5.3 Concentration Changes Over Time

Reaction order tells you how the rate depends on concentration of a reactant (zero-, first-, second-order for this topic). To figure it out from concentration vs time data, try the integrated-rate plots from the CED: - Zeroth order: [A] vs time is linear. Slope = -k (so k = -slope). - First order: ln[A] vs time is linear. Slope = -k. Also half-life is constant: t1/2 = 0.693/k (useful for radioactive decay examples). - Second order: 1/[A] vs time is linear. Slope = +k. So take your concentration data, plot [A], ln[A], and 1/[A] vs time. The plot that gives a straight line tells you the order and its slope gives k (matching CED 5.3.A.1–5.3.A.4). For exam free-response, show the linear fit and use the slope to calculate k and (if first order) t1/2. For extra practice, check the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and thousands of practice problems (https://library.fiveable.me/practice/ap-chemistry).

Look at how concentration changes with time and make three plots: [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. Whichever plot is most linear tells the order for A: - Zeroth order: [A] vs. t is linear and follows [A]t − [A]0 = −kt (slope = −k). - First order: ln[A] vs. t is linear and follows ln[A]t − ln[A]0 = −kt (slope = −k). Also half-life is constant: t1/2 = 0.693/k. - Second order: 1/[A] vs. t is linear and follows 1/[A]t − 1/[A]0 = kt (slope = k). Use R² (or how straight the line looks) to pick the best fit. If you only have initial-rate data, use the method of initial rates to find the exponent. For practice, check the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and try problems at Fiveable’s practice page (https://library.fiveable.me/practice/ap-chemistry).

Because the integrated rate laws for different reaction orders give different mathematical relationships between [A] and t, you can’t reliably tell order or get k from the raw [A] vs t curve alone. For a first-order reaction ln[A] vs t is linear (ln[A]t − ln[A]0 = −kt) and for a second-order reaction 1/[A] vs t is linear (1/[A]t − 1/[A]0 = kt). Plotting those transformed quantities straightens the curved concentration data so you can (1) identify the order by which plot is linear and (2) get k directly from the slope (CED 5.3.A.2–5.3.A.4). First-order half-life is constant (t1/2 = 0.693/k), which also follows only when you’ve confirmed first order. For a quick refresher and practice on making these plots, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and try more problems at (https://library.fiveable.me/practice/ap-chemistry).

Integrated rate laws just rewrite the rate law so concentration vs time is explicit for different reaction orders. Use these forms from the CED to identify order and get k from the slope: - Zeroth order: [A]t = [A]0 − kt → plot [A] vs t; slope = −k. - First order: ln[A]t = ln[A]0 − kt → plot ln[A] vs t; slope = −k. Half-life is constant: t1/2 = 0.693/k (useful on the exam). - Second order: 1/[A]t = 1/[A]0 + kt → plot 1/[A] vs t; slope = +k. Which one to use: try all three linear plots—the one that’s straight tells you the order (CED 5.3.A.1–5.3.A.4). Slopes give k and let you predict concentrations at any t. For reactions that behave like first order because one reactant is in large excess, use a pseudo-first-order approximation (see Topic 5.3 study guide: https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).

When they say “the slope gives you the rate constant,” they mean that for the integrated rate-law plot that is linear for a given reaction order, the slope of that straight line equals (±) the numerical value of k. For the common cases in the CED: - Zeroth order: plot [A] vs. t → slope = d[A]/dt = −k (so slope is −k). - First order: plot ln[A] vs. t → slope = −k (so the line’s slope = −k). - Second order: plot 1/[A] vs. t → slope = +k. So you identify the order by which plot is linear (CED 5.3.A.1–3), fit a line, and take the magnitude (and sign) of its slope to get k (CED 5.3.A.4). For first-order reactions you can also use t1/2 = 0.693/k to check k (CED 5.3.A.5). For more worked examples and practice, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and the AP practice problems (https://library.fiveable.me/practice/ap-chemistry).

For a first-order reaction the integrated rate law is ln[A]t = ln[A]0 − kt (CED 5.3.A.2). To get the half-life, set [A]t = 1/2 [A]0 and solve: ln(1/2 [A]0) = ln[A]0 − kt1/2 ln(1/2) = −kt1/2 t1/2 = −ln(1/2)/k = ln(2)/k ≈ 0.693/k Because ln(2) is just a constant and the initial concentration [A]0 cancels out, t1/2 depends only on k, not on [A]0—so every equal-k first-order reaction has the same half-life regardless of starting amount. That’s why radioactive decay (a common AP example) shows a constant half-life (CED 5.3.A.6). For more practice and a short Topic 5.3 review, see the concentration changes study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and the AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

Radioactive decay is always treated as first order because each nucleus has a constant probability per unit time to decay—that probability (k) doesn’t depend on how many other nuclei are around. That gives the differential rate law: rate = −d[N]/dt = k[N]. Integrating yields ln[N]t − ln[N]0 = −kt, so a plot of ln[N] vs. time is linear (CED 5.3.A.2). A key consequence is a constant half-life: t1/2 = 0.693/k (CED 5.3.A.5), which matches experimental decay data. This is why radioactive processes are the canonical example of first-order kinetics on the AP exam (CED 5.3.A.6). If you want a quick refresher on integrated rate laws and practice problems, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and Fiveable’s AP Chemistry practice set (https://library.fiveable.me/practice/ap-chemistry).

First-order vs second-order shows up in which transformed plot is linear. - First order: [A] vs t is an exponential decay (curved). A plot of ln[A] vs t is linear with slope = -k (integrated law: ln[A]t − ln[A]0 = -kt). Half-life is constant and t1/2 = 0.693/k (so repeated half-lives are equal)—useful on the AP (CED 5.3.A.2, 5.3.A.5). - Second order: [A] vs t also curves, but more steeply as [A] gets small. A plot of 1/[A] vs t is linear with slope = +k (integrated law: 1/[A]t − 1/[A]0 = kt) (CED 5.3.A.3). On the exam, identify order by which transformed plot (ln[A] or 1/[A]) gives a straight line and use its slope to find k (CED 5.3.A.4). For a quick review, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Don’t try to rote-memorize every algebraic rearrangement—learn the three forms and how to recognize them from plots and slopes. Quick checklist to memorize and use on exam: - Zeroth order: [A] vs t is linear → [A]t − [A]0 = −kt (slope = −k). - First order: ln[A] vs t is linear → ln[A]t − ln[A]0 = −kt (slope = −k). Half-life constant: t1/2 = 0.693/k. - Second order: 1/[A] vs t is linear → 1/[A]t − 1/[A]0 = kt (slope = +k). Study strategy: practice reading which plot is linear, then write the matching integrated form and identify slope sign. Make a one-line cheat: “0 → [A], 1 → ln[A], 2 → 1/[A].” Do lots of quick graph/fit problems so recognition becomes automatic (AP cares about identifying order from plots and using slope to get k per CED 5.3.A.1–4). For focused review, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and drill practice problems (https://library.fiveable.me/practice/ap-chemistry).

You measure concentration vs. time by repeatedly recording a property that changes as the reactant (or product) changes, then use integrated rate laws to get order and k. Common methods: - Spectrophotometry/colorimetry: measure absorbance at set times and convert to [A] with Beer's law—great for colored or UV-active species. - Titration samples: withdraw small aliquots at times, quench if needed, titrate to find [A]. - Conductivity or ion-selective electrodes: for ionic reactants/products. - Gas collection or pressure: measure product gas volume or pressure vs. time. - Initial-rate or pseudo-first-order: keep one reactant in large excess so one concentration is effectively constant. Plot your data per the CED: [A] vs. t (zeroth), ln[A] vs. t (first; linear; use t1/2 = 0.693/k), and 1/[A] vs. t (second). That determines order and k. For lab review and practice problems, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ), Unit 5 overview (https://library.fiveable.me/ap-chemistry/unit-5), and lots of practice questions (https://library.fiveable.me/practice/ap-chemistry).

Because the half-life formula t1/2 = 0.693/k comes directly from the first-order integrated rate law. For a first-order reaction the integrated law is ln[A]t − ln[A]0 = −kt, which rearranges to [A]t = [A]0 e^(−kt). Setting [A]t = 1/2[A]0 gives 1/2 = e^(−kt1/2) → t1/2 = ln2 / k (≈0.693/k). The key point: the fraction remaining (1/2) is independent of [A]0 because concentration decays exponentially for first order, so t1/2 is constant. For other orders the half-life depends on the initial concentration. Example: zeroth order t1/2 = [A]0/(2k) and second order t1/2 = 1/(k[A]0). Because those depend on [A]0, there’s no single constant half-life formula like the first-order one. This is exactly why AP CED emphasizes first-order ln[A] vs. time linearity and the constant half-life relation (see 5.3.A.2 and 5.3.A.5). For a quick review, check the Topic 5.3 study guide on Fiveable (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Pick the plot that would give a straight line for the possible integrated rate law. Test three options and see which is linear: - Zeroth order: plot [A] vs t. If linear with slope = −k, reaction is zero order (use [A]t − [A]0 = −kt). - First order: plot ln[A] vs t. If linear with slope = −k, reaction is first order (ln[A]t − ln[A]0 = −kt). Also first-order half-life is constant: t1/2 = 0.693/k. - Second order: plot 1/[A] vs t. If linear with slope = +k, reaction is second order (1/[A]t − 1/[A]0 = kt). AP CED expects you to identify order from these integrated plots and use the slope to get k (5.3.A.1–5.3.A.4). Practically, transform your concentration data into all three plots and pick the one with the best straight line (highest R²). For extra practice, see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and more problems (https://library.fiveable.me/practice/ap-chemistry).

For a zero-order reaction the concentration vs. time plot is a straight line with a constant negative slope. The integrated rate law is [A]t − [A]0 = −kt, so [A]t = [A]0 − kt—a linear decrease in [A] over time with slope = −k. The reaction goes to zero at t = [A]0/k. Unlike first order, the half-life depends on the initial concentration (t1/2 = [A]0/(2k)), so it’s not constant. On the AP exam you should recognize a zero order from a linear [A] vs. t graph and use its slope to find k (CED 5.3.A.1 and 5.3.A.4). For a quick refresher check the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Start from the differential rate law for a first-order reaction: rate = −d[A]/dt = k[A]. Separate variables and integrate: d[A]/[A] = −k dt ∫ from [A]0 to [A]t (d[A]/[A]) = −∫0 to t k dt ln[A]t − ln[A]0 = −kt Rearrange: ln[A]t = −kt + ln[A]0. That’s the equation of a straight line (y = mx + b) with y = ln[A], x = t, slope m = −k, and intercept b = ln[A]0. So if the reaction is truly first order, plotting ln[A] vs. time gives a straight line whose slope equals −k. This is exactly why AP asks you to use ln[A] vs time to identify first order (CED 5.3.A.2) and to find k (CED 5.3.A.4). Also useful: first-order half-life is constant, t1/2 = 0.693/k (CED 5.3.A.5). For extra practice, check the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and more problems (https://library.fiveable.me/practice/ap-chemistry).

Use the integrated-rate plots and the slope of the linear fit to read k directly: - Zeroth order: plot [A] vs t. Integrated form: [A]t − [A]0 = −kt. The slope of [A] vs t = −k, so k = −(slope). Units: M·s⁻¹ (or M·min⁻¹). - First order: plot ln[A] vs t. Integrated form: ln[A]t − ln[A]0 = −kt. The slope of ln[A] vs t = −k, so k = −(slope). Units: s⁻¹ (or min⁻¹). Useful AP fact: half-life t1/2 = 0.693/k for first order. - Second order: plot 1/[A] vs t. Integrated form: 1/[A]t − 1/[A]0 = kt. The slope of 1/[A] vs t = +k, so k = slope. Units: M⁻¹·s⁻¹. Make sure you fit a straight line (least-squares) and check which plot is linear to identify order first (CED 5.3.A.1–A.4). For practice and examples see the Topic 5.3 study guide (https://library.fiveable.me/ap-chemistry/unit-5/concentration-changes-over-time/study-guide/Owtcz3zbxFZXsPI7YfwJ) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).