Questions on rate equations, orders of reaction and the rate constant

Question 1

Describe the rate–concentration graph and the concentration–time graph for a reactant that is zero order, first order and second order, and explain the shape of each concentration–time graph.

Accepted Answer

Indicative content

zero order: rate against concentration is a horizontal line, because the rate does not depend on the concentration
zero order: concentration against time is a straight line with a constant negative gradient, because the reactant is used at a steady rate
first order: rate against concentration is a straight line through the origin with a gradient of k
first order: concentration against time is a curve with a constant half-life, because the rate falls in proportion to the concentration
second order: rate against concentration is a curve rising ever more steeply from the origin, rate proportional to concentration squared
second order: concentration against time is a curve whose half-lives double in turn, because the rate falls with the square of the concentration

Levels

Level 3 (5–6 marks) all six graphs are described correctly and the shape of each concentration–time graph is explained from how the rate changes with concentration, including the constant half-life of first order and the lengthening half-lives of second order
Level 2 (3–4 marks) the three rate–concentration graphs are right and two concentration–time graphs are explained; the third is not, or half-lives are never used
Level 1 (1–2 marks) one or two graphs are described, often with the first-order concentration–time graph drawn as a straight line falling to zero

Question 2

State what is meant by the order of reaction with respect to a reactant and by the overall order of a reaction, and give the units of the rate constant for reactions that are first order and second order overall.

Accepted Answer

B1 the order with respect to a reactant is the power to which its concentration is raised in the rate equation B1 the overall order is the sum of the powers of all the concentration terms in the rate equation B1 first order overall: s⁻¹, since (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³) leaves s⁻¹ B1 second order overall: mol⁻¹ dm³ s⁻¹; accept the units in any order, such as dm³ mol⁻¹ s⁻¹

Question 3

Deduce the order with respect to A, the rate equation and the value of k, with its units, from the initial rates in the table. A is the only reactant.

Accepted Answer

B1 the rate is the same whatever [A], so the reaction is zero order with respect to A B1 rate = k, or rate = k[A]⁰ M1 k equals the rate, since [A]⁰ = 1 A1 k = 2.5 × 10⁻⁴ mol dm⁻³ s⁻¹, the units of a rate

Question 4

Calculate the rate constant, with its units, for a reaction with rate = k[A][B], given a rate of 4.5 × 10⁻⁵ mol dm⁻³ s⁻¹ when [A] = 0.040 mol dm⁻³ and [B] = 0.025 mol dm⁻³. Then calculate the rate immediately after the mixture is diluted with an equal volume of water.

Accepted Answer

M1 k = rate ÷ ([A][B]) = 4.5 × 10⁻⁵ ÷ (0.040 × 0.025) = 4.5 × 10⁻⁵ ÷ 1.0 × 10⁻³ A1 k = 0.045 mol⁻¹ dm³ s⁻¹, to two significant figures M1 dilution halves both concentrations, so the rate is multiplied by 0.5 × 0.5 = 0.25 A1 new rate = 0.25 × 4.5 × 10⁻⁵ = 1.1 × 10⁻⁵ mol dm⁻³ s⁻¹, to two significant figures

Question 5

Explain why changing the concentration of a zero-order reactant has no effect on the rate, and why increasing the concentration of a first-order reactant increases the rate but does not change k.

Accepted Answer

B1 the zero-order reactant does not take part in the rate-determining step or in any step before it B1 it reacts in a fast step after the slow step, so the rate is limited by the slow step whatever its concentration B1 for a first-order reactant the rate is k multiplied by the concentration, so a higher concentration gives a proportionally higher rate B1 k is fixed at a given temperature and changes only with temperature or a catalyst; reject 'k increases with concentration'

Question 6

Describe how the rate constant is found from a graph for a reaction with rate = k[A], and for one with rate = k[A]², and explain why the second needs a different plot.

Accepted Answer

B1 for rate = k[A], rate against [A] is a straight line through the origin B1 the gradient of that line is k B1 for rate = k[A]², rate against [A] is a curve with no single gradient to measure B1 rate against [A]² has the form y = mx, a straight line through the origin whose gradient is k

Question 7

Compare the graphs expected for a reactant that is first order with those for a reactant that is second order, for rate against concentration and for concentration against time.

Accepted Answer

B1 rate against concentration: a straight line through the origin for first order, but a curve rising ever more steeply from the origin for second order B1 doubling the concentration doubles the rate for first order but multiplies it by four for second order B1 concentration against time: both are curves that fall steeply then level off, but the second-order curve has the longer tail, because its rate falls with the square of the concentration B1 the half-life is constant for first order, whereas each half-life is double the previous one for second order

Question 8

Evaluate a student's claim that the reaction 2NO(g) + O₂(g) → 2NO₂(g), whose rate equation is rate = k[NO]²[O₂], shows that the orders in a rate equation can be taken from the coefficients.

Accepted Answer

B1 for this reaction the orders do equal the coefficients, 2 for NO and 1 for O₂ B1 but that agreement was found by experiment and could not have been predicted from the equation B1 a counter-example is 2N₂O₅ → 4NO₂ + O₂, which is first order although the coefficient of N₂O₅ is 2 B1 so the claim is wrong: one case of agreement is a coincidence of mechanism, and orders must always be measured

Question 9

Calculate the rate constant, with its units, for a reaction with A alone in the rate equation, after showing from the data in the table that it is second order in A.

Accepted Answer

B1 from 0.20 to 0.40 [A] doubles and the rate is multiplied by 2.4 × 10⁻³ ÷ 6.0 × 10⁻⁴ = 4, so the reaction is second order in A M1 k = rate ÷ [A]², or the gradient of rate against [A]²: 6.0 × 10⁻⁴ ÷ 0.040, the same for every pair of values M1 units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² leaves mol⁻¹ dm³ s⁻¹ A1 k = 0.015 mol⁻¹ dm³ s⁻¹, to two significant figures

Question 10

Calculate the initial rate of H₂(g) + I₂(g) → 2HI(g), rate = k[H₂][I₂], and the initial rate of formation of HI, using the data in the table.

Accepted Answer

M1 [H₂] = 0.010 ÷ 2.00 = 0.0050 and [I₂] = 0.020 ÷ 2.00 = 0.010 mol dm⁻³ M1 rate = 2.4 × 10⁻² × 0.0050 × 0.010; error carried forward from M1 A1 rate = 1.2 × 10⁻⁶ mol dm⁻³ s⁻¹, two significant figures A1 HI forms at twice that rate, 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹, two significant figures

Question 11

Calculate the value of the rate constant, with its units, for a reaction with rate = k[A]²[B], given that the rate is 3.6 × 10⁻⁴ mol dm⁻³ s⁻¹ when [A] = 0.030 mol dm⁻³ and [B] = 0.020 mol dm⁻³.

Accepted Answer

M1 k = rate ÷ ([A]²[B]) = 3.6 × 10⁻⁴ ÷ (0.030² × 0.020) = 3.6 × 10⁻⁴ ÷ 1.8 × 10⁻⁵ M1 units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ leaves two lots of mol dm⁻³ underneath A1 k = 20 mol⁻² dm⁶ s⁻¹, to two significant figures

Question 12

Calculate the new rate when the volume of a mixture of nitrogen monoxide and oxygen is halved at constant temperature. The rate equation is rate = k[NO]²[O₂] and the rate before the change is 1.5 × 10⁻³ mol dm⁻³ s⁻¹.

Accepted Answer

M1 halving the volume doubles the concentration of both gases M1 the rate is multiplied by 2² for NO and by 2 for O₂, a factor of 8 A1 new rate = 8 × 1.5 × 10⁻³ = 1.2 × 10⁻² mol dm⁻³ s⁻¹, to two significant figures

Question 13

Explain why the orders in a rate equation cannot be deduced from the balanced equation for the reaction.

Accepted Answer

B1 the balanced equation shows only the overall amounts of each substance that react, not how the reaction happens B1 most reactions take place in a series of steps, and the rate depends on the slowest step and the steps before it B1 the particles in those steps need not match the coefficients, so the orders must be found by experiment

Question 14

Explain how a concentration–time graph for a single reactant can show that the reaction is first order with respect to that reactant.

Accepted Answer

B1 read the time for the concentration to fall to half, then to a quarter, then to an eighth of its starting value B1 for a first-order reactant these successive half-lives are equal B1 because the rate is proportional to the concentration, so halving any concentration takes the same time

Question 15

Justify the factor by which the rate of S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq), rate = k[S₂O₈²⁻][I⁻], changes when [I⁻] is tripled, and identify the error in a student's answer of 9.

Accepted Answer

B1 I⁻ is first order in the rate equation, so tripling [I⁻] triples the rate: a factor of 3 B1 the student squared the factor because I⁻ has a coefficient of 2 in the balanced equation B1 orders come from the rate equation, found by experiment, not from the coefficients; the second I⁻ reacts in a step after the slow one

Question 16

Identify the order with respect to A in each case: (a) doubling [A] doubles the rate; (b) tripling [A] multiplies the rate by 9; (c) halving [A] leaves the rate unchanged.

Accepted Answer

B1 (a) first order, since 2¹ = 2 B1 (b) second order, since 3² = 9 B1 (c) zero order, since the rate does not depend on [A]

Question 17

Suggest why ammonia decomposing on a hot tungsten wire is zero order with respect to ammonia at high pressure, but first order at very low pressure.

Accepted Answer

B1 the reaction happens only on the surface of the tungsten, which has a fixed number of sites B1 at high pressure every site is occupied, so extra ammonia cannot react any faster and the rate does not depend on its concentration B1 at very low pressure most sites are empty, so the rate is proportional to how often ammonia molecules reach them, which is proportional to the concentration

Question 18

Suggest why H⁺ appears in the rate equation for the reaction of propanone with iodine, rate = k[CH₃COCH₃][H⁺], although it does not appear in the overall equation, and why iodine does not appear in the rate equation although it is a reactant.

Accepted Answer

B1 H⁺ is a catalyst: it is used in one step and regenerated in a later one, so it cancels from the overall equation B1 it takes part in the rate-determining step or a step before it, so its concentration affects the rate B1 iodine reacts only in a fast step after the rate-determining step, so it is zero order

Question 19

Deduce the order of reaction with respect to A, and the value and units of the rate constant, from the graph of initial rate against the initial concentration of A. The concentrations of the other reactants were the same in every run.

Accepted Answer

B1 the graph is a straight line through the origin, so rate is proportional to [A] and the reaction is first order with respect to A M1 k is the gradient: rate ÷ [A] for any point, such as 0.016 ÷ 0.40 A1 k = 0.040 s⁻¹; the unit must be s⁻¹, from mol dm⁻³ s⁻¹ ÷ mol dm⁻³

Question 20

Explain how raising the temperature and adding a catalyst each affect the value of the rate constant, given that neither changes any concentration.

Accepted Answer

B1 raising the temperature increases k, because a larger fraction of collisions have at least the activation energy B1 a catalyst increases k, because it provides an alternative route with a lower activation energy B1 the rate rises in both cases although every concentration is unchanged, so the change must be in k

Question 21

Calculate the rate of formation of NO₂ and the rate of formation of O₂ in the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g), at a moment when the concentration of N₂O₅ is falling at 4.0 × 10⁻⁶ mol dm⁻³ s⁻¹.

Accepted Answer

M1 the mole ratio N₂O₅ : NO₂ : O₂ is 2 : 4 : 1, so NO₂ forms at twice the rate N₂O₅ is used and O₂ at half that rate A1 rate of formation of NO₂ = 2 × 4.0 × 10⁻⁶ = 8.0 × 10⁻⁶ mol dm⁻³ s⁻¹ A1 rate of formation of O₂ = 0.5 × 4.0 × 10⁻⁶ = 2.0 × 10⁻⁶ mol dm⁻³ s⁻¹, to two significant figures

Question 22

Calculate the new rate of a reaction with rate = k[X][Y]², whose rate is 2.0 × 10⁻⁵ mol dm⁻³ s⁻¹, when [X] is doubled and [Y] is tripled at the same temperature.

Accepted Answer

M1 the rate is multiplied by 2 for X and by 3² = 9 for Y, a factor of 2 × 9 = 18; reject adding the factors, 2 + 9 A1 new rate = 18 × 2.0 × 10⁻⁵ = 3.6 × 10⁻⁴ mol dm⁻³ s⁻¹, to two significant figures

Quantity	Value
Amount of H₂	0.010 mol
Amount of I₂	0.020 mol
Volume of the vessel	2.00 dm³
Rate constant, k	2.4 × 10⁻² mol⁻¹ dm³ s⁻¹