The graphical method determines reaction order by testing which form of the concentration data produces a straight line when plotted as a function of time. This works because the integrated rate laws for zeroth-, first-, and second-order reactions can each be written in the linear form
\[ y=mx+b \]In each case, time \(t\) is plotted on the horizontal axis. The quantity plotted on the vertical axis depends on the proposed reaction order.
For a reaction that is zeroth order in \(A\), the differential rate law is
\[ -\frac{d[A]}{dt}=k \]Integration gives
\[ [A]=[A]_0-kt \]Comparing this equation with
\[ y=mx+b \]gives the following graphical relationship:
\[ \underbrace{[A]}_{y} = \underbrace{(-k)}_{m} \underbrace{t}_{x} + \underbrace{[A]_0}_{b} \]Therefore, if a plot of \([A]\) versus \(t\) is linear, the data are consistent with zeroth-order kinetics. The slope is \(-k\), and the vertical intercept is \([A]_0\).
For a reaction that is first order in \(A\),
\[ -\frac{d[A]}{dt}=k[A] \]Integration gives
\[ \ln[A]=\ln[A]_0-kt \]In linear form,
\[ \underbrace{\ln[A]}_{y} = \underbrace{(-k)}_{m} \underbrace{t}_{x} + \underbrace{\ln[A]_0}_{b} \]Therefore, if a plot of \(\ln[A]\) versus \(t\) is linear, the data are consistent with first-order kinetics. The slope is \(-k\), and the vertical intercept is \(\ln[A]_0\).
For a reaction that is second order in \(A\),
\[ -\frac{d[A]}{dt}=k[A]^2 \]Integration gives
\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]In linear form,
\[ \underbrace{\frac{1}{[A]}}_{y} = \underbrace{k}_{m} \underbrace{t}_{x} + \underbrace{\frac{1}{[A]_0}}_{b} \]Therefore, if a plot of \(1/[A]\) versus \(t\) is linear, the data are consistent with second-order kinetics. In this case, the slope is \(+k\), and the vertical intercept is \(1/[A]_0\).
The same concentration–time data are transformed in three different ways and tested for linearity:
| Proposed order | Quantity plotted against time | Slope | Intercept |
|---|---|---|---|
| Zeroth order | \([A]\) | \(-k\) | \([A]_0\) |
| First order | \(\ln[A]\) | \(-k\) | \(\ln[A]_0\) |
| Second order | \(\displaystyle \frac{1}{[A]}\) | \(+k\) | \(\displaystyle \frac{1}{[A]_0}\) |
The plot that most closely follows a straight line identifies the reaction order. Once the correct plot has been selected, the rate constant is obtained from the magnitude of its slope.
For zeroth- and first-order plots,
\[ k=-\text{slope} \]while for a second-order plot,
\[ k=\text{slope} \]A graph may appear approximately linear even when the fit is imperfect. A linear regression provides an equation of the form
\[ y=mx+b \]along with a measure of how closely the data follow the fitted line. The coefficient of determination, \(R^2\), is often used for this purpose. A value closer to 1 indicates a more nearly linear relationship.
The correct reaction order should generally produce the most linear plot and the largest \(R^2\) value. However, \(R^2\) should not be used blindly: the data points and residual deviations from the fitted line should also be examined, especially when the possible fits have very similar \(R^2\) values.
The units of the slope provide another useful check:
| Plot | Slope units | Units of \(k\) |
|---|---|---|
| \([A]\) versus \(t\) | \(\mathrm{M\,s^{-1}}\) | \(\mathrm{M\,s^{-1}}\) |
| \(\ln[A]\) versus \(t\) | \(\mathrm{s^{-1}}\) | \(\mathrm{s^{-1}}\) |
| \(1/[A]\) versus \(t\) | \(\mathrm{M^{-1}\,s^{-1}}\) | \(\mathrm{M^{-1}\,s^{-1}}\) |
Big picture: The integrated rate laws convert nonlinear concentration–time behavior into testable linear relationships. By determining whether \([A]\), \(\ln[A]\), or \(1/[A]\) varies linearly with time, the reaction order can be identified and the rate constant can be obtained directly from the slope of the appropriate graph.
The concentration of a reactant \(A\) was measured as a function of time:
| Time (s) | \([A]\) (M) |
|---|---|
| 0 | 0.800 |
| 20 | 0.536 |
| 40 | 0.359 |
| 60 | 0.241 |
| 80 | 0.162 |
| 100 | 0.108 |
Use the graphical method to determine whether the reaction is zeroth, first, or second order. Then determine the rate constant.
To test the three possible reaction orders, calculate \(\ln[A]\) and \(1/[A]\) for each concentration.
| Time (s) | \([A]\) (M) | \(\ln[A]\) | \(1/[A]\) (\(\mathrm{M^{-1}}\)) |
|---|---|---|---|
| 0 | 0.800 | \(-0.223\) | 1.25 |
| 20 | 0.536 | \(-0.623\) | 1.87 |
| 40 | 0.359 | \(-1.02\) | 2.79 |
| 60 | 0.241 | \(-1.42\) | 4.15 |
| 80 | 0.162 | \(-1.82\) | 6.17 |
| 100 | 0.108 | \(-2.23\) | 9.26 |
The integrated rate laws predict the following linear plots:
| Reaction order | Plot that should be linear |
|---|---|
| Zeroth order | \([A]\) versus \(t\) |
| First order | \(\ln[A]\) versus \(t\) |
| Second order | \(1/[A]\) versus \(t\) |
The concentration \([A]\) does not decrease by a constant amount during each 20-second interval, so a plot of \([A]\) versus time is not linear. Likewise, the values of \(1/[A]\) do not increase by a constant amount.
The values of \(\ln[A]\), however, decrease by approximately \(0.400\) during each 20-second interval:
\[ -0.223,\ -0.623,\ -1.02,\ -1.42,\ -1.82,\ -2.23 \]Therefore, a plot of \(\ln[A]\) versus time produces a straight line. The reaction is first order in \(A\).
\[ \boxed{\text{Rate}=k[A]} \]The integrated first-order rate law is
\[ \ln[A]=\ln[A]_0-kt \]Comparing this equation with \(y=mx+b\), the slope of the graph is
\[ m=-k \]Using the first and last data points to calculate the slope,
\[ m = \frac{\ln[A]_{100}-\ln[A]_0} {100\ \mathrm{s}-0\ \mathrm{s}} \] \[ m = \frac{-2.23-(-0.223)} {100\ \mathrm{s}} \] \[ m=-2.00\times10^{-2}\ \mathrm{s^{-1}} \]Because \(m=-k\),
\[ k=-m \] \[ \boxed{k=2.00\times10^{-2}\ \mathrm{s^{-1}}} \]The vertical intercept of the first-order plot is
\[ b=\ln[A]_0 \]From the linear relationship,
\[ b\approx-0.223 \]Therefore,
\[ [A]_0=e^{-0.223}=0.800\ \mathrm{M} \]which agrees with the measured initial concentration.
Physical interpretation: A first-order reaction does not lose the same concentration during each equal time interval. Instead, it loses the same fraction of the reactant. Taking the natural logarithm of the concentration converts this exponential decay into a linear relationship whose slope is \(-k\).
The following concentration–time data were collected for reactant \(A\). Select a graphical relationship and generate the corresponding plot. Use the linearity of the plots to determine the reaction order.
1. Choose a relationship to plot:
2. What is the order of the reaction?
| Reaction order | Linear plot | Slope | Intercept |
|---|---|---|---|
| Zeroth order | \([A]\) versus \(t\) | \(-k\) | \([A]_0\) |
| First order | \(\ln[A]\) versus \(t\) | \(-k\) | \(\ln[A]_0\) |
| Second order | \(\displaystyle \frac{1}{[A]}\) versus \(t\) | \(+k\) | \(\displaystyle \frac{1}{[A]_0}\) |
| Reaction order | Units of slope (and \(k\)) |
|---|---|
| Zeroth order | \(\mathrm{M\,s^{-1}}\) |
| First order | \(\mathrm{s^{-1}}\) |
| Second order | \(\mathrm{M^{-1}\,s^{-1}}\) |
Big picture: The graphical method transforms concentration–time data into simple linear relationships. Rather than guessing the reaction order, each possible integrated rate law is tested directly. The graph that produces the best straight line identifies the reaction order, while the slope and intercept provide the rate constant and the initial concentration.