Skip to content
Marlbridge
Free trial

Flashcards · AP® Chemistry · Unit 5

Unit 5: Kinetics flashcards

72 cards on topics 5.1 to 5.11. Pick a topic or study the whole unit.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 72 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

5.1 Reaction Rates

What is a reaction rate, and why is it always positive?
The change in concentration (or amount) of a species per unit time. For a reactant, rate = −Δ[A]/Δt; the minus sign cancels the negative Δ[A]. For a product, rate = +Δ[C]/Δt. [A] falling from 0.500 M to 0.350 M in 25 s gives 0.0060 M s⁻¹.
How are the rates of different species in one reaction related?
By the coefficients. For aA + bB → cC, rate of reaction = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt. In 2SO₂ + O₂ → 2SO₃, if SO₃ forms at 0.040 M s⁻¹, O₂ is used at 0.020 M s⁻¹ and the rate of reaction is 0.020 M s⁻¹.
How do you find average, instantaneous and initial rates from a concentration–time graph?
Average rate: slope of the line joining two points (belongs to that interval only). Instantaneous rate: slope of the tangent at one time. Initial rate: the instantaneous rate at t = 0.
Why does the rate of a reaction usually fall as it proceeds?
Reactant concentrations fall, so reacting particles collide less often. The rate is not constant during a run.
What factors increase reaction rate?
Higher concentration (or gas pressure), higher temperature, larger surface area of a solid, and a catalyst. For temperature, say that more collisions have enough energy, not just that there are more collisions.
Why does powdered zinc react with acid faster than one lump of the same mass?
More surface is exposed, so acid particles collide with zinc atoms more often. Adding more solid does not raise its "concentration"; for solids, surface area matters.
How can a reaction rate be measured experimentally?
Follow a property that changes: gas volume, mass loss, colour (absorbance), pressure, conductivity or pH. Convert an amount rate to a concentration rate with moles per second ÷ volume (L) = M s⁻¹. Keep temperature constant.

5.2 Introduction to Rate Law

What is a rate law?
rate = k[A]ᵐ[B]ⁿ. Each exponent is the order with respect to that reactant; their sum is the overall order. Orders are found by experiment, not read from the coefficients of the overall equation.
How do you find the order in a reactant from initial-rate data?
Choose two experiments where only that reactant's concentration changes. Rate ratio = (concentration ratio)ᵐ, so m = log(rate ratio) ÷ log(conc. ratio). Doubling [A]: rate × 1 means zero order, × 2 first order, × 4 second order.
How do you work out the units of k?
Units of k = M¹⁻ⁿ s⁻¹ where n is the overall order. Zero order: M s⁻¹; first: s⁻¹; second: M⁻¹ s⁻¹; third: M⁻² s⁻¹.
What happens to k when concentrations change, and what makes k change?
k is the same for every experiment at one temperature; changing concentration changes the rate, not k. k changes only when the temperature changes (or a catalyst is added).
For rate = k[A][B]², how does the rate change if [B] triples, or if [A] halves and [B] doubles?
[B] tripled: rate × 3² = × 9. [A] halved and [B] doubled: × ½ × 4 = × 2.
Why are initial rates used to find rate laws?
At t = 0 the concentrations are known exactly and products have not built up. The slope of the concentration–time curve at t = 0 gives the initial rate.
How do you calculate k from one experiment?
Substitute the rate and concentrations into the rate law. A first-order reaction with rate 8.0 × 10⁻³ M s⁻¹ at [A] = 0.20 M: k = 8.0 × 10⁻³ ÷ 0.20 = 0.040 s⁻¹.

5.3 Concentration Changes Over Time

How do you find the order of a reaction from concentration–time data?
Plot [A], ln[A] and 1/[A] against t. Whichever is a straight line gives the order: [A] vs t → zero; ln[A] vs t → first; 1/[A] vs t → second. With equal time steps, the column with constant differences is the straight one.
What are the integrated rate laws for zero, first and second order?
Zero: [A]ₜ − [A]₀ = −kt. First: ln[A]ₜ − ln[A]₀ = −kt. Second: 1/[A]ₜ − 1/[A]₀ = kt. Use the natural log, not log₁₀.
How do you get k from the straight-line plot?
Zero order: slope of [A] vs t = −k. First order: slope of ln[A] vs t = −k. Second order: slope of 1/[A] vs t = +k. k is always positive.
What is half-life, and for which order is it constant?
The time for [A] to fall to half. Only first order has a constant half-life: t½ = 0.693/k. Zero-order half-lives get shorter; second-order half-lives get longer.
What fraction of a first-order reactant remains after n half-lives?
(½)ⁿ. After two half-lives one quarter is left (not none); after three, one eighth (0.125).
How do you convert between k and half-life for a first-order reaction?
t½ = 0.693/k and k = 0.693/t½. t½ = 45 s gives k = 0.0154 s⁻¹. A ln[A] slope of −0.0350 min⁻¹ gives t½ = 19.8 min. Match time units to those of k.
What order is radioactive decay?
First order: each isotope has a fixed half-life, independent of how much is present.

5.4 Elementary Reactions

What is an elementary reaction?
A single molecular event: one particle reacts on its own, or two (rarely three) particles collide and react in that collision. An overall reaction is the sum of one or more elementary steps.
How do you write the rate law of an elementary step?
Each reactant's coefficient is its order, because rate is proportional to how often the particles meet. Products do not appear. NO + NO₃ → 2NO₂ gives rate = k[NO][NO₃]. This works only for elementary steps.
What is molecularity?
The number of reactant particles in an elementary step. A → products: unimolecular, k[A]. A + B or 2A → products: bimolecular, k[A][B] or k[A]². 2A + B → products: termolecular, k[A]²[B], k in M⁻² s⁻¹. Count particles, not substances.
Why are termolecular steps rare?
Three particles must meet at the same instant with the right energy and orientation. They are rare, not impossible. A step with four or more reactant particles cannot be a single step.
How do you test whether an overall reaction could be a single elementary step?
Compare the experimental rate law with the rate law from the coefficients. Different → not one step (2X + Y → Z with rate = k[X][Y] cannot be one step). Same → one step is possible but not proved.
For the elementary step A + 2B → products, how does the rate change if [B] doubles, or both double?
Rate = k[A][B]². [B] doubled: × 4. Both doubled: × 2 × 4 = × 8.

5.5 Collision Model

What two conditions must a collision meet to produce products?
Enough energy (at least the activation energy, Eₐ) and a suitable orientation, so the atoms that must bond actually meet. In most reactions only a small fraction of collisions succeed.
Why does increasing concentration increase rate?
It increases collision frequency in proportion. It does not change the energy or orientation of collisions. For A + B → products, collision frequency ∝ [A][B]; doubling both gives × 4.
Why does a small temperature rise increase the rate so much?
Mainly because a much larger fraction of collisions have energy ≥ Eₐ. Collision frequency rises only slightly. Eₐ itself does not change with temperature.
What does the area under a Maxwell–Boltzmann curve beyond Eₐ represent?
The particles, and so the share of collisions, with at least the activation energy. On heating this area grows a lot while the total area stays the same.
How does a Maxwell–Boltzmann curve change at higher temperature?
Lower, wider and shifted right, with the same total area. Do not draw the higher-temperature peak taller.
Why can a fast collision between two reactant molecules fail to react?
The orientation is wrong: the atoms that must bond do not meet. Larger, less symmetrical particles have a smaller fraction of suitably oriented collisions, so k is smaller if Eₐ is similar.
What does the rate constant k contain in the collision model?
The energy fraction (collisions with energy ≥ Eₐ) and the orientation fraction. Collision frequency is covered by the concentration terms: rate = k[A][B].

5.6 Reaction Energy Profile

What does a reaction energy profile for one elementary step look like?
Energy against reaction coordinate (progress of the rearrangement, not time): from the reactant level up to a single peak, the transition state, then down to the product level. One step has one hump.
What is a transition state?
The highest-energy arrangement along the path, with old bonds partly broken and new bonds partly formed. It cannot be isolated, so it is not an intermediate.
How do you read activation energy and ΔH from an energy profile?
Eₐ(forward) = E(transition state) − E(reactants), always positive. Eₐ(reverse) = E(transition state) − E(products). ΔH = E(products) − E(reactants): negative if products are lower (exothermic).
How are forward and reverse activation energies related?
Eₐ(reverse) = Eₐ(forward) − ΔH. With Eₐ(forward) = 75 kJ mol⁻¹ and ΔH = −30 kJ mol⁻¹, Eₐ(reverse) = 105 kJ mol⁻¹.
What does the Arrhenius equation tell you qualitatively?
k = A e^(−Eₐ/RT), with T in kelvin. e^(−Eₐ/RT) is linked to the fraction of collisions that can reach the transition state; A holds collision frequency and orientation. For similar A, the step with the larger Eₐ has its rate rise by the larger factor when heated.
Does an exothermic reaction have to be fast?
No. ΔH is about the energy difference between reactants and products; speed depends on Eₐ. An exothermic step can be slow.
Does the energy profile change with temperature?
No. The profile, including the peak height, is a property of the reaction; heating changes the distribution of particle energies, not Eₐ.

5.7 Introduction to Reaction Mechanisms

What is a reaction mechanism?
A series of elementary steps, each a single molecular event, by which a reaction occurs. Adding the steps and cancelling species that appear on both sides must give the overall balanced equation.
How do you tell an intermediate from a catalyst in a mechanism?
Intermediate: made in an early step, used in a later one. Catalyst: used in an early step, made back in a later one. Neither appears in the overall equation. In A₂ + C → A₂C, A₂C + C → A₂ + C₂: A₂C is the intermediate, A₂ the catalyst; overall 2C → C₂.
How does the concentration of an intermediate change over time?
It starts at zero, builds up a little while the reaction runs, and returns to zero. It is a real species that can, in principle, be detected.
How is evidence used to support a proposed mechanism?
A mechanism is a model. If an intermediate that only one proposed mechanism predicts is detected, that supports it over the others, but does not prove it. Several mechanisms can add up to the same equation.
What mistakes are common when adding mechanism steps?
Forgetting a multiplier (an intermediate made twice but used once means the later step happens twice); cancelling unequal amounts; treating the overall equation as one collision; calling a transition state an intermediate.
Do the coefficients of an overall equation tell you how many particles collide?
No. Only the coefficients of an elementary step show its molecularity. The overall equation is a summary of all steps.

5.8 Reaction Mechanism and Rate Law

What is the rate-limiting step?
The step that is much slower than the rest. It sets the rate of the whole reaction.
How do you get the rate law when the first step is slow?
The rate law of the reaction is the rate law of that elementary step: each species raised to its coefficient in the step. Step 1 (slow) J₂ + L → J₂L gives rate = k[J₂][L].
Do reactants used only in fast steps after the slow step appear in the rate law?
No. They are zero order, even though they are used up in the reaction.
Can a catalyst or an intermediate appear in an experimental rate law?
Intermediates never appear in a rate law compared with experiment. A catalyst can, if it reacts in the slow step: in J₂ + L → J₂L (slow), J₂L + L → J₂ + L₂, the catalyst J₂ is in the rate law.
When is a proposed mechanism consistent with the evidence?
When its steps add up to the overall equation, the steps are reasonable (uni- or bimolecular), and it predicts the experimental rate law. Consistent is not proven; two mechanisms can both fit.
How does the rate change when [A] changes by a factor f?
By fⁿ, where n is the order in A. Order 2 with [A] tripled: × 9. Overall order is the sum of exponents; k[A][B] is second order with k in M⁻¹ s⁻¹.

5.9 Pre-Equilibrium Approximation

Why can't you use a later slow step's rate law directly?
It usually contains an intermediate, whose concentration you cannot set in an experiment. An intermediate must not appear in the final rate law.
What is the pre-equilibrium approximation?
When the step before the slow step is fast and reversible, assume it stays at equilibrium: its forward rate equals its reverse rate. Use this to express the intermediate's concentration in terms of reactants.
What are the steps for deriving a rate law with a fast pre-equilibrium?
Write the slow step's rate law from its molecularity. Identify the intermediate. Set forward rate = reverse rate for the fast step and solve for [intermediate]. Substitute and combine constants into one k.
What rate law does A + B ⇌ I (fast), I + B → P (slow) give?
[I] = (k₁/k₋₁)[A][B], so rate = k₂[I][B] = (k₂k₁/k₋₁)[A][B]². K₁ = k₁/k₋₁, with the forward constant on top.
What rate law does X₂ ⇌ 2X (fast), X + Y → XY (slow) give?
[X] = (K₁[X₂])^½, so rate = k₂K₁^½[X₂]^½[Y]. Do not lose the square root. Multiplying [X₂] by 9 triples the rate.
How do you calculate K₁ for a fast pre-equilibrium step?
K₁ = k₁ / k₋₁. With k₁ = 6.0 × 10³ M⁻¹ s⁻¹ and k₋₁ = 2.0 × 10⁵ s⁻¹, K₁ = 0.030 M⁻¹.

5.10 Multistep Reaction Energy Profile

How do you read a multistep reaction energy profile?
Each peak is a transition state, one per elementary step. Each valley between peaks is an intermediate. Count peaks, not valleys, to get the number of steps; a chain of steps has (steps − 1) intermediates.
How do you find the activation energy of a step in a multistep profile?
Its peak minus the valley where that step starts, not the overall reactants. The reverse Eₐ of a step is its peak minus the valley where the step ends (Eₐ − ΔH).
How do you identify the rate-limiting step from an energy profile?
It is the step with the largest Eₐ, measured from its own starting valley. The biggest drop is ΔH, not a barrier, so it says nothing about speed.
How do you find overall ΔH for a multistep reaction?
Overall ΔH = products − reactants = sum of the steps' ΔH. It depends only on reactants and products, not the route.
Step 1 has Eₐ = 40, ΔH = −10; step 2 has Eₐ = 70, ΔH = +25 kJ mol⁻¹. What are the key energies?
With reactants at 0: intermediate at −10 kJ mol⁻¹; TS 2 at −10 + 70 = 60 kJ mol⁻¹; overall ΔH = +15 kJ mol⁻¹. Step 2 is rate-limiting (barrier 70).
What profile shapes match a slow first step and a fast pre-equilibrium?
Slow first step: the first climb is the largest; the rate law comes from step 1. Fast pre-equilibrium then slow step: a low first hump, an intermediate that can easily slide back, then a much larger climb.

5.11 Catalysis

What is a catalyst, and how does it speed up a reaction?
A substance that increases the rate and is not used up overall. It increases the number of effective collisions by opening a new path with lower activation energy and/or helping reactants meet with better orientation. It does not give particles more energy.
How does a catalyst appear in a mechanism?
It is used up in one step and made again later, through new elementary steps and intermediates. In Q + R → QR, QR + T → V + R, R is the catalyst and QR the intermediate. If it reacts in the slow step it appears in the rate law.
Does a catalyst change ΔH or affect only the forward reaction?
Neither. Reactants and products keep the same energies, so ΔH is unchanged. The reverse barrier drops by the same amount as the forward barrier. Ea 90 → 55 kJ mol⁻¹ with ΔH = −25 kJ mol⁻¹ lowers the reverse barrier from 115 to 80 kJ mol⁻¹.
What are the three types of catalysis in the course?
Binding catalysts such as enzymes (a catalyst–reactant complex forms); acid–base catalysis (a proton is gained or lost); surface catalysis (reactants bond to a solid surface).
Why is powdered nickel a better hydrogenation catalyst than a nickel bar?
In surface catalysis only surface atoms take part. Powder has more surface atoms, so more sites where H₂ and the alkene can bind.
How do you draw a catalysed reaction profile?
Same reactant and product levels as the uncatalysed path, but a lower highest transition state. A catalysed path usually has extra humps with a dip for the new intermediate, not just one lower hump.
What happens to the rate if you triple the concentration of a catalyst consumed in the slow first step?
The rate triples, if that step is first order in the catalyst, because rate ∝ [catalyst].

Study Unit 5 further

AP® and Advanced Placement® are trademarks registered by the College Board, which is not affiliated with, and does not endorse, this website. Marlbridge is an independent tuition provider. It is not an authorized AP course provider, an AP school or an exam centre, and these resources are not official College Board materials. Reading them or doing simulations does not meet the laboratory requirement of an AP science course.