First-order Integrated Rate Law: Equations And Examples

Chemical kinetics measures how fast reactants turn into products, but knowing the speed isn't enough if you can't predict concentrations over time. The first-order integrated rate law connects reactant concentration, rate constants, and elapsed time into a single, straightforward equation you can solve algebraically.

In a first-order reaction, the rate depends linearly on the concentration of a single reactant. If you double the reactant concentration, the reaction rate doubles right along with it. While the differential rate law tells you how fast the reaction goes at any exact micro-second, the integrated version lets you calculate how much reactant remains after minutes, hours, or days.

Deriving the First-Order Rate Equation

To track concentration over time, chemists use calculus to integrate the differential rate law. For a generic reactant $A$ turning into products, the differential rate law is written as:

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

Here, $[A]$ represents the molar concentration of the reactant, $t$ stands for time, and $k$ is the first-order rate constant with units of inverse time ($s^{-1}$ or $min^{-1}$).

Rearranging this expression to group the concentration terms on one side and time on the other gives:

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

Integrating both sides from time zero (where concentration is $[A]_0$) to a later time $t$ (where concentration is $[A]_t$) yields the natural logarithm form of the integrated rate law:

$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$

You can also write this equation by separating the logarithmic terms:

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

Notice how this matches the standard linear equation format $y = mx + b$. If you plot $\ln[A]_t$ on the y-axis against time $t$ on the x-axis, you get a straight line where the slope equals $-k$ and the y-intercept is $\ln[A]_0$. This graphical test is one of the easiest ways to prove experimentally whether an unknown reaction follows first-order kinetics.

Exponential Decay and Concentration Profiles

Because natural logarithms can be cumbersome for quick mental checks, chemists often convert the logarithmic equation into an exponential form by taking the inverse natural log of both sides:

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

This formula reveals that reactant concentration decays exponentially. Unlike zero-order reactions, which consume constant amounts of reactant per unit of time, a first-order reaction consumes a larger absolute amount early on when concentration is high, and progressively smaller amounts as the reactant depletes. The reaction never technically reaches absolute zero concentration in finite time, though it approaches it asymptotically.

Calculating Half-Life for First-Order Reactions

The half-life ($t_{1/2}$) of a reaction is the time required for the concentration of a reactant to drop to exactly half of its initial value. For first-order processes, half-life has a unique and powerful property: it doesn't depend on the starting concentration at all.

To find the half-life formula, substitute $[A]_t = \frac{1}{2}[A]0$ and $t = t{1/2}$ into the integrated rate law:

$\ln\left(\frac{\frac{1}{2}[A]_0}{[A]0}\right) = -k , t{1/2}$

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

Since $\ln(0.693) \approx -0.693$, the negative signs cancel out, leaving:

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

This independence makes first-order half-lives exceptionally useful. Whether you start with a 1.0 M solution or a 0.001 M solution, it will take the exact same amount of time for the concentration to cut in half. Radioactive decay, pharmaceutical drug clearance in the human body, and many thermal decomposition reactions rely on this constant half-life property.

Step-by-Key-Step Problem Solving Example

Let's walk through a practical calculation to see how these equations work in practice.

Imagine a thermal decomposition reaction that follows first-order kinetics with a rate constant $k = 3.46 \times 10^{-3} , s^{-1}$. If your initial reactant concentration is $0.500 , M$, what concentration remains after 120 seconds?

1. Identify the known variables: $[A]_0 = 0.500 , M$ $k = 0.00346 , s^{-1}$ $t = 120 , s$

2. Select the correct integrated rate law: $\ln[A]_t = -kt + \ln[A]_0$

3. Substitute the known values into the equation: $\ln[A]_t = -(0.00346 , s^{-1})(120 , s) + \ln(0.500)$

4. Calculate the numerical values: $-0.00346 \times 120 = -0.4152$ $\ln(0.500) = -0.6931$ $\ln[A]_t = -0.4152 - 0.6931 = -1.1083$

5. Exponentiate to solve for $[A]_t$: $[A]_t = e^{-1.1083} \approx 0.330 , M$

After two minutes, about $0.330 , M$ of the original reactant remains unreacted.

Common Pitfalls and Experimental Mistakes

Students and laboratory researchers often make a few avoidable errors when working with first-order integrated rate laws:

• Unit mismatches: Ensure your time units for $k$ match your time units for $t$. If $k$ is given in $min^{-1}$, do not plug seconds into your equation without converting first.
• Confusing reaction orders: Do not use the natural log equation for zero-order or second-order reactions. Zero-order reactions use linear concentration drops, while second-order reactions require inverse concentrations ($\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$).
• Forgetting temperature dependence: Rate constants are only constant at a fixed temperature. Changing the temperature alters $k$ according to the Arrhenius equation.

Frequently Asked Questions

How do I know if a reaction is first-order?

You can determine reaction order by testing experimental data against different integrated rate law equations. If plotting $\ln[A]$ versus time yields a straight line with a negative slope, the reaction is first-order with respect to that reactant.

Can a first-order rate constant be negative?

No. Rate constants ($k$) are always positive values. If your calculation yields a negative rate constant, check your logarithmic signs and make sure you subtracted initial and final concentrations in the correct order.

Why is radioactive decay always first-order?

Radioactive decay is a probabilistic nuclear event where individual unstable nuclei disintegrate independently of their neighbors or external physical conditions like temperature and pressure. Because the rate of decay depends strictly on the number of radioactive atoms present at any given moment, it inherently follows first-order kinetics.

How does temperature affect the first-order rate constant?

Raising the temperature increases kinetic energy, causing molecules to collide more frequently and with greater energy. This drastically increases the rate constant $k$, which in turn speeds up the reaction and shortens the half-life.