5.3 Concentration Changes Over Time

1. AGraph 1 Correct
   In a zero-order reaction the rate does not depend on [X], so [X] falls by the same amount in each equal time interval: the graph of [X] against time is a straight line with slope −k until X is used up.
2. BGraph 2
   A student who thinks every reaction slows down as its reactant is used up picks this curve. That is true when the rate depends on [X] (first or second order), but in a zero-order reaction the rate stays the same, so the graph is a straight line, not a curve that levels off.
3. CGraph 3
   A student who reads 'zero order' as 'no change in [X]' picks this flat line. Zero order means that the rate does not depend on [X]; X is still used up, at a constant rate, so [X] falls steadily.
4. DGraph 4
   A student who thinks a reaction speeds up as it proceeds picks this curve, which gets steeper with time. In a zero-order reaction the rate, and so the slope of the graph, stays the same.

Working Zero order: rate = k, independent of [X]. [X]_t = [X]₀ − kt, so [X] falls along a straight line of slope −k until X is used up. The key is the graph that is a straight, falling line.

CED 5.3.A.1 · Read this in Fix

1. AMeasure [A] at regular times in one trial, then plot [A] against time to see whether the plot of the data is a straight line.
   A student who thinks a straight-line graph of [A] against time shows a first-order reaction picks this. A linear [A]-time graph shows zero order; first- and second-order reactions both give curved [A]-time graphs, so this plot cannot tell them apart.
2. BWrite the balanced equation for the decomposition and take the coefficient of A in it as the order with respect to A.
   A student who thinks the order equals the coefficient in the balanced equation picks this. The order with respect to a reactant is found from experimental data, not from the coefficients of the overall equation.
3. CRun two trials with different volumes of the same solution of A and compare how long each one takes to react.
   A student who thinks the rate depends on the amount of A rather than its concentration picks this. Different volumes of the same solution have the same [A], so they react at the same rate and give no information about the order.
4. DMeasure [A] at regular times in one trial, then plot ln[A] and the reciprocal of [A] against time to find the linear plot. Correct
   For a first-order reaction a plot of ln[A] against time is linear; for a second-order reaction a plot of 1/[A] against time is linear. Plotting both from one set of [A]-time data shows which order fits.

CED 5.3.A.2 · Read this in Fix

1. AA plot of 1/[A] against time; k is −1 times its slope
   A student who thinks every integrated-rate-law plot has a slope of −k picks this. That is true of the zero-order and first-order plots, which fall with time; 1/[A] rises as A is used up, and its slope is +k.
2. BA plot of 1/[A] against time; k equals its slope Correct
   For a second-order reaction 1/[A]_t − 1/[A]₀ = kt, so 1/[A] = kt + 1/[A]₀: a plot of 1/[A] against time is a straight line whose slope is k, positive because 1/[A] rises as A is used up.
3. CA plot of ln[A] against time; k equals −1 times its slope
   A student who mixes up the two transformed plots picks this. A linear ln[A]-time plot, with slope −k, is the test for a first-order reaction; for a second-order reaction it is the 1/[A] plot that is linear.
4. DA plot of [A]² against time; k is the slope of that plot
   A student who thinks the quantity to plot is [A] raised to the order of the reaction picks this. A second-order reaction gives a straight line when 1/[A], not [A]², is plotted against time.

Working Second order: 1/[A]_t − 1/[A]₀ = kt, so 1/[A]_t = kt + 1/[A]₀. A plot of 1/[A] against t is linear with slope +k.

CED 5.3.A.3 · Read this in Fix

1. A0.035 s⁻¹
   A student who thinks a straight [X]-time graph shows a first-order reaction picks this, using ln([X]₀/[X]_t)/t = ln(0.80/0.20)/(40 s) = 0.035 s⁻¹. A first-order reaction gives a straight line only when ln[X] is plotted; a straight line for [X] itself means zero order.
2. B0.094 M⁻¹ s⁻¹
   A student who takes the order from the coefficient 2 in the equation treats the reaction as second order and uses (1/[X]_t − 1/[X]₀)/t = (5.0 − 1.25) M⁻¹/(40 s) = 0.094 M⁻¹ s⁻¹. Orders come from experimental data, and these data give a straight line for [X], not for 1/[X].
3. C0.015 M s⁻¹ Correct
   The graph of [X] against time is a straight line, so the reaction is zero order in X and [X]_t − [X]₀ = −kt. The slope is (0.20 M − 0.80 M)/(40 s) = −0.015 M s⁻¹, so k = 0.015 M s⁻¹; M s⁻¹ is the unit of a zero-order rate constant.
4. D0.0050 M s⁻¹
   A student who reads the rate constant from one point, dividing the concentration there by the time, picks this: 0.20 M/(40 s) = 0.0050 M s⁻¹. The rate constant comes from the slope, the change in [X] divided by the change in time: 0.60 M/(40 s).

Working The graph of [X] against time is a straight line, so the reaction is zero order in X: [X]_t − [X]₀ = −kt. Slope = (0.20 M − 0.80 M)/(40 s − 0 s) = −0.015 M s⁻¹, so k = 0.015 M s⁻¹. Distractors: first order assumed, k = ln(0.80/0.20)/(40 s) = 0.035 s⁻¹; second order assumed from the coefficient 2, k = (1/0.20 − 1/0.80) M⁻¹/(40 s) = 0.094 M⁻¹ s⁻¹; one point read as concentration ÷ time, 0.20 M/40 s = 0.0050 M s⁻¹.

CED 5.3.A.4 · Read this in Fix

1. AThe half-life by itself, with no other data from the trial Correct
   Falling to one-quarter of the starting value takes two half-lives (1 → 1/2 → 1/4) whatever the starting concentration, because the half-life of a first-order reaction is constant. The time is 2 × t₁/2.
2. BThe half-life together with the starting concentration of A
   A student who thinks the half-life depends on the starting concentration picks this. For a first-order reaction t₁/2 = 0.693/k does not involve [A]₀, so reaching one-quarter takes two half-lives from any starting concentration.
3. CThe half-life and how the half-life changes as [A] falls
   A student who thinks each half-life is longer than the one before, because the reaction slows as A is used up, picks this, expecting to need to know how much longer the second half-life is. For a first-order reaction every half-life is the same, so the time to reach one-quarter is exactly two half-lives.
4. DThe half-life and the total volume of the mixture
   A student who thinks the amount of A, rather than its concentration, controls how fast it reacts picks this. The volume changes neither [A] nor k, so it does not affect the time.

Working First order: the half-life is constant. [A]₀ → [A]₀/2 → [A]₀/4 is two half-lives, so time = 2 × t₁/2, whatever [A]₀ or the volume; the second half-life equals the first.

CED 5.3.A.5 · Read this in Fix

1. ADiagram 1
   A student who thinks a half-life is half the time needed for the whole sample to decay picks this, expecting no radioactive nuclei after two half-lives. Each half-life removes half of the radioactive nuclei still present, so 3 of the 12 remain after 12 hours.
2. BDiagram 2 Correct
   12 hours is two half-lives. After the first, half of the 12 radioactive nuclei remain (6); after the second, half of those remain (3). The 9 nuclei that decayed are still present as nuclei of the stable isotope, so the portion holds 3 radioactive and 9 stable nuclei.
3. CDiagram 3
   A student who thinks decayed nuclei disappear picks this, which shows the 3 remaining radioactive nuclei and nothing else. Each nucleus that decays becomes a nucleus of the stable isotope, so 9 stable nuclei should also be present.
4. DDiagram 4
   A student who thinks each half-life is longer than the one before, because decay slows as nuclei are used up, picks this, with more than 3 radioactive nuclei left. The decay rate does fall, but in proportion to the number of radioactive nuclei left, so every half-life is 6 hours and 3 remain.

Working 12 h = 2 × 6 h = two half-lives. Radioactive nuclei: 12 → 6 → 3. Each decayed nucleus becomes a stable product nucleus: 12 − 3 = 9. The key shows 3 radioactive and 9 stable nuclei.

CED 5.3.A.6 · Read this in Fix

1. AThe half-life is twice as long, and the initial rate doubles.
   A student who thinks a larger starting amount takes longer to fall to half picks this. The initial rate does double, but so does the amount that must react, so the time to halve is unchanged: t₁/2 = 0.693/k.
2. BThe half-life is unchanged, and the initial rate does not change.
   A student who treats the rate as the rate constant picks this. k is unchanged at the same temperature, but the rate, k[A], depends on [A], so doubling [A] doubles the initial rate.
3. CThe half-life is unchanged, and the initial rate doubles. Correct
   For a first-order reaction t₁/2 = 0.693/k, which does not involve [A], and k is unchanged at the same temperature, so the half-life is the same. The rate law is rate = k[A], so doubling [A] doubles the initial rate: twice as much A reacts twice as fast, which is why halving takes the same time.
4. DThe half-life is twice as long, and the initial rate does not change.
   A student who thinks a reaction runs at a fixed speed, whatever the amount of reactant, picks this: twice as much A at the same rate would take twice as long to halve. For a first-order reaction the rate is proportional to [A], so the initial rate doubles and the half-life is unchanged.

Working Rate law: rate = k[A]; same temperature, so same k. Initial rate ratio = k(0.40 M)/k(0.20 M) = 2. Half-life t₁/2 = 0.693/k does not involve [A]₀, so it is unchanged.

CED 5.3.A.5 · Read this in Fix

1. AFirst order in D: the absorbance is proportional to [D], so a plot of ln[D] against time is also linear. Correct
   With the wavelength and path length fixed, A = εbc makes the absorbance proportional to [D], so ln(absorbance) = ln[D] + a constant. A linear ln(absorbance)-time plot therefore means that ln[D] is linear in time, the test for a first-order reaction; the slope is −k.
2. BSecond order in D: the absorbance is proportional to [D], so a plot of ln[D] against time is also linear.
   A student who mixes up the transformed plots picks this. The absorbance argument is right, but a linear ln[D]-time plot shows a first-order reaction; a second-order reaction needs a linear 1/[D]-time plot.
3. CZero order in D: a straight-line plot against time shows that D is used up at a constant rate.
   A student who thinks any straight-line plot against time means a constant rate picks this. The plotted quantity is a logarithm: a constant slope of ln(absorbance) means that a constant fraction of D reacts each minute, so the rate falls as D is used up.
4. DNo conclusion: a plot made from absorbance values rather than from [D] gives no information about the order.
   A student who thinks only a plot of the concentration itself can show the order picks this. Because the absorbance is proportional to [D] (A = εbc with ε and b fixed), ln(absorbance) differs from ln[D] by a constant, so the plot has the shape and slope of the ln[D] plot.

CED 5.3.A.2 · Read this in Fix

1. A0.16 M⁻¹ s⁻¹
   A student who reads the rate constant from a single point, dividing the plotted value by the time, picks this: 4.0 M⁻¹/(25 s) = 0.16 M⁻¹ s⁻¹. k is the slope, the change in 1/[E] divided by the change in time: (4.0 − 2.0) M⁻¹/(25 s).
2. B0.028 s⁻¹
   A student who links a linear 1/[E] plot with a first-order reaction picks this, converting to [E] = 0.50 M and 0.25 M and using ln([E]₀/[E]_t)/t = ln 2/(25 s) = 0.028 s⁻¹. A linear ln[E] plot shows first order; a linear 1/[E] plot shows second order.
3. C0.010 M s⁻¹
   A student who treats the rate constant as the rate picks this: [E] falls from 0.50 M to 0.25 M in 25 s, an average rate of 0.010 M s⁻¹. That rate changes as [E] falls; k is the constant slope of the linear 1/[E] plot.
4. D0.080 M⁻¹ s⁻¹ Correct
   The plot of 1/[E] against time is a straight line, so the reaction is second order in E and 1/[E]_t − 1/[E]₀ = kt. The slope is (4.0 − 2.0) M⁻¹/(25 s) = 0.080 M⁻¹ s⁻¹, which is k; M⁻¹ s⁻¹ is the unit of a second-order rate constant.

Working The plot of 1/[E] against time is a straight line, so the reaction is second order in E: 1/[E]_t − 1/[E]₀ = kt. Slope = (4.0 − 2.0) M⁻¹/(25 s) = 0.080 M⁻¹ s⁻¹ = k. Distractors: plotted value ÷ time, 4.0 M⁻¹/(25 s) = 0.16 M⁻¹ s⁻¹; first order assumed with [E] = 0.50 M and 0.25 M, ln(0.50/0.25)/(25 s) = 0.028 s⁻¹; average rate taken as k, (0.50 − 0.25) M/(25 s) = 0.010 M s⁻¹.

CED 5.3.A.4 · Read this in Fix

1. AIt is a straight line whose slope equals −k throughout.
   A student who thinks a first-order reaction gives a straight line on a graph of [X] against time picks this. A straight [X]-time line, with slope −k, is the sign of a zero-order reaction; for a first-order reaction it is ln[X] against time that is linear.
2. BIt is a curve whose slope becomes steeper and steeper as time passes.
   A student who thinks a reaction speeds up as it proceeds picks this. For a first-order reaction the rate, shown by the steepness of the curve, is greatest at the start, when [X] is largest.
3. CIt is a curve that reaches [X] = 0 at the end of two half-lives.
   A student who thinks a half-life is half the time the reaction takes to finish picks this. Each half-life halves the [X] that remains, so after two half-lives one-quarter of the starting [X] is left.
4. DIt is a curve whose slope becomes less steep as time passes. Correct
   For a first-order reaction the rate is k[X], so the rate falls as X is used up: [X] drops quickly at first and then more and more slowly, giving a curve that levels off toward zero.

CED 5.3.A.1 · Read this in Fix

1. AThe initial rate will be doubled.
   A student who links a linear reciprocal plot with a first-order reaction picks this. Equal steps in 1/[G] mean a linear 1/[G]-time plot, which shows a second-order reaction; its rate depends on [G]², so the rate quadruples.
2. BThe initial rate will be quadrupled. Correct
   Only the 1/[G] column changes by equal amounts in equal times (1.00 M⁻¹ every 10 s), so a plot of 1/[G] against time is linear and the reaction is second order: rate = k[G]². Doubling [G] multiplies the rate by 2² = 4.
3. CThe initial rate will be unchanged.
   A student who thinks a quantity that changes by equal steps in equal times shows a constant rate picks this, treating the reaction as zero order. The equal steps are in 1/[G], not in [G]; [G] itself falls by smaller amounts in each interval.
4. DThe initial rate will be multiplied by 8.
   A student who takes the order from the coefficient 3 in the equation treats the reaction as third order and multiplies the rate by 2³ = 8. Orders come from experimental data: only the 1/[G] column changes by equal steps, so the reaction is second order and the rate quadruples.

Working Differences over each 10 s: [G] −0.167, −0.083, −0.050 (not constant); ln[G] −0.41, −0.29, −0.22 (not constant); 1/[G] +1.00, +1.00, +1.00 M⁻¹ (constant). So 1/[G] is linear in time: second order, rate = k[G]² (k = 0.100 M⁻¹ s⁻¹). Doubling [G]: rate × 2² = 4. Distractors: first order (doubled); zero order (unchanged); order taken from the coefficient 3, 2³ = 8.

CED 5.3.A.3 · Read this in Fix

1. AZero order, because the half-life does not depend on [A]₀ in these trials.
   A student who thinks any quantity that does not depend on concentration points to zero order picks this. Zero order means that the rate does not depend on [A]; here it is the half-life that does not depend on [A]₀, which is the sign of a first-order reaction.
2. BZero order, because each trial reacts at the same rate as the other two.
   A student who thinks equal times mean equal rates picks this. In the same 50 s the 0.40 M trial loses 0.20 M of A and the 0.10 M trial only 0.05 M, so its rate is four times as large: the rate is proportional to [A], as in a first-order reaction.
3. CFirst order, because the half-life does not depend on [A]₀ in these trials. Correct
   For a first-order reaction t₁/2 = 0.693/k, which does not involve [A]₀, so the same half-life at every starting concentration is evidence that the reaction is first order in A.
4. DNo order can be assigned, because every reaction has a half-life that is constant.
   A student who thinks every reaction has a constant half-life picks this. A half-life that is the same at every starting concentration is a property of first-order reactions, so these data do identify the order.

CED 5.3.A.5 · Read this in Fix

1. AThe time to fall from N/2 to N/4 equals the time to fall from N to N/2. Correct
   Radioactive decay is first order: the number of nuclei decaying per day is proportional to the number present, so the time to lose half of whatever is present, the half-life, is 5.0 days at every stage.
2. BFalling from N/2 to N/4 takes longer than falling from N to N/2.
   A student who thinks each half-life is longer than the one before, because the decay slows as nuclei are used up, picks this. The decay rate does fall, but in proportion to the number of nuclei left, so halving N/2 takes as long as halving N.
3. CFalling from N/2 to N/4 takes less time than falling from N to N/2.
   A student who thinks a larger amount takes longer to fall to half picks this, expecting the smaller number, N/2, to halve sooner. Half as many nuclei decay at half the rate, so the time to halve is the same.
4. DIt falls by N/2 every 5.0 days until no radioactive nuclei remain.
   A student who thinks decay proceeds at a fixed rate, whatever the number of nuclei left, picks this: losing N/2 every 5.0 days would leave none after 10 days. The decay rate falls as nuclei are used up; after 10 days N/4 remain.

CED 5.3.A.6 · Read this in Fix

1. A0.49 M Correct
   For a first-order reaction ln[A]_t − ln[A]₀ = −kt. Here kt = (0.020 s⁻¹)(25 s) = 0.50, so ln[A]_t = ln(0.80) − 0.50 = −0.723 and [A]_t = 0.49 M.
2. B0.30 M
   A student who thinks [A] falls along a straight line in a first-order reaction picks this, using [A]_t = [A]₀ − kt = 0.80 − (0.020)(25) = 0.30. That equation describes a zero-order reaction; for first order it is ln[A] that falls linearly with time.
3. C0.57 M
   A student who links the reciprocal equation with a first-order reaction picks this: 1/[A]_t = 1/0.80 + (0.020)(25) = 1.75 M⁻¹, so [A]_t = 0.57 M. The 1/[A] equation applies to a second-order reaction; the first-order equation uses ln[A].
4. D0.40 M
   A student who thinks the reaction keeps a fixed speed picks this, using the initial rate, (0.020 s⁻¹)(0.80 M) = 0.016 M s⁻¹, for the whole 25 s: 0.80 − 0.40 = 0.40. In a first-order reaction the rate falls as [A] falls, so less than 0.40 M reacts.

Working First order: ln[A]_t − ln[A]₀ = −kt, so ln[A]_t = ln(0.80) − (0.020 s⁻¹)(25 s) = −0.223 − 0.50 = −0.723, and [A]_t = e^−0.723 = 0.49 M (equivalently 0.80 M × e^−0.50). Distractors: zero-order equation, 0.80 − (0.020)(25) = 0.30; second-order equation, 1/[A]_t = 1/0.80 + (0.020)(25) = 1.75, [A]_t = 0.57; initial rate (0.020 × 0.80 = 0.016 M s⁻¹) kept for 25 s, 0.80 − 0.40 = 0.40.

CED 5.3.A.4 · Read this in Fix

1. A0.12 s⁻¹ Correct
   [A] halves from 0.60 M to 0.30 M in the first 6.0 s and halves again, to 0.15 M, in the next 6.0 s. A half-life that stays the same as [A] falls shows a first-order reaction, for which k = 0.693/t₁/2 = 0.693/(6.0 s) = 0.12 s⁻¹.
2. B0.046 s⁻¹
   A student who takes a half-life to be half of the time the whole reaction lasts picks this, reading about 30 s for A to be used up and using 0.693/(15 s) = 0.046 s⁻¹. The half-life is the time for [A] to fall to half of its value, which the graph shows is 6.0 s.
3. C0.050 M s⁻¹
   A student who treats the rate constant as the rate picks this, from how fast [A] falls at the start: (0.60 M − 0.30 M)/(6.0 s) = 0.050 M s⁻¹. That rate changes as [A] falls; k for this first-order reaction comes from the constant half-life.
4. D0.28 M⁻¹ s⁻¹
   A student who takes the order from the coefficient 2 in the equation treats the reaction as second order and uses (1/[A]_t − 1/[A]₀)/t = (3.33 − 1.67) M⁻¹/(6.0 s) = 0.28 M⁻¹ s⁻¹. Orders come from experimental data, and the constant half-life on the graph shows first order.

Working [A] falls from 0.60 M to 0.30 M in the first 6.0 s and from 0.30 M to 0.15 M in the next 6.0 s: the half-life is constant, so the reaction is first order in A. k = 0.693/t₁/2 = 0.693/(6.0 s) = 0.12 s⁻¹. Distractors: half-life taken as half of the 30 s the reaction appears to last, 0.693/(15 s) = 0.046 s⁻¹; average rate over the first 6.0 s taken as k, (0.60 − 0.30) M/(6.0 s) = 0.050 M s⁻¹; second order assumed from the coefficient 2, (1/0.30 − 1/0.60) M⁻¹/(6.0 s) = 0.28 M⁻¹ s⁻¹.

CED 5.3.A.5 · Read this in Fix