17.5Integrated Rate Laws
Section · module m68791
What this needs
Learn these first — this section will not make sense without them.
- Needs4.2Linear Functions
The text writes ln[A]ₜ = −kt + ln[A]₀ directly over y = mx + b (and does the same for 1/[A]ₜ and for [A]ₜ) and asks students to read k off as the negative of a plotted slope and the initial concentration off as the y-intercept; a student who cannot recognize a rearranged equation as a disguised line has no way to know which two quantities on a data plot correspond to slope and intercept, making the graphical-determination examples unworkable rather than just harder.
The graphical-determination example reads k off as −slope and [A]₀ off as the y-intercept of a ln[A] vs. t plot, attaching physical units (h⁻¹) to the slope — this is §4.3's own objective (‘graph and interpret applications of slope-intercept form,’ ‘always attach units to slope and intercept’) rather than §4.2's mechanics of writing a line's equation.
Rearranging the first-order law to t = ln([A]₀/[A]ₜ)·(1/k) depends on treating ln([A]₀/[A]ₜ) as one quantity rather than trying to split the quotient inside the log apart by hand; a student who has not internalized the log-of-a-quotient identity commonly mishandles the sign and gets a negative time or a negative k.
Every half-life and time-to-decompose problem (‘how long until 80% has decomposed?’) is solved by isolating t in [A]ₜ = [A]₀e^(−kt) via natural log — the exponential-equation-solving skill itself; a student who cannot move between the exponential and log forms cannot finish these problems even though the chemistry setup (x and 0.200x) is trivial.
What this touches
Algebra that turns up here. Not a blocker, but this is where you will see it used.
- Applies1.7Rational Expressions
Deriving the second-order half-life combines 1/(½[A]₀) − 1/[A]₀ into 2/[A]₀ − 1/[A]₀ = 1/[A]₀ — ordinary combining of rational expressions over a common denominator, just written in concentration units instead of x's; skipping the common-denominator step is where students lose the factor of 2.
- Illustrates4.4Fitting Linear Models to Data
The chapter's central move — plotting [A], ln[A], or 1/[A] against t and checking which one straightens out — is a cleaner demonstration that choosing a transformation before fitting a line is itself a mathematical decision than algebra §4.4, which never transforms a scatter plot at all: its cricket-chirp data is already roughly straight before you touch it, and its exam-scores-by-age plot is set aside as showing no trend rather than reshaped into one — the section's whole counsel on nonlinear data is that a correlation coefficient is meaningless for it, never that a transformation might straighten it. Watching the same nonlinear data get linearized three different ways makes ‘why transform first’ concrete in a way neither of those examples can.
The integrated first-order law [A]ₜ = [A]₀e^(−kt) is algebra §6.8's continuous decay model A = A₀e^(rt) with r = −k under a different variable name, and once a student has solved 6.8's exponential decay problems, every half-life and time-to-decompose example in this section is the identical calculation with concentration standing in for amount.
The figure caption explicitly classifies one curve as ‘exponentially decaying’ (first order, quartz surface) and the other as ‘linear’ (zero order, tungsten surface) from the plot shape alone, before any fitting is done — exactly the scatter-diagram shape-recognition (linear vs. exponential vs. logarithmic) that is §6.9's first objective, not the choose-a-transformation move already bridged to §4.4.
- Applies9.4Geometric Sequences
The H₂O₂ decay data (1.000 M → 0.500 → 0.250 → 0.125 → 0.0625 M over equal 6-hour steps) is a geometric sequence with common ratio ½ sitting inside a chemistry table; recognizing it as one lets a student predict the next entry without recomputing the integrated rate law from scratch.