Integrated Rate Law: First Order

Radioactive decay, drug metabolism, and the decomposition of many gas-phase molecules share one mathematical fingerprint: an exponential decline in concentration over time. If a reaction is first order, the rate depends on only one concentration raised to the first power. This single fact forces the concentration to fall exponentially and forces the half-life to stay constant, no matter how much material you start with. This regularity is not a coincidence. It follows directly from solving a simple differential equation, and it gives chemists a powerful diagnostic. If a plot of $\ln[A]$ against time gives a straight line, the reaction is first order, and the slope hands you the rate constant immediately. This article derives the first-order integrated rate law from the differential rate law. It shows why the half-life is constant. It works through two full numerical examples. It also flags the common errors students make when they mix up first-order and other-order behavior.

Key takeaways

A first-order reaction obeys $-d[A]/dt = k[A]$ , giving exponential decay $[A]_t = [A]_0 e^{-kt}$ .
Taking logarithms linearizes the law: $\ln[A]_t = \ln[A]_0 - kt$ , a straight line of slope $-k$ .
The half-life of a first-order reaction is $t_{1/2} = \ln 2 / k$ , independent of the starting concentration.
Units of $k$ for a first-order process are inverse time (s⁻¹, min⁻¹, yr⁻¹, etc.).
A straight $\ln[A]$ vs. $t$ plot is the standard experimental test for first-order kinetics.
Key idea: first-order kinetics means the fractional loss of reactant per unit time is constant, which mathematically forces exponential decay and a concentration-independent half-life.

The core insight

Because the rate of a first-order reaction is directly proportional to the single remaining concentration, integrating the rate law produces an exponential decay curve whose logarithm is perfectly linear in time and whose half-life never changes, however much reactant you start with.

The physical picture

Consider a reaction A → products where experiments show the rate depends only on $[A]$ to the first power:

-\frac{d[A]}{dt} = k[A]

This differential statement says something intuitive: the instantaneous rate of disappearance is proportional to however much A is currently present. Halve the concentration, and the rate immediately halves too. This is fundamentally different from, say, a zero-order process (constant rate, a straight-line concentration decay) or a second-order process (rate falls off much faster at first, then levels out more gradually). Because the rate law is a differential equation, not yet an explicit function of time, chemists integrate it to get a formula for $[A]$ at any moment. The result — worked out in full below — is an exponential decay curve. On a normal concentration-vs-time plot this curve looks like a smooth downward curve that never quite reaches zero. On a semi-log plot (natural log of concentration vs. time) that same curve becomes a perfectly straight line. This transformation is the workhorse diagnostic of first-order kinetics: chemists routinely plot $\ln[A]$ against $t$ and check for linearity before trusting a rate constant. A second essential feature falls straight out of the exponential form: the time required for the concentration to fall to half its current value is always the same number, regardless of what that current value is. Whether you start at 1.0 M or 0.001 M, the first-order half-life clock ticks at the identical rate. This constant half-life is the single most distinctive experimental signature separating first-order from all other simple orders, and it is explored in depth in the follow-on article on half-life order-dependence.

Figure

Concentration decay and its logarithmic linearization for a first-order reaction with k = 0.10 s⁻¹

The core relations

-\frac{d[A]}{dt} = k[A]

where: $k$ is the first-order rate constant (units s⁻¹), $[A]$ is the reactant concentration at time $t$ .

[A]_t = [A]_0 \, e^{-kt}

where: $[A]_0$ is the initial concentration at $t = 0$ , $[A]_t$ is the concentration after elapsed time $t$ .

\ln[A]_t = \ln[A]_0 - kt

where: this is the linearized form; a plot of $\ln[A]_t$ against $t$ is a straight line of slope $-k$ and intercept $\ln[A]_0$ .

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

where: $t_{1/2}$ is the half-life, the time for $[A]$ to fall to half its current value, constant for a given $k$ .

Derivation AFrom the differential rate law to the ln-linear plot

Goal: integrate the first-order differential rate law to obtain an explicit expression for $[A]_t$ , then linearize it and extract the half-life.

Step 1. Separate the variables. Starting from the rate law,

-\frac{d[A]}{dt} = k[A]

rearrange so that every term involving $[A]$ sits on one side and every term involving $t$ sits on the other:

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

This step is legitimate because $k$ is a constant (temperature fixed) and $[A]$ appears only through its own differential — a classic separable first-order ordinary differential equation.

Step 2. Integrate both sides. Integrate the left side from $[A]_0$ at $t=0$ to $[A]_t$ at time $t$ , and the right side from $0$ to $t$ :

\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt

The left integral is a standard logarithmic form; the right integral is trivial. Carrying out both:

\ln[A]_t - \ln[A]_0 = -kt

Step 3. Rearrange into the linear (ln) form. Adding $\ln[A]_0$ to both sides gives the linearized integrated rate law:

\ln[A]_t = \ln[A]_0 - kt

This is the equation of a straight line, $y = mx + b$ , with $y = \ln[A]_t$ , $x = t$ , slope $m = -k$ , and intercept $b = \ln[A]_0$ . This immediately explains why plotting $\ln[A]$ against $t$ is the standard first-order diagnostic: linearity confirms the order, and the slope delivers $k$ without any further fitting.

Step 4. Exponentiate to recover the concentration-time law. Exponentiating both sides of Step 3 undoes the logarithm:

[A]_t = [A]_0\, e^{-kt}

This is the explicit exponential decay law. Because $e^{-kt}$ is always positive, $[A]_t$ never reaches exactly zero in finite time — it only asymptotically approaches it, matching the physical expectation that reaction rate slows as reactant is depleted but a tiny amount always remains in principle.

Step 5. Derive the half-life from the exponential law. By definition, the half-life $t_{1/2}$ is the time at which $[A]_t = \tfrac{1}{2}[A]_0$ . Substitute into Step 4:

\frac{1}{2}[A]_0 = [A]_0\, e^{-k t_{1/2}}

Cancel $[A]_0$ (valid since it is nonzero) and take the natural log of both sides:

\ln\!\left(\frac{1}{2}\right) = -k\, t_{1/2}

Since $\ln(1/2) = -\ln 2$ , this becomes:

t_{1/2} = \frac{\ln 2}{k}

Notice that $[A]_0$ has completely disappeared from the final expression. This is the algebraic origin of the constant half-life: because the original differential equation is linear in $[A]$ , halving the starting concentration simply rescales the whole exponential curve without changing how long it takes to halve again.

\boxed{t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}, \quad \text{independent of } [A]_0}

Worked examples

Example 1 — Foundational: finding $k$ and predicting a later concentration. A first-order gas-phase decomposition A → products is monitored at constant temperature. At $t = 0$ , $[A]_0 = 0.500$ M. After $t = 120$ s, $[A] = 0.220$ M. Step 1: Find $k$ using the ln-linear form.

\ln[A]_t = \ln[A]_0 - kt \implies k = \frac{\ln[A]_0 - \ln[A]_t}{t}

Compute the logarithms:

\ln(0.500) = -0.6931, \qquad \ln(0.220) = -1.5141

k = \frac{(-0.6931) - (-1.5141)}{120\ \text{s}} = \frac{0.8210}{120\ \text{s}} = 6.842\times10^{-3}\ \text{s}^{-1}

Step 2: Sanity-check the units. Since first order requires $k$ in s⁻¹, and we divided a dimensionless log ratio by seconds, the units check out. Step 3: Predict $[A]$ at $t = 300$ s using the exponential form.

[A]_{300} = [A]_0\, e^{-kt} = 0.500\ \text{M} \times e^{-(6.842\times10^{-3}\,\text{s}^{-1})(300\,\text{s})}

= 0.500\ \text{M} \times e^{-2.053} = 0.500\ \text{M} \times 0.1284 = 0.0642\ \text{M}

Step 4: Find the half-life.

t_{1/2} = \frac{0.693}{6.842\times10^{-3}\ \text{s}^{-1}} = 101.3\ \text{s}

Sanity check: after two half-lives (202.6 s) concentration should be $0.500/4 = 0.125$ M. Using the exponential law at $t=202.6$ s: $0.500 \times e^{-(6.842\times10^{-3})(202.6)} = 0.500 \times e^{-1.386} = 0.500\times 0.2500 = 0.125$ M. Consistent.

Example 2 — Applied: pharmacokinetic dosing interval (harder, multi-step). A drug is cleared from the bloodstream by a first-order process with half-life $t_{1/2} = 6.0$ h. A patient receives an initial dose producing a plasma concentration of $C_0 = 8.0\ \mu\text{g/mL}$ . The drug is considered sub-therapeutic (ineffective) once concentration drops below $1.0\ \mu\text{g/mL}$ . A physician wants to know the maximum interval between doses to keep the patient always above the threshold, and also wants the effective rate constant. Step 1: Convert half-life to rate constant.

k = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{6.0\ \text{h}} = 0.1155\ \text{h}^{-1}

Step 2: Use the integrated law to find the time to fall from $8.0\ \mu\text{g/mL}$ to $1.0\ \mu\text{g/mL}$ .

\ln[A]_t = \ln[A]_0 - kt \implies t = \frac{\ln[A]_0 - \ln[A]_t}{k}

\ln(8.0) = 2.0794, \qquad \ln(1.0) = 0

t = \frac{2.0794 - 0}{0.1155\ \text{h}^{-1}} = 18.0\ \text{h}

Step 3: Cross-check using the half-life directly, since $8.0 \to 4.0 \to 2.0 \to 1.0\ \mu\text{g/mL}$ is exactly three half-lives:

3 \times t_{1/2} = 3 \times 6.0\ \text{h} = 18.0\ \text{h}

Both methods agree exactly, confirming internal consistency (the concentration ratio $8.0/1.0 = 8 = 2^3$ makes the half-life shortcut exact here). Step 4: Interpret clinically. The physician should schedule doses no more than every 18 hours to guarantee the plasma level never falls below the therapeutic floor before the next dose, assuming each new dose restores concentration to $8.0\ \mu\text{g/mL}$ and clearance restarts from that point (a simplification; real pharmacokinetics often adds absorption and multi-compartment effects, but the elimination phase alone is well described by this first-order law).

Limiting cases & common traps

Situation	What actually happens	Common mistake
Very long times	$[A]_t \to 0$ but never reaches exactly zero mathematically	Assuming the reaction "finishes" at a specific finite time
Comparing to zero order	Zero order gives a straight $[A]$ vs. $t$ line, not $\ln[A]$ vs. $t$	Plotting $[A]$ vs. $t$ and calling any straight-looking curve first order
Comparing to second order	Second order linearizes as $1/[A]$ vs. $t$ , and its half-life depends on $[A]_0$	Using $t_{1/2} = 0.693/k$ for a reaction that is not actually first order
Units of $k$	Always inverse time (s⁻¹, min⁻¹) for first order only	Quoting $k$ with concentration units, which signals order was misassigned
Multiple half-lives	Fraction remaining after $n$ half-lives is $(1/2)^n$	Forgetting that half-lives multiply fractionally rather than subtracting linearly

The chemist behind it

The mathematical form of first-order kinetics — exponential decay governed by a proportionality constant — was first rigorously connected to chemical transformation through the work on inversion of cane sugar and, more broadly, through the systematic kinetic studies pioneered in the late nineteenth century. The formalism was cemented by chemists studying radioactive decay and unimolecular gas reactions, where the same exponential law appears with total universality. Louis Jacques Thénard and Jacobus van 't Hoff both contributed early quantitative treatments of reaction rates, but it was van 't Hoff's systematic classification of reaction order — building directly on the differential rate-law framework — that gave chemists the language to recognize first-order behavior as a distinct, universal kinetic class rather than a coincidence of one particular reaction.

Origins · 1 minJacobus Henricus van 't Hoff and how this was discovered

Jacobus Henricus van 't Hoff

1852–1911 · Dutch physical chemist who systematized reaction order and rate laws, later the first Nobel laureate in Chemistry (1901)

:::quiz[If [A]0 = 2.0 M and the reaction is first order, what is [A] after exactly two half-lives?]

1.0 M
0.50 M ✓
0.25 M
1.5 M

:::faq[Does a straight ln[A] vs t plot prove the reaction is first order?] It provides strong supporting evidence but is not an absolute proof in isolation; a well-designed kinetics study checks that this linearity persists across multiple half-lives and multiple starting concentrations, and often cross-validates with other diagnostics such as constant half-life or the method of initial rates, covered in the article on determining reaction order.

Ask the tutor: How does the constant half-life of first order break down for second-order or zero-order reactions, and how do their integrated laws differ?

Why is the rate constant k for a first-order reaction always in units of inverse time?

Because the rate law is $-d[A]/dt = k[A]$ , and rate always carries units of concentration per time (M/s, for example). Since $[A]$ on the right side already carries units of concentration, $k$ must supply only the inverse-time part to make the units balance. This is a quick, reliable check: if you compute a rate constant and it comes out with concentration units still attached, the reaction is not actually first order and you should recheck the order before trusting the number.

How is the first-order integrated law different from radioactive decay equations?

It is not different at all — radioactive decay is the textbook example of first-order kinetics, just with the rate constant renamed the decay constant $\lambda$ and the half-life computed the identical way, $t_{1/2} = \ln 2/\lambda$ . Nuclear decay is first order because each nucleus decays independently of all others, exactly mirroring a unimolecular chemical reaction where each molecule reacts independently, giving rate proportional only to how many are present.

Can I use the half-life shortcut instead of the full integrated law?

Yes, whenever the elapsed time is a whole-number multiple of the half-life, because the fraction remaining after $n$ half-lives is exactly $(1/2)^n$ . For any other elapsed time you need the full integrated law, $\ln[A]_t = \ln[A]_0 - kt$ , or equivalently $[A]_t = [A]_0 e^{-kt}$ , since the half-life relation alone only pins down concentration at discrete time points.

What happens to the first-order law if the reaction is reversible?

Once a reverse reaction becomes significant, the simple exponential decay to zero breaks down because the system approaches a nonzero equilibrium concentration rather than complete depletion. The integrated law is modified to decay exponentially toward that equilibrium value instead of toward zero, a case treated fully in the article on reversible, opposing reactions.