Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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In preparation: 0 of 13 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R2.2.1 Rate of reaction
Rate of reaction
The change in concentration of a particular reactant or product per unit time, usually in mol dm⁻³ s⁻¹. For a reactant the concentration falls, and the rate is quoted as a positive value (rate = −Δ[reactant]/Δt); for a product it is +Δ[product]/Δt. In practice the progress of a reaction can also be followed by another measurable quantity that changes with it, such as the volume of gas produced (rate in cm³ s⁻¹) or the loss in mass of a reaction flask (rate in g s⁻¹).
Rate from the tangent to a graph
The rate at a particular instant is the gradient of the tangent to a graph of concentration, volume or mass against time at that instant. A straight line is drawn touching the curve at that time, two well-separated points are read from the line, and gradient = (change in y)/(change in x), with units of the y-quantity divided by time (e.g. cm³ s⁻¹). The initial rate is the gradient of the tangent at t = 0. For most reactions the gradient, and so the rate, decreases as the reactants are used up.
Students often think The rate of a reaction is the time it takes for the reaction to finish, so a larger time means a larger rate. In fact No. Rate is a change in concentration (or another measured quantity) per unit time. A shorter reaction time goes with a greater rate, so time and rate are inversely related, not the same thing.
Students often think A faster reaction makes more product, so the rate can be judged from how much product is formed in total. In fact Not necessarily. The rate says how quickly product forms; the total amount of product depends on the amounts of reactants (the limiting reactant). A faster reaction reaches the same final amount sooner.
R2.2.2 Collision theory
Collision theory
The model in which reacting species must collide in order to react, and a collision leads to reaction only if the particles have a combined kinetic energy equal to or greater than the activation energy, Ea, and collide with a proper orientation. Only a small fraction of all collisions meet both conditions; those that do are called successful collisions.
Collision geometry (orientation)
The relative orientation of particles at the moment they collide. For most reactions a collision is successful only if the atoms that form the new bond meet; collisions with an unsuitable orientation do not lead to reaction however much energy they have. Single atoms react in any orientation, but larger molecules often react only when they collide at a particular site.
Kinetic energy and absolute temperature
The average (mean) kinetic energy of the particles in a substance is directly proportional to its absolute temperature in kelvin (T/K = θ/°C + 273). Doubling the kelvin temperature doubles the mean kinetic energy; a rise from 20 °C to 40 °C increases it only by the factor 313/293 ≈ 1.07. At a given temperature the particles have a wide range of kinetic energies, and all gases at the same temperature have the same mean kinetic energy per particle.
Students often think Whether a collision leads to reaction depends only on how much energy it has (or simply on how often particles collide); the orientation of the particles does not matter. In fact No. A collision must also have a proper orientation, so that the atoms that form the new bonds meet.
Students often think The mean kinetic energy of the particles is proportional to the Celsius temperature, so doubling the temperature in °C doubles it. In fact No. It is proportional to the absolute temperature in kelvin; 0 °C is 273 K, not zero energy.
R2.2.3 Collision frequency
Collision frequency
The number of collisions between reactant particles per unit time in a given volume. It increases with concentration (or, for gases, pressure), with the surface area of a solid reactant, and slightly with temperature. Increasing collision frequency increases the rate, because the number of successful collisions per unit time rises even though the fraction of collisions that succeed is unchanged.
Effect of concentration and pressure
Increasing the concentration of a reactant in solution, or the pressure of a gaseous reactant (by adding more gas or by reducing the volume), puts more particles in each unit of volume. Collisions become more frequent, so there are more successful collisions per unit time and the rate increases. The energy of each collision, and so the fraction of collisions with E ≥ Ea, is not changed.
Effect of surface area
Only the particles at the surface of a solid can collide with the other reactant. Dividing a solid into smaller pieces or a powder increases its total surface area for the same mass, so collisions with the other reactant are more frequent and the rate increases. The total amount of product is unchanged if the amounts of reactants are unchanged.
Effect of temperature
Raising the temperature increases the mean kinetic energy of the particles. A much larger fraction of collisions then have energy equal to or greater than Ea, which is the main reason the rate increases; collisions also become slightly more frequent. For many reactions a rise of 10 K near room temperature roughly doubles the rate. Raising the temperature increases the rate of both endothermic and exothermic reactions.
Students often think A bigger lump has a bigger surface, so it has the larger surface area and reacts faster. In fact No. Breaking a solid into smaller pieces exposes more surface; the same mass of powder has a far larger total surface area than a single lump.
Students often think Crowding the particles together, or breaking up a solid, makes the particles collide harder, so a larger fraction of collisions have enough energy to react. In fact No. These changes make collisions more frequent. The energy of each collision depends on temperature, so the fraction of collisions with E ≥ Ea is unchanged.
R2.2.4 Activation energy, Ea
Activation energy, Ea
The minimum energy that colliding particles need for a successful collision leading to a reaction, in kJ mol⁻¹. On an energy profile it is the energy difference between the reactants and the transition state. Ea is fixed for a given reaction pathway; it is not changed by temperature or concentration, and it is different from the enthalpy change, ΔH.
Maxwell–Boltzmann energy distribution
A graph of the number (or fraction) of particles against kinetic energy for a sample at one temperature. It starts at the origin, rises to a peak (the most probable energy), and falls to a long tail that approaches but does not reach the energy axis. The area under the curve represents the total number of particles. The area beyond Ea represents the particles with enough energy to react. At a higher temperature the curve has the same area but is flatter and broader: the peak is lower and moves to higher energy, and the area beyond Ea increases greatly.
Students often think Heating gives every particle the same extra energy, so the whole energy distribution shifts to higher energy without changing shape, and every particle ends up more energetic than before. In fact No. The particles have a spread of energies before and after heating. The mean energy increases and the distribution becomes flatter and broader; it does not simply move along the energy axis.
Students often think Every change that increases the rate, including raising the temperature or adding a catalyst, works mainly by making the particles collide more often. In fact No. A rise of 10 K near room temperature increases the collision frequency by less than 2%, but it greatly increases the fraction of collisions with E ≥ Ea. A catalyst does not change the collision frequency at all; it lowers Ea.
R2.2.5 Catalyst
Catalyst
A substance that increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy; it is not consumed in the overall reaction. A catalyst does not change the energies of the reactants or products, so ΔH is unchanged, and it lowers the activation energy of the forward and reverse reactions by the same amount.
Energy profiles with and without a catalyst
An energy profile plots potential energy against reaction progress. For an exothermic reaction the products are lower than the reactants; for an endothermic reaction they are higher. The maximum of the curve is the transition state, and Ea is the height of the maximum above the reactants. The profile for the catalysed pathway starts and ends at the same energies but has a lower maximum, so Ea is smaller and ΔH is the same.
Catalysts and the Maxwell–Boltzmann distribution
At a fixed temperature the Maxwell–Boltzmann curve is the same with or without a catalyst. Marking the lower Ea of the catalysed pathway on the energy axis shows a larger area of the curve beyond Ea, so a greater fraction of collisions have enough energy to react and the rate increases.
Enzyme
A biological catalyst, usually a protein, that increases the rate of a reaction in living organisms by providing a pathway with a lower activation energy. For example, catalase catalyses the decomposition of hydrogen peroxide into water and oxygen.
Students often think A reaction cannot take place at all unless a catalyst is present, so a catalyst is what starts the reaction. In fact Yes. A catalyst only provides a faster pathway; the uncatalysed reaction still occurs, usually more slowly.
Students often think A catalyst lowers the energy of the products (or lowers ΔH), making the reaction more exothermic, and that is why less energy is needed to react. In fact No. A catalyst changes the pathway, not the energies of the reactants or products, so ΔH is unchanged.
R2.2.6 Elementary step and reaction mechanism HL
Elementary step and reaction mechanism
An elementary step is a single molecular event, such as one collision, in which bonds are broken and/or formed. A reaction mechanism is the sequence of elementary steps by which a reaction occurs. The steps must add up to the overall stoichiometric equation, and the mechanism must be consistent with the experimentally determined rate equation.
Rate-determining step
The slowest elementary step in a mechanism; it limits the rate of the overall reaction. The rate equation contains the species taking part in the rate-determining step and in any steps before it. The rate-determining step need not be the first step: when a fast step comes first, the rate equation includes the reactants of that step as well.
Reaction intermediate
A species formed in one elementary step and used up in a later step, so it does not appear in the overall equation. On an energy profile an intermediate lies in a minimum (a dip) between two maxima; it has a real, if often short, lifetime and can in principle be detected.
Transition state
The highest-energy arrangement of atoms in an elementary step, at a maximum on the energy profile, in which bonds are partly broken and partly formed. A transition state exists only momentarily and cannot be isolated. A multistep reaction has one transition state for each step.
Students often think The first step of a mechanism is always the slow, rate-determining step, and its reactants (with its coefficients) give the rate equation. In fact No. The rate-determining step is the slowest step, which can come after one or more fast steps; then the rate equation includes the reactants of those earlier steps as well.
Students often think A proposed mechanism only has to give the correct rate equation; it does not matter whether the steps add up to the overall equation. In fact No. A mechanism must be consistent with both the kinetic data (the rate equation) and the stoichiometric data (the steps must add up to the overall equation).
R2.2.7 Energy profile of a multistep reaction HL
Energy profile of a multistep reaction
A profile with one maximum (transition state) for each elementary step and a minimum for each intermediate between them. The activation energy of each step is measured from the energy of the species entering that step (the reactants or an intermediate) up to the transition state of that step. The rate-determining step is the slowest step, the one with the largest activation energy.
Students often think The activation energy of any step is the height of its maximum above the original reactants. In fact No. The activation energy of a step is measured from the species entering that step (for a later step, the intermediate) up to that step's transition state.
R2.2.8 Molecularity HL
Molecularity
The number of reacting particles (molecules, atoms, ions or radicals) that take part in an elementary step. A unimolecular step involves one particle (e.g. N₂O₄ → 2NO₂), a bimolecular step two particles (e.g. 2NO₂ → NO₃ + NO), and a termolecular step three (e.g. 2NO(g) + O₂(g) → 2NO₂(g), if it occurs in a single three-particle collision). Termolecular steps are rare because a simultaneous three-particle collision is unlikely. Molecularity applies to elementary steps only, not to an overall equation.
Students often think Molecularity (or order) is the number of different reactant species, so a step or rate equation with only one kind of reactant is unimolecular (or first order). In fact The number of particles. In 2NO → N₂O₂ two NO molecules collide, so the step is bimolecular even though only one kind of species reacts.
Students often think Molecularity counts all the particles in the step, the products as well as the reactants. In fact No. Molecularity counts only the reacting particles in the elementary step; the products are not counted.
R2.2.9 Rate equation HL
Rate equation
An expression of the form rate = k[A]^m[B]^n relating the rate to the concentrations of reactants, where m and n are the orders with respect to A and B. It depends on the mechanism and can only be determined experimentally; the orders are not deduced from the coefficients of the balanced equation, and products do not normally appear in it.
Initial rates method
A way to determine a rate equation from experimental data. The initial rate is measured in several experiments in which the concentration of one reactant is changed while the others are kept constant. If doubling [A] leaves the rate unchanged, doubles it or quadruples it, the order with respect to A is 0, 1 or 2 respectively.
Students often think The orders in the rate equation are the coefficients of the reactants in the balanced equation. In fact No. Rate equations depend on the mechanism and can only be determined experimentally; the orders need not equal the stoichiometric coefficients.
Students often think A rate equation is written like the expression for K, with the concentrations of the products divided by those of the reactants. In fact No. A rate equation contains the concentrations of the species in the rate-determining step (and steps before it), raised to experimentally determined orders; products do not normally appear, and there is no 'products over reactants' form.
R2.2.10 Order of reaction HL
Order of reaction
The order with respect to a reactant is the exponent to which its concentration is raised in the rate equation; it can describe the number of particles of that reactant taking part in the rate-determining step (and steps before it). The overall order is the sum of the orders with respect to each reactant. Only integer orders (0, 1, 2) are assessed.
Graphs for zero-, first- and second-order reactions
Concentration–time graphs: zero order, a straight line with a negative gradient; first order, a curve whose gradient decreases, with the concentration falling by the same fraction in equal time intervals; second order, a curve that falls more steeply at first and then flattens more than the first-order curve. Rate–concentration graphs: zero order, a horizontal line; first order, a straight line through the origin; second order, an upward curve (a parabola) through the origin.
Students often think The rate is always directly proportional to the concentration of each reactant, so every reactant can be treated as first order. In fact No. Only for a first-order reactant. For a zero-order reactant the rate does not depend on its concentration; for a second-order reactant the rate is proportional to its concentration squared.
Students often think A reactant that is zero order is not used up (or does not take part in the reaction), because it does not appear in the rate equation. In fact Yes. A zero-order reactant still reacts and its concentration falls; its concentration simply does not affect the rate, often because it is not involved in the rate-determining step.
R2.2.11 Rate constant, k HL
Rate constant, k
The constant of proportionality in the rate equation. For a given reaction its value depends on temperature (it increases as temperature increases) but not on the concentrations of the reactants. It is found by rearranging the rate equation, for example k = rate/([A]^m[B]^n).
Units of the rate constant
The units of k follow from the overall order, because k = rate/(concentration terms): zero order, mol dm⁻³ s⁻¹; first order, s⁻¹; second order, mol⁻¹ dm³ s⁻¹; third order, mol⁻² dm⁶ s⁻¹.
Students often think The rate constant always has the same units as the rate, mol dm⁻³ s⁻¹. In fact Only for a zero-order reaction. The units of k depend on the overall order, because k = rate/(concentration terms).
Students often think To isolate k (or A), the other side of the equation is multiplied by the remaining terms rather than divided by them. In fact By dividing: k = rate/([A]^m[B]^n), and A = k/e^(−Ea/RT) = k × e^(Ea/RT).
R2.2.12 Arrhenius equation HL
Arrhenius equation
k = Ae^(−Ea/RT), where k is the rate constant, A the Arrhenius factor, Ea the activation energy in J mol⁻¹, R the gas constant (8.31 J K⁻¹ mol⁻¹) and T the absolute temperature in K. The term e^(−Ea/RT) is the fraction of collisions with energy equal to or greater than Ea, so k increases as T increases, and increases more steeply for a reaction with a larger Ea.
Linear form of the Arrhenius equation
Taking natural logarithms gives ln k = −Ea/RT + ln A. A graph of ln k (y-axis) against 1/T (x-axis) is a straight line with gradient −Ea/R and y-intercept ln A. Hence Ea = −gradient × R, and A = e^(intercept).
Students often think k is a constant for a given reaction, so it does not change with temperature. In fact No. k is constant for a given reaction only at a fixed temperature; it increases as temperature increases, as described by the Arrhenius equation.
Students often think The gradient of ln k against 1/T is −Ea (or ln(k₂/k₁) divided by (1/T₁ − 1/T₂) gives Ea directly), so R need not be included. In fact No. The gradient is −Ea/R, so Ea = −gradient × R, with R = 8.31 J K⁻¹ mol⁻¹.
R2.2.13 Arrhenius factor, A HL
Arrhenius factor, A
The pre-exponential factor in the Arrhenius equation. It takes into account the frequency of collisions with proper orientations, so it is smaller for reactions whose particles must collide in a particular orientation. A has the same units as k.
Determining Ea and A from experimental data
Ea is found from the gradient of a graph of ln k against 1/T (Ea = −gradient × R), or from rate constants at two temperatures: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Temperatures must be in kelvin, natural logarithms must be used, and because R is in J K⁻¹ mol⁻¹ the result is in J mol⁻¹. A is found from ln A = ln k + Ea/RT, or from the intercept of the graph (A = e^intercept).
Students often think Temperatures can be substituted into the Arrhenius equation in °C, as they are given in the question. In fact No. T in the Arrhenius equation must be the absolute temperature in kelvin.
Students often think log₁₀ and ln are interchangeable in the Arrhenius equation, so either logarithm key on a calculator can be used. In fact No. The linear form uses natural logarithms: ln k = −Ea/RT + ln A. Using log₁₀ without the conversion factor (ln x = 2.303 log₁₀ x) gives wrong values of Ea and A.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Hydrogen peroxide decomposes: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Which quantity is the rate of this reaction at a particular moment?
Answer and reasoning
The total time taken for all of the H₂O₂ in the sample to decompose — A student who treats rate and time as the same thing picks this. Time taken is not a rate; a reaction that takes longer has a smaller rate. A rate is a change divided by a time.
The total volume of O₂ given off by the time the reaction has stopped — A student who confuses how fast with how much picks this. The total volume of O₂ depends on the amount of H₂O₂ used, not on how quickly it decomposes, and it is not divided by any time.
The decrease in the concentration of H₂O₂ per unit time at that moment — Rate is the change in concentration of a reactant or product per unit time, in mol dm⁻³ s⁻¹. At a particular moment it is the gradient of the tangent to the concentration–time curve, quoted as a positive value for the decrease in [H₂O₂].
The concentration of H₂O₂ at that moment divided by the time elapsed so far — A student who divides the y-value by the x-value picks this. The rate is a change in concentration per unit time; [H₂O₂]/t is not even an average rate, and the rate at an instant needs the gradient of the tangent.
2 Two reactant molecules in a gas collide with a combined kinetic energy greater than the activation energy, Ea, of the reaction, but they do not react. Which statement explains this?
Answer and reasoning
No catalyst was present, and a reaction cannot take place at all without one. — A student who thinks reactions need a catalyst picks this. A catalyst only provides a faster pathway; without one, collisions with E ≥ Ea and a proper orientation still react.
They collided with an orientation that did not bring the reacting atoms together. — A successful collision needs both energy equal to or greater than Ea and a proper orientation. With enough energy but the wrong collision geometry, the atoms that must bond do not meet, so no reaction occurs.
Their combined energy was less than the size of the enthalpy change of reaction, ΔH. — A student who confuses Ea with ΔH picks this. The energy a collision needs is Ea, the barrier to the transition state; ΔH is the energy difference between reactants and products and is not the threshold.
No spark or heating had first supplied the activation energy to start the reaction. — A student who thinks activation energy is a one-off 'start-up' energy from a spark or flame picks this. Ea is the minimum energy each colliding pair must have, and this collision already had more than Ea, so no outside energy was missing. The collision failed because of its orientation.
3 Two experiments use equal masses of calcium carbonate and the same volume of excess hydrochloric acid of the same concentration at the same temperature. In experiment 1 the calcium carbonate is a single lump; in experiment 2 it is a fine powder. Which statement is correct?
Answer and reasoning
The powder reacts faster, as more of its particles are exposed, so collisions with acid are more frequent. — The powder has a much larger total surface area than the lump of the same mass, so more calcium carbonate particles can be hit by acid particles. Collisions are more frequent, so there are more successful collisions per second.
The lump reacts faster, as its greater size gives it the larger surface area for the acid particles to attack. — A student who compares one lump with one grain picks this. For equal masses, breaking the solid up exposes far more surface in total, so the powder has the larger surface area and reacts faster.
The powder reacts faster, as its exposed particles collide with the acid particles with more energy. — A student who thinks surface area changes collision energy picks this. The energies of collisions depend on temperature, which is the same; the powder reacts faster because collisions are more frequent.
The powder gives off more carbon dioxide in total, as its faster reaction goes further to completion. — A student who links a faster reaction with more product picks this. The masses of calcium carbonate are equal and the acid is in excess, so both give the same volume of CO₂; the powder simply gives it off sooner.
4 A Maxwell–Boltzmann distribution is plotted for a sample of gas at 300 K, with number of particles on the y-axis and kinetic energy on the x-axis. The curve starts at the origin, rises to a peak and falls to a long tail that approaches the energy axis. Which describes the curve for the same sample at 400 K?
Answer and reasoning
The peak is higher and at a higher energy, as more particles move fast, and the area is larger. — A student who thinks a higher temperature gives 'more' of everything picks this. The area represents the number of particles, which does not change; because the distribution spreads out, the peak must be lower.
The whole curve moves to higher energy by the same amount, keeping its shape and height. — A student who thinks every particle gains the same energy picks this. The distribution changes shape: it becomes flatter and broader, with the peak lower and at a higher energy.
The peak is lower and at a higher energy, the tail is higher, and the area under the curve is unchanged. — The number of particles is unchanged, so the area under the curve is the same. At the higher temperature the energies spread out to higher values, so the curve is flatter: the peak is lower and moves to the right, and more particles are in the high-energy tail.
The curve is unchanged, but the activation energy becomes lower, so more particles exceed it. — A student who thinks heating lowers Ea picks this. Ea is fixed for the pathway; it is the curve that changes with temperature, so that a larger area lies beyond the same Ea.
5 Which statement correctly distinguishes a reaction intermediate from a transition state in a multistep reaction? HL
Answer and reasoning
A transition state can be isolated if the reaction is stopped quickly enough, just as an intermediate can. — A student who treats a transition state as a real species picks this. A transition state is at an energy maximum and exists only momentarily; it cannot be isolated however the reaction is stopped.
An intermediate is present at the start and is re-formed at the end; a transition state is used up. — A student who confuses intermediates with catalysts picks this. A catalyst is present at the start and re-formed; an intermediate is produced in one step and consumed in a later step.
An intermediate lies at an energy minimum between steps; a transition state is at a maximum and cannot be isolated. — An intermediate is formed in one step and used in a later one, and sits in a dip on the energy profile, so it has a finite lifetime and can in principle be detected. A transition state sits at a peak, with bonds partly broken and partly formed, and cannot be isolated.
An intermediate appears in the overall equation for the reaction; a transition state does not. — A student who does not cancel species when adding steps picks this. An intermediate is produced and then consumed, so it cancels and does not appear in the overall equation.
6 A reaction occurs in two steps. Its energy profile, with energies relative to the reactants, shows: reactants 0 kJ mol⁻¹; a first maximum at +45 kJ mol⁻¹; a minimum at +15 kJ mol⁻¹; a second maximum at +80 kJ mol⁻¹; products −20 kJ mol⁻¹. What is the activation energy of the rate-determining step? HL
Answer and reasoning
+65 kJ mol⁻¹ — The minimum is the intermediate and the two maxima are transition states, so step 2 starts from the intermediate. Ea(step 1) = 45 − 0 = 45 kJ mol⁻¹; Ea(step 2) = 80 − 15 = 65 kJ mol⁻¹. Step 2 has the larger activation energy and the highest transition state, so it is rate-determining, with Ea = +65 kJ mol⁻¹.
+45 kJ mol⁻¹ — A student who assumes the first step is rate-determining picks this. Step 2 has the larger activation energy (80 − 15 = 65 kJ mol⁻¹), so it is the slow step.
+80 kJ mol⁻¹ — A student who measures every activation energy from the original reactants picks this. Step 2 starts from the intermediate at +15 kJ mol⁻¹, so its Ea is 80 − 15 = 65 kJ mol⁻¹.
−35 kJ mol⁻¹ — A student who confuses Ea with ΔH picks this: −35 kJ mol⁻¹ is ΔH for step 2 (−20 − 15). Ea is the rise from the intermediate to transition state 2, +65 kJ mol⁻¹, and is always positive.
Working Ea(step 1) = E(TS1) − E(reactants) = 45 − 0 = 45 kJ mol⁻¹. Ea(step 2) = E(TS2) − E(intermediate) = 80 − 15 = 65 kJ mol⁻¹. Step 2 has the larger Ea (and TS2 is the highest point), so step 2 is rate-determining: Ea = +65 kJ mol⁻¹.
7 Which statement about molecularity is correct? HL
Answer and reasoning
2NO(g) → N₂O₂(g) is a unimolecular step. — A student who counts kinds of species picks this. The coefficient 2 means two NO molecules collide, so the step is bimolecular.
N₂O₄(g) → 2NO₂(g) is a termolecular step. — A student who counts product particles as well picks this. Only one N₂O₄ molecule reacts, so the step is unimolecular.
2NO₂(g) → NO₃(g) + NO(g) is a bimolecular step. — Two NO₂ molecules collide in this elementary step, so two reacting particles take part and it is bimolecular. The products are not counted.
Overall, 2H₂(g) + O₂(g) → 2H₂O(g) is termolecular. — A student who reads an overall equation as a single collision picks this. Molecularity applies only to elementary steps; the overall equation shows stoichiometry, and this reaction occurs by a series of steps, not a single collision of three molecules. A genuine termolecular elementary step is, for example, 2NO(g) + O₂(g) → 2NO₂(g) if it occurs in a single three-particle collision.
8 A reaction is zero order with respect to reactant X. Which statement about a graph for this reaction is correct? HL
Answer and reasoning
The concentration–time graph for X is a straight line with a negative gradient. — For a zero-order reactant the rate does not depend on [X], so X is used up at a constant rate and its concentration falls in a straight line until it is used up.
The concentration–time graph for X is a horizontal line, as X is not used up in the reaction. — A student who thinks a zero-order reactant does not react picks this. X is consumed; only its effect on the rate is zero, so [X] falls at a constant rate.
The rate–concentration graph for X is a straight line sloping downwards, with a negative gradient. — A student who carries the concentration–time shape over to the rate–concentration graph picks this. For zero order the rate is the same at every [X], so the rate–concentration graph is a horizontal line.
The rate–concentration graph for X is a straight line through the origin, as rate rises with [X]. — A student who assumes rate is always proportional to concentration picks this. A straight line through the origin is the first-order shape; for zero order the rate does not change with [X], so the line is horizontal.
9 The rate equation for a reaction is rate = k[A]², with concentrations in mol dm⁻³ and the rate in mol dm⁻³ s⁻¹. What are the units of k? HL
Answer and reasoning
mol dm⁻³ s⁻¹, as the rate constant has the same units as the rate — A student who thinks k always has the units of rate picks this. That is true only for a zero-order reaction; for rate = k[A]² the units are mol⁻¹ dm³ s⁻¹.
mol⁻¹ dm³ s⁻¹, as k = rate/[A]² for a second-order reaction — k = rate/[A]², so its units are (mol dm⁻³ s⁻¹)/(mol dm⁻³)² = mol⁻¹ dm³ s⁻¹, the units for any reaction that is second order overall.
s⁻¹, as only one reactant, A, appears in the rate equation — A student who counts reactant species instead of reading the exponent picks this. [A]² makes the reaction second order overall, so k has units of mol⁻¹ dm³ s⁻¹, not the first-order unit s⁻¹.
mol³ dm⁻⁹ s⁻¹, as k = rate × [A]² when the rate equation is rearranged — A student who multiplies instead of dividing picks this. Rearranging rate = k[A]² gives k = rate/[A]², so the units are mol⁻¹ dm³ s⁻¹.
10 The rate constant of a reaction varies with absolute temperature according to the Arrhenius equation, k = Ae^(−Ea/RT). A student plots ln k (y-axis) against 1/T (x-axis) for one reaction over a range of temperatures. Which describes the graph? HL
Answer and reasoning
A horizontal line, as k is a constant for a given reaction — A student who thinks k does not change with temperature picks this. k is constant only at a fixed temperature; it increases as T rises, so ln k changes with 1/T.
A straight line with a negative gradient, equal to −Ea/R — Taking natural logarithms gives ln k = −(Ea/R)(1/T) + ln A, which has the form y = mx + c. The graph is a straight line with gradient −Ea/R and y-intercept ln A. Because the gradient is negative, ln k increases as 1/T decreases: k increases with temperature.
A straight line with a negative gradient, equal to −Ea — A student who forgets R picks this. The exponent is −Ea/RT, so the gradient of ln k against 1/T is −Ea/R, and Ea = −gradient × R.
A straight line with a positive gradient, as k rises with T — A student who carries 'k rises with T' straight onto the graph picks this. The x-axis is 1/T, which decreases as T increases, so ln k is largest at small 1/T and the gradient, −Ea/R, is negative.
Read the ones marked not yet in Learn, then Verify.
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28 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Magnesium ribbon reacts with excess hydrochloric acid, and the hydrogen is collected in a gas syringe. A graph of the volume of hydrogen (y-axis, cm³) against time (x-axis, s) is a curve that is steep at first, becomes less steep and levels off at 60 cm³ at 100 s. The curve passes through the point (20 s, 34 cm³). A tangent drawn to the curve at t = 20 s passes through the points (0 s, 10 cm³) and (40 s, 58 cm³). What is the rate of reaction at t = 20 s?
Answer and reasoning
1.70 cm³ s⁻¹ — A student who divides the coordinates of the point gets 34 cm³ ÷ 20 s = 1.70 cm³ s⁻¹. That is the average rate over the first 20 s, which is larger than the rate at 20 s because the reaction has been slowing down. Use the tangent: 48 cm³ ÷ 40 s = 1.20 cm³ s⁻¹.
0.60 cm³ s⁻¹ — A student who assumes the rate is the same throughout divides the final volume by the total time: 60 cm³ ÷ 100 s = 0.60 cm³ s⁻¹. That is the average over the whole reaction; the rate at 20 s is the tangent's gradient, 1.20 cm³ s⁻¹.
0.83 cm³ s⁻¹ — A student who calculates the gradient upside down gets 40 s ÷ 48 cm³ = 0.83 (in s cm⁻³, which is not a rate unit). Gradient = Δy/Δx = 48 cm³ ÷ 40 s = 1.20 cm³ s⁻¹.
1.20 cm³ s⁻¹ — The rate at 20 s is the gradient of the tangent: (58 − 10) cm³ ÷ (40 − 0) s = 48 cm³ ÷ 40 s = 1.20 cm³ s⁻¹.
Working Gradient of the tangent at t = 20 s = Δ(volume)/Δ(time) = (58 cm³ − 10 cm³)/(40 s − 0 s) = 48 cm³/40 s = 1.20 cm³ s⁻¹.
2 The concentration of a reactant, R, is followed at constant temperature. A graph of [R] (y-axis, mol dm⁻³) against time (x-axis, s) is a curve that starts at 0.0500 mol dm⁻³, becomes less steep and reaches zero at 1500 s. The curve passes through the point (300 s, 0.0220 mol dm⁻³). A tangent drawn to the curve at t = 300 s passes through the points (0 s, 0.0370 mol dm⁻³) and (600 s, 0.0070 mol dm⁻³). What is the rate of reaction at t = 300 s?
Answer and reasoning
7.33 × 10⁻⁵ mol dm⁻³ s⁻¹ — A student who divides the value of y at the point by the time gets 0.0220 mol dm⁻³ ÷ 300 s = 7.33 × 10⁻⁵ mol dm⁻³ s⁻¹. The concentration remaining is not a change in concentration, so this is not a rate. Use the tangent: 0.0300 mol dm⁻³ ÷ 600 s = 5.00 × 10⁻⁵ mol dm⁻³ s⁻¹.
5.00 × 10⁻⁵ mol dm⁻³ s⁻¹ — The rate at 300 s is the gradient of the tangent at that time, quoted as a positive value for a reactant: (0.0370 − 0.0070) mol dm⁻³ ÷ (600 − 0) s = 0.0300 ÷ 600 = 5.00 × 10⁻⁵ mol dm⁻³ s⁻¹.
3.33 × 10⁻⁵ mol dm⁻³ s⁻¹ — A student who assumes the reaction goes at one steady rate divides the total change by the total time: 0.0500 mol dm⁻³ ÷ 1500 s = 3.33 × 10⁻⁵ mol dm⁻³ s⁻¹. The curve becomes less steep, so the rate changes; the rate at 300 s is the gradient of the tangent, 5.00 × 10⁻⁵ mol dm⁻³ s⁻¹.
9.33 × 10⁻⁵ mol dm⁻³ s⁻¹ — A student who takes the average rate so far as the rate at 300 s gets (0.0500 − 0.0220) mol dm⁻³ ÷ 300 s = 9.33 × 10⁻⁵ mol dm⁻³ s⁻¹. That is the average over the first 300 s, when the reaction was faster; the rate at 300 s itself is the gradient of the tangent, 5.00 × 10⁻⁵ mol dm⁻³ s⁻¹.
Working Gradient of the tangent at t = 300 s = Δ[R]/Δt = (0.0070 − 0.0370) mol dm⁻³/(600 − 0) s = −0.0300 mol dm⁻³/600 s = −5.00 × 10⁻⁵ mol dm⁻³ s⁻¹. The rate is quoted as a positive value for the decrease in [R]: rate = 5.00 × 10⁻⁵ mol dm⁻³ s⁻¹.
3 Sample P is neon gas at 40 °C. Sample Q contains twice as many krypton atoms as P and is at 20 °C. Which statement comparing the kinetic energies of the atoms in the two samples is correct?
Answer and reasoning
The mean kinetic energy of the atoms in P is greater than that in Q, but only by about 7%. — Mean kinetic energy is proportional to the absolute temperature: 40 °C = 313 K and 20 °C = 293 K, so the ratio is 313/293 = 1.07. The size of the sample does not affect the mean energy per atom. Neither the number of atoms nor their mass affects the mean kinetic energy per atom.
The mean kinetic energy of the atoms in P is twice the mean kinetic energy in Q. — A student who uses the Celsius temperatures picks this, since 40 °C is twice 20 °C. Mean kinetic energy is proportional to the kelvin temperature: 313 K/293 K = 1.07, an increase of only about 7%.
The atoms in Q have the greater mean kinetic energy, as krypton atoms are heavier. — A student who thinks heavier particles have more kinetic energy at a given temperature picks this. At the same temperature every gas has the same mean kinetic energy per particle, and krypton atoms just move more slowly than neon atoms. Q is colder, so the mean kinetic energy of its atoms is lower.
The atoms in Q have the greater mean kinetic energy, as Q contains more atoms. — A student who thinks a larger sample, holding more heat, has more energetic particles picks this. Q has more total energy because it has more atoms, but the mean kinetic energy per atom depends only on temperature, and Q is colder.
4 Hydrogen iodide decomposes: 2HI(g) → H₂(g) + I₂(g). A sample of hydrogen iodide is compressed to half its volume at constant temperature. What is the effect on the rate of reaction, and why?
Answer and reasoning
It is unchanged: there are equal amounts of gas on both sides of the equation, so pressure has no effect. — A student who applies Le Châtelier's principle to rates picks this. Equal amounts of gas on each side matter for an equilibrium position, not for the rate. Compressing the gas increases the concentration of HI, so the rate increases.
It increases: the compressed molecules collide more energetically, so more collisions exceed Ea. — A student who thinks crowding makes collisions harder picks this. At constant temperature the energy distribution is unchanged; compression increases how often molecules collide, not how energetically.
It increases: a larger amount of H₂ and I₂ can now form from the same amount of HI as before. — A student who links a faster rate with more product picks this. The amount of HI is unchanged, so compressing it does not change the amount of H₂ and I₂ that can form; it changes how quickly they form.
It increases: collisions become more frequent, and the fraction with E ≥ Ea is unchanged. — Halving the volume doubles the concentration of HI molecules, so they collide more often. The temperature, and so the fraction of collisions with E ≥ Ea, is unchanged; more collisions per second means more successful collisions per second.
5 For a reaction with Ea = 50 kJ mol⁻¹, raising the temperature from 300 K to 310 K almost doubles the rate. Which statement gives the main reason for this large increase?
Answer and reasoning
A much larger fraction of all the collisions now have energy equal to or greater than Ea. — A 10 K rise increases the mean kinetic energy by only about 3%, but because Ea lies far out in the tail of the Maxwell–Boltzmann distribution, the area beyond Ea grows greatly. Far more collisions are successful, which is the main reason the rate nearly doubles.
The particles move faster, so they collide more often, and more collisions give more reactions. — A student who explains every rate increase by collision frequency picks this. Collisions do become slightly more frequent, but by less than 2% for this 10 K rise, far too little to double the rate. The main effect is the larger fraction with E ≥ Ea.
The activation energy decreases as the temperature rises, so more of the collisions succeed. — A student who thinks heating lowers Ea picks this. Ea stays at 50 kJ mol⁻¹; the temperature rise changes the energy distribution so that more collisions have at least that energy.
Every particle gains the same amount of energy, which carries many more of them past Ea. — A student who pictures every particle gaining the same energy picks this. Particles have a spread of energies that is redistributed by collisions; the distribution becomes broader and flatter rather than moving along as a block.
6 Catalase is an enzyme that catalyses the exothermic decomposition of hydrogen peroxide: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). The energy profile of the uncatalysed reaction rises from the reactants to a maximum and falls to products that are lower in energy than the reactants. How does the profile differ when catalase is present?
Answer and reasoning
The maximum is lower and the products are lower in energy, so the reaction becomes more exothermic. — A student who thinks a catalyst lowers the whole profile picks this. The products are the same substances with the same energy, so ΔH is unchanged; only the maximum is lower.
The maximum is lower, while the energies of the reactants and products, and so ΔH, are unchanged. — An enzyme is a biological catalyst: it provides an alternative pathway with a lower activation energy. It does not change the energies of the reactants or products, so the profile starts and ends at the same energies and ΔH is unchanged.
The reactants are raised to a higher energy, closer to the top of an unchanged maximum. — A student who thinks a catalyst gives particles energy picks this. Catalase does not add energy to the reactants; it lowers the maximum by providing a different pathway.
The forward Ea is lower but the reverse Ea is unchanged, as only the forward reaction speeds up. — A student who thinks a catalyst affects only the forward reaction picks this. Lowering the maximum lowers the forward and reverse activation energies by the same amount, so ΔH stays the same.
7 A Maxwell–Boltzmann distribution is drawn for a reaction mixture at constant temperature. The activation energy of the uncatalysed reaction is marked as a vertical line far out in the tail of the curve; the activation energy of the catalysed pathway is marked as a second vertical line at a lower energy. Which statement explains why the catalysed reaction is faster?
Answer and reasoning
The catalyst gives the particles extra energy, moving the curve to higher energies past the Ea line. — A student who thinks a catalyst supplies energy picks this. The catalyst does not change the particles' energies; the curve is unchanged at constant temperature and it is the Ea line that moves to lower energy.
The catalyst makes the particles collide more often, while the fraction with E ≥ Ea stays the same. — A student who explains every rate increase by collision frequency picks this. A catalyst does not make particles collide more often; it lowers Ea, which increases the fraction of collisions with enough energy.
The catalyst lowers ΔH of the reaction, so particles need less energy to react when they collide. — A student who thinks a catalyst changes ΔH picks this. ΔH is unchanged by a catalyst; what is lowered is the activation energy of the alternative pathway.
The curve is unchanged, but a larger area lies beyond the lower Ea, so more collisions succeed. — At constant temperature the energy distribution is the same with or without the catalyst. The catalysed pathway has a lower Ea, so a larger area under the curve, and so a larger fraction of collisions, has at least the required energy.
8 For a hypothetical endothermic reaction, the uncatalysed forward activation energy is 280 kJ mol⁻¹ and ΔH = +110 kJ mol⁻¹. A catalyst lowers the forward activation energy to 230 kJ mol⁻¹. What is the activation energy of the catalysed reverse reaction?
Answer and reasoning
+170 kJ mol⁻¹ — A student who thinks a catalyst lowers only the forward activation energy keeps the uncatalysed reverse value, 280 − 110 = 170 kJ mol⁻¹. The catalyst lowers the single maximum shared by both directions, so the reverse barrier also falls by 50 kJ mol⁻¹, to 120 kJ mol⁻¹.
−110 kJ mol⁻¹ — A student who equates activation energy with ΔH takes the reverse activation energy to be ΔH of the reverse reaction, −110 kJ mol⁻¹. An activation energy is the rise from the species that react up to the maximum, so it is positive; here it is 230 − 110 = 120 kJ mol⁻¹.
+120 kJ mol⁻¹ — The products lie 110 kJ mol⁻¹ above the reactants and the catalysed maximum lies 230 kJ mol⁻¹ above the reactants, so the reverse barrier is 230 − 110 = 120 kJ mol⁻¹. The catalyst lowers the forward and reverse activation energies by the same 50 kJ mol⁻¹.
+340 kJ mol⁻¹ — A student who adds the size of ΔH to the forward barrier, as for an exothermic profile, gets 230 + 110 = 340 kJ mol⁻¹. This reaction is endothermic, so the products lie above the reactants and the reverse barrier is smaller than the forward one: 230 − 110 = 120 kJ mol⁻¹.
Working On the energy profile the products lie 110 kJ mol⁻¹ above the reactants. The catalysed maximum lies 230 kJ mol⁻¹ above the reactants, so it lies 230 − 110 = 120 kJ mol⁻¹ above the products: Ea(reverse, catalysed) = +120 kJ mol⁻¹. Check: the catalyst lowers both barriers by the same 280 − 230 = 50 kJ mol⁻¹; the uncatalysed reverse Ea is 280 − 110 = 170 kJ mol⁻¹, and 170 − 50 = 120 kJ mol⁻¹.
9 Nitrogen monoxide reacts with hydrogen: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g). The experimentally determined rate equation is rate = k[NO]²[H₂]. Which proposed mechanism is consistent with both the rate equation and the overall equation? HL
Answer and reasoning
2NO → N₂O₂ (slow); N₂O₂ + H₂ → N₂ + H₂O₂; H₂O₂ + H₂ → 2H₂O (both fast) — A student who assumes the first step is rate-determining picks this. The steps add up correctly to the overall equation, but with a slow first step the rate equation would be rate = k[NO]², with no H₂ term, which does not match the data.
2NO ⇌ N₂O₂ (fast); N₂O₂ + H₂ → N₂ + H₂O₂ (slow); H₂O₂ → 2OH (fast) — A student who checks only the rate equation picks this. The slow step, after the fast first step, gives rate = k[NO]²[H₂], but the steps add up to 2NO + H₂ → N₂ + 2OH, not the overall equation, so the mechanism is inconsistent with the stoichiometry.
2NO + 2H₂ → N₂ + 2H₂O as a single step, in one collision of all four molecules — A student who reads the overall equation as one collision picks this. A simultaneous collision of four molecules is extremely unlikely, and a single step would give rate = k[NO]²[H₂]², not the observed rate equation.
2NO ⇌ N₂O₂ (fast); N₂O₂ + H₂ → N₂O + H₂O (slow); N₂O + H₂ → N₂ + H₂O (fast) — The steps add up to 2NO + 2H₂ → N₂ + 2H₂O once the intermediates N₂O₂ and N₂O cancel. The slow second step involves N₂O₂ (formed from 2NO in the fast first step) and H₂, so the rate equation is rate = k[NO]²[H₂]. Here the rate-determining step is not the first step.
10 Nitrogen dioxide reacts with carbon monoxide: NO₂(g) + CO(g) → NO(g) + CO₂(g), ΔH = −226 kJ mol⁻¹. The experimentally determined rate equation is rate = k[NO₂]². The proposed mechanism is step 1: 2NO₂(g) → NO₃(g) + NO(g); step 2: NO₃(g) + CO(g) → NO₂(g) + CO₂(g). Which statement about the energy profile for this mechanism follows from these data? HL
Answer and reasoning
The first maximum is the highest point on the profile, as step 1 is the rate-determining step. — A student who measures every activation energy from the original reactants concludes that the rate-determining step must have the highest maximum. Step 2 starts from the intermediate, so its activation energy is the rise from the minimum; if the intermediate lies well above the reactants, the second maximum can be the higher point even though step 2 has the smaller activation energy. The data fix only which rise is larger.
The rise from the reactants to the first maximum exceeds that from the minimum to the second, as step 1 is slow. — The rate equation contains NO₂ to the second power, matching the two NO₂ molecules that collide in step 1, and does not contain CO, which reacts only in step 2; so step 1 is the rate-determining step. The rate-determining step is the step with the largest activation energy, so the rise from the reactants to the first transition state is larger than the rise from the intermediate (NO₃ + NO, with CO) to the second. The profile has two maxima, a minimum for the intermediate, and ends 226 kJ mol⁻¹ below the reactants.
NO₃ and NO sit at the top of the first maximum, as they are the species formed in step 1. — A student who treats a transition state as a real species, like an intermediate, places NO₃ and NO at the first maximum. NO₃ is the intermediate: it is formed in step 1 and used in step 2, so it sits in the minimum between the two maxima. The first maximum is a transition state, with bonds partly broken and partly formed, which cannot be isolated.
The first maximum is the transition state of a unimolecular step, as only NO₂ reacts in step 1. — A student who counts kinds of reactant particle calls step 1 unimolecular because only NO₂ reacts. Two NO₂ molecules collide in step 1, so it is bimolecular, which is also why the rate equation is second order with respect to NO₂.
11 A reaction has the overall equation 2A + B → C + D. A student states that its rate equation must be rate = k[A]²[B]. Which response is correct? HL
Answer and reasoning
The student is correct, as orders are the coefficients of the reactants in the balanced equation. — A student who writes the rate equation from the balanced equation picks this. The orders must be found by experiment; the balanced equation shows only the overall stoichiometry.
The rate equation can only be found by experiment, and its orders need not match the coefficients. — Rate equations depend on the mechanism and can only be determined experimentally. The orders may happen to equal the coefficients, but they cannot be deduced from the balanced equation.
The rate equation must include the products, written as rate = k[C][D]/[A]²[B], like K. — A student who writes the rate equation like an equilibrium expression picks this. A rate equation gives the rate in terms of reactant concentrations raised to experimental orders; it is not a ratio of products to reactants.
The orders are the coefficients of the reactants in the first step of the mechanism. — A student who assumes the first step is rate-determining picks this. The rate equation reflects the slowest step (and any steps before it), which need not be the first step, and is confirmed only by experiment.
12 The initial rate of the reaction A + 2B → C was measured at constant temperature. Experiment 1: [A] = 0.010 mol dm⁻³, [B] = 0.020 mol dm⁻³, initial rate = 1.5 × 10⁻⁵ mol dm⁻³ s⁻¹. Experiment 2: [A] = 0.020 mol dm⁻³, [B] = 0.020 mol dm⁻³, initial rate = 6.0 × 10⁻⁵ mol dm⁻³ s⁻¹. Experiment 3: [A] = 0.020 mol dm⁻³, [B] = 0.060 mol dm⁻³, initial rate = 1.8 × 10⁻⁴ mol dm⁻³ s⁻¹. What is the rate equation? HL
Answer and reasoning
rate = k[A]¹[B]² — A student who takes the orders from the coefficients in A + 2B → C picks this. The data show the rate quadruples when [A] doubles (second order in A) and triples when [B] triples (first order in B).
rate = k[A]⁴[B]³ — A student who takes the order as the factor by which the rate changes picks this. The order is the exponent: 2ⁿ = 4 gives n = 2 for A, and 3ⁿ = 3 gives n = 1 for B.
rate = k[A]¹[B]¹ — A student who assumes the rate is directly proportional to every concentration picks this. Doubling [A] quadruples the rate, so the order in A is 2, not 1.
rate = k[A]²[B]¹ — Experiments 1 and 2: [A] doubles with [B] constant and the rate increases by 6.0 ÷ 1.5 = 4 = 2², so the order in A is 2. Experiments 2 and 3: [B] triples with [A] constant and the rate increases by 18 ÷ 6.0 = 3 = 3¹, so the order in B is 1.
Working Exp 1 → 2: [A] × 2, [B] constant, rate × (6.0 × 10⁻⁵)/(1.5 × 10⁻⁵) = 4 = 2², so order in A = 2. Exp 2 → 3: [B] × 3, [A] constant, rate × (1.8 × 10⁻⁴)/(6.0 × 10⁻⁵) = 3 = 3¹, so order in B = 1. Rate = k[A]²[B]; overall order 3.
13 The rate equation for a reaction is rate = k[A][B]². At constant temperature, the concentrations of A and B are both doubled. By what factor does the rate increase? HL
Answer and reasoning
6 — A student who adds the separate effects gets 2 + 4 = 6. The concentration terms in the rate equation are multiplied, so the factors multiply: 2 × 4 = 8.
3 — A student who takes the overall order as the factor by which the rate changes gets 3. The overall order is an exponent: doubling every concentration multiplies the rate by 2³ = 8.
8 — The factors multiply: doubling [A] (first order) multiplies the rate by 2¹ = 2, and doubling [B] (second order) multiplies it by 2² = 4, so the rate increases by 2 × 4 = 8, which is 2 raised to the overall order, 3.
4 — A student who treats each reactant as first order gets 2 × 2 = 4. B is second order, so doubling [B] alone quadruples the rate; the total factor is 2 × 4 = 8.
Working New rate/old rate = (2[A])(2[B])²/([A][B]²) = 2¹ × 2² = 8.
14 Reactant A is used up in two separate reactions at constant temperature. Reaction 1 is first order with respect to A and reaction 2 is second order with respect to A. Which statement about graphs for these reactions is correct? HL
Answer and reasoning
For reaction 2, a graph of rate against [A] curves up from the origin, as doubling [A] quadruples the rate. — For reaction 2, rate = k[A]², so the rate–concentration graph is a parabola through the origin: doubling [A] quadruples the rate, and the gradient increases as [A] increases. A graph of rate against [A]² would be a straight line through the origin.
For reaction 2, a graph of rate against [A] is a straight line through the origin, as for reaction 1. — A student who treats every reactant as first order picks this. A straight line through the origin is the first-order shape; for reaction 2, rate = k[A]², so doubling [A] quadruples the rate and the graph curves upwards.
For reaction 1, a graph of rate against [A] is a curve that falls towards zero, like the [A]–time graph. — A student who carries the concentration–time shape over to the rate–concentration graph picks this. For reaction 1, rate = k[A], so the rate is directly proportional to [A]: the graph is a straight line through the origin, rising as [A] increases.
For reaction 1, a graph of [A] against time is a straight line, as A is used up steadily. — A student who thinks a reaction goes at one steady rate picks this. Only a zero-order reactant is used up at a constant rate. For reaction 1 the rate falls as [A] falls, so the [A]–time graph is a curve whose gradient decreases, with [A] falling by the same fraction in equal time intervals.
15 The rate equation for a reaction is rate = k[A]²[B]. When [A] = 0.020 mol dm⁻³ and [B] = 0.050 mol dm⁻³, the rate is 3.0 × 10⁻⁵ mol dm⁻³ s⁻¹. What is the value of k? HL
Answer and reasoning
0.030 mol⁻² dm⁶ s⁻¹ — A student who treats A as first order divides by [A][B] = 1.0 × 10⁻³ and gets 0.030. The rate equation has [A]², so divide by (0.020)² × 0.050 = 2.0 × 10⁻⁵.
1.5 × 10⁻² mol⁻² dm⁶ s⁻¹ — A student who calculates [A]² as 2 × [A] divides by 2 × 0.020 × 0.050 = 2.0 × 10⁻³ and gets 1.5 × 10⁻². (0.020)² = 4.0 × 10⁻⁴, so k = 1.5.
6.0 × 10⁻¹⁰ mol⁻² dm⁶ s⁻¹ — A student who multiplies instead of dividing gets 3.0 × 10⁻⁵ × 2.0 × 10⁻⁵ = 6.0 × 10⁻¹⁰. k = rate/([A]²[B]) = 1.5 mol⁻² dm⁶ s⁻¹.
16 A plot of ln k (y-axis) against 1/T (x-axis, in K⁻¹) for a reaction is a straight line with a gradient of −6.00 × 10³ K. Using ln k = −Ea/RT + ln A and R = 8.31 J K⁻¹ mol⁻¹, what is the activation energy of the reaction? HL
Answer and reasoning
49.9 kJ mol⁻¹ — The gradient is −Ea/R, so Ea = −gradient × R = 6.00 × 10³ K × 8.31 J K⁻¹ mol⁻¹ = 4.99 × 10⁴ J mol⁻¹ = 49.9 kJ mol⁻¹. The negative gradient shows that k increases as T increases.
6.00 × 10³ J mol⁻¹ — A student who takes the gradient as −Ea, leaving out R, gets Ea = 6.00 × 10³ J mol⁻¹. That number is Ea/R, in K; the gradient is −Ea/R, so multiply by R = 8.31 J K⁻¹ mol⁻¹ to obtain Ea = 49.9 kJ mol⁻¹.
0.722 kJ mol⁻¹ — A student who divides by R when rearranging gets 6.00 × 10³ ÷ 8.31 = 722, read as 722 J mol⁻¹ = 0.722 kJ mol⁻¹. From gradient = −Ea/R, multiplying both sides by −R gives Ea = −gradient × R = 49.9 kJ mol⁻¹.
4.99 × 10⁴ kJ mol⁻¹ — A student who reads the result of 6.00 × 10³ × 8.31 as kJ mol⁻¹ gets 4.99 × 10⁴ kJ mol⁻¹. With R in J K⁻¹ mol⁻¹ the result is in J mol⁻¹: 4.99 × 10⁴ J mol⁻¹ = 49.9 kJ mol⁻¹.
Working ln k = −(Ea/R)(1/T) + ln A has the form y = mx + c, so gradient = −Ea/R. Ea = −gradient × R = −(−6.00 × 10³ K) × 8.31 J K⁻¹ mol⁻¹ = 4.99 × 10⁴ J mol⁻¹ = 49.9 kJ mol⁻¹.
17 The rate constant of a first-order reaction is 2.00 × 10⁻³ s⁻¹ at 25.0 °C and 8.00 × 10⁻³ s⁻¹ at 45.0 °C. Using ln k = −Ea/RT + ln A and R = 8.31 J K⁻¹ mol⁻¹, what is the activation energy of the reaction? HL
Answer and reasoning
0.65 kJ mol⁻¹ — A student who uses the Celsius temperatures gets 1/25 − 1/45 = 0.0178 and Ea = 8.31 × 1.386/0.0178 = 648 J mol⁻¹ = 0.65 kJ mol⁻¹. T must be in kelvin: 298 K and 318 K.
23.7 kJ mol⁻¹ — A student who uses log₁₀ instead of ln gets log₁₀ 4 = 0.602 and Ea = 8.31 × 0.602/(2.111 × 10⁻⁴) = 23.7 kJ mol⁻¹. The Arrhenius equation uses natural logarithms: ln 4 = 1.386.
54.6 kJ mol⁻¹ — T₁ = 298 K, T₂ = 318 K. Subtracting the two equations: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). ln 4 = 1.386 and 1/298 − 1/318 = 2.111 × 10⁻⁴ K⁻¹, so Ea = 8.31 × 1.386/(2.111 × 10⁻⁴) = 5.46 × 10⁴ J mol⁻¹ = 54.6 kJ mol⁻¹.
6.57 kJ mol⁻¹ — A student who leaves out R gets 1.386/(2.111 × 10⁻⁴) = 6.57 × 10³, read as 6.57 kJ mol⁻¹. That value is Ea/R in K; multiply by R = 8.31 J K⁻¹ mol⁻¹ to obtain Ea = 54.6 kJ mol⁻¹.
18 A first-order reaction has Ea = 100 kJ mol⁻¹, and its rate constant at 127 °C is 3.0 × 10⁻³ s⁻¹. Using k = Ae^(−Ea/RT) and R = 8.31 J K⁻¹ mol⁻¹, what is the Arrhenius factor, A? HL
Answer and reasoning
3.1 × 10⁻³ s⁻¹ — A student who substitutes Ea = 100 (in kJ mol⁻¹) with R in J K⁻¹ mol⁻¹ gets Ea/RT = 0.0301 and A = 3.0 × 10⁻³ × e^0.0301 = 3.1 × 10⁻³ s⁻¹. Convert Ea to 100 000 J mol⁻¹ first.
4.2 × 10³⁸ s⁻¹ — A student who uses T = 127 instead of 400 K gets Ea/RT = 100 000/(8.31 × 127) = 94.8 and A = 3.0 × 10⁻³ × e^94.8 = 4.2 × 10³⁸ s⁻¹. T must be in kelvin: 400 K.
3.6 × 10²⁷ s⁻¹ — A student who uses log₁₀ in place of ln writes log₁₀ A = log₁₀ k + Ea/RT = −2.52 + 30.08 = 27.56 and gets A = 10^27.56 = 3.6 × 10²⁷ s⁻¹. With natural logarithms, ln A = ln k + Ea/RT = −5.81 + 30.08 = 24.27, so A = e^24.27 = 3.5 × 10¹⁰ s⁻¹.
3.5 × 10¹⁰ s⁻¹ — T = 127 + 273 = 400 K and Ea = 100 000 J mol⁻¹, so Ea/RT = 100 000/(8.31 × 400) = 30.08. A = k/e^(−Ea/RT) = 3.0 × 10⁻³ × e^30.08 = 3.0 × 10⁻³ × 1.16 × 10¹³ = 3.5 × 10¹⁰ s⁻¹.
Working T = 127 + 273 = 400 K; Ea = 100 kJ mol⁻¹ = 100 000 J mol⁻¹. Ea/RT = 100 000/(8.31 × 400) = 30.08. A = k/e^(−Ea/RT) = k × e^(Ea/RT) = 3.0 × 10⁻³ s⁻¹ × e^30.08 = 3.0 × 10⁻³ × 1.16 × 10¹³ = 3.5 × 10¹⁰ s⁻¹ (A has the same units as k). Equivalently ln A = ln k + Ea/RT = −5.81 + 30.08 = 24.27, and A = e^24.27.
19 Reaction 1 occurs between two gaseous atoms. Reaction 2 occurs between two large gaseous molecules and happens only if they collide at one particular site on each molecule. Assume the two reactions have the same Ea and the same collision frequency at a given temperature. Which reaction has the larger Arrhenius factor, A, and why? HL
Answer and reasoning
Reaction 1, as a greater fraction of its collisions have a suitable orientation for reaction. — A takes into account the frequency of collisions with proper orientations. Atoms can react whichever way they meet, but the large molecules react only when their reactive sites meet, so a smaller fraction of their collisions are properly oriented and A is smaller for reaction 2.
Neither: A depends only on the fraction of collisions with E ≥ Ea, which is the same. — A student who confuses A with the energy term picks this. The fraction with E ≥ Ea is e^(−Ea/RT), which is indeed the same here; A describes the frequency of properly oriented collisions, which differs.
Neither: A depends only on how often the particles collide, which is the same for both. — A student who ignores collision geometry picks this. A counts collisions with proper orientations, not all collisions; most collisions between the large molecules miss the reactive site.
Reaction 2, as A is larger when collisions must have one particular orientation to react. — A student who reads 'A takes orientation into account' as 'A is bigger when orientation matters more' picks this. A counts only the collisions that are properly oriented, so the strict orientation requirement of reaction 2 makes its A smaller, not larger.
20 A proposed mechanism for the reaction H₂(g) + I₂(g) → 2HI(g) is: step 1: I₂(g) ⇌ 2I(g) (fast); step 2: H₂(g) + 2I(g) → 2HI(g) (slow). Which statement about molecularity in this mechanism is correct? HL
Answer and reasoning
Step 2 is bimolecular, as it involves two different reacting species, H₂ and I. — A student who counts the kinds of species picks this. Molecularity counts particles, not kinds: the coefficient 2 means two I atoms take part alongside one H₂ molecule, so three particles collide and the step is termolecular.
The slow step is termolecular, as one H₂ molecule and two I atoms must all collide together in it. — Molecularity is the number of reacting particles in an elementary step. In step 2 three particles, one H₂ molecule and two I atoms, react in one collision, so the step is termolecular. A simultaneous three-particle collision with the correct orientation is improbable, which is consistent with step 2 being the slow step.
Step 1 is termolecular, as three particles, one I₂ molecule and two I atoms, appear in it. — A student who counts the products as well as the reactants picks this. Only the reacting particles count: a single I₂ molecule breaks apart in step 1, so the forward step is unimolecular.
The overall reaction is bimolecular, as one H₂ molecule collides with one I₂ molecule. — A student who reads the overall equation as a single collision picks this. Molecularity applies only to elementary steps; the overall equation gives the stoichiometry, and in the proposed mechanism H₂ never collides with I₂ but with two I atoms.
21 The graph shows Maxwell–Boltzmann energy distributions, P and Q, for the same sample of a gas at two different temperatures. The activation energy, Ea, of a reaction of the gas is marked. Which statement is correct?
Answer and reasoning
P is at the higher temperature, because its higher peak shows more of its particles have energy above Ea. — A student who thinks the hotter curve has the higher peak picks this. The area under each curve is the same (same number of particles), so the hotter sample, spread over more energies, must have the lower peak. P's high peak lies at low energy; beyond Ea it is Q that has the larger area.
Q is at the higher temperature, because every one of its particles has more energy than any particle in P. — A student who thinks heating gives every particle the same extra energy picks this. The curves overlap: Q still has many low-energy particles and P has some high-energy ones. Heating changes the shape of the distribution, increasing the fraction of particles above Ea, not every particle's energy.
Q is at the higher temperature, and more of its particles have energies of at least Ea. — At a higher temperature the distribution flattens and spreads to higher energies, so the hotter sample has the lower, broader peak: that is Q. Beyond the Ea line the area under Q is larger than under P, so a larger fraction of Q's particles have energy of at least Ea and a larger fraction of collisions can react.
Q is at the higher temperature, which lowers Ea for Q, so more of its particles can react. — A student who thinks heating lowers the activation energy picks this. Ea is fixed by the reaction pathway, so the one Ea line applies to both samples; what changes with temperature is the fraction of particles whose energy is at least Ea.
22 The diagram shows the energy profile of an exothermic reaction without a catalyst (solid line) and with a catalyst (dashed line). Which arrow represents the activation energy of the catalysed reverse reaction?
Answer and reasoning
Arrow Y — A student who thinks a catalyst lowers only the forward activation energy picks this. The catalysed pathway has one lower maximum that both directions pass over, so the reverse activation energy is also lowered: it is Z, from the products to the catalysed maximum. Y is the uncatalysed reverse activation energy.
Arrow W — A student who measures every activation energy from the reactants picks this. W is the activation energy of the catalysed forward reaction. The reverse reaction starts from the products, so its activation energy is measured from the products level: Z.
Arrow X — A student who confuses activation energy with ΔH picks this. X is the energy difference between reactants and products, ΔH, which the catalyst does not change. The activation energy of the catalysed reverse reaction is the rise from the products to the catalysed maximum, Z.
Arrow Z — The reverse reaction starts from the products, and with the catalyst present it passes over the lower, catalysed maximum. Its activation energy is the rise from the products level to that maximum, which is Z.
23 The graph shows how the concentration of a reactant, A, changes with time at constant temperature. A tangent to the curve has been drawn at t = 100 s. What is the rate of reaction at t = 100 s?
Answer and reasoning
6.0 × 10⁻⁴ mol dm⁻³ s⁻¹ — A student who divides the concentration at the point by the time gets 0.060 ÷ 100 = 6.0 × 10⁻⁴ mol dm⁻³ s⁻¹. The concentration remaining is not a change in concentration. Use the tangent: it drops 0.090 mol dm⁻³ over 300 s, giving 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
3.0 × 10⁻⁴ mol dm⁻³ s⁻¹ — The rate at 100 s is the gradient of the tangent there. Reading where the tangent cuts the axes, it falls from 0.090 mol dm⁻³ at 0 s to 0 at 300 s: 0.090 ÷ 300 = 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹ (quoted as a positive rate of decrease of [A]).
4.0 × 10⁻⁴ mol dm⁻³ s⁻¹ — A student who takes the average rate over the first 100 s gets (0.100 − 0.060) ÷ 100 = 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹. The reaction was faster before 100 s than at 100 s, so this average is too large; the rate at 100 s is the tangent's gradient, 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
2.0 × 10⁻⁴ mol dm⁻³ s⁻¹ — A student who assumes one steady rate throughout divides the total change by the total time: 0.100 ÷ 500 = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. The curve becomes less steep, so the rate changes; at 100 s it is the gradient of the tangent, 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
Working The rate at 100 s is the gradient of the tangent at that point. The tangent crosses the [A] axis at 0.090 mol dm⁻³ and the time axis at 300 s, so gradient = (0.000 − 0.090) mol dm⁻³ ÷ (300 − 0) s = −3.0 × 10⁻⁴ mol dm⁻³ s⁻¹. The rate is quoted as a positive value for the decrease in [A]: rate = 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
24 The graph shows the volume of hydrogen collected against time when the same mass of magnesium ribbon reacts with an excess of hydrochloric acid in two experiments, P and Q. One experiment used 1.0 mol dm⁻³ acid and the other 2.0 mol dm⁻³ acid; all other conditions were identical. Which statement is correct?
Answer and reasoning
P used the 2.0 mol dm⁻³ acid, since its reaction continues for longer and so has the higher rate. — A student who equates a longer reaction time with a higher rate picks this. A high rate means a large change in a short time: Q reaches the final volume sooner, so Q has the higher rate and used the more concentrated acid.
The rates of P and Q are the same, since both experiments produce the same final volume of hydrogen. — A student who judges rate from the amount of product picks this. The final volume depends on the amount of the limiting reactant, magnesium, which is the same in both; the rate is shown by the steepness of the curve, and Q is steeper.
Q used the 2.0 mol dm⁻³ acid, since more frequent collisions give it the steeper curve and higher rate. — Q is steeper, so its rate is higher: the more concentrated acid gives more frequent collisions between H⁺ ions and the magnesium surface, so more successful collisions per second. Both curves level off at the same volume because magnesium is the limiting reactant and its mass is the same in both experiments.
Q used the 2.0 mol dm⁻³ acid, since its more crowded particles collide with more energy, raising the rate. — A student who thinks concentration changes the energy of collisions picks this. Q did use the more concentrated acid, but the collisions are no more energetic: the temperature is the same. Higher concentration means more frequent collisions, so more successful collisions per second.
25 The graph shows how the initial rate of the reaction A + B → C varies with the initial concentration of A, with [B] and the temperature kept constant. Which statement is correct? HL
Answer and reasoning
The order with respect to A is zero, so [A] does not appear in the rate equation, although A is consumed. — A horizontal rate–concentration graph means the rate is independent of [A]: the order with respect to A is zero, so [A]⁰ = 1 and A is omitted from the rate equation, rate = k[B]ⁿ. A is still a reactant and is used up as the reaction proceeds.
A takes no part in the reaction, since changing its concentration has no effect on the rate. — A student who thinks a zero-order reactant does not react picks this. A is consumed, as the equation shows; it is zero order because it is not involved in the rate-determining step, so its concentration does not affect how fast the reaction goes.
A graph of [A] against time for this reaction would also be a horizontal line, as the rate does not depend on [A]. — A student who carries the shape of the rate–concentration graph over to the concentration–time graph picks this. Because the rate is constant, [A] falls at a constant rate: the [A]–time graph is a straight line with a negative gradient, not a horizontal line.
The order with respect to A is one, since doubling [A] multiplies the rate by a factor of one. — A student who takes the order to be the factor by which the rate changes picks this. Order is the exponent of [A] in the rate equation: doubling [A] leaves the rate unchanged, which means 2ⁿ = 1 and n = 0. For first order the rate would double.
26 The graph shows how the concentration of A changes with time for the reaction 2A(g) → B(g) + C(g) at constant temperature. Which statement is correct? HL
Answer and reasoning
The reaction is second order with respect to A, as the coefficient of A in the equation is 2. — A student who takes orders from the coefficients of the balanced equation picks this. Orders can only be found experimentally, from data such as this graph. The constant half-life of 50 s shows first order; for second order the half-life would double each time the concentration halved.
The reaction is zero order with respect to A, as A is being used up at one steady rate. — A student who assumes a reaction proceeds at one steady rate picks this. The graph is a curve, not a straight line: [A] falls by 0.040 in the first 50 s but only by 0.020 in the next 50 s, so the rate falls as [A] falls. A zero-order reactant would give a straight line.
A graph of rate against [A] for this reaction would be a curve of the same shape, falling to zero. — A student who carries the concentration–time shape over to the rate–concentration graph picks this. The constant half-life shows first order, rate = k[A], so a graph of rate against [A] is a straight line through the origin, rising as [A] increases.
The reaction is first order with respect to A, as [A] halves in equal time intervals. — The concentration falls from 0.080 to 0.040, to 0.020 and to 0.010 mol dm⁻³ in equal 50 s intervals: a constant half-life. A constant half-life, independent of the starting concentration, is the signature of a first-order reaction, rate = k[A].
27 The graph shows ln k plotted against 1/T for a reaction; two points on the line are marked. Using ln k = −Ea/RT + ln A and R = 8.31 J K⁻¹ mol⁻¹, what is the activation energy of the reaction? HL
Answer and reasoning
8.0 kJ mol⁻¹ — A student who takes the gradient as −Ea, leaving out R, reads 8.0 × 10³ as Ea in J mol⁻¹ and writes 8.0 kJ mol⁻¹. The gradient is −Ea/R, so Ea = 8.0 × 10³ K × 8.31 J K⁻¹ mol⁻¹ = 66 kJ mol⁻¹.
66 kJ mol⁻¹ — Between the marked points ln k falls by 8.0 while 1/T rises by 1.0 × 10⁻³ K⁻¹, so the gradient is −8.0 × 10³ K. The gradient equals −Ea/R, so Ea = 8.0 × 10³ K × 8.31 J K⁻¹ mol⁻¹ = 6.65 × 10⁴ J mol⁻¹ ≈ 66 kJ mol⁻¹.
0.96 kJ mol⁻¹ — A student who divides the gradient by R gets 8.0 × 10³ ÷ 8.31 = 963 J mol⁻¹ = 0.96 kJ mol⁻¹. Rearranging gradient = −Ea/R by multiplying both sides by −R gives Ea = −gradient × R = 66 kJ mol⁻¹.
0.066 kJ mol⁻¹ — A student who ignores the 10⁻³ in the axis label takes Δ(1/T) as 1.0 K⁻¹, giving a gradient of −8.0 K and Ea = 8.0 × 8.31 = 66 J mol⁻¹ = 0.066 kJ mol⁻¹. The axis numbers are 1/T in units of 10⁻³ K⁻¹, so Δ(1/T) = 1.0 × 10⁻³ K⁻¹, the gradient is −8.0 × 10³ K and Ea = 66 kJ mol⁻¹.
Working Reading the marked points: (1/T = 2.5 × 10⁻³ K⁻¹, ln k = −2.0) and (1/T = 3.5 × 10⁻³ K⁻¹, ln k = −10.0). Gradient = Δ(ln k)/Δ(1/T) = (−10.0 − (−2.0)) ÷ ((3.5 − 2.5) × 10⁻³ K⁻¹) = −8.0 ÷ 1.0 × 10⁻³ = −8.0 × 10³ K. Since gradient = −Ea/R, Ea = −gradient × R = 8.0 × 10³ K × 8.31 J K⁻¹ mol⁻¹ = 6.65 × 10⁴ J mol⁻¹ = 66 kJ mol⁻¹ (2 s.f.).
28 The diagram shows the energy profile of a reaction that occurs in two elementary steps. Which arrow represents the activation energy of the rate-determining step? HL
Answer and reasoning
Arrow W — A student who assumes the first step is always rate-determining picks this. The slowest step has the largest activation energy: the rise from the intermediate to the second maximum, X, is larger than W, so step 2 is rate-determining.
Arrow Y — A student who measures every activation energy from the original reactants picks this. Step 2 begins at the intermediate, not the reactants, so its activation energy is the rise from the minimum to the second maximum, X.
Arrow Z — A student who confuses activation energy with ΔH picks this. Z is the overall enthalpy change, reactants to products. Activation energies are rises to a transition state; the rate-determining step's is X.
Arrow X — The minimum is the intermediate and each maximum is a transition state. Step 2 starts from the intermediate and climbs to the second, higher maximum: comparing the two rises, X (minimum to second maximum) is larger than W, so step 2 is the slow, rate-determining step and X is its activation energy.
That was your twenty minutes. Real practice on R2.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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