Question

Provide a conceptual rationale for the differences in the half-lives of zero-, first-, and second-order reactions. Explain.

Short answer

Differences in half-life expressions are showing why half-life expectations are different depending on the order of the reaction.

Step-by-step solution

Understanding the concept of Half-life.

The half-life of a reaction is the amount of time needed for the concentration of a reactant to decrease by half from the initial one. Here we have a zero-order kinetic rate law, with [A] being the current concentration, initial concentration, ‘k’ reaction rate constant, and time ‘t’.

$\left[\text{A}\right]={\left[\text{A}\right]}_0-kt$

Understanding the concept of half-life for zero-order kinetics.

The half-life for zero-order kinetics, therefore, depends on both initial concentration and reaction rate constant:

$t_{1/2}=\frac{{\left[\text{A}\right]}_0}{2\text{k}}$

Understanding the concept of half-life for first-order kinetics.

When we take the first-order reaction into consideration and substitute [A] with ${\left[\text{A}\right]}_0$what we can see is that for the first-order reaction, the half-life depends solely on the reaction rate constant.

role="math" localid="1663752974925" $\begin{array}{rcl}\left[\text{A}\right]& =& {\left[\text{A}\right]}_0{e}^{-kt}\\ t_{1/2}& =& \frac{\ln 2}{k}\end{array}$

Understanding the concept of half-life for second-order kinetics.

When we then look at the second-order reaction and isolate time, we can see that half-life in this case, just like in the case of zero-order reaction, depends on both initial concentration and reaction rate constant, but in a different way.

$\begin{array}{rcl}\left[\text{A}\right]& =& {\left[\text{A}\right]}_0{e}^{-kt}\\ \frac{1}{\left[\text{A}\right]}& =& kt+\frac{1}{{\left[\text{A}\right]}_0}\\ t_{1/2}& =& \frac{1}{k{\left[\text{A}\right]}_0}\end{array}$