Orders of reaction
Consider a series of reactions of the form
\[\mathrm{A} \longrightarrow \mathrm{B}\] with a rate constant at unity (k = 1). The reactions follow 0th, 1st, or 2nd-order kinetics giving the following rate laws
The concentrations, [A], and rates (0th, 1st, and 2nd) are given as
[A] rate0 rate1 rate2
1.00 1.00 1.00 1.00
0.90 1.00 0.90 0.81
0.80 1.00 0.80 0.64
0.70 1.00 0.70 0.49
0.60 1.00 0.60 0.36
0.50 1.00 0.50 0.25
0.40 1.00 0.40 0.16
0.30 1.00 0.30 0.09
0.20 1.00 0.20 0.04
0.10 1.00 0.10 0.01
0.00 1.00 0.00 0.00
Reproduce the rates (rate0, rate1, rate2) by setting k = 1 and using [A] across a range of 1.00 to 0.00 in Excel.
Plotting the rates of reaction vs. concentration gives
Important features:
For the same set of data, we plot the concentration of reactant vs. time. We do this by implementing the integrated rate laws for a 0th, 1st, and 2nd order reactions given as
\[[\mathrm{A}]_t = -kt + [\mathrm{A}]_0 ~~~ \longrightarrow ~~~ t = \dfrac{\left ( [\mathrm{A}]_t - [\mathrm{A}]_0\right )}{-k} \]
\[\ln[\mathrm{A}]_{t} = -kt + \ln\mathrm[\mathrm{A}]_0 ~~~ \longrightarrow ~~~ t = \dfrac{\left ( \ln[\mathrm{A}]_{t} - \ln\mathrm[\mathrm{A}]_0 \right ) }{-k}\]
\[\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0} ~~~ \longrightarrow ~~~ t = \dfrac{\left ( \dfrac{1}{[\mathrm{A}]_t}- \dfrac{1}{[\mathrm{A}]_0} \right )}{k}\]
Note that the slope for each of these linear equations is equal to the rate constant.
[A] time0 time1 time2
1.00 0.00 0.00 0.00
0.90 0.10 0.11 0.11
0.80 0.20 0.22 0.25
0.70 0.30 0.36 0.43
0.60 0.40 0.51 0.67
0.50 0.50 0.69 1.00
0.40 0.60 0.92 1.50
0.30 0.70 1.20 2.33
0.20 0.80 1.61 4.00
0.10 0.90 2.30 9.00
Reproduce the times (time0, time1, and time2) in Excel for each order of reaction using the appropriate integrated rate laws.
Plotting the reactant concentration vs. time gives
To obtain plots with straight lines for each order, transform the y-axis appropriately for the order of reaction.
0th order 1st order 2nd order
[A] time0 ln([A]) time1 1/[A] time2
1.00 0.00 0.00 0.00 1.00 0.00
0.90 0.10 -0.11 0.11 1.11 0.11
0.80 0.20 -0.22 0.22 1.25 0.25
0.70 0.30 -0.36 0.36 1.43 0.43
0.60 0.40 -0.51 0.51 1.67 0.67
0.50 0.50 -0.69 0.69 2.00 1.00
0.40 0.60 -0.92 0.92 2.50 1.50
0.30 0.70 -1.20 1.20 3.33 2.33
0.20 0.80 -1.61 1.61 5.00 4.00
0.10 0.90 -2.30 2.30 10.00 9.00
Reproduce the y-axis values ([A], ln([A]), and 1/[A]) for each order of reaction in Excel using the appropriate integrated rate laws.
The slope is the rate constant, k, and the y-intercept is the initial reactant concentration (here [A]0 = 1).
One important concept to keep in mind is regarding integrated rate laws is, if we know the order of the reaction, we can predict how much reactant is left after a certain amount of time, t, has passed.
So far we have considered a rate constant, k, of 1. What happens to our rate plots as we change k? Taking our reaction data from above, k is varied (0.5, 1, and 2).
We will see that as k gets bigger, the rate increase and the time of reaction for [A] to go to 0 gets shorter.
0th order
1st order
2nd order
Half-life (\(t_{1/2}\)) is defined as the amount of time required for exactly one-half of an initial reactant concentration (\(\frac{1}{2}[\mathrm{A}]_0\)) to be consumed. The half-life equations (0th, 1st, and 2nd) are directly derived from their respective integrated rate laws (see derivations).
The half-life equations are
\[t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k}\]
\[t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k}\]
\[ t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0} \]
Given the reactant data from above (where k = 1)
[A] time0 time1 time2
1.00 0.00 0.00 0.00
0.90 0.10 0.11 0.11
0.80 0.20 0.22 0.25
0.70 0.30 0.36 0.43
0.60 0.40 0.51 0.67
0.50 0.50 0.69 1.00
0.40 0.60 0.92 1.50
0.30 0.70 1.20 2.33
0.20 0.80 1.61 4.00
0.10 0.90 2.30 9.00
what would the half-life be (in arbitrary time units) for the 0th, 1st, and 2nd order reactions?
If we assume our initial concentration to be
\[[\mathrm{A}]_0 = 1~M\]
then, the half-life for our 0th order reaction would be
\[t_{1/2} = \dfrac{[\mathrm{A}]_0}{2k} = \dfrac{1}{2(1)} = 0.5\]
This means 0.5 time units must pass for \([\mathrm{A}]\) to equal 0.5 M.
For our 1st order reaction, the half life would be
\[t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{1} = 0.693\]
For our 2nd order reaction, the half-life would be
\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0} = \dfrac{1}{(1)(1)} = 1\]
For attribution, please cite this work as
Dornshuld (2022, Feb. 14). Rate Plots. Retrieved from https://dornshuld.chemistry.msstate.edu/notes/ch12/kinetics.html
BibTeX citation
@misc{dornshuld2022rate, author = {Dornshuld, Eric Van}, title = {Rate Plots}, url = {https://dornshuld.chemistry.msstate.edu/notes/ch12/kinetics.html}, year = {2022} }