Integrated Rate Law for a First-Order Reaction
Imagine you have a reaction where the rate depends only on the concentration of one reactant — double the concentration, double the rate. That’s a first-order reaction. The question is: if you start with some amount of A, how does its concentration actually decay over time? The differential rate law tells you the instantaneous speed, but the integrated rate law tells you the path the concentration follows.
The Intuition
Think of a radioactive sample. Every atom has the same fixed probability of decaying per second, regardless of how many atoms are left. So if you start with 1000 atoms, in the first second maybe 100 decay. In the next second, you have 900 left, so only 90 decay. The number decaying per second keeps dropping because there are fewer atoms to decay. The result is not a straight-line drop in concentration — it’s a curve that falls steeply at first, then flattens out. That curve is an exponential decay.
For a first-order chemical reaction, the same logic holds: the rate at any moment is proportional to how much reactant is still there. So the concentration doesn’t fall by equal amounts in equal time intervals; it falls by equal fractions in equal time intervals.
The Derivation (Short)
Start with the differential rate law for a first-order reaction:
−dtd[A]=k[A]
Rearrange so that all [A] terms are on one side and dt on the other:
[A]d[A]=−kdt
Integrate from time 0 (initial concentration [A]0) to time t (concentration [A]):
∫[A]0[A][A]d[A]=−k∫0tdt
The left side integrates to ln[A]−ln[A]0, and the right side gives −kt. So:
ln[A]−ln[A]0=−kt
Or equivalently:
ln[A]0[A]=−kt
Convert to base-10 logarithms (since lnx=2.303log10x):
log[A]0[A]=−2.303kt
Rearranging gives the form you usually see in exams:
k=t2.303log[A][A]0
The Linear Plot — The Key Insight
The equation ln[A]=ln[A]0−kt is of the form y=mx+c, where y=ln[A], m=−k, x=t, and c=ln[A]0. So if you plot ln[A] (or log[A]) against time, you get a straight line with slope −k (or −k/2.303 for base-10 logs).
This is the hallmark of a first-order reaction. If your experimental data gives a straight line when you plot log[reactant] vs. time, the reaction is first order. If it curves, it’s not.
You don’t need to know [A]0 to find k from the plot — just measure the slope. The intercept gives ln[A]0, which is a useful check.
What This Means Practically
- The half-life (t1/2) is constant for a first-order reaction. Set [A]=[A]0/2 in the integrated law: ln(1/2)=−kt1/2, so t1/2=kln2≈k0.693. It doesn’t depend on how much you started with. …