Master AP Chemistry Topic 5.3 with this free practice quiz. Test your skills on integrated rate laws, linear graphs, and first order half life to prepare for your AP exam.
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The Three Integrated Rate Laws
An integrated rate law connects concentration directly to time, so you can calculate how much reactant remains at any moment. Each order has its own form, and each form is a straight line in disguise.
Zero order: [A]ₜ = [A]₀ − kt. Concentration falls by the same amount in every equal time interval.
First order: ln[A]ₜ = ln[A]₀ − kt. Concentration falls by the same fraction in every equal time interval.
Second order: 1/[A]ₜ = kt + 1/[A]₀. The reciprocal of concentration grows linearly with time.
Here [A]₀ is the starting concentration, [A]ₜ the concentration at time t, and k the rate constant in the units belonging to that order: M s⁻¹ for zero order, s⁻¹ for first order, M⁻¹ s⁻¹ for second order.
Half-Lives
The half-life t₁/₂ is the time needed for the concentration to fall to half its value at the start of that interval. Each order behaves differently, and the behaviour is diagnostic.
First order: t₁/₂ = ln 2 / k = 0.693 / k. There is no concentration term, so the half-life is constant throughout the reaction. Every half-life halves what is left: after n half-lives the fraction remaining is (1/2)ⁿ. Losing 75% of the reactant means reaching one quarter, which is exactly two half-lives.
Second order: t₁/₂ = 1 / (k[A]₀). The half-life depends on the starting concentration: halve [A]₀ and the half-life doubles. Successive half-lives get longer as the reaction proceeds and the concentration drops.
Zero order: t₁/₂ = [A]₀ / (2k). Starting with less reactant shortens the half-life, because the constant consumption rate reaches the halfway point sooner.
Reading the Graphs
Plot the data three ways and find which plot is straight. That plot names the order.
[A] against time is linear for zero order, with slope −k. ln[A] against time is linear for first order, with slope −k. 1/[A] against time is linear for second order, with slope +k. Only one of the three plots is straight for a given reaction; the other two curve.
The intercepts carry information too. On the first-order plot, the intercept is ln[A]₀, so [A]₀ = e^(intercept). On the second-order plot the intercept is 1/[A]₀. The slope of the correct plot gives k directly, with its sign: negative for the zero-order and first-order plots, positive for the second-order plot.
Worked Examples
Example 1: first order, two routes to the same answer
A first-order reaction has k = 0.0347 s⁻¹ and [A]₀ = 0.80 M.
Half-life: t₁/₂ = 0.693 / 0.0347 = 20.0 s.
After 60.0 s, find [A]. Method one, half-lives: 60.0 s is 3 half-lives, so [A] = 0.80 × (1/2)³ = 0.80 / 8 = 0.10 M. Method two, integrated law: ln[A] = ln(0.80) − (0.0347)(60.0) = −0.223 − 2.082 = −2.305, so [A] = e^(−2.305) = 0.10 M. The two methods agree, as they must.
Example 2: second order
A second-order reaction has [A]₀ = 0.20 M and k = 0.50 M⁻¹ s⁻¹. Find [A] after 10 s.
1/[A] = kt + 1/[A]₀ = (0.50)(10) + 1/0.20 = 5.0 + 5.0 = 10, so [A] = 0.10 M.
Check with the half-life: t₁/₂ = 1 / (k[A]₀) = 1 / (0.50 × 0.20) = 10 s. After exactly one half-life the concentration should be halved, and 0.10 M is half of 0.20 M. Consistent.
Traps Specific to This Topic
The first-order half-life formula t₁/₂ = 0.693 / k belongs to first order only. Using it for a second-order or zero-order reaction is wrong even when a k value is supplied, because those half-lives contain [A]₀.
Sign of the slope separates second order from the others. The 1/[A] plot rises, so its slope is +k. Reporting k as negative from a second-order plot, or taking the slope of the ln[A] plot as +k, flips the value.
The integrated first-order law uses natural logarithms. Substituting log₁₀ without converting makes k too small by a factor of 2.303, because ln x = 2.303 log₁₀ x.
Percentages need translating into fractions remaining. 75% consumed is 25% remaining, one quarter, two half-lives for first order, not 0.75 of a half-life and not three half-lives.
A constant half-life is evidence, not an assumption. If data show equal halving times, the reaction is first order; if the halving time doubles as concentration falls, it is second order. Let the pattern identify the order before reaching for a formula.