First-Order Integrated Rate Law and Half-Life

Integrating rate = k[A] over time gives a logarithmic relationship. Written as ln[A]t = -kt + ln[A]0 it is in y = mx + b form, so the graph of ln[A] against t is linear with slope -k and intercept ln[A]0.

Here the working convention is −d[A]/dt = k[A]. This is the disappearance-rate law for A; if a different text defines rate by dividing by A's stoichiometric coefficient, its reported k can differ by that factor. Match the convention before combining data and formulas. Separating variables gives d[A]/[A] = −k dt; integrating between the measured endpoints gives ln([A]t/[A]₀) = −kt. The ratio is dimensionless, and kt must also be dimensionless.

Subtracting the intercept gives ln([A]t/[A]0) = -kt. This version is convenient whenever a question gives a fraction or percentage remaining rather than absolute concentrations, because the units cancel.

Choose the rearrangement from the unknown: [A]t = [A]₀e^−kt, t = ln([A]₀/[A]t)/k, or k = ln([A]₀/[A]t)/t. For 0.800 M falling to 0.200 M in 60.0 s, k = ln(4)/60.0 = 0.0231 s⁻¹. Then t₁/₂ = ln(2)/k = 30.0 s, consistent with two halvings in 60.0 s. Use unrounded k in later calculations.

When the question gives percent consumed, subtract it from 100% first. At 90% conversion the fraction remaining is 0.10, so kt = −ln(0.10) = 2.303. Putting 0.90 into the logarithm instead would calculate the time for only 10% conversion.

After n half-lives the fraction remaining is (1/2)ⁿ: 50 percent, 25 percent, 12.5 percent, 6.25 percent. When the elapsed time is a whole number of half-lives, counting is faster and safer than logarithms.

The half-life is a fixed time for halving the amount currently present, not for losing half the original amount each time. Starting with 80 mg, successive equal intervals leave 40, 20, 10 and 5 mg. The losses are 40, 20, 10 and 5 mg. After n = t/t₁/₂ half-lives, the remaining fraction is 2⁻ⁿ, even when n is not an integer. First-order decay approaches zero continuously; it does not predict an exact finite time at which concentration becomes mathematically zero.

Radioactive decay is strictly first order. So are many decompositions such as N₂O₅ and the elimination of many drugs from the bloodstream, which is why dosing intervals are quoted in half-lives.

A straight ln[A] versus t plot must be supported over a useful concentration range, with slope units matching the time axis. A slope of −0.0200 min⁻¹ gives k = 0.0200 min⁻¹, not s⁻¹. A straight base-10 log[A] plot has slope −k/2.303. Apparent first-order behavior can also arise from a multireactant process with another reactant in excess; the graph establishes the observed time dependence, not a unique molecular mechanism.