4.2 First-order integrated rate law

First-Order Reactions

Derivation of the First-Order Integrated Rate Law

The differential rate law for a first-order reaction states that the rate of disappearance of reactant A is directly proportional to its concentration:

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

where $[A]$ is the concentration of reactant A and $k$ is the rate constant. To get a useful equation that relates concentration to time, you need to integrate this expression. Here's how:

1. Separate variables by dividing both sides by $[A]$ and multiplying by $dt$: $\frac{d[A]}{[A]} = -k \, dt$

2. Integrate both sides. The left side runs from the initial concentration $[A]_0$ to $[A]$ at time $t$, and the right side runs from $0$ to $t$: $\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = \int_{0}^{t} -k \, dt$

3. Evaluate the integrals: $\ln[A] - \ln[A]_0 = -kt$

4. Rearrange to get the two standard forms of the first-order integrated rate law:

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

$\ln\frac{[A]}{[A]_0} = -kt$

The first form is especially useful because it has the shape of $y = mx + b$, which connects directly to graphical analysis.

Application of the First-Order Rate Law

Once you know $k$ and $[A]_0$, you can find the concentration of A at any time $t$. Start with:

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

Plug in your known values for $k$, $t$, and $[A]_0$, then solve for $\ln[A]$. To get $[A]$ itself, exponentiate both sides:

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

This exponential form makes it clear that first-order reactions show exponential decay. The concentration drops quickly at first, then tapers off as less reactant remains.

Example: Suppose $[A]_0 = 0.80 \, M$ and $k = 0.045 \, s^{-1}$. To find $[A]$ at $t = 10 \, s$:

$[A] = 0.80 \, e^{-(0.045)(10)} = 0.80 \, e^{-0.45} = 0.80 \times 0.638 = 0.51 \, M$

You can also rearrange to solve for $t$ if you know the starting and final concentrations.

Characteristics in ln(Concentration) vs. Time Plots

Because $\ln[A] = -kt + \ln[A]_0$ has the form $y = mx + b$, a plot of $\ln[A]$ versus time produces a straight line if the reaction is first-order.

• The slope of that line equals $-k$
• The y-intercept equals $\ln[A]_0$

This is the primary graphical test for first-order kinetics. If you plot your data and the $\ln[A]$ vs. time graph is linear, you've confirmed first-order behavior. If it curves, the reaction follows a different order, and you'd need to test second-order or zero-order plots instead.

Rate Constant Calculation from Data

To determine $k$ experimentally:

1. Collect concentration vs. time data for the reaction.
2. Calculate $\ln[A]$ for each data point.
3. Plot $\ln[A]$ on the y-axis versus time on the x-axis.
4. Check whether the plot is linear. If it is, the reaction is first-order.
5. Find the slope of the best-fit line. The rate constant is the negative of that slope:

$k = -\text{slope}$

You can calculate the slope from any two points on the line using $\frac{\Delta(\ln[A])}{\Delta t}$, though a linear regression across all data points gives a more reliable result.

Half-Life for First-Order Reactions

The half-life ($t_{1/2}$) is the time it takes for the reactant concentration to drop to half its current value. For first-order reactions, you can derive it by setting $[A] = \frac{[A]_0}{2}$ in the integrated rate law:

$\ln\frac{[A]_0 / 2}{[A]_0} = -kt_{1/2}$

$\ln\frac{1}{2} = -kt_{1/2}$

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

Notice that $[A]_0$ canceled out entirely. This means the half-life of a first-order reaction is constant regardless of how much reactant you start with. Whether you begin with 1.0 M or 0.01 M, it takes the same amount of time to lose half. This constant half-life is a signature feature of first-order kinetics and is the same principle behind radioactive decay.

You can also rearrange to find $k$ from a known half-life:

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

This relationship gives you a quick way to convert between $k$ and $t_{1/2}$ without needing concentration data at all.