First-Order Reactions
First-Order Reactions
An equation relating the rate constant k to the initial concentration [A]0 and the concentration [A]t present after any given time t can be derived for a first-order reaction and shown to be:
\(\text{ln}\left(\cfrac{{\left[A\right]}_{t}}{{\left[A\right]}_{0}}\right)=\text{−}kt\)
or
\(\text{ln}\left(\cfrac{{\left[A\right]}_{0}}{{\left[A\right]}_{t}}\right)=kt\)
or
\(\left[A\right]={\left[A\right]}_{0}{e}^{-kt}\)
Example
The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, C4H8 at 500 °C is 9.2 \(×\) 10−3 s−1:
\({\text{C}}_{4}{\text{H}}_{8}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2C}}_{2}{\text{H}}_{4}\)
How long will it take for 80.0% of a sample of C4H8 to decompose?
Solution
We use the integrated form of the rate law to answer questions regarding time:\(\text{ln}\left(\cfrac{{\left[A\right]}_{0}}{\left[A\right]}\right)=kt\)
There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.
The initial concentration of C4H8, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:
\(\begin{array}{cc}\hfill t& =\text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{\left[x\right]}{\left[0.200x\right]}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{k}\hfill \\ & =\text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{0.100\phantom{\rule{0.2em}{0ex}}{\text{mol L}}^{-1}}{0.020\phantom{\rule{0.2em}{0ex}}{\text{mol L}}^{-1}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{9.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}}\hfill \\ & =1.609\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{9.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}}\hfill \\ & =1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{s}\hfill \end{array}\)
We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:
\(\begin{array}{ccc}\hfill \text{ln}\left[A\right]& =& \left(\text{−}k\right)\left(t\right)+\text{ln}{\left[A\right]}_{0}\hfill \\ \hfill y& =& mx+b\hfill \end{array}\)
A plot of ln[A] versus t for a first-order reaction is a straight line with a slope of −k and an intercept of ln[A]0. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.
Example
Determination of Reaction Order by Graphing
Show that the data in this lesson can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the rate of decomposition of H2O2 from this data.
Solution
The data from this lesson with the addition of values of ln[H2O2] are given in the figure below.
The linear relationship between the ln[H2O2] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.
| Trial | Time (h) | [H2O2] (M) | ln[H2O2] |
|---|---|---|---|
| 1 | 0 | 1.000 | 0.0 |
| 2 | 6.00 | 0.500 | −0.693 |
| 3 | 12.00 | 0.250 | −1.386 |
| 4 | 18.00 | 0.125 | −2.079 |
| 5 | 24.00 | 0.0625 | −2.772 |
The plot of ln[H2O2] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.
The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H2O2] versus time where:
\(\text{slope}=\phantom{\rule{0.1em}{0ex}}\cfrac{\text{change in}\phantom{\rule{0.2em}{0ex}}y}{\text{change in}\phantom{\rule{0.2em}{0ex}}x}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{\text{Δ}y}{\text{Δ}x}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{\text{Δln}\left[{\text{H}}_{2}{\text{O}}_{2}\right]}{\text{Δ}t}\)
In order to determine the slope of the line, we need two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:
\(\begin{array}{ccc}\hfill \text{slope}& =& \cfrac{-1.386-\left(-0.693\right)}{\text{12.00 h}-\text{6.00 h}}\hfill \\ & =& \cfrac{-0.693}{\text{6.00 h}}\hfill \\ & =& -1.155\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{-1}\hfill \\ \hfill k& =& -\text{slope}=-\left(-1.155\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{-1}\right)=1.155\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{h}}^{-1}\hfill \end{array}\)
This lesson is part of:
Chemical Kinetics