All first order reactions have a constant half life, t½, which means that the time taken for the concentration of the reactant to fall to half its initial value is constant and it is independent of the initial concentration.
We can determine the half life for a first order reaction from a graph of concentration against time. The decomposition of dinitrogen pentoxide, N2O5, to form dinitrogen oxide and oxygen is first order:
rate = k [N2O5]
The concentration of the reactant decays exponentially with time (this is a common feature of all first order reactions). We can calculate the half life by finding the time taken for the concentration of N2O5 to half from 2.0 mol dm-3 to 1.0 mol dm-3 (shown on the graph below by the orange lines). A second half life is calculated by finding the time taken for the concentration of N2O5 to half from 1.0 mol dm-3 to 0.5 mol dm-3 (shown on the graph below by the green lines). The values are the same because the half life is constant.
We could also use this graph to find a value for k, the rate constant. The first step is to find the rate of the reaction at a specific concentration by drawing a tangent to the curve at that concentration.
The second step is to use these values to find k, the rate constant. The tangent was drawn at the point on the curve at which the concentration of N2O5 was 1.27 mol dm-3 so this is the value for concentration that we use in the equation:
However, this is a fairly crude method of determining k since drawing an accurate tangent on a curved line is notoriously difficult!
The good news is that there is another way 😊. The bad news its that this other way requires us to use the integrated rate equation for a first order reaction 😖.
I am going to run through the maths for deriving the equation for finding the value for k, given that we know the half life for the reaction, because I’ve noticed that the text books simply pull this equation out of thin air and expect you to learn and use it (which is never an educationally sound method of teaching anything!).
A basic understanding of natural logarithms and the laws of logarithms can be found here and the maths behind deriving an integrated rate equation for a first order reaction can be found here. Both posts are written for students not doing Maths A level, so don’t be scared!
The integrated rate equation for a first order reaction looks like so:
Since one half life is the time taken for the concentration of N2O5 to halve
Let’s substitute that into the integrated rate equation and rearrange for k
Now, in text books this equation is often written as
The reason being that …
So, to finish …
There is also a third way to find the value of k, the rate constant … we can rearrange the integrated rate equation into the form y = m𝑥 + c and so linearise our data. The result will be a straight line graph of ln [N2O5] against time, and the gradient will be -k.
The advantage of this method is that we don’t need to know the half life for the reaction (we can find k and then use the method above to calculate the half life).
Practice question
The decarboxylation of 2,4,6-trinitrobenzoic acid is a first order reaction.
ln [C7H3O8N3]t = – kt + ln [C7H3O8N3]0
The decrease in the concentration of 2,4,6-trinitrobenzoic acid was followed over time and the results are shown below:
| Time / mins | [C7H3O8N3] x 10-4 / mol dm-3 |
|---|---|
| 0 | 2.78 |
| 20 | 2.29 |
| 30 | 2.07 |
| 55 | 1.59 |
| 80 | 1.25 |
| 160 | 0.56 |
Find a value for the rate constant, k, and hence determine the half life for this reaction.
Answer
k is the gradient of the graph of ln [C7H3O8N3]t against time. Once we know k, we can use the equation t½ = 0.693 / k to calculate the half life.
| Time / mins | [C7H3O8N3] x 10-4 / mol dm-3 | ln [C7H3O8N3] |
|---|---|---|
| 0 | 2.78 | -8.18 |
| 20 | 2.29 | -8.38 |
| 30 | 2.07 | -8.48 |
| 55 | 1.59 | -8.75 |
| 80 | 1.25 | -8.99 |
| 160 | 0.56 | -9.79 |
Remember to plug in the entire number (2.78 x 10-4) when plugging the concentrations into your calculator to find their natural logarithms. Exam questions are sneaky like that!
t½ = 0.693 / k = 0.693 / 0.0100 = 69.3 mins