First-Order Integrated Rate Law

Introduction

For a first-order disappearance, rate = k[A]. As [A] falls, the absolute rate slows in direct proportion, but the same fraction of remaining A disappears during equal time intervals. Integration yields exponential decay, a linear ln[A] plot and a constant half-life under fixed conditions.

Core explanation

Start with −d[A]/dt = k[A]. Rearranging gives d[A]/[A] = −k dt. Integrating from initial concentration [A]0 to [A]t yields ln[A]t − ln[A]0 = −kt, or ln([A]t/[A]0)=−kt. The ratio inside the logarithm is dimensionless, which is the formally clean expression. Rearranged, [A]t=[A]0e^(−kt).

Plotting ln[A] against time gives a straight line with slope −k and intercept ln[A]0 when concentration is expressed relative to a stated unit or standard concentration. In routine school graphs, ln of numerical concentration values is used as shorthand. A direct [A] versus time plot curves downward. The rate constant k has units inverse time, such as s⁻¹, because kt must be dimensionless in the exponential.

For half-life, set [A]t/[A]0=1/2. Then ln(1/2)=−kt₁/₂ and t₁/₂=ln2/k≈0.693/k. The initial concentration cancels. Thus if a first-order reaction takes 10 min to fall from 0.80 M to 0.40 M, it takes another 10 min to fall from 0.40 M to 0.20 M, assuming the same k and conditions.

After n half-lives, [A]=[A]0/2ⁿ. This is a useful quick estimate, but a non-integer time requires the exponential or logarithmic equation. For example, after three half-lives, one eighth remains, not one third. “Constant fraction” should not be mistaken for “constant amount.” The amount removed in each successive half-life halves.

First-order decay appears in many unimolecular decompositions and radioactive decay, though nuclear decay is not a chemical reaction. A first-order rate law does not prove a one-step unimolecular mechanism in chemical kinetics; pseudo-first-order conditions or more complex mechanisms can produce the same observed law.

At equilibrium in a reversible reaction, reactant concentration need not decay to zero. A simple irreversible first-order equation would then be inadequate. Likewise, a catalyst change or temperature drift changes k, invalidating a single constant-k fit. Plotting and checking residuals helps test the model rather than assuming it.

Step-by-step reasoning

1. Confirm rate = k[A] over the data range. 2. Use ln([A]t/[A]0)=−kt or [A]t=[A]0e^(−kt). 3. For graph data, obtain k from the negative ln[A] slope. 4. For half-life, use 0.693/k. 5. Keep time units consistent with k and check model assumptions.

Visual explanation

Draw a curved [A] versus time graph with marked concentrations 1, 1/2, 1/4 and 1/8 at equal time gaps. Next to it draw a straight ln[A] versus time line descending with slope −k. The paired plots show why a log transform identifies first order.

Real-world analogy

If a fixed percentage of a remaining balance is removed every month, the absolute deduction shrinks while the fraction stays the same. First-order decay likewise removes a constant fraction in each equal interval.

Real-world example

Some chemical decompositions show a straight ln(concentration) versus time graph. A laboratory can estimate k from that slope and compare runs at different temperatures, provided the same mechanism and measurement calibration apply.

Why?

Why is first-order half-life independent of initial concentration? Rate grows in proportion to concentration, exactly compensating for the larger amount that must be removed when starting with more reactant.

Common misconception

“A constant half-life means the same molar amount disappears each interval.” It means the same fraction disappears; the absolute amount gets smaller after each half-life.

Worked example

A first-order reaction has k=0.0231 min⁻¹ and starts at 0.800 M. Half-life is 0.693/0.0231=30.0 min. After 60.0 min, two half-lives have passed, so [A]=0.800/4=0.200 M. The exponential calculation gives 0.800e^(−0.0231×60)≈0.200 M, consistent within rounding.

Quick check

1. Which plot is linear for ideal first-order disappearance? Answer: ln[A] against time, with slope −k.

Exam focus

Write both log and exponential forms, give k in inverse-time units and use constant half-life correctly. Distinguish fractional loss from constant absolute loss.

Advanced insight

For reversible A ⇌ B under suitable first-order forward and reverse laws, the approach of concentration to equilibrium can be exponential, but the exponent involves the sum of forward and reverse rate constants. A plateau requires a modified model rather than irreversible decay to zero.

Summary

First-order disappearance produces exponential concentration decay and a straight ln[A] plot. k has inverse-time units, and t₁/₂=0.693/k is independent of initial concentration when k remains constant.

Practice questions

1. If k=0.10 min⁻¹, what is half-life? Answer: 0.693/0.10=6.93 min. 2. What fraction remains after three half-lives? Answer: (1/2)³=1/8 of the starting amount. 3. Why does the direct concentration-time plot curve? Answer: The absolute disappearance rate falls as concentration falls, even though the fractional rate stays constant.