reactants: powdered solids react faster than lumps because more surface is exposed.
- Radiation (light) in photochemical reactions (e.g. H₂ + Cl₂ → 2HCl) supplies the activation energy.
- Orientation of the colliding molecules (the steric factor P in collision theory) and the activation energy Ea itself.
| Factor | Effect on rate | Reason |
|---|---|---|
| Concentration ↑ | ↑ | More collisions per second |
| Temperature ↑ | ↑ (≈ ×2 per 10 K) | More molecules with E ≥ Ea |
| Catalyst | ↑ | Lower Ea |
| Surface area ↑ | ↑ | More effective collisions on surface |
Extra depth. Besides concentration, temperature and catalyst, the nature of the reactants matters: ionic reactions such as AgNO₃ + NaCl occur instantly, whereas reactions requiring bond cleavage are slow. For heterogeneous reactions, the surface area is critical (a powdered solid reacts faster than a lump), and in photochemical reactions the intensity of light.
Conclusion: Concentration (pressure), temperature and catalyst are the principal factors, along with nature of reactants, surface area and radiation.
Exercise 3.6 — A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Rate law. Rate = k[A]². Let the initial rate be r = k[A]².
(i) Doubled. [A]′ = 2[A]: r′ = k(2[A])² = 4k[A]² = 4r — the rate becomes four times.
(ii) Halved. [A]′ = [A]/2: r′ = k([A]/2)² = k[A]²/4 = r/4 — the rate becomes one-fourth.
Extra depth. Second-order reactions are very sensitive to concentration: halving the concentration decreases the rate fourfold. Compare first-order reactions, where halving the concentration halves the rate; this difference allows the order to be determined experimentally by the method of initial rates.
Conclusion: For a second-order reaction: doubling [A] quadruples the rate; halving [A] reduces the rate to ¼ of its value.
Exercise 3.7 — What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Effect. The rate constant increases with temperature; for most reactions it nearly doubles for every 10 °C rise (e.g. N₂O₅ decomposition: t½ = 12 min at 50 °C, 5 h at 25 °C, 10 days at 0 °C). A rise in temperature increases the fraction of molecules having energy ≥ Ea (the Maxwell–Boltzmann curve shifts and broadens, Fig. 3.9).
Quantitative representation — Arrhenius equation:
k = A e^(−Ea/RT)
A = Arrhenius (frequency/pre-exponential) factor, related to the collision frequency; Ea = activation energy (J mol⁻¹); R = 8.314 J K⁻¹ mol⁻¹; T = temperature in kelvin; e^(−Ea/RT) = fraction of molecules with energy ≥ Ea.
Logarithmic forms: ln k = ln A − Ea/RT, log k = log A − Ea/(2.303 RT): a plot of ln k against 1/T is a straight line of slope −Ea/R and intercept ln A (Fig. 3.10). For two temperatures T₁ and T₂:
log (k₂/k₁) = (Ea / 2.303R) × (1/T₁ − 1/T₂) = (Ea / 2.303R) × (T₂ − T₁)/(T₁T₂)
Extra depth. The Arrhenius equation was proposed by van’t Hoff and interpreted physically by Arrhenius; the pre-exponential factor A includes the collision frequency and a steric factor, and e^(−Ea/RT) gives the fraction of molecules with sufficient energy. Plotting ln k against 1/T gives a straight line with slope −Ea/R, the standard way of finding Ea experimentally.
Conclusion: k increases exponentially with T according to k = Ae^(−Ea/RT).
Exercise 3.8 — In a pseudo first order reaction in water, the following results were obtained: t/s: 0, 30, 60, 90; [A]/mol L⁻¹: 0.55, 0.31, 0.17, 0.085. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Concept. Average rate = −Δ[A]/Δt over the interval.
r_av = −(0.17 − 0.31) mol L⁻¹ ÷ (60 − 30) s = 0.14 ÷ 30 = 4.67 × 10⁻³ mol L⁻¹ s⁻¹
(For comparison, from 0 to 30 s the average rate is (0.55 − 0.31)/30 = 8.0 × 10⁻³ mol L⁻¹ s⁻¹ — the rate falls as [A] falls, as expected for a first-order process.)
Extra depth. For a pseudo first-order reaction in water the concentration of water is practically constant, so the rate depends only on [A]; the average rate decreases with time (8.0 × 10⁻³ in the first 30 s and 4.67 × 10⁻³ in the second 30 s). The ratio 0.55 : 0.31 : 0.17 : 0.085 shows that [A] roughly halves every 30 s, characteristic of first-order kinetics with half-life ≈ 30 s.
Conclusion: Average rate between 30 s and 60 s = 4.67 × 10⁻³ mol L⁻¹ s⁻¹.
Exercise 3.9 — A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?
(i) Differential rate equation:
Rate = −d[R]/dt = k[A][B]²
(ii) [B] tripled. r′ = k[A](3[B])² = 9k[A][B]² = 9r — the rate increases nine times.
(iii) [A] and [B] both doubled. r′ = k(2[A])(2[B])² = 2 × 4 k[A][B]² = 8r — the rate increases eight times.
Overall order = 1 + 2 = 3 (third order), so doubling all concentrations multiplies the rate by 2³ = 8.
Extra depth. The reaction is third order overall and the rate law shows how the order of a reaction can be 1 for A and 2 for B; the overall effect of doubling both reactants (eight-fold) is 2³. Such a third-order reaction is rare; most reactions are first or second order, and termolecular elementary steps are uncommon because the collision of three bodies at once is very improbable.
Conclusion: (i) Rate = k[A][B]²; (ii) 9 times; (iii) 8 times.
Exercise 3.10 — In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given: A/mol L⁻¹: 0.20, 0.20, 0.40; B/mol L⁻¹: 0.30, 0.10, 0.05; r₀/mol L⁻¹ s⁻¹: 5.07 × 10⁻⁵, 5.07 × 10⁻⁵, 1.43 × 10⁻⁴. What is the order of the reaction with respect to A and B?
Let r₀ = k[A]ˣ[B]ʸ.
Order in B. Experiments 1 and 2 have the same [A] = 0.20 M; [B] changes from 0.30 to 0.10 M (÷ 3) but the rate stays 5.07 × 10⁻⁵. So (0.30/0.10)ʸ = 1 ⇒ y = 0 — zero order in B.
Order in A. Since B does not affect the rate, compare experiments 2 and 3: [A] doubles (0.20 → 0.40) and the rate changes from 5.07 × 10⁻⁵ to 1.43 × 10⁻⁴.
(r₃/r₂) = 1.43 × 10⁻⁴ ÷ 5.07 × 10⁻⁵ = 2.82 = (0.40/0.20)ˣ = 2ˣ ⇒ x = log 2.82 ÷ log 2 = 1.5
2^1.5 = 2.83 ✔. The rate law is r = k[A]^1.5[B]⁰ and the overall order is 1.5.
Extra depth. The method of initial rates compares experiments in which only one concentration is varied at a time; a fractional order (1.5) can result from a mechanism with a pre-equilibrium or chain steps. Always verify with all the data: the fit k = r/[A]^1.5 should give the same k (here about 5.67 × 10⁻⁴ L^½ mol^−½ s⁻¹) for each experiment.
Conclusion: Order with respect to A = 3/2; with respect to B = 0 (overall order 1.5).
Exercise 3.11 — The following results have been obtained during the kinetic studies of the reaction 2A + B → C + D. Experiment I: [A] = 0.1, [B] = 0.1, rate of formation of D = 6.0 × 10⁻³; II: 0.3, 0.2, 7.2 × 10⁻²; III: 0.3, 0.4, 2.88 × 10⁻¹; IV: 0.4, 0.1, 2.40 × 10⁻² (mol L⁻¹ min⁻¹). Determine the rate law and the rate constant for the reaction.
Let rate = k[A]ˣ[B]ʸ.
Order in B — experiments II and III keep [A] = 0.3 constant while [B] doubles (0.2 → 0.4): the rate becomes 2.88 × 10⁻¹ ÷ 7.2 × 10⁻² = 4 = 2ʸ ⇒ y = 2.
Order in A — experiments I and IV keep [B] = 0.1 constant while [A] increases four times (0.1 → 0.4): the rate becomes 2.40 × 10⁻² ÷ 6.0 × 10⁻³ = 4 = 4ˣ ⇒ x = 1.
Rate law: rate = k[A][B]² (third order overall)
Rate constant from experiment I: k = rate ÷ ([A][B]²) = 6.0 × 10⁻³ ÷ (0.1 × 0.01) = 6.0 mol⁻² L² min⁻¹. Check with II: 6.0 × 0.3 × 0.04 = 7.2 × 10⁻² ✔; III: 6.0 × 0.3 × 0.16 = 0.288 ✔; IV: 6.0 × 0.4 × 0.01 = 2.4 × 10⁻² ✔.
| Experiment | [A] | [B] | Observed rate | k[A][B]² with k = 6.0 |
|---|---|---|---|---|
| I | 0.1 | 0.1 | 6.0 × 10⁻³ | 6.0 × 10⁻³ |
| II | 0.3 | 0.2 | 7.2 × 10⁻² | 7.2 × 10⁻² |
| III | 0.3 | 0.4 | 2.88 × 10⁻¹ | 2.88 × 10⁻¹ |
| IV | 0.4 | 0.1 | 2.40 × 10⁻² | 2.40 × 10⁻² |
Extra depth. This is the standard technique of finding a rate law from tabulated initial rates; verifying that all four experiments give the same k = 6.0 mol⁻² L² min⁻¹ is the check on the derived order. The rate of formation of D is used as the rate; if the rate of disappearance of A were used the value would be twice as large because of the coefficient 2.
Conclusion: Rate law: rate = k[A][B]²; k = 6.0 mol⁻² L² min⁻¹.
Exercise 3.12 — The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment I: [A] = 0.1, [B] = 0.1, rate 2.0 × 10⁻²; II: [A] = —, [B] = 0.2, rate 4.0 × 10⁻²; III: [A] = 0.4, [B] = 0.4, rate —; IV: [A] = —, [B] = 0.2, rate 2.0 × 10⁻².
Rate law. First order in A, zero order in B: rate = k[A][B]⁰ = k[A]. From experiment I: k = 2.0 × 10⁻² ÷ 0.1 = 0.2 min⁻¹.
| Experiment | [A] (mol L⁻¹) | [B] (mol L⁻¹) | Initial rate (mol L⁻¹ min⁻¹) | Working |
|---|---|---|---|---|
| I | 0.1 | 0.1 | 2.0 × 10⁻² | k = 2.0 × 10⁻²/0.1 = 0.2 min⁻¹ |
| II | 0.2 | 0.2 | 4.0 × 10⁻² | [A] = rate/k = 4.0 × 10⁻²/0.2 = 0.2 |
| III | 0.4 | 0.4 | 8.0 × 10⁻² | rate = 0.2 × 0.4 = 8.0 × 10⁻² |
| IV | 0.1 | 0.2 | 2.0 × 10⁻² | [A] = 2.0 × 10⁻²/0.2 = 0.1 |
Since the rate does not depend on [B] (zero order), experiments I and IV give the same rate for the same [A] = 0.1 even though [B] differs.
Extra depth. Because the order in B is zero, doubling [B] has no effect; the rate depends only on [A], as in an enzyme-catalysed reaction at saturation. Filling in tables is a standard exam format and tests the understanding of orders more than the arithmetic.
Conclusion: II: [A] = 0.2 mol L⁻¹; III: rate = 8.0 × 10⁻² mol L⁻¹ min⁻¹; IV: [A] = 0.1 mol L⁻¹.
Exercise 3.13 — Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s⁻¹ (ii) 2 min⁻¹ (iii) 4 years⁻¹.
Formula. For a first order reaction t½ = 0.693/k.
(i) t½ = 0.693 ÷ 200 s⁻¹ = 3.47 × 10⁻³ s
(ii) t½ = 0.693 ÷ 2 min⁻¹ = 0.3465 min ≈ 0.35 min (≈ 20.8 s)
(iii) t½ = 0.693 ÷ 4 years⁻¹ = 0.173 years (≈ 63 days)
Extra depth. Half-lives span a huge range: 3.47 × 10⁻³ s for k = 200 s⁻¹ (a very fast reaction) to about 63 days for k = 4 year⁻¹. The relation t½ = 0.693/k is only for first-order reactions; for zero-order reactions t½ = [R]₀/2k and for second-order t½ = 1/(k[R]₀).
Conclusion: (i) 3.47 × 10⁻³ s; (ii) 0.35 min; (iii) 0.173 year.
Exercise 3.14 — The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Concept. Radioactive decay is first order. k = 0.693/t½ = 0.693 ÷ 5730 years = 1.209 × 10⁻⁴ year⁻¹.
Taking the initial amount as 100 and the present amount as 80:
t = (2.303/k) log([R]₀/[R]) = (2.303 ÷ 1.209 × 10⁻⁴) × log (100/80) = 19046 × 0.09691 ≈ 1845 years
Using t = 2.303/1.209 × 10⁻⁴ = 1.905 × 10⁴; log 1.25 = 0.0969.
Extra depth. Radiocarbon dating works because living organisms continuously take in ¹⁴C so the ratio ¹⁴C/¹²C is constant, but after death the ¹⁴C decays with half-life 5730 years. An artifact with 80 % of the original ¹⁴C is about 1845 years old, i.e. roughly one-third of a half-life (log 1.25 ÷ log 2 = 0.322 half-lives).
Conclusion: The artifact is about 1845 years old.
Exercise 3.15 — The experimental data for decomposition of N₂O₅ [2N₂O₅ → 4NO₂ + O₂] in gas phase at 318 K are given: t/s: 0, 400, 800, 1200, 1600, 2000, 2400, 2800, 3200; 10² × [N₂O₅]/mol L⁻¹: 1.63, 1.36, 1.14, 0.93, 0.78, 0.64, 0.53, 0.43, 0.35. (i) Plot [N₂O₅] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N₂O₅] and t. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
(i) Plot of [N₂O₅] against t. The points fall on a smooth, concave-up decaying curve (the concentration falls by about the same fraction in equal times), characteristic of a first order reaction.
(ii) Half-life from the graph. Initial concentration = 1.63 × 10⁻² M, so half = 0.815 × 10⁻² M. Reading the smooth curve at this concentration gives t½ ≈ 1440 s (the experimental points between 1200 s (0.93) and 1600 s (0.78) show a little scatter; the best-fit curve crosses 0.815 at ≈ 1440 s).
(iii) Graph of log[N₂O₅] against t. Values of log[N₂O₅] (with [N₂O₅] in mol L⁻¹):
| t / s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
|---|---|---|---|---|---|---|---|---|---|
| [N₂O₅] × 10² | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
| log [N₂O₅] | −1.788 | −1.866 | −1.943 | −2.032 | −2.108 | −2.194 | −2.276 | −2.367 | −2.456 |
The points lie on a straight line: slope = −(2.456 − 1.788)/3200 ≈ −2.08 × 10⁻⁴ s⁻¹ (least-squares slope ≈ −2.08 × 10⁻⁴ s⁻¹). A straight line of log c against t proves a first-order reaction.
(iv) Rate law. Rate = k[N₂O₅] (first order).
(v) Rate constant. k = −2.303 × slope = 2.303 × 2.08 × 10⁻⁴ ≈ 4.8 × 10⁻⁴ s⁻¹. Check with the first-order formula using the first and last points: k = (2.303/3200) log (1.63/0.35) = 7.197 × 10⁻⁴ × 0.6682 = 4.81 × 10⁻⁴ s⁻¹; the individual values at each time lie between 4.5 × 10⁻⁴ and 4.8 × 10⁻⁴ s⁻¹.
k ≈ 4.8 × 10⁻⁴ s⁻¹
(vi) Half-life from k. t½ = 0.693/k = 0.693 ÷ 4.8 × 10⁻⁴ = 1444 s ≈ 1.44 × 10³ s, in agreement with the graphical value of about 1440 s.
Extra depth. The best way to prove first-order kinetics is the straight-line plot of log c against t; the near-constant values of k calculated at different times (4.5–4.8 × 10⁻⁴ s⁻¹) support this. The half-life from the graph (≈ 1440 s) and from k (1444 s) are consistent. The scatter in the individual values of k (4.5–4.8 × 10⁻⁴ s⁻¹) arises from experimental errors in the measured concentrations, which is why the slope of the best-fit line is used.
Conclusion: First order: rate = k[N₂O₅], k ≈ 4.8 × 10⁻⁴ s⁻¹, t½ ≈ 1440 s (graph) = 1444 s (from k).
Exercise 3.16 — The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Method 1 — formula. t = (2.303/k) log([R]₀/[R]) with [R] = [R]₀/16.
t = (2.303 ÷ 60 s⁻¹) × log 16 = 0.03838 × 1.2041 = 4.62 × 10⁻² s
Method 2 — half-lives. 1/16 = (½)⁴, so four half-lives are needed: t½ = 0.693/60 = 1.155 × 10⁻² s; t = 4 × 1.155 × 10⁻² = 4.62 × 10⁻² s ✔.
Extra depth. A first-order reaction needs the time of exactly four half-lives to fall to 1/16; the very high k (60 s⁻¹) means that the reaction is complete in less than a tenth of a second — typical of very fast reactions in solution. The general formula for n half-lives is [R] = [R]₀/2ⁿ.
Conclusion: Time required = 4.62 × 10⁻² s (≈ 0.046 s).
Exercise 3.17 — During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 mg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Concept. Radioactive decay is first order: k = 0.693/t½ = 0.693 ÷ 28.1 = 2.467 × 10⁻² year⁻¹; [R] = [R]₀ e^(−kt), or log([R]₀/[R]) = kt/2.303.
After 10 years: kt = 0.2467; [R] = 1 mg × e^(−0.2467) = 0.781 mg. (Using log form: log(1/[R]) = 0.2467/2.303 = 0.1071 ⇒ [R] = 0.781 mg.)
After 60 years: kt = 1.480; [R] = 1 mg × e^(−1.480) = 0.228 mg. (Equivalent: 60/28.1 = 2.135 half-lives ⇒ 2^(−2.135) = 0.228.)
| Time (years) | kt | Fraction left | Mass of ⁹⁰Sr left |
|---|---|---|---|
| 10 | 0.2467 | 0.781 | 0.781 mg |
| 60 | 1.480 | 0.228 | 0.228 mg |
Extra depth. ⁹⁰Sr (t½ = 28.1 years) is a dangerous radioactive fission product because it replaces calcium in bones; once absorbed, it stays there for decades. The numerical results show that after 10 years 78 % remains and after 60 years 23 % remains, so its biological hazard persists for generations.
Conclusion: 0.781 mg remains after 10 years and 0.228 mg after 60 years.
Exercise 3.18 — For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Concept. First order: t = (2.303/k) log([R]₀/[R]). Let [R]₀ = 100.
For 99 % completion: [R] = 100 − 99 = 1.
t₉₉ = (2.303/k) log (100/1) = (2.303/k) × 2 = 4.606/k
For 90 % completion: [R] = 100 − 90 = 10.
t₉₀ = (2.303/k) log (100/10) = (2.303/k) × 1 = 2.303/k
t₉₉ ÷ t₉₀ = (4.606/k) ÷ (2.303/k) = 2 ⇒ t₉₉ = 2 t₉₀
In general, completion of 99.9 % needs 3 × (2.303/k) = 6.909/k = 10 half-lives (Example 3.8).
Extra depth. The result generalises: for first-order reactions the time for 99.9 % completion is 3 times that for 90 %, and the ratio of the times for completion 1 − 10⁻ⁿ is n : 1. In the same way t₉₉.₉ = 10 t½ (Example 3.8), which is why first-order processes are called “complete” after about ten half-lives.
Conclusion: Hence the time for 99 % completion is exactly twice the time for 90 % completion — shown.
Exercise 3.19 — A first order reaction takes 40 min for 30% decomposition. Calculate t½.
Step 1 — k. [R]₀ = 100, [R] = 100 − 30 = 70 after t = 40 min.
k = (2.303/40 min) log (100/70) = 0.057575 × 0.1549 = 8.918 × 10⁻³ min⁻¹
Step 2 — half-life.
t½ = 0.693 ÷ k = 0.693 ÷ 8.918 × 10⁻³ min⁻¹ = 77.7 min
Extra depth. The half-life of 77.7 min here (k = 8.92 × 10⁻³ min⁻¹) is much longer than the time of 40 min used for 30 % decomposition, which is correct, because 30 % decomposition needs less time than the 50 % of a half-life. Always use the fraction remaining (70 %), not the fraction decomposed, in the logarithm.
Conclusion: t½ = 77.7 min.
Exercise 3.20 — For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained: t (s): 0, 360, 720; P (mm Hg): 35.0, 54.0, 63.0. Calculate the rate constant.
Reaction. (CH₃)₂CHN=NCH(CH₃)₂(g) → C₆H₁₄(g) + N₂(g), i.e. A(g) → B(g) + C(g). Initial pressure p₀ = 35.0 mm Hg; if the pressure of A falls by x, the total pressure is p_t = (p₀ − x) + x + x = p₀ + x, so x = p_t − p₀ and p_A = p₀ − x = 2p₀ − p_t.
At t = 360 s: p_A = 2(35.0) − 54.0 = 16.0 mm Hg.
k = (2.303/360 s) log (35.0/16.0) = 6.397 × 10⁻³ × 0.3400 = 2.175 × 10⁻³ s⁻¹
At t = 720 s: p_A = 70.0 − 63.0 = 7.0 mm Hg.
k = (2.303/720 s) log (35.0/7.0) = 3.199 × 10⁻³ × 0.6990 = 2.236 × 10⁻³ s⁻¹
The two values agree (confirming first-order kinetics); the mean is k ≈ 2.2 × 10⁻³ s⁻¹.
Extra depth. For gas-phase reactions in which the number of moles increases, the total pressure rises with time and the partial pressure of the reactant is 2p₀ − p_t (for A → B + C). Agreement between k values at 360 s and 720 s (2.18 and 2.24 × 10⁻³ s⁻¹) confirms first order; azoisopropane decomposes into hexane and nitrogen via free radicals.
Conclusion: k ≈ 2.2 × 10⁻³ s⁻¹.
Exercise 3.21 — The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume: SO₂Cl₂(g) → SO₂(g) + Cl₂(g). Experiment 1: time 0 s, total pressure 0.5 atm; Experiment 2: time 100 s, total pressure 0.6 atm. Calculate the rate of the reaction when total pressure is 0.65 atm.
Step 1 — find k. A(g) → B(g) + C(g): p_A = 2p₀ − p_t, with p₀ = 0.5 atm. At t = 100 s, p_t = 0.6 atm ⇒ p_A = 1.0 − 0.6 = 0.4 atm.
k = (2.303/100 s) log (0.5/0.4) = 0.02303 × 0.09691 = 2.23 × 10⁻³ s⁻¹
Step 2 — pressure of SO₂Cl₂ when p_t = 0.65 atm. p_A = 2(0.5) − 0.65 = 0.35 atm.
Step 3 — rate. For a first-order reaction, rate = k × p_A.
Rate = 2.23 × 10⁻³ s⁻¹ × 0.35 atm = 7.8 × 10⁻⁴ atm s⁻¹
Extra depth. The decomposition SO₂Cl₂ → SO₂ + Cl₂ is a classic gas-phase first-order reaction (Example 3.6 has a similar treatment for N₂O₅). The rate in atm s⁻¹ is the product of k (s⁻¹) and the partial pressure of the reactant (atm).
Conclusion: The rate when p_t = 0.65 atm is 7.8 × 10⁻⁴ atm s⁻¹ (k = 2.23 × 10⁻³ s⁻¹).
Exercise 3.22 — The rate constant for the decomposition of N₂O₅ at various temperatures is given: T/°C: 0, 20, 40, 60, 80; 10⁵ × k/s⁻¹: 0.0787, 1.70, 25.7, 178, 2140. Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50 °C.
Step 1 — convert data. T (K) = 273, 293, 313, 333, 353; 1/T and ln k:
| T / °C | T / K | 10³/T (K⁻¹) | k / s⁻¹ | ln k |
|---|---|---|---|---|
| 0 | 273 | 3.663 | 7.87 × 10⁻⁷ | −14.06 |
| 20 | 293 | 3.413 | 1.70 × 10⁻⁵ | −10.98 |
| 40 | 313 | 3.195 | 2.57 × 10⁻⁴ | −8.27 |
| 60 | 333 | 3.003 | 1.78 × 10⁻³ | −6.33 |
| 80 | 353 | 2.833 | 2.14 × 10⁻² | −3.84 |
Step 2 — Arrhenius equation. ln k = ln A − (Ea/R)(1/T): the slope of the best straight line = −Ea/R. The least-squares line has slope ≈ −1.21 × 10⁴ K and intercept ln A ≈ 30.3.
Ea = −slope × R = 1.21 × 10⁴ K × 8.314 J K⁻¹ mol⁻¹ ≈ 1.0 × 10⁵ J mol⁻¹ ≈ 100 kJ mol⁻¹
ln A ≈ 30.3 ⇒ A ≈ 1.4 × 10¹³ s⁻¹
(Using only the end points: Ea = 2.303R log(k₈₀/k₀) ÷ (1/273 − 1/353) = 102 kJ mol⁻¹ — the result of the graph depends slightly on how the line is drawn; the accepted value for N₂O₅ is about 103 kJ mol⁻¹.)
Step 3 — predicted rate constants. ln k = 30.3 − 12103/T:
At 30 °C (303 K): ln k = 30.3 − 39.94 = −9.64 ⇒ k ≈ 6.6 × 10⁻⁵ s⁻¹
At 50 °C (323 K): ln k = 30.3 − 37.47 = −7.17 ⇒ k ≈ 7.8 × 10⁻⁴ s⁻¹
These lie between the values for 20 °C (1.70 × 10⁻⁵) and 40 °C (2.57 × 10⁻⁴), and between 40 °C and 60 °C (1.78 × 10⁻³), respectively, as expected.
Extra depth. Arrhenius plots are the standard way of finding Ea and A: the slope −Ea/R of ln k versus 1/T gives Ea ≈ 100 kJ mol⁻¹ and the intercept gives A ≈ 10¹³ s⁻¹, typical for unimolecular decompositions where A is of the order of a molecular vibration frequency. The values for rates at 30 °C and 50 °C are obtained by interpolation along the line.
Conclusion: Ea ≈ 100 kJ mol⁻¹, A ≈ 1.4 × 10¹³ s⁻¹; k(30 °C) ≈ 6.6 × 10⁻⁵ s⁻¹, k(50 °C) ≈ 7.8 × 10⁻⁴ s⁻¹.
Exercise 3.23 — The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Concept. k = A e^(−Ea/RT) ⇒ log k = log A − Ea/(2.303RT) ⇒ log A = log k + Ea/(2.303RT).
Ea/(2.303RT) = 179900 ÷ (2.303 × 8.314 × 546) = 179900 ÷ 10454.2 = 17.2086. log k = log (2.418 × 10⁻⁵) = −4.6165.
log A = −4.6165 + 17.2086 = 12.592 ⇒ A = 3.9 × 10¹² s⁻¹
Since k is in s⁻¹, A has the same units (s⁻¹), as the reaction is first order.
Extra depth. A pre-exponential factor of about 10¹² s⁻¹ is typical for unimolecular reactions and is close to the frequency of molecular vibrations (10¹²–10¹⁴ s⁻¹); for bimolecular reactions A is a collision frequency in units of L mol⁻¹ s⁻¹ (10⁹–10¹¹). The high Ea (179.9 kJ mol⁻¹) accounts for the high temperature (546 K) at which the reaction is studied.
Conclusion: Pre-exponential factor A ≈ 3.9 × 10¹² s⁻¹.
Exercise 3.24 — Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.
Concept. The unit of k (s⁻¹) shows first order: [A] = [A]₀ e^(−kt), or log([A]₀/[A]) = kt/2.303.
kt = 2.0 × 10⁻² s⁻¹ × 100 s = 2.0
[A] = 1.0 × e^(−2.0) = 0.135 mol L⁻¹
(Equivalent: log ([A]₀/[A]) = 2.0 ÷ 2.303 = 0.8686 ⇒ [A]₀/[A] = 7.39 ⇒ [A] = 0.135 mol L⁻¹.) About 86.5 % has reacted.
Extra depth. The unit of k tells us the order, so the integrated first-order equation is used; the answer 0.135 mol L⁻¹ shows that 86.5 % of A has reacted in 100 s, i.e. 2 / ln 2 = 2.9 half-lives (t½ = 34.7 s). Such a calculation is used in pharmacokinetics for drug elimination.
Conclusion: [A] after 100 s = 0.135 mol L⁻¹.
Exercise 3.25 — Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t½ = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Step 1 — k. k = 0.693/t½ = 0.693 ÷ 3.00 h = 0.231 h⁻¹.
Step 2 — fraction left. log([R]₀/[R]) = kt/2.303 = 0.231 × 8 ÷ 2.303 = 0.8025 ⇒ [R]₀/[R] = 6.34.
Fraction remaining = [R]/[R]₀ = 1/6.34 = 0.158 (≈ 0.157)
(Equivalent: 8 h ÷ 3 h = 2.667 half-lives ⇒ fraction = 2^(−2.667) = 0.1575.) The reaction is C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (inversion of cane sugar; pseudo first order).
Extra depth. The inversion of cane sugar (sucrose → glucose + fructose) is a pseudo first-order reaction because water is in large excess; it is named “inversion” because the optical rotation changes from positive (sucrose, +66.5°) to negative (the equimolar mixture of glucose and fructose is laevorotatory). Its rate can therefore be followed with a polarimeter.
Conclusion: About 0.157 (15.7 %) of the sucrose remains after 8 hours.
Exercise 3.26 — The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s⁻¹) e^(−28000 K/T). Calculate Ea.
Concept. Compare with the Arrhenius equation k = A e^(−Ea/RT): the exponent −28000 K/T corresponds to −Ea/(RT), so Ea/R = 28000 K.
Ea = 28000 K × 8.314 J K⁻¹ mol⁻¹ = 2.328 × 10⁵ J mol⁻¹ = 232.8 kJ mol⁻¹
(A = 4.5 × 10¹¹ s⁻¹ is the pre-exponential factor.)
Extra depth. Ea/R = 28000 K gives Ea = 232.8 kJ mol⁻¹, a large value typical of C–C bond-breaking in the cracking of hydrocarbons; the high Ea explains why very high temperatures (700–1100 K) are needed. The pre-exponential factor 4.5 × 10¹¹ s⁻¹ has the dimensions of k for a first-order reaction.
Conclusion: Ea = 232.8 kJ mol⁻¹.
Exercise 3.27 — The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 − 1.25 × 10⁴ K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
(a) Ea. Compare log k = log A − Ea/(2.303RT) with the given equation: Ea/(2.303R) = 1.25 × 10⁴ K.
Ea = 1.25 × 10⁴ × 2.303 × 8.314 = 2.393 × 10⁵ J mol⁻¹ = 239.3 kJ mol⁻¹
(b) Temperature for t½ = 256 min. k = 0.693/t½ = 0.693 ÷ (256 × 60 s) = 4.51 × 10⁻⁵ s⁻¹ (the equation gives k in s⁻¹, as in NCERT’s answer). log k = log (4.51 × 10⁻⁵) = −4.346.
−4.346 = 14.34 − 1.25 × 10⁴/T ⇒ 1.25 × 10⁴/T = 18.686 ⇒ T = 668.9 K ≈ 669 K
(If k were taken in min⁻¹ — 2.707 × 10⁻³ min⁻¹ — the temperature would come out as 739 K; the standard answer treats k in s⁻¹, which is the usual unit for A = 10¹⁴·³⁴ s⁻¹.)
Extra depth. The given linear equation log k = 14.34 − 1.25 × 10⁴/T is exactly of the form log k = log A − Ea/(2.303RT), so Ea = 239.3 kJ mol⁻¹ and A = 10¹⁴·³⁴ s⁻¹ ≈ 2.2 × 10¹⁴ s⁻¹; the temperature at which t½ = 256 min is found from k = 0.693/256 min and the equation. The decomposition of H₂O₂ is catalysed by I⁻, Fe³⁺ and enzymes (catalase), which lower Ea enormously.
Conclusion: Ea = 239.3 kJ mol⁻¹; t½ = 256 min at about 669 K (≈ 396 °C).
Exercise 3.28 — The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10 °C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?
Data. T₁ = 10 + 273 = 283 K, k₁ = 4.5 × 10³ s⁻¹, k₂ = 1.5 × 10⁴ s⁻¹, Ea = 60000 J mol⁻¹.
log (k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)
log (1.5 × 10⁴ ÷ 4.5 × 10³) = log 3.333 = 0.5229. Ea/(2.303R) = 60000 ÷ 19.147 = 3133.7 K.
1/T₁ − 1/T₂ = 0.5229 ÷ 3133.7 = 1.669 × 10⁻⁴ ⇒ 1/T₂ = 3.534 × 10⁻³ − 1.669 × 10⁻⁴ = 3.367 × 10⁻³ ⇒ T₂ = 297 K
Extra depth. A reaction with Ea = 60 kJ mol⁻¹ has its k increased 3.33-fold by a rise from 283 K to only 297 K; the large temperature sensitivity is why chemical reactions are slowed by refrigeration. The result in kelvin (297 K) should be converted to 24 °C for a clear physical interpretation.
Conclusion: k becomes 1.5 × 10⁴ s⁻¹ at about 297 K (24 °C).
Exercise 3.29 — The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s⁻¹. Calculate k at 318 K and Ea.
Step 1 — ratio of rate constants. For first order t = (2.303/k) log([R]₀/[R]).
10 % completion: t₁ = (2.303/k₁) log (100/90) = 0.10536/k₁ (using ln). 25 % completion: t₂ = (2.303/k₂) log (100/75) = 0.28768/k₂. Given t₁ = t₂:
k₂/k₁ = 0.28768 ÷ 0.10536 = 2.73
Step 2 — Ea. log (k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂) with T₁ = 298 K, T₂ = 308 K:
log 2.73 = 0.4362 = Ea × 10 ÷ (19.147 × 298 × 308) ⇒ Ea = 0.4362 × 19.147 × 9178.4 = 7.67 × 10⁴ J mol⁻¹ ≈ 76.7 kJ mol⁻¹
Step 3 — k at 318 K from k = A e^(−Ea/RT):
log k = log (4 × 10¹⁰) − 76664 ÷ (2.303 × 8.314 × 318) = 10.602 − 12.591 = −1.989 ⇒ k₃₁₈ = 1.02 × 10⁻² s⁻¹
Check: k at 298 K = 4 × 10¹⁰ × e^(−76664/(8.314 × 298)) = 1.46 × 10⁻³ s⁻¹ and at 308 K = 3.98 × 10⁻³ s⁻¹ (ratio 2.73 ✔); the times t₁ = 0.10536/1.46 × 10⁻³ and t₂ = 0.28768/3.98 × 10⁻³ are both ≈ 72.3 s ✔.
Extra depth. The problem combines first-order kinetics with the Arrhenius equation: equal times for different extents of reaction at two temperatures give the ratio of rate constants, then Ea = 76.7 kJ mol⁻¹, and k at 318 K from the given A. The check — k at 298 K and 308 K gives both times ≈ 72 s — verifies the solution.
Conclusion: Ea ≈ 76.7 kJ mol⁻¹ and k(318 K) ≈ 1.02 × 10⁻² s⁻¹.
Exercise 3.30 — The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Data. k₂/k₁ = 4 (rate ∝ k at fixed concentrations), T₁ = 293 K, T₂ = 313 K.
log 4 = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂) = Ea × 20 ÷ (19.147 × 293 × 313)
log 4 = 0.6021; 19.147 × 293 × 313 = 1.756 × 10⁶; so Ea = 0.6021 × 1.756 × 10⁶ ÷ 20.
Ea = 5.29 × 10⁴ J mol⁻¹ = 52.9 kJ mol⁻¹ (52.86 kJ mol⁻¹)
Extra depth. A fourfold increase in rate for a 20 K rise (293 → 313 K) corresponds to Ea = 52.9 kJ mol⁻¹, similar to the “rate doubles every 10 K” rule: the rate constant doubles per 10 K gives 4 for 20 K. Notice that this calculation assumes Ea is independent of temperature, which is a good approximation over a short range.
Conclusion: Ea ≈ 52.9 kJ mol⁻¹.