Questions on the Arrhenius equation and the rate-determining step

Question 1

Explain how the rate equations for the hydrolysis of 1-bromobutane and of 2-bromo-2-methylpropane by hydroxide ions show that they react by different mechanisms, and describe each mechanism.

Accepted Answer

Indicative content

1-bromobutane has rate = k[C₄H₉Br][OH⁻], so both the halogenoalkane and hydroxide are in the rate-determining step
2-bromo-2-methylpropane has rate = k[(CH₃)₃CBr], so only the halogenoalkane is in the rate-determining step
SN2: the hydroxide lone pair attacks the δ+ carbon from the side opposite the bromine
SN2: the C–Br bond breaks as the C–O bond forms, in one step through a single transition state
SN1: the C–Br bond breaks heterolytically in a slow first step to give a tertiary carbocation and Br⁻
SN1: the hydroxide ion bonds to the carbocation in a fast second step, so hydroxide is zero order

Levels

Level 3 (5–6 marks) both rate equations are read correctly to say which species are in the rate-determining step, and both mechanisms are described in full, the one-step SN2 attack from the side opposite the bromine with a single transition state, and the two-step SN1 with the slow heterolysis to a tertiary carbocation and the fast attack by hydroxide; the answer links each mechanism back to its rate equation
Level 2 (3–4 marks) the rate equations are matched to the two mechanisms and both are outlined; the answer does not say which species are in the rate-determining step, or one mechanism lacks its key step
Level 1 (1–2 marks) one mechanism is named or the rate equations are said to differ; the overall order is commonly taken as the number of steps in the mechanism

Question 2

Explain how the mechanism of the acid-catalysed iodination of propanone accounts for its rate equation, rate = k[CH₃COCH₃][H⁺], and identify the intermediates and the catalyst in it.

Accepted Answer

Indicative content

in the first, fast step a lone pair on the oxygen of propanone accepts H⁺, giving protonated propanone, (CH₃)₂C=OH⁺
the second step, loss of H⁺ from a carbon atom of the protonated propanone to give the enol, is slow and rate-determining
the protonated propanone is an intermediate, so it cannot appear in the rate equation; the one propanone and one H⁺ that made it take its place
iodine reacts with the C=C bond of the enol only in the fast third step, after the rate-determining step, so it is zero order
the protonated propanone and the enol are both intermediates: each is made in one step and used up in a later one
H⁺ is the catalyst: it is used in step 1 and given back in step 2, so it is in the rate equation but not in the overall equation

Levels

Level 3 (5–6 marks) all three steps are described with their relative speeds, the rate equation is derived from the slow step with the intermediate replaced by the species that formed it, the zero order in iodine is explained from its place after the rate-determining step, and both intermediates and the catalyst are identified with the reasons
Level 2 (3–4 marks) the slow step is identified and the rate equation explained from it, with the intermediates or the catalyst named; why iodine is zero order, or why an intermediate cannot appear in the rate equation, is not explained
Level 1 (1–2 marks) H⁺ is called the catalyst or the slow step said to set the rate; iodine is usually put in the rate equation because it is a reactant in the overall equation

Question 3

Calculate the activation energy of the reaction between sodium thiosulfate and hydrochloric acid from the results in the table, obtained by timing the disappearance of a cross at four temperatures with the same concentrations in every run.

Accepted Answer

M1 temperatures converted to kelvin and 1/T calculated: 3.356 × 10⁻³ K⁻¹ at 298 K and 3.049 × 10⁻³ K⁻¹ at 328 K M1 ln(1/t) calculated: −5.74 for 310 s and −3.37 for 29 s M1 gradient of ln(1/t) against 1/T = (−3.37 − (−5.74)) ÷ (3.049 × 10⁻³ − 3.356 × 10⁻³) = −7.7 × 10³ K M1 Ea = −gradient × R = 7.7 × 10³ × 8.31; error carried forward from M3 A1 Ea = 64 kJ mol⁻¹, two significant figures; allow 62 to 66 kJ mol⁻¹ from a best-fit line through all four points

Question 4

Calculate the activation energy of a reaction from the rate constants in the table.

Accepted Answer

M1 ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), with ln(1.0 × 10⁻² ÷ 2.5 × 10⁻³) = ln 4.0 = 1.386 M1 1/298 − 1/318 = 2.11 × 10⁻⁴ K⁻¹ M1 Ea = 8.31 × 1.386 ÷ 2.11 × 10⁻⁴ = 5.46 × 10⁴ J mol⁻¹ A1 Ea = 55 kJ mol⁻¹, to two significant figures as the rate constants are given; accept 5.5 × 10⁴ J mol⁻¹

Question 5

Calculate the pre-exponential factor A for a first-order reaction with Ea = 75.0 kJ mol⁻¹ and k = 3.2 × 10⁻⁴ s⁻¹ at 300 K.

Accepted Answer

M1 rearrange: A = k ÷ e^(−Ea/RT) = k × e^(Ea/RT) M1 Ea/RT = 75 000 ÷ (8.31 × 300) = 30.08 M1 e^30.08 = 1.16 × 10¹³, so A = 3.2 × 10⁻⁴ × 1.16 × 10¹³ A1 A = 3.7 × 10⁹ s⁻¹, to two significant figures, with the units of a first-order k

Question 6

Outline how the activation energy of a reaction could be found using a clock reaction at several temperatures.

Accepted Answer

B1 run the clock reaction with the same concentrations at five or six temperatures, measuring each temperature in the mixture B1 record the time t at each, and use 1/t as a measure of the rate, which is proportional to k at fixed concentrations B1 plot ln(1/t) against 1/T with T in kelvin and draw the best-fit straight line B1 the gradient equals −Ea/R, so Ea = −gradient × 8.31 J K⁻¹ mol⁻¹

Question 7

Compare the SN1 and SN2 mechanisms for the hydrolysis of halogenoalkanes by hydroxide ions.

Accepted Answer

B1 SN2 takes place in one step, whereas SN1 takes place in two steps B1 SN2 passes through a transition state only, whereas SN1 forms a carbocation intermediate B1 SN2 has rate = k[RX][OH⁻], second order, whereas SN1 has rate = k[RX], first order B1 SN2 is typical of primary halogenoalkanes, whereas SN1 is typical of tertiary ones

Question 8

Evaluate the mechanism X + X → X₂ (slow), then X₂ + Y → X₂Y (fast), proposed for the reaction 2X + Y → X₂Y, whose rate equation is rate = k[X][Y].

Accepted Answer

B1 the steps add up to the overall equation, so the mechanism passes that test B1 its slow step involves two X and no Y, so it predicts rate = k[X]², which is not the rate equation measured B1 the mechanism is therefore rejected B1 a consistent mechanism is X + Y → XY (slow), then XY + X → X₂Y (fast)

Question 9

State what is meant by the rate-determining step of a reaction. Then give the rate equation, and its overall order, for a reaction whose mechanism is A + B → AB (fast), then AB + C → D (slow).

Accepted Answer

B1 the slowest step in the reaction mechanism, which limits the overall rate; reject 'the first step' B1 rate = k[A][B][C], since the intermediate AB is replaced by the A and B that formed it B1 the reaction is third order overall; error carried forward from the rate equation given

Question 10

Calculate the activation energy, in kJ mol⁻¹, of the reaction whose graph of ln k against 1/T is shown. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 gradient taken from two widely separated points on the line, such as (3.00, −2.50) and (3.50, −5.60): −3.10 ÷ (0.50 × 10⁻³) = −6.20 × 10³ K, the 10⁻³ from the axis unit included M1 gradient = −Ea/R, so Ea = 6.20 × 10³ × 8.31 = 5.15 × 10⁴ J mol⁻¹ A1 Ea = +51.5 kJ mol⁻¹, to three significant figures; accept 50 to 53 kJ mol⁻¹ from a gradient read off the graph

Question 11

Calculate the rate constant at 310 K for the reaction described in the table.

Accepted Answer

M1 ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (50 000 ÷ 8.31) × (1/300 − 1/310) M1 = 6017 × 1.075 × 10⁻⁴ = 0.647, so k₂/k₁ = e^0.647 = 1.91 A1 k₂ = 1.91 × 4.0 × 10⁻³ = 7.6 × 10⁻³ s⁻¹, to two significant figures

Question 12

Explain, using the Arrhenius equation, why a small rise in temperature causes a large increase in the rate constant.

Accepted Answer

B1 e^(−Ea/RT) is the fraction of collisions with energy equal to or greater than the activation energy B1 raising T makes −Ea/RT less negative, and because the dependence is exponential this fraction rises steeply B1 so a much larger proportion of collisions succeed and k, and the rate, rise by far more than the temperature does

Question 13

Explain why tertiary halogenoalkanes are hydrolysed by an SN1 mechanism while primary halogenoalkanes are hydrolysed by SN2.

Accepted Answer

B1 a tertiary carbocation is stabilised by the three alkyl groups, which push electron density towards the positive carbon B1 the three alkyl groups also crowd the carbon atom and hinder attack by the hydroxide ion from the side opposite the halogen B1 a primary carbocation is unstable and a primary carbon is open to attack, so the one-step SN2 route has the lower activation energy

Question 14

Suggest a two-step mechanism for 2NO(g) + O₂(g) → 2NO₂(g), whose rate equation is rate = k[NO]²[O₂], given that collisions of three molecules at once are very unlikely.

Accepted Answer

B1 step 1, fast: two NO molecules combine, 2NO → N₂O₂ B1 step 2, slow: N₂O₂ + O₂ → 2NO₂ B1 the steps add to the overall equation, and the slow step uses O₂ and an intermediate made from two NO, giving [NO]²[O₂]

Question 15

Suggest which species take part in the rate-determining step of the reaction H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l), whose rate equation is rate = k[H₂O₂][I⁻], and why H⁺ does not appear.

Accepted Answer

B1 one H₂O₂ molecule and one I⁻ ion react in the rate-determining step B1 the H⁺ ions and the second I⁻ ion react only in fast steps after the slow step B1 species that react only after the rate-determining step do not affect the rate, so H⁺ is zero order

Question 16

Calculate the factor by which the rate constant of a reaction at 300 K increases when a catalyst lowers its activation energy from 95.0 kJ mol⁻¹ to 80.0 kJ mol⁻¹, assuming A is unchanged.

Accepted Answer

M1 k(catalysed) ÷ k(uncatalysed) = e^(−80 000/RT) ÷ e^(−95 000/RT) = e^(15 000/RT) M1 15 000 ÷ (8.31 × 300) = 6.02 A1 factor = e^6.02 = 4.1 × 10², to two significant figures, with no units as it is a ratio

Question 17

Explain why, in the thiosulfate and acid method for the activation energy, a graph of ln(1/t) against 1/T has the same gradient as a graph of ln k against 1/T.

Accepted Answer

B1 the cross disappears when a fixed amount of sulfur has formed, so 1/t is proportional to the rate B1 the starting concentrations are the same in every run, so the rate, and therefore 1/t, is a constant multiple of k B1 ln(1/t) = ln k + a constant, so the constant moves the line up or down without changing its gradient, −Ea/R

Question 18

Show that raising the temperature from 310 K to 320 K roughly doubles the rate constant of a reaction whose activation energy is 60.0 kJ mol⁻¹. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 from k = Ae^(−Ea/RT), ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂), with Ea in J mol⁻¹ M1 = (60 000 ÷ 8.31) × (1/310 − 1/320) = 7220 × 1.008 × 10⁻⁴ = 0.728 A1 k₂ ÷ k₁ = e^0.728 = 2.07, so the rate constant roughly doubles; allow 2.1

Question 19

Give the Arrhenius equation, and the units in which Ea and T must be used in it when R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

B1 k = Ae^(−Ea/RT), where A is the pre-exponential factor B1 Ea must be in J mol⁻¹ and T in kelvin

Question 20

Explain, using the Arrhenius equation, why adding a catalyst increases the rate constant of a reaction.

Accepted Answer

B1 the catalyst provides an alternative route with a lower activation energy B1 a smaller Ea makes e^(−Ea/RT) larger at the same temperature, so k increases

Question 21

Explain why an intermediate formed before the rate-determining step never appears in the rate equation, and state what appears in its place.

Accepted Answer

B1 an intermediate is made and used up during the reaction, so its concentration cannot be set independently; it depends on the concentrations of the reactants that formed it B1 the reactants that formed it in the earlier step appear in the rate equation in its place

Question 22

Identify the catalyst and the intermediate in this mechanism for the destruction of ozone: Cl + O₃ → ClO + O₂, then ClO + O → Cl + O₂.

Accepted Answer

B1 Cl is the catalyst: it is used in the first step and regenerated in the second B1 ClO is the intermediate: it is made in the first step and used up in the second

Temperature / °C	Time for the cross to disappear / s
25	310
35	134
45	61
55	29

Quantity	Value
Activation energy, Ea	50.0 kJ mol⁻¹
Rate constant at 300 K	4.0 × 10⁻³ s⁻¹
Temperature required	310 K