First-Order Reactions

A zero-order law means rate is insensitive to the specified concentration over a particular range at fixed conditions. It need not remain zero order when reactants become scarce or a surface is no longer saturated.

First order means the exponents in the empirical power-law rate law sum to one. A common single-reactant disappearance law is −d[A]/dt=k[A]. It does not mean that the balanced equation contains only one reactant; overall order is established from kinetic evidence.

$r = k[A]$

In a First-Order reaction, the units of k will be $s^{-1}$, or $1/s$. If you are ever given a $k$, then you would be able to tell the overall reaction order based on the units $k$ has. This doesn’t always mean you have enough information to use the rate equation, as you still need to know which reactant the reaction rate relies upon.

This means that as the concentration of the reactant decreases, it will progress more slowly. Many first-order reactions are decomposition reactions, where one compound decomposes into daughter compounds:

$A\rightarrow B+C$

A first-order disappearance model does not require products to escape. If the reverse reaction becomes important, use a reversible model rather than this simple irreversible law. For hydrogen peroxide, the observed law depends on catalyst, temperature and other conditions; the balanced decomposition equation alone cannot establish first order.

Now when you are working with a reaction that is in first-order, you should expect to see a relationship between concentration and time that is non-linear. Instead, you will have a reaction that is initially very quick, however as the concentration drops off, the reaction slows, meaning that the concentration will take longer to drop.

For the specified law −d[A]/dt=k[A], a plot of ln([A]/1 M) versus time is linear. The logarithm uses a dimensionless ratio. The slope is −k, and k has inverse-time units.

To figure out why this graph is linear, we will have to derive the Integrated First-Order Rate Equation. Once again, this involves calculus to perform, but it should solve our question of why we need to use the $\ln[A]$ vs $t$ graph to find a linear relationship.

Now we have two forms for our Integrated First-Order Rate Equation. The first form explains why we see a linear relationship develop when we graph $\ln[A]$ vs. $t$.

For y=ln([A]/1 M), the integrated line is y=ln([A]₀/1 M)−kt. Its slope is −k and its intercept is ln([A]₀/1 M), not the initial concentration itself.

One last thing before we move on is the Half-Life of a First-Order reaction.

Let’s derive the half-life equation, starting with the $ln[A]$ form of the First-Order rate equation:

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

$[A]=\frac12[A]_0$

$\ln\frac{[A]_0}2=ln[A]_0-kt$

$\ln[A]_0-\ln2=\ln[A]_0-kt$

$ln2=kt$

$t_\frac12=\frac{\ln2}k$

Perhaps surprisingly, this half-life equation is independent of the initial concentration, unlike the Zeroth-Order reaction’s half-life. This means that you can determine the half-life of a first order reaction by only knowing the $k$ of the reaction.

One great application is that a first order reaction has a constant half-life, meaning you can find the ‘quarter-life’ by merely doubling the half-life of the reaction. This is because you would start with a concentration of $[A]_0$, and after one half-life passes, you are down to $\frac{[A]_0}2$. Should you allow another half-life to pass, you’ll be at $\frac12\cdot\frac{[A]_0}2$, or $\frac{[A]_0}4$.

$t_{\frac1{2^n}}=n\cdot\frac{ln2}k$

Looking at a normalized graph, you can see this relationship in action. If you’re unfamiliar with normalization, that merely means that instead of looking at $[A]$, we are looking at $[A]/[A]_o$, meaning that if the y-axis was 1, then $[A]=[A]_0$.