Rate Plots

Orders of reaction

Eric Van Dornshuld https://dornshuld.chemistry.msstate.edu (Mississippi State University)https://chemistry.msstate.edu
2022-02-14

Rate vs. Concentration

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:


Concentration vs. Time

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


Concentration vs. Time (linear forms)

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).


0th Order


1st Order


2nd Order

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.


Effect of rate constant, k

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

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?


Half-life Example

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

\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0} = \dfrac{1}{(1)(1)} = 1\]


Citation

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}
}