5.2 Introduction to Rate Law

1. AWeigh the tightly sealed flask every 10 seconds, starting when the reactants are mixed.
   A student who thinks the mass falls whenever a gas is produced picks this. In a sealed flask the CO₂ cannot leave, so the mass reading stays the same throughout and gives no information about the rate.
2. BRecord the absorbance of the mixture at regular time intervals after mixing. Correct
   Br₂ is the only colored species, so the absorbance of the mixture is proportional to [Br₂]. Readings at regular times show how quickly [Br₂] is changing at different stages of the trial, which is what a rate at several times requires.
3. CMeasure how strongly the mixture absorbs light once its color has stopped changing.
   A student who takes the amount that has reacted as the rate picks this. One reading at the end shows only that the Br₂ has been used up; it does not show how quickly [Br₂] was changing at any time.
4. DTime how long it takes, after mixing, for the color of the mixture to disappear.
   A student who treats the rate as the time the reaction takes picks this. One overall time cannot show how the rate differs from one moment to another during the trial.

CED 5.2.A.1 · Read this in Fix

1. A12
   A student who thinks the rate changes by the same factor as each concentration picks this: 4 × 3 = 12. That is right for B, which is first order, but the reaction is second order in A, so making [A] 4 times as great makes the rate 4² = 16 times as great.
2. B24
   A student who multiplies the concentration factor by the order picks this: (2 × 4) for A and (1 × 3) for B, giving 8 × 3 = 24. The order is an exponent, so the factor for A is 4² = 16, not 2 × 4 = 8.
3. C19
   A student who adds the separate effects of the two changes picks this: 16 for A plus 3 for B. The concentration terms in a rate law are multiplied, so the two factors are multiplied: 16 × 3 = 48.
4. D48 Correct
   With rate = k[A]²[B] and the same k, the rate changes by (4)² = 16 because of A and by 3 because of B. The concentration terms are multiplied in the rate law, so the factors multiply: 16 × 3 = 48.

Working Rate = k[A]²[B], and k is the same in both trials. rate₂/rate₁ = (4)²(3) = 16 × 3 = 48. Distractors: each concentration factor used without its order, 4 × 3 = 12; order multiplied by the concentration factor, (2 × 4)(1 × 3) = 24; the two effects added, 16 + 3 = 19.

CED 5.2.A.2 · Read this in Fix

1. AThe rate does not depend on [B]: the order with respect to B is zero. Correct
   A rate law contains the concentration of each reactant raised to its order. [B] is absent, which means [B]⁰ = 1: changing [B] does not change the rate, although B is still consumed in the reaction.
2. BThe rate law is incomplete: a reactant has an order of at least one.
   A student who thinks every reactant in the equation must appear in the rate law picks this. A reactant can be zero order: it is needed for the reaction, but the rate does not depend on its concentration.
3. CThe reactant B is a catalyst: a catalyst is left out of a rate law.
   A student who thinks a substance missing from the rate law is a catalyst picks this. B is a reactant in the balanced equation and is used up, so it is not a catalyst; it is zero order.
4. DThe reactant B is in excess: the limiting reactant sets the rate.
   A student who thinks the limiting reactant alone decides the rate picks this. The limiting reactant decides how much product forms. The rate law was found by varying [B] as well as [A], and the rate did not respond to [B].

CED 5.2.A.3 · Read this in Fix

1. A2.0 × 10⁻² M⁻² s⁻¹
   A student who takes the orders from the coefficients in the equation picks this, using rate = k[X][Y]²: k = (1.0 × 10⁻³)/((0.20)(0.50)²) = 2.0 × 10⁻² M⁻² s⁻¹. The data show the opposite orders: the rate is 9 times as great when [X] is tripled and 3 times as great when [Y] is tripled.
2. B5.0 × 10⁻² M⁻² s⁻¹ Correct
   Tripling [X] alone (trials 1 and 2) makes the rate 9 times as great, so the reaction is second order in X; tripling [Y] alone (trials 1 and 3) triples the rate, so it is first order in Y. With rate = k[X]²[Y], trial 1 gives k = (1.0 × 10⁻³ M s⁻¹)/((0.20 M)²(0.50 M)) = 5.0 × 10⁻² M⁻² s⁻¹.
3. C1.0 × 10⁻² M⁻¹ s⁻¹
   A student who thinks the rate is directly proportional to each concentration picks this, using rate = k[X][Y]: k = (1.0 × 10⁻³)/((0.20)(0.50)) = 1.0 × 10⁻² M⁻¹ s⁻¹. Trials 1 and 2 show that tripling [X] makes the rate 9 times as great, so the reaction is second order in X, and k has the units of a third-order rate constant.
4. D2.5 × 10⁻¹ M⁻³ s⁻¹
   A student who finds an order by dividing the rate factor by the concentration factor picks this: 9 ÷ 3 = 3 for X and 3 ÷ 3 = 1 for Y, so k = (1.0 × 10⁻³)/((0.20)³(0.50)) = 2.5 × 10⁻¹ M⁻³ s⁻¹. The order is an exponent: 3² = 9, so the reaction is second order in X.

Working Trials 1 and 2: [X] is 3 times as great with [Y] the same, and the rate is 9 times as great; 3² = 9, so second order in X. Trials 1 and 3: [Y] is 3 times as great with [X] the same, and the rate is 3 times as great, so first order in Y. Rate = k[X]²[Y]. From trial 1: k = (1.0 × 10⁻³ M s⁻¹)/((0.20 M)²(0.50 M)) = (1.0 × 10⁻³)/(0.020) M⁻² s⁻¹ = 5.0 × 10⁻² M⁻² s⁻¹. Distractors: orders from coefficients, k = (1.0 × 10⁻³)/((0.20)(0.50)²) = 2.0 × 10⁻² M⁻² s⁻¹; each reactant taken as first order, k = (1.0 × 10⁻³)/((0.20)(0.50)) = 1.0 × 10⁻² M⁻¹ s⁻¹; order in X taken as 9 ÷ 3 = 3, k = (1.0 × 10⁻³)/((0.20)³(0.50)) = 2.5 × 10⁻¹ M⁻³ s⁻¹.

CED 5.2.A.4 · Read this in Fix

1. A1.2 × 10⁻² M s⁻¹
   A student who thinks the rate changes by the same factor as each concentration picks this: 3 × 2 × 2.0 × 10⁻³ = 1.2 × 10⁻² M s⁻¹. Trials 1 and 2 show that the reaction is second order in A, so tripling [A] makes the rate 9 times as great.
2. B3.6 × 10⁻² M s⁻¹ Correct
   Doubling [A] alone (trials 1 and 2) makes the rate 4 times as great, so the reaction is second order in A; doubling [B] alone (trials 1 and 3) doubles the rate, so it is first order in B. Compared with trial 1, trial 4 has 3 times the [A] and 2 times the [B]: rate = 3² × 2 × 2.0 × 10⁻³ M s⁻¹ = 3.6 × 10⁻² M s⁻¹.
3. C2.4 × 10⁻² M s⁻¹
   A student who takes the orders from the coefficients picks this, using rate = k[A][B]²: 3 × 2² × 2.0 × 10⁻³ = 2.4 × 10⁻² M s⁻¹. The data show second order in A and first order in B, the reverse of the coefficients.
4. D5.0 × 10⁻³ M s⁻¹
   A student who thinks the rate is proportional to the total reactant concentration picks this: the total goes from 0.20 M in trial 1 to 0.50 M in trial 4, a factor of 2.5, giving 5.0 × 10⁻³ M s⁻¹. Each concentration enters the rate law separately, raised to its own order.

Working Trials 1 and 2: [A] doubles with [B] the same and the rate is 4 times as great: second order in A. Trials 1 and 3: [B] doubles with [A] the same and the rate doubles: first order in B. Rate = k[A]²[B]. Trial 4 compared with trial 1: [A] is 3 times as great and [B] is 2 times as great, so rate = (3)²(2)(2.0 × 10⁻³ M s⁻¹) = 18 × 2.0 × 10⁻³ = 3.6 × 10⁻² M s⁻¹. Distractors: each reactant taken as first order, 3 × 2 × 2.0 × 10⁻³ = 1.2 × 10⁻²; orders from coefficients, 3 × 2² × 2.0 × 10⁻³ = 2.4 × 10⁻²; rate proportional to the total concentration, (0.50/0.20) × 2.0 × 10⁻³ = 5.0 × 10⁻³.

CED 5.2.A.5 · Read this in Fix

1. AOrder 4
   A student who takes each order from the coefficient in the balanced equation picks this: 2 + 2 = 4. The graphs show second order in X, but the rate does not change at all with [Y], so the order in Y is zero, not 2.
2. BOrder 5
   A student who takes each order to be the factor by which the rate changes picks this: the rate is 4 times as great when [X] doubles ('fourth order') and changes by a factor of 1 when [Y] doubles ('first order'), giving 4 + 1 = 5. The order is the exponent: 2² = 4 gives 2 for X, and 2⁰ = 1 gives 0 for Y.
3. COrder 2 Correct
   In graph 1 the rate is 4 times as great when [X] doubles and 9 times as great when [X] triples, so the reaction is second order in X. In graph 2 the rate does not change with [Y], so it is zero order in Y. The overall order is the sum of the powers: 2 + 0 = 2.
4. DOrder 3
   A student who thinks every reactant must have an order of at least one picks this, taking second order in X from graph 1 and adding 1 for Y. Graph 2 shows that the rate does not depend on [Y]: the order in Y is zero.

Working Graph 1: when [X] doubles from 0.10 M to 0.20 M the rate goes from 1 to 4 (× 10⁻³ M s⁻¹), a factor of 4 = 2², and from 0.10 M to 0.30 M it goes from 1 to 9 = 3²: second order in X. Graph 2: the rate is the same at every [Y]: zero order in Y. Overall order = 2 + 0 = 2. Distractors: coefficients, 2 + 2 = 4; order taken as the rate factor, 4 + 1 = 5; Y given an order of at least one, 2 + 1 = 3.

CED 5.2.A.3 · Read this in Fix

1. AThe rate falls to one-half when [X] falls to one-half of its starting value. Correct
   In the first 20 s, 4 of the 8 X atoms react; in the next 20 s, when [X] starts at half its original value, 2 react. The amount reacting in equal times is halved when [X] is halved, so the rate is proportional to [X].
2. BThe rate falls to one-fourth when [X] falls to one-half of its starting value.
   A student who takes the order from the coefficient 2 in the equation picks this, expecting rate = k[X]². The diagrams show 4 X atoms reacting in the first 20 s and 2 in the next 20 s: the rate is halved, not quartered, when [X] is halved.
3. CThe rate stays the same while [X] falls to one-half of its starting value.
   A student who treats the rate as the time taken picks this, because each stage shown lasts 20 s. The rate is the amount reacting per unit time: 4 X atoms react in the first 20 s but only 2 in the next 20 s.
4. DThe rate rises to 1.5 times when [X₂] rises from its value at 20 s to its value at 40 s.
   A student who takes the amount of product present as the rate picks this, because the number of X₂ molecules goes from 2 to 3. The rate is how quickly the amounts change: only 1 more X₂ forms in the second 20 s, compared with 2 in the first 20 s, so the reaction has slowed.

CED 5.2.A.2 · Read this in Fix

1. AThe exponents in the rate law become greater on heating, so the 25°C rate law does not fit.
   A student who thinks a faster reaction has higher orders picks this. An order describes how the rate responds to a concentration; the quantity in the rate law that depends on temperature is the rate constant.
2. BThe rate constant depends on [A] and [B], so it changed with the new concentrations.
   A student who thinks the rate constant changes with concentration picks this. At a fixed temperature k is the same whatever the concentrations; it is the change in temperature that changed k.
3. CThe rate constant is fixed for a reaction, so the measured rate has to contain an error.
   A student who thinks the rate constant has the same value under all conditions picks this. A rate constant is constant only at one temperature; at 45°C it is larger than at 25°C, so the measured rate is not an error.
4. DThe rate constant is greater at 45°C, so the value measured at 25°C gives too low a rate. Correct
   The value of a rate constant depends on temperature. The k measured at 25°C does not apply at 45°C, where k is larger, so a prediction made with the 25°C value is too low.

CED 5.2.A.4 · Read this in Fix

1. ASupported: the rate changed by the same factor as [A] did, so the order in A equals one.
   A student who assigns the whole change in rate to the reactant being asked about picks this. [B] doubled too, so the data cannot show what A alone does: the result would also fit zero order in A and first order in B.
2. BNot supported: the coefficient of A in the equation is two, so the order is two.
   A student who takes the order from the coefficient in the balanced equation picks this. Orders come from experimental data, not from the coefficients of the overall equation, and these two trials cannot isolate the order in A because [B] changed as well.
3. CNot supported: [B] changed as well, so the effect of [A] on the rate is not isolated. Correct
   Both concentrations doubled, so the doubling of the rate reflects the orders in A and in B together (it shows only that the two orders add up to 1). To find the order in A, trials are needed in which [A] differs and [B] is the same.
4. DNot supported: the rate became two times as great, so the data show an order of two.
   A student who takes the order to be the factor by which the rate changes picks this. The order is an exponent that links the concentration factor to the rate factor, and here two concentrations changed, so no single order can be read from the two trials.

CED 5.2.A.5 · Read this in Fix

1. A6.0 × 10⁻² M s⁻¹ Correct
   The initial rate is the size of the slope of the tangent to the curve at time 0. The tangent falls from 1.20 M at 0 s to 0 M at 20 s, so the rate is (1.20 M)/(20 s) = 6.0 × 10⁻² M s⁻¹.
2. B3.0 × 10⁻² M s⁻¹
   A student who takes the initial rate from the first two plotted points on the curve picks this: (1.20 M − 0.60 M)/(20 s) = 3.0 × 10⁻² M s⁻¹. That is the average rate over the first 20 s; the curve is steepest at the start, so the initial rate, given by the tangent, is greater.
3. C1.5 × 10⁻² M s⁻¹
   A student who gives the reaction a single rate, the total change divided by the total time, picks this: (1.20 M − 0.30 M)/(60 s) = 1.5 × 10⁻² M s⁻¹. That is the average rate over 60 s; the rate at time 0 is the slope of the tangent there.
4. D5.0 × 10⁻³ M s⁻¹
   A student who divides a concentration by the time at which it is measured picks this: (0.30 M)/(60 s) = 5.0 × 10⁻³ M s⁻¹. A rate is a change in concentration divided by a time interval; the initial rate is the slope of the tangent at time 0.

Working The initial rate is the size of the slope of the tangent at time 0. The tangent runs from (0 s, 1.20 M) to (20 s, 0 M): slope = (0 − 1.20 M)/(20 s − 0 s) = −0.060 M s⁻¹, so the initial rate is 6.0 × 10⁻² M s⁻¹. Distractors: first two plotted points on the curve, (1.20 − 0.60)/20 = 3.0 × 10⁻²; whole graph, (1.20 − 0.30)/60 = 1.5 × 10⁻²; last point read as concentration ÷ time, 0.30/60 = 5.0 × 10⁻³.

CED 5.2.A.1 · Read this in Fix

1. A[X] = 0.20 M and [Y] = 0.60 M
   A student who thinks the rate is directly proportional to each concentration picks this, because [X][Y] = 0.12 is the largest product. The rate law has [X] squared, and (0.20)²(0.60) = 0.024 is less than (0.50)²(0.20) = 0.050.
2. B[X] = 0.10 M and [Y] = 0.80 M
   A student who thinks the rate depends on the total reactant concentration picks this, because 0.90 M is the largest total. Each concentration enters the rate law separately: (0.10)²(0.80) = 0.0080, the smallest value of the four.
3. C[X] = 0.30 M and [Y] = 0.30 M
   A student who thinks the reactant present in the smaller concentration alone sets the rate picks this, because its smaller concentration, 0.30 M, is the largest of the four. The rate depends on both concentrations: (0.30)²(0.30) = 0.027, less than 0.050.
4. D[X] = 0.50 M and [Y] = 0.20 M Correct
   The rate is proportional to [X]²[Y]. For this mixture [X]²[Y] = (0.50)²(0.20) = 0.050, larger than for the others: (0.30)²(0.30) = 0.027, (0.20)²(0.60) = 0.024 and (0.10)²(0.80) = 0.0080.

CED 5.2.A.2 · Read this in Fix

1. A4
   A student who thinks the rate changes by the same factor as each concentration picks this: 2 × 2 = 4. That is right for Y, which is first order, but the reaction is second order in X, so doubling [X] multiplies the rate by 2² = 4, and the overall factor is 4 × 2 = 8.
2. B6
   A student who adds the separate effects of the two changes picks this: 4 for X plus 2 for Y. The concentration terms in a rate law are multiplied, so the two factors are multiplied: 4 × 2 = 8.
3. C1
   A student who thinks the rate depends on the amounts of X and Y present picks this, because no gas was added or removed. The rate law contains concentrations, and halving the volume doubles both [X] and [Y].
4. D8 Correct
   Halving the volume at constant temperature doubles the concentration of each gas, and k does not change. With rate = k[X]²[Y], the rate is multiplied by 2² = 4 because of X and by 2 because of Y: 4 × 2 = 8.

Working Concentration = n/V. Halving V with the same amounts doubles both [X] and [Y]; k is unchanged because the temperature is the same. rate₂/rate₁ = (2)²(2) = 8. Distractors: each concentration factor used without its order, 2 × 2 = 4; the two effects added, 2² + 2 = 6; amounts unchanged so rate taken as unchanged, 1.

CED 5.2.A.2 · Read this in Fix

1. A5.0 × 10⁻⁷ M s⁻¹
   A student who divides the concentration at 200 s by the time picks this: (1.0 × 10⁻⁴ M)/(200 s) = 5.0 × 10⁻⁷ M s⁻¹. A rate is a change in concentration divided by the time interval: [D] changes by 6.0 × 10⁻⁵ M in the first 200 s.
2. B3.0 × 10⁻⁷ M s⁻¹ Correct
   The absorbance falls from 0.80 to 0.50 in the first 200 s. Dividing by 5.0 × 10³ M⁻¹ gives [D] = 1.6 × 10⁻⁴ M and 1.0 × 10⁻⁴ M, so the average rate is (6.0 × 10⁻⁵ M)/(200 s) = 3.0 × 10⁻⁷ M s⁻¹.
3. C2.5 × 10⁻⁷ M s⁻¹
   A student who gives the reaction a single rate, the total change divided by the total time followed, picks this: (1.6 × 10⁻⁴ M − 6.0 × 10⁻⁵ M)/(400 s) = 2.5 × 10⁻⁷ M s⁻¹. The curve is steeper in the first 200 s than in the second, so the average rate over the first 200 s is greater than the average over 400 s.
4. D6.0 × 10⁻⁵ M s⁻¹
   A student who takes the amount of reactant used up as the rate picks this: [D] falls by 6.0 × 10⁻⁵ M in the first 200 s. A rate is that change divided by the time it takes, 200 s.

Working From the graph, A = 0.80 at 0 s and A = 0.50 at 200 s. [D] = A/(5.0 × 10³ M⁻¹): 1.6 × 10⁻⁴ M at 0 s and 1.0 × 10⁻⁴ M at 200 s. Average rate = −Δ[D]/Δt = (1.6 × 10⁻⁴ M − 1.0 × 10⁻⁴ M)/(200 s) = 3.0 × 10⁻⁷ M s⁻¹. Distractors: concentration at 200 s ÷ time, (1.0 × 10⁻⁴ M)/(200 s) = 5.0 × 10⁻⁷; whole graph, (1.6 × 10⁻⁴ M − 6.0 × 10⁻⁵ M)/(400 s) = 2.5 × 10⁻⁷; amount of D used up in the first 200 s given as the rate, 6.0 × 10⁻⁵.

CED 5.2.A.1 · Read this in Fix