3.4 Integrated Rate Law
Concentration vs. Time
At the beginning of this page, we saw some concentration vs. time plots and then transitioned into rate vs. concentration plots when analyzing rate laws. We now transition back to concentration vs. time plots for different orders of reactions. These plots are a product of integrated rate laws which show the linear relationship of reactant concentration vs. time and have the form of y=mx+b. There is an integrated rate law for each order (zeroth- first- and second-) of reaction. These are derived from the corresponding rate laws (derivations not shown).
3.4.1 First-order Integrated Rate Law
Consider the following concentration vs. time plot for a first-order reaction.
Figure 3.12: Concentration vs. time plot for a first-order reaction.
This plot uses, once again, the numerical values of the rate constants and corresponding temperatures found above. The concentration of A decreases exponentially with time.
We can obtain a linear relationship by transforming the data in order to extract the rate constant from the slope. To do this for a first-order reaction, take the natural log of the concentration!
The first-order integrated rate law is
\[\ln\mathrm{[A]}_t = -kt + \ln\mathrm{[A]_0}\] and gives a linear correlation between the natural log of the concentration and time (e.g. reaction progress). The plot below shows the transformed data ([A] → ln[A]).
Figure 3.13: Linear integrated rate law plot for a first-order reaction.
The slope of each line is the is related to the rate constant.
\[\mathrm{slope} = -k\]
First-order reactions give linear relationships when the natural log of the reactant concentration is plotted vs. time.
3.4.2 Zeroth-order Integrated Rate Law
Consider the following concentration vs. time plot for a zeroth-order reaction. Again, I use the numerical values of the rate constants and corresponding temperatures found above.
Figure 3.14: Integrated rate law plot for a zeroth-order reaction.
The concentration of A decreases linearly with time. Since the relationship is already in a linear form, there is no need to transform the data! The rate constant can be obtained directly from the slope.
\[\mathrm{slope} = -k\]
The zeroth-order integrated rate law is
\[\mathrm{[A]}_t = -kt + \mathrm{[A]_0}\]
Zeroth-order reactions give linear relationships when the reactant concentration is plotted vs. time.
3.4.3 Second-order Integrated Rate Law
Consider the following concentration vs. time plot for a second-order reaction. Again, I use the numerical values of the rate constants and corresponding temperatures found above.
Figure 3.15: Concentration vs. time plot for a second-order reaction.
Once again, we see an exponentially decreasing relationship between reactant concentration and time. We can transform this data, once again, to obtain a linear relationship allowing us to extract the rate constant from the slope. We do this by taking the inverse of the reactant concentration ([A] → 1/[A]).
The rate constant can be obtained directly from the slope.
\[\mathrm{slope} = k\]
The second-order integrated rate law is
\[\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}\] and gives a linear correlation between the inverse of the concentration and time (e.g. reaction progress). The plot below shows the transformed data.
Figure 3.16: Integrated rate law plot for a second-order reaction.
Second-order reactions give linear relationships when the inverse of the reactant concentration is plotted vs. time.
Practice
Consider the following second-order reaction. The reaction has a rate constant of 0.3387 M–1 s–1 (at 326.86 °C or 600 K) and begins with 5.0 M NO2.
\[2\mathrm{NO_2}(g) \longrightarrow \mathrm{2NO}(g) + \mathrm{O_2}(g)\] How much time (in s) will it take for 1.0 M of the reactant to be consumed?
Solution
Since the problem is asking for time, we realize that we must solve the integrated rate law for this second-order reaction (remember, we know the reaction order by analyzing the units of the given rate constant).
We initially begin with 5.0 M NO2 and we are finding the time required for 1.0 M of reactant to be consumed. Therefore, [NO2]0 = 5.0 and [NO2]t = 4.0
Rearrange the equation for t and solve.
\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} ~~\longrightarrow \\[1.5ex] t &= \dfrac{\left ( \frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} \right )}{k} \\[1.5ex] &= \dfrac{\left ( \frac{1}{4.0~M} - \frac{1}{5.0~M} \right )}{0.3387~M^{-1}~\mathrm{s}~^{-1}} \\[1.5ex] &= 0.14762~\mathrm{s} \end{align*}\]
Math String: ((1/4) - (1/5))/0.3387
In about one-fifteenth of a second, 1 M NO2 is transformed to products. This is a fast reaction!
Extend
How much time will it take for the reaction to go from 2.0 M NO2 to 1.0 M NO2?
Remember, the reaction is second-order and, as we determined from the practice problem above, the reaction is slowing down. We would expect that the time required for this will be longer than 0.15 seconds!
Solve
\[\begin{align*} \dfrac{1}{[\mathrm{A}]_t} &= kt + \dfrac{1}{[\mathrm{A}]_0} ~~\longrightarrow \\[1.5ex] t &= \dfrac{\left ( \frac{1}{[\mathrm{A}]_t} - \frac{1}{[\mathrm{A}]_0} \right )}{k} \\[1.5ex] &= \dfrac{\left ( \frac{1}{1.0~M} - \frac{1}{2.0~M} \right )}{0.3387~M^{-1}~\mathrm{s}~^{-1}} \\[1.5ex] &= 1.476~\mathrm{s} \end{align*}\]
Math String: ((1/1) - (1/2))/0.3387
We see that the reaction takes much longer (about 10 times as long) at this stage in the reaction!
Plot
We’ve determined two points along this reaction numerically. Let’s go ahead and plot the data from 5.0 M down to 0 M and see what it looks like!
Figure 3.17: Concentration vs. time plot for this second-order reaction.
Here we see the second-order reaction starts out really fast (reactant is being quickly consumed) but as the NO2 concentration decreases, the reaction rate increases significantly.
Extend Again
Let’s make the concentration vs. time relationship linear! We can get the rate constant from the slope of the line! For a second-order reaction, we simply plot the inverse concentration (i.e. 1/[NO2]) vs. time.
Figure 3.18: Integrated rate plot for this second-order reaction.
The equation for the resulting line is
\[y = 0.3387x + 0.2\]
The given rate constant of reaction is 0.3387 s–1 which exactly matches the slope of our integrated rate plot for a second-order reaction!