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CBSE · Class 12 · Chemistry

Unit 3: Chemical Kinetics

5 concepts · 40 practice questions with answers · Free to read

Rate of a reaction: average and instantaneous

Link to Std 11 : In Std 11 you balanced equations and learnt that a reaction goes towards equilibrium; kinetics asks how fast it goes.

From daily life

A firecracker reacts in a flash, iron rusts in months, and ripening of a mango takes days: the same idea, different speeds.

Simple definition

Rate of reaction is the change in the concentration of a reactant or product per unit time. For R → P the average rate = −Δ[R] ÷ Δt = Δ[P] ÷ Δt over an interval; the instantaneous rate is the rate at one moment (slope of the tangent). Units: mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹).

Concentration of reactant falls with time; rate is the slopetime →[R]average rate (chord)instantaneous rate(tangent)rate = −Δ[R] ÷ Δt = +Δ[P] ÷ Δt (slope is always taken as positive)

In the figure : The curve gets flatter with time because reactant is being used up. The chord measures the average rate over an interval; the tangent measures the rate at one instant.

Rule 1: Average rate — Average rate = (change in concentration) ÷ (time interval). The rate is positive; since the reactant concentration decreases, a minus sign is placed before Δ[R].

  • Easy : [R] falls from 0.080 to 0.056 mol L⁻¹ in 20 s: rate = 0.024 ÷ 20 = 1.2 × 10⁻³ mol L⁻¹ s⁻¹.
  • Medium : A product rises from 0.00 to 0.045 mol L⁻¹ in 30 s: rate = 0.045 ÷ 30 = 1.5 × 10⁻³ mol L⁻¹ s⁻¹.
  • Exam-type : The average rate between two close points is nearly the instantaneous rate at the middle point, so a shorter interval gives a better value.

Rule 2: Rate and stoichiometry — For a A + b B → c C + d D, rate of reaction = −(1/a) Δ[A]/Δt = −(1/b) Δ[B]/Δt = (1/c) Δ[C]/Δt = (1/d) Δ[D]/Δt. Dividing by the coefficient makes the rate the same for every species.

  • Easy : For 2A → B, if [A] falls from 0.50 to 0.38 mol L⁻¹ in 60 s: −Δ[A]/Δt = 2.0 × 10⁻³; reaction rate = ½ × that = 1.0 × 10⁻³ mol L⁻¹ s⁻¹.
  • Medium : N₂ + 3H₂ → 2NH₃: if NH₃ forms at 0.60 mol L⁻¹ s⁻¹, the reaction rate is 0.60 ÷ 2 = 0.30 and H₂ is used at 3 × 0.30 = 0.90 mol L⁻¹ s⁻¹.
  • Exam-type : For 4NH₃ + 5O₂ → 4NO + 6H₂O, if O₂ is used at 0.25 mol L⁻¹ s⁻¹, rate = 0.25 ÷ 5 = 0.050 and water forms at 6 × 0.050 = 0.30 mol L⁻¹ s⁻¹.

Rule 3: Factors that affect the rate — Rate depends on the concentration of reactants, temperature, catalyst, surface area of solids, and (for some reactions) light. Higher concentration or temperature usually increases the rate.

  • Easy : Powdered marble reacts faster with acid than a lump: more surface area.
  • Medium : Food spoils faster in summer than in a refrigerator: higher temperature, faster reaction.
  • Exam-type : Two reactions: one at 0.1 M and one at 0.2 M of the same reactant; usually the second is faster. How much faster depends on the order, covered in the next section.
Remember

Rate is always positive. Divide by the coefficient to compare rates of different species.

Common mistake

Forgetting to divide by the coefficient when comparing rates of disappearance and formation.

Shortcut

Memory clue: "Rate = change ÷ coefficient ÷ time" — divide by the stoichiometric number to get the common rate.

Spot the difference

PointAverage rateInstantaneous rate
Time intervalOver a period ΔtAt one instant
Found fromChord (two points)Tangent at a point
UseRough speedExact speed at that moment

Practice questions

Easy
  1. 1Write the unit of reaction rate.
  2. 2Why is a minus sign used for the rate of change of a reactant?
  3. 3Name two factors that increase rate.
Medium
  1. 4[R] changes from 0.080 to 0.056 mol L⁻¹ in 20 s. Find the average rate.
  2. 5For 2A → B, [A] falls from 0.50 to 0.38 mol L⁻¹ in 60 s. Find the rate of the reaction.
  3. 6N₂ + 3H₂ → 2NH₃: rate of formation of NH₃ is 0.60 mol L⁻¹ s⁻¹. Find the rate of disappearance of H₂.
Challenging
  1. 7For 4NH₃ + 5O₂ → 4NO + 6H₂O, O₂ is consumed at 0.25 mol L⁻¹ s⁻¹. Find the rate of formation of H₂O.
  2. 8Why does the instantaneous rate of most reactions fall with time?
Show answers (try them yourself first)
  1. 1mol L⁻¹ s⁻¹.
  2. 2Δ[R] is negative (the reactant concentration decreases) but a rate is positive.
  3. 3Concentration and temperature (also catalyst, surface area).
  4. 4Rate = 0.024 ÷ 20 = 1.2 × 10⁻³ mol L⁻¹ s⁻¹.
  5. 5Rate of reaction = ½ × (0.12 ÷ 60) = 1.0 × 10⁻³ mol L⁻¹ s⁻¹.
  6. 6Rate of reaction = 0.60 ÷ 2 = 0.30; rate of H₂ disappearance = 3 × 0.30 = 0.90 mol L⁻¹ s⁻¹.
  7. 7Reaction rate = 0.25 ÷ 5 = 0.050; H₂O forms at 6 × 0.050 = 0.30 mol L⁻¹ s⁻¹.
  8. 8Reactant is consumed, so its concentration falls, and fewer collisions occur per second.

Rate law, order, rate constant and molecularity

Link to Std 11 : You know that equilibrium constants depend on balanced equations; the rate law, however, comes from experiments.

From daily life

Doubling the number of customers at a counter does not always double the speed of the queue; the way the rate responds to a change is the order.

Simple definition

The rate law expresses the rate as a power function of the concentrations: rate = k[A]ˣ[B]ʸ. The powers x, y are the orders with respect to A and B, and the overall order is x + y. k is the rate constant (the rate when all concentrations are 1 mol L⁻¹). Molecularity is the number of species colliding in an elementary step.

Rate against concentration for different orders[A] →ratezero orderfirst ordersecond orderrate = k[A]ⁿ (n is found from experiments, not from the equation)

In the figure : For zero order the rate does not change with concentration; for first order the rate is proportional to concentration; for second order the rate rises with the square.

Rule 1: Order and units of k — Order can be 0, a whole number or a fraction. Unit of k for order n = (mol L⁻¹)^(1−n) s⁻¹: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹.

  • Easy : Rate = k[A]²[B] has order 2 + 1 = 3.
  • Medium : If rate = k[A]² and [A] is doubled, the rate becomes 2² = 4 times.
  • Exam-type : Rate = k[A]^(1/2)[B]²: order = 2.5; the unit of k is mol^(−1.5) L^(1.5) s⁻¹.

Rule 2: Finding the order from data — Compare two experiments in which only one concentration changes. If doubling a concentration doubles the rate, order 1; if it makes the rate 4 times, order 2; if the rate does not change, order 0.

  • Easy : Exp 1: [A] = 0.10, [B] = 0.10, rate = 2.0 × 10⁻³. Exp 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10⁻³. Doubling A made the rate 4 times, so order in A = 2.
  • Medium : Exp 3: [A] = 0.10, [B] = 0.20, rate = 4.0 × 10⁻³. Doubling B doubled the rate, so order in B = 1.
  • Exam-type : So rate = k[A]²[B]. k = 2.0 × 10⁻³ ÷ (0.10² × 0.10) = 2.0 L² mol⁻² s⁻¹; for [A] = 0.30 and [B] = 0.15 the rate is 2.0 × 0.09 × 0.15 = 0.027 mol L⁻¹ s⁻¹.

Rule 3: Molecularity and mechanism — Molecularity counts molecules in an elementary step (1, 2 or rarely 3) and is a whole number; it is defined only for elementary reactions. Order is experimental and can be zero or fractional. For a multistep reaction the slowest step, the rate-determining step, controls the overall rate.

  • Easy : For H₂O₂ + 3I⁻ + 2H⁺ → 2H₂O + I₃⁻ the experimental rate is k[H₂O₂][I⁻], second order overall, although the balanced equation has six reactant particles; the slow step decides the order.
  • Medium : An elementary step A + B → products has molecularity 2 and order 2.
  • Exam-type : Order and molecularity are equal only for an elementary step; for a complex reaction order may differ from the sum of coefficients.

Rule 4: Pseudo first order reactions — If one reactant is present in large excess, its concentration hardly changes, so a second-order reaction behaves like a first-order one (pseudo first order).

  • Easy : Hydrolysis of ethyl acetate in a large excess of water has rate = k[CH₃COOC₂H₅] only.
  • Medium : Inversion of cane sugar in water follows a first-order rate law with a large excess of water.
  • Exam-type : A reaction A + B → C with rate = k[A][B] is run with [B] = 2.0 M (much larger than [A]); then rate = k′[A] with k′ = 2.0 k, a first-order law.
Remember

Order comes from experiments, molecularity from the mechanism. Order may be 0 or a fraction; molecularity never is.

Common mistake

Taking the stoichiometric coefficients of the balanced equation as the orders.

Shortcut

Memory clue: "Double it, watch the rate" — doubles → first order, quadruples → second order, no change → zero order.

Spot the difference

PointOrderMolecularity
Found fromExperimentMechanism (elementary step)
Possible values0, 1, 2, 3, fractions1, 2, (3)
Applies toAny reactionElementary reactions only

Practice questions

Easy
  1. 1Define overall order of a reaction.
  2. 2What is the unit of k for a first-order reaction?
  3. 3Can molecularity be zero? Give a reason.
Medium
  1. 4If rate = k[A]² and [A] is tripled, by what factor does the rate change?
  2. 5Find the order and the unit of k when rate = k[A]^(1/2)[B]².
  3. 6Find k if rate = 2.0 × 10⁻³ mol L⁻¹ s⁻¹ when [A] = [B] = 0.10 M and rate = k[A]²[B].
Challenging
  1. 7Using Exp 1 ([A] = 0.10, [B] = 0.10, rate = 2.0 × 10⁻³) and k = 2.0, predict the rate at [A] = 0.30 and [B] = 0.15 (rate = k[A]²[B]).
  2. 8Why is hydrolysis of ethyl acetate with excess water first order even though two reactants are involved?
Show answers (try them yourself first)
  1. 1The sum of the powers of the concentration terms in the rate law.
  2. 2s⁻¹.
  3. 3No; an elementary step needs at least one species, so molecularity is a whole number of 1 or more.
  4. 4It becomes 3² = 9 times.
  5. 5Order = 0.5 + 2 = 2.5; unit of k = mol^(−1.5) L^(1.5) s⁻¹.
  6. 6k = 2.0 × 10⁻³ ÷ (0.10² × 0.10) = 2.0 L² mol⁻² s⁻¹.
  7. 7Rate = 2.0 × 0.30² × 0.15 = 0.027 mol L⁻¹ s⁻¹.
  8. 8Water is in large excess, so its concentration is essentially constant and the rate depends only on the ester concentration.

Integrated rate equations and half-life

Link to Std 11 : In Std 11 you used logarithms and exponential decay of radioactive nuclei; first-order reactions follow the same law.

From daily life

A first-order change is like the foam on a drink: each minute the same fraction disappears, no matter how much foam is left.

Simple definition

Zero order: [R] = [R]₀ − kt, half-life t½ = [R]₀ ÷ 2k. First order: ln([R]₀ ÷ [R]) = kt, i.e. k = (2.303 ÷ t) log([R]₀ ÷ [R]), half-life t½ = 0.693 ÷ k (does not depend on [R]₀).

Zero-order and first-order plotst →[R]Zero orderstraight line, slope = −kt →log[R]First orderlog[R] vs t: slope = −k ÷ 2.303First-order half-life = 0.693 ÷ k (independent of concentration)
Live figure

Integrated rate equations and half-life

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In the figure : Zero-order: [R] against t is a straight line. First-order: log[R] against t is a straight line with slope −k ÷ 2.303.

Rule 1: Zero-order reactions — The rate is independent of the concentration: rate = k. This happens when the rate is limited by something else, such as the surface of a catalyst. [R] falls linearly and the reaction ends at t = [R]₀ ÷ k.

  • Easy : k = 2.0 × 10⁻³ mol L⁻¹ s⁻¹ and [R]₀ = 0.10 M: the reaction ends at t = 0.10 ÷ 2.0 × 10⁻³ = 50 s.
  • Medium : Half-life = [R]₀ ÷ 2k = 25 s; half-life depends on the starting concentration.
  • Exam-type : Decomposition of NH₃ on a hot platinum surface is zero order at high pressure.

Rule 2: First-order reactions — k = (2.303 ÷ t) log([R]₀ ÷ [R]). Unit of k is s⁻¹. After n half-lives the fraction left is (½)ⁿ. Time for 90 % completion: t = 2.303 ÷ k.

  • Easy : [R]₀ = 0.80 M and [R] = 0.40 M after 100 s: t½ = 100 s, k = 0.693 ÷ 100 = 6.93 × 10⁻³ s⁻¹.
  • Medium : After 3 half-lives (300 s) the concentration is 0.80 ÷ 8 = 0.1 M.
  • Exam-type : Time for 90 % completion with this k: t = 2.303 ÷ (6.93 × 10⁻³) = 332.0 s.

Rule 3: Half-life and finding k — For first order t½ = 0.693 ÷ k, independent of [R]₀; for zero order t½ = [R]₀ ÷ 2k; for second order t½ = 1 ÷ k[R]₀. The order can be told from how t½ changes when [R]₀ changes.

  • Easy : If t½ of a reaction stays 100 s when [R]₀ is halved, it is first order.
  • Medium : [R] falls from 0.50 to 0.35 M in 40 min: k = (2.303 ÷ 40) × log(0.50 ÷ 0.35) = 8.9 × 10⁻³ min⁻¹.
  • Exam-type : A radioactive nucleus with t½ = 5 h: after 15 h, fraction left = (½)³ = 1/8, an example of first-order decay.
Remember

First order: half-life is constant. Zero order: half-life is proportional to [R]₀.

Common mistake

Using ln with 2.303 together, or log with the factor 0.693; use 2.303 with log and 0.693 for t½.

Shortcut

Memory clue: "Half is constant for first" — each half-life removes the same fraction, so t½ stays fixed.

Spot the difference

PointZero orderFirst order
Integrated law[R] = [R]₀ − ktln([R]₀ ÷ [R]) = kt
Half-life[R]₀ ÷ 2k0.693 ÷ k
Straight-line plot[R] vs tlog[R] vs t

Practice questions

Easy
  1. 1Write the half-life formula for a first-order reaction.
  2. 2What is the unit of k for a zero-order reaction?
  3. 3What fraction is left after 3 half-lives?
Medium
  1. 4Find k of a first-order reaction with t½ = 100 s.
  2. 5[R] falls from 0.50 to 0.35 M in 40 min. Find k.
  3. 6How long does a zero-order reaction with k = 2.0 × 10⁻³ mol L⁻¹ s⁻¹ and [R]₀ = 0.10 M take to finish?
Challenging
  1. 7For k = 6.93 × 10⁻³ s⁻¹ (first order), find the time for 90 % completion.
  2. 8Show that for a first-order reaction t½ does not depend on the initial concentration.
Show answers (try them yourself first)
  1. 1t½ = 0.693 ÷ k.
  2. 2mol L⁻¹ s⁻¹.
  3. 3(½)³ = 1/8.
  4. 4k = 0.693 ÷ 100 = 6.93 × 10⁻³ s⁻¹.
  5. 5k = (2.303 ÷ 40) × log(0.50 ÷ 0.35) = 0.0089 min⁻¹.
  6. 6t = 0.10 ÷ 2.0 × 10⁻³ = 50 s.
  7. 7t = 2.303 ÷ (6.93 × 10⁻³) = 332.0 s.
  8. 8At t½, [R] = [R]₀ ÷ 2, so k = (2.303 ÷ t½) log 2 = 0.693 ÷ t½; [R]₀ cancels.

Temperature dependence and the Arrhenius equation

Link to Std 11 : In Std 11 you saw that heating increases molecular motion; here the same idea gives a quantitative law for the rate constant.

From daily life

Milk spoils in hours on a warm day but lasts days in a fridge; a 10 K change in temperature often doubles or triples a reaction rate.

Simple definition

The Arrhenius equation is k = A e^(−Ea/RT), where A is the pre-exponential (frequency) factor, Ea the activation energy (minimum extra energy needed by colliding molecules), R = 8.314 J K⁻¹ mol⁻¹ and T the temperature in kelvin. For two temperatures: log(k₂ ÷ k₁) = [Ea ÷ (2.303 R)] × (T₂ − T₁) ÷ (T₁T₂).

Fraction of molecules with energy above Ea rises with temperaturekinetic energy →fraction of moleculesEaTT + 10more molecules beyond Eawhen T is higherk = A e^(−Ea/RT) · log k = log A − Ea ÷ (2.303 R T)
Live figure

Temperature dependence and the Arrhenius equation

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In the figure : Raising the temperature moves the curve to the right, so a larger fraction of molecules has energy above Ea. Since the curve is nearly flat there, a small rise in T gives a large increase in the number of effective molecules.

Rule 1: Reading the equation — Higher T or lower Ea gives a larger k. Plotting ln k (or log k) against 1/T gives a straight line of slope −Ea ÷ R (or −Ea ÷ 2.303R) and intercept ln A (log A).

  • Easy : A reaction with Ea = 0 would have k = A at every temperature.
  • Medium : The unit of A is the same as the unit of k.
  • Exam-type : From a straight-line graph of log k vs 1/T with slope −2620 K: Ea = 2620 × 2.303 × 8.314 = 50.2 kJ mol⁻¹.

Rule 2: Using two temperatures — Use the two-temperature form to compare rates or to find Ea. Temperatures are in kelvin.

  • Easy : Ea = 50 kJ mol⁻¹, 300 K → 310 K: log(k₂ ÷ k₁) = (50000 ÷ 19.147) × (10 ÷ 93000) = 0.281, so k₂ ÷ k₁ = 1.91.
  • Medium : If the rate doubles from 300 to 310 K: Ea = 2.303 × 8.314 × log 2 × (300 × 310 ÷ 10) = 53.6 kJ mol⁻¹.
  • Exam-type : A high Ea means the rate is very sensitive to temperature.

Rule 3: Activation energy and the energy distribution — Only collisions between molecules with energy ≥ Ea can react. The fraction of such molecules is e^(−Ea/RT). A 10 K rise roughly doubles the fraction, which is why rates roughly double for every 10 K increase.

  • Easy : Raising T from 300 K to 310 K increases the fraction e^(−Ea/RT) for Ea = 50 kJ mol⁻¹ by a factor of 1.91.
  • Medium : The average kinetic energy changes only a little for a 10 K rise, but the fraction of high-energy molecules grows a lot.
  • Exam-type : At 300 K the fraction e^(−50000 ÷ (8.314 × 300)) = 1.97e-09; this tiny fraction is why not every collision reacts.
Remember

Rate constant increases with temperature; the effect is bigger for reactions with larger activation energy.

Common mistake

Using °C instead of kelvin in the Arrhenius equation.

Shortcut

Memory clue: "Hot and low-barrier is fast" — high temperature or low Ea makes k large.

Spot the difference

PointLow EaHigh Ea
Rate at a given TFastSlow
Sensitivity to temperatureSmallLarge
Fraction e^(−Ea/RT)LargerSmaller

Practice questions

Easy
  1. 1State the Arrhenius equation.
  2. 2What is activation energy?
  3. 3Why must temperatures be in kelvin?
Medium
  1. 4A reaction with Ea = 50 kJ mol⁻¹ goes from 300 K to 310 K. By what factor does k change?
  2. 5If the rate doubles between 300 K and 310 K, find Ea.
  3. 6A graph of log k vs 1/T has slope −2620 K. Find Ea.
Challenging
  1. 7Explain, using the energy distribution curve, why a 10 K rise greatly increases the rate.
  2. 8A reaction has Ea = 100 kJ mol⁻¹. Compare its k at 400 K and 410 K.
Show answers (try them yourself first)
  1. 1k = A e^(−Ea/RT).
  2. 2The minimum extra energy that colliding molecules must have to form products.
  3. 3Because T appears in the exponent in the form of RT; the kelvin scale starts at absolute zero.
  4. 4k₂ ÷ k₁ = 1.91.
  5. 5Ea = 53.6 kJ mol⁻¹.
  6. 6Ea = 2620 × 2.303 × 8.314 = 50.2 kJ mol⁻¹.
  7. 7Raising T flattens the distribution so that a much larger fraction of molecules has energy above Ea, although the average energy changes only a little.
  8. 8log(k₂ ÷ k₁) = (100000 ÷ 19.147) × (10 ÷ (400 × 410)) = 0.318; k₂ ÷ k₁ = 2.08.

Catalysis and collision theory

Link to Std 11 : In Std 11 you learnt that a catalyst does not change the position of equilibrium; here you see why it speeds up the reaction.

From daily life

A match head does not light by itself in air but does when struck: the strike supplies the activation energy.

Simple definition

A catalyst increases the rate by providing an alternative path with lower activation energy; it is not used up and it does not change ΔH or K. Collision theory: a reaction occurs when molecules collide with enough energy (≥ threshold energy) and the right orientation; k = P Z(AB) e^(−Ea/RT), where P is the steric factor and Z(AB) is the collision frequency.

Energy profile with and without a catalystprogress of reaction →energyred: no catalyst (Ea)green: with catalyst (Ea′)reactantsproductsCatalyst lowers Ea; ΔH and the final equilibrium do not change
Live figure

Catalysis and collision theory

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In the figure : The red path has a high hill (Ea); the catalyst gives a green path with a lower hill. Reactants and products stay at the same levels, so ΔH is unchanged.

Rule 1: How a catalyst works — Catalyst lowers Ea, so a bigger fraction of collisions are effective. It speeds up forward and reverse reactions equally and it does not alter the equilibrium constant.

  • Easy : Lowering Ea from 75 to 50 kJ mol⁻¹ at 300 K multiplies k by e^(25000 ÷ (8.314 × 300)) = 2.3e+04.
  • Medium : A catalyst makes equilibrium reach sooner, but the equilibrium composition is the same.
  • Exam-type : MnO₂ speeds up decomposition of H₂O₂; Pt speeds up the oxidation of SO₂ in the contact process.

Rule 2: Collision theory — Not every collision leads to reaction. The colliding molecules must have enough energy (threshold energy = Ea + average energy of reactants) and a proper orientation. Rate = P × Z(AB) × e^(−Ea/RT).

  • Easy : Two molecules colliding with energy lower than Ea bounce apart.
  • Medium : In CH₃Br + OH⁻ → CH₃OH + Br⁻ the OH⁻ must hit the C atom from the side opposite to Br for reaction; otherwise there is no reaction.
  • Exam-type : The steric factor P (between 0 and 1 for most reactions) accounts for orientation; a smaller P means a more demanding orientation.

Rule 3: Activation energy and reaction path — Activation energy is the energy hill between reactants and the activated complex (transition state). A reaction can be exothermic and still slow if Ea is high.

  • Easy : Burning of coal is exothermic but needs a spark to start, because Ea is high.
  • Medium : Raising temperature, increasing concentration (more collisions) and adding a catalyst (lower Ea) all increase rate, but only the catalyst changes Ea.
  • Exam-type : For the same reaction compare two paths: path 1 Ea = 80 kJ mol⁻¹, path 2 Ea = 60 kJ mol⁻¹; path 2 is faster at the same temperature.
Remember

A catalyst lowers Ea for both directions; it does not change ΔH, K or the final yield.

Common mistake

Saying a catalyst changes the equilibrium constant or the energy of the reactants and products.

Shortcut

Memory clue: "Catalyst cuts the hill, not the valleys" — Ea falls, but reactant and product energies stay fixed.

Spot the difference

PointWithout catalystWith catalyst
EaHigherLower
ΔH and KSameSame
RateSlowerFaster

Practice questions

Easy
  1. 1What is a catalyst?
  2. 2Does a catalyst change ΔH of a reaction?
  3. 3What is the steric factor?
Medium
  1. 4Ea falls from 75 to 50 kJ mol⁻¹ on adding a catalyst. By what factor does k change at 300 K?
  2. 5Why are only some collisions effective?
  3. 6Why is combustion of coal slow at room temperature although it is exothermic?
Challenging
  1. 7Draw (describe) the energy profile for a catalysed and an uncatalysed exothermic reaction and label Ea, Ea′ and ΔH.
  2. 8A catalyst is added to a reaction at equilibrium. What happens to the equilibrium composition?
Show answers (try them yourself first)
  1. 1A substance that increases the rate of a reaction by lowering the activation energy without being used up.
  2. 2No.
  3. 3The factor P in k = P Z e^(−Ea/RT) that accounts for the orientation of the colliding molecules.
  4. 4k₂ ÷ k₁ = e^(25000 ÷ (8.314 × 300)) = 2.3e+04.
  5. 5Only collisions with enough energy and the correct orientation lead to products.
  6. 6Its Ea is high at room temperature, so very few collisions are energetic enough; a spark supplies the energy to start it.
  7. 7Both profiles start at the reactant level and end at a lower product level (same ΔH); the uncatalysed path has a high hill (Ea), the catalysed path a lower hill (Ea′).
  8. 8Nothing: the composition is the same, but equilibrium is reached sooner.