Second-Order Reactions
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:
\(\text{Rate}=k{\left[A\right]}^{2}\)
For these second-order reactions, the integrated rate law is:
\(\cfrac{1}{\left[A\right]}\phantom{\rule{0.1em}{0ex}}=kt+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}\)
where the terms in the equation have their usual meanings as defined earlier.
Example
The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas (C4H6) with itself produces C8H12 gas as follows:
\({\text{2C}}_{4}{\text{H}}_{\text{6}}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{\text{12}}\left(g\right)\)
The reaction is second order with a rate constant equal to 5.76 \(×\) 10−2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?
Solution
We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:\(\cfrac{1}{\left[A\right]}\phantom{\rule{0.1em}{0ex}}=kt+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}\)
We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 \(×\) 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:
\(\begin{array}{ccc}\hfill \cfrac{1}{\left[A\right]}& =& \left(5.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{L mol}}^{-1}\phantom{\rule{0.2em}{0ex}}{\mathrm{min}}^{-1}\right)\phantom{\rule{0.2em}{0ex}}\left(10\phantom{\rule{0.2em}{0ex}}\text{min}\right)+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{0.200\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-1}}\hfill \\ \hfill \cfrac{1}{\left[A\right]}& =& \left(5.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{L mol}}^{-1}\right)+5.00\phantom{\rule{0.2em}{0ex}}{\text{L mol}}^{-1}\hfill \\ \hfill \cfrac{1}{\left[A\right]}& =& 5.58\phantom{\rule{0.2em}{0ex}}{\text{L mol}}^{-1}\hfill \\ \hfill \left[A\right]& =& 1.79\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{mol L}}^{-1}\hfill \end{array}\)
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
The integrated rate law for our second-order reactions has the form of the equation of a straight line:
\(\begin{array}{ccc}\hfill \cfrac{1}{\left[A\right]}& =& kt+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}\hfill \\ \hfill y& =& mx+b\hfill \end{array}\)
A plot of \(\cfrac{1}{\left[A\right]}\) versus t for a second-order reaction is a straight line with a slope of k and an intercept of \(\cfrac{1}{{\left[A\right]}_{0}}.\) If the plot is not a straight line, then the reaction is not second order.
Example: Determination of Reaction Order by Graphing
The data below are for the same reaction described in the example above. Test these data to confirm that this dimerization reaction is second-order.
Solution
| Trial | Time (s) | [C4H6] (M) |
|---|---|---|
| 1 | 0 | 1.00 \(×\) 10−2 |
| 2 | 1600 | 5.04 \(×\) 10−3 |
| 3 | 3200 | 3.37 \(×\) 10−3 |
| 4 | 4800 | 2.53 \(×\) 10−3 |
| 5 | 6200 | 2.08 \(×\) 10−3 |
In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C4H6] versus t and compare it with a plot of \(\cfrac{\text{1}}{\left[{\text{C}}_{4}{\text{H}}_{6}\right]}\) versus t. The values needed for these plots follow.
| Time (s) | \(\cfrac{1}{\left[{\text{C}}_{4}{\text{H}}_{6}\right]}\phantom{\rule{0.4em}{0ex}}\left({M}^{-1}\right)\) | ln[C4H6] |
|---|---|---|
| 0 | 100 | −4.605 |
| 1600 | 198 | −5.289 |
| 3200 | 296 | −5.692 |
| 4800 | 395 | −5.978 |
| 6200 | 481 | −6.175 |
The plots are shown in the figure below. As you can see, the plot of ln[C4H6] versus t is not linear, therefore the reaction is not first order. The plot of \(\cfrac{1}{\left[{\text{C}}_{4}{\text{H}}_{6}\right]}\) versus t is linear, indicating that the reaction is second order.
These two graphs show first- and second-order plots for the dimerization of C4H6. Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.
This lesson is part of:
Chemical Kinetics