Order rate equations

A chemist watching a reaction does not need to keep sampling it forever. Once the order is known, a single equation predicts the exact concentration left at any future instant — or how long a given amount will take to react — without a fresh experiment each time. That predictive machinery is the integrated rate equation. ISC Class 12 restricts this to two cases: zero order, where the rate never changes with concentration, and first order, where the rate falls in step with concentration. Both are derived from the differential rate law by straightforward calculus, and each carries its own signature: a straight-line graph, a distinct unit for kk, and a distinct behaviour of half-life. Recognising which signature a data set shows is exactly what board numericals test.

The core insight

Integrating the differential rate law over concentration and time turns a rate expression into a formula for concentration-at-time-tt, giving [A]=[A]0−kt[A]=[A]_0-kt for zero order and k=2.303tlog⁡[A]0[A]k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]} for first order, each with its own units of kk and its own half-life behaviour.

The core idea

Why integration is needed. The differential rate law only gives the instantaneous rate at one concentration; it says nothing about how long a reaction takes to reach a given extent. Integrating that rate law with respect to time converts "rate now" into "concentration at any later time" — the whole point of the integrated equations.

Zero-order case. A zero-order reaction has Rate=k[A]0=k\text{Rate}=k[A]^0=k, a constant that does not depend on how much AA is left. Writing Rate=−d[A]dt\text{Rate}=-\dfrac{d[A]}{dt} and separating variables:

−d[A]dt=k  ⟹  −d[A]=k dt-\frac{d[A]}{dt}=k \implies -d[A]=k\,dt

where: [A][A] is the concentration of reactant AA at time tt, and kk is the rate constant. Integrating from [A]0[A]_0 at t=0t=0 to [A][A] at time tt gives the working formula:

[A]=[A]0−kt[A]=[A]_0-kt

A graph of [A][A] against tt is a straight line falling with slope −k-k, and it genuinely reaches zero at a finite time. Units of kk: mol L−1s−1\text{mol L}^{-1}\text{s}^{-1} — identical to the units of rate itself, because [A]0=1[A]^0=1 carries no concentration units.

First-order case. A first-order reaction has Rate=k[A]\text{Rate}=k[A], so:

−d[A]dt=k[A]  ⟹  −d[A][A]=k dt-\frac{d[A]}{dt}=k[A] \implies -\frac{d[A]}{[A]}=k\,dt

Integrating from [A]0[A]_0 at t=0t=0 to [A][A] at time tt gives:

k=1tln⁡[A]0[A]=2.303tlog⁡10[A]0[A]k=\frac{1}{t}\ln\frac{[A]_0}{[A]}=\frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]}

where: log⁡10\log_{10} is the base-10 logarithm (what calculators give by default), and the factor 2.3032.303 converts it to the natural log ln⁡\ln used in the calculus step (ln⁡x=2.303log⁡10x\ln x=2.303\log_{10}x). Units of kk: s−1\text{s}^{-1} (or min−1\text{min}^{-1}, generally time−1\text{time}^{-1}) — notice kk carries no concentration units at all, since [A]0/[A][A]_0/[A] is a dimensionless ratio.

Reading the graphs. For zero order, [A][A] versus tt is linear. For first order, [A][A] versus tt curves (exponential decay), but log⁡[A]\log[A] versus tt is the straight line, with slope −k/2.303-k/2.303. This graphical test is the fastest way to identify an unknown order from data without doing any arithmetic first.

Figure
Two rate order graphs side by side, a straight declining line for zero order concentration against time reaching zero, and a curved exponential decay for first order concentration against time that only approaches zero
Zero order versus first order — how concentration falls with time Zero order: [A] = [A]0 minus kt time t [A] [A]0 reaches zero at finite t First order: [A] = [A]0 e to the minus kt time t [A] [A]0 approaches zero, never quite reaches it

Half-life. Half-life t1/2t_{1/2} is the time for [A][A] to fall to [A]0/2[A]_0/2. Substituting into the zero-order equation gives t1/2=[A]02kt_{1/2}=\dfrac{[A]_0}{2k} — proportional to the starting concentration. Substituting into the first-order equation gives kt1/2=2.303log⁡2=0.693kt_{1/2}=2.303\log 2=0.693, so:

t1/2=0.693kt_{1/2}=\frac{0.693}{k}

where: t1/2t_{1/2} depends only on kk, not on [A]0[A]_0 — this concentration-independence is the fastest experimental fingerprint of a first-order reaction, and it is why radioactive decay (always first order) has a fixed half-life regardless of how much sample you start with.

Derivation AThe first-order integrated rate equation

The goal of this derivation is to turn the differential statement "the rate of disappearance of AA is proportional to how much AA is currently present" into an explicit algebraic formula linking the rate constant kk to the initial and final concentrations and the elapsed time, using only separation of variables and a definite integral — nothing beyond what is already used for the zero-order case, but with a 1/[A]1/[A] term that changes the outcome from a straight line to a logarithm.

Step 1. Write the differential rate law for a first-order reaction: Rate=−d[A]dt=k[A]\text{Rate}=-\dfrac{d[A]}{dt}=k[A]. The negative sign reflects that [A][A] is decreasing while kk and [A][A] are both positive.

Step 2. Separate the variables so that everything involving concentration sits on one side and everything involving time sits on the other: −d[A][A]=k dt-\dfrac{d[A]}{[A]}=k\,dt. This step is only valid because [A][A] appears to the first power — it is precisely what makes the first-order case different from zero order, where no such division was needed.

Step 3. Integrate both sides as a definite integral, the left from [A]0[A]_0 (at t=0t=0) to [A][A] (at time tt), the right from 00 to tt: −∫[A]0[A]d[A][A]=k∫0tdt-\displaystyle\int_{[A]_0}^{[A]}\frac{d[A]}{[A]}=k\int_0^t dt, which gives −(ln⁡[A]−ln⁡[A]0)=kt-\big(\ln[A]-\ln[A]_0\big)=kt, i.e. ln⁡[A]0[A]=kt\ln\dfrac{[A]_0}{[A]}=kt.

Step 4. Convert the natural log to base-10 log for practical calculator use, using ln⁡x=2.303log⁡10x\ln x=2.303\log_{10}x, and rearrange to isolate kk: k=1tln⁡[A]0[A]=2.303tlog⁡10[A]0[A]k=\dfrac{1}{t}\ln\dfrac{[A]_0}{[A]}=\dfrac{2.303}{t}\log_{10}\dfrac{[A]_0}{[A]}. This is the working first-order formula: given any two of kk, tt, and the concentration ratio, the third can be found directly.

Worked example

Question. The decomposition of a certain first-order reaction has a rate constant k=4.606×10−2 min−1k=4.606\times10^{-2}\ \text{min}^{-1}. If the initial concentration of reactant is 0.80 mol L−10.80\ \text{mol L}^{-1}, find (a) the concentration remaining after 20 minutes, and (b) the time needed for the concentration to fall to 0.10 mol L−10.10\ \text{mol L}^{-1}.

Step 1. Rearrange the first-order equation for the log ratio: log⁡[A]0[A]=kt2.303\log\dfrac{[A]_0}{[A]}=\dfrac{kt}{2.303}.

Step 2. For part (a), substitute t=20 mint=20\ \text{min}: log⁡[A]0[A]=(4.606×10−2)(20)2.303=0.92122.303=0.400\log\dfrac{[A]_0}{[A]}=\dfrac{(4.606\times10^{-2})(20)}{2.303}=\dfrac{0.9212}{2.303}=0.400.

Step 3. Undo the log: [A]0[A]=100.400=2.512\dfrac{[A]_0}{[A]}=10^{0.400}=2.512, so [A]=0.802.512=0.318 mol L−1[A]=\dfrac{0.80}{2.512}=0.318\ \text{mol L}^{-1}.

Step 4. For part (b), substitute [A]=0.10 mol L−1[A]=0.10\ \text{mol L}^{-1}: log⁡0.800.10=log⁡8=0.903\log\dfrac{0.80}{0.10}=\log 8=0.903, so t=2.303×0.9034.606×10−2=2.0804.606×10−2=45.2 mint=\dfrac{2.303\times0.903}{4.606\times10^{-2}}=\dfrac{2.080}{4.606\times10^{-2}}=45.2\ \text{min}.

Final answer. After 20 minutes, about 0.318 mol L⁻¹ of reactant remains, and it takes about 45.2 minutes for the concentration to drop to 0.10 mol L⁻¹ — consistent with a first-order reaction whose concentration only ever approaches zero, never exactly reaching it in finite time.

Competitive stretch

Cross-check with half-life. For this reaction, t1/2=0.693k=0.6934.606×10−2=15.05 mint_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{4.606\times10^{-2}}=15.05\ \text{min}. Starting from 0.80 mol L−10.80\ \text{mol L}^{-1}, after two half-lives (30.1 min30.1\ \text{min}) the concentration should be 0.80/4=0.20 mol L−10.80/4=0.20\ \text{mol L}^{-1}, and after roughly three half-lives (45.2 min45.2\ \text{min}) it should be near 0.80/8=0.10 mol L−10.80/8=0.10\ \text{mol L}^{-1} — matching Step 4 exactly, since 45.2 min45.2\ \text{min} is almost precisely 3×15.05 min3\times15.05\ \text{min}. This is a fast sanity check for any first-order numerical: convert the target ratio to a power of 22 and compare against whole multiples of t1/2t_{1/2}.

Common traps

  • Mixing up ln⁡\ln and log⁡10\log_{10} without the conversion factor 2.3032.303 — always write ln⁡x=2.303log⁡10x\ln x=2.303\log_{10}x explicitly rather than trusting memory.
  • Applying the first-order log formula to a zero-order problem, or the zero-order linear formula to a first-order problem — check the stated order before choosing an equation.
  • Assuming [A][A] reaches exactly zero for a first-order reaction; only zero-order concentration reaches zero in finite time, first-order concentration only approaches it.
  • Writing first-order kk with concentration units, or zero-order kk without them — zero order is mol L−1time−1\text{mol L}^{-1}\text{time}^{-1}, first order is time−1\text{time}^{-1} only.

Ask the tutor: if picking between the zero-order and first-order formulas, or converting between log and ln, is slowing you down, ask your AI tutor to generate a fresh numerical and walk through which equation applies and why.

Why does zero-order concentration reach exactly zero but first-order concentration never does, mathematically?

The zero-order equation [A]=[A]0−kt[A]=[A]_0-kt is linear, so it crosses zero at the finite time t=[A]0/kt=[A]_0/k and the reaction is genuinely over. The first-order equation is exponential, [A]=[A]0e−kt[A]=[A]_0e^{-kt}, and an exponential only ever gets arbitrarily small — it approaches zero as tt grows but never reaches it for any finite tt. In practice, the reactant becomes so dilute after several half-lives that it is treated as "finished," but the pure mathematics keeps it strictly positive.

How do I tell from raw experimental data whether a reaction is zero or first order, without knowing k in advance?

Plot the data two ways. If concentration against time gives a straight line, the reaction is zero order and the slope is −k-k. If concentration against time curves but log of concentration against time gives a straight line, the reaction is first order, with slope −k/2.303-k/2.303. A quicker check once you suspect first order is to see whether the half-life stays the same as the reaction proceeds — constant half-life across different starting points is the signature of first order, since it does not depend on [A]0[A]_0.

Why is the number 2.303 always showing up, and can I just skip it?

It converts natural logarithm to base-ten logarithm, since ln⁡x=2.303log⁡10x\ln x=2.303\log_{10}x. The calculus derivation naturally produces ln⁡\ln, but calculators and log tables in exams typically use log⁡10\log_{10}, so the conversion factor is unavoidable if you want a numerical answer in base-ten log form. Skipping it, or using it inconsistently between two steps of the same problem, is one of the most common arithmetic slips in kinetics numericals — always check whether a given kk or a computed log value already assumes base-ten or natural log before combining terms.

Before this

Where this leads