12 Integrated Rate Laws; Application of Pseudo-First Order (Flooding) Techniques (M13Q5)

Integrated Rate Laws

The rate laws we have seen so far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.

Please note: The integrated rate laws discussed in this text are derived for a reaction with reactant stoichiometric coefficient of 1, as in A → Product. If the coefficient was different, such as 2A → Product, the integrated rate laws would be slightly different.

First-Order Reactions

For the first-order reaction:

A   →   Product

the rate is:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]

 Because this is a first-order reaction, the rate law is:

rate  =  k[A]

These two expressions can be combined:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]  =  k[A]

 

Calculus can used to integrate this expression to give us a useful equation relating the rate constant, k, with [A]₀, the initial concentration, and [A]_t, the concentration after time, t, has passed.

ln [A]_t = –kt + ln [A]₀    or      [A]_t  =  [A]₀e^–kt

Example 1

The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, C₄H₈ at 500 °C is 9.2 × 10⁻³ s⁻¹:

C₄H₈(g)   →   2 C₂H₄(g)

How long will it take for 80.0% of a sample of C₄H₈ to decompose?

Solution
We use the integrated form of the rate law to answer questions regarding time:

ln[latex](\frac{[A]_{t}}{[A]_{0}})[/latex]  =  –kt

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]₀, [A]_t, and k, and need to find t.

The initial concentration of C₄H₈, [A]₀, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:

t = ln[latex](\frac{[A]_{t}}{[A]_{0}})\;\times\;[/latex] – [latex]\frac{1}{k}[/latex]

t = ln[latex](\frac{0.200x}{x})\;\times\;[/latex]  – [latex]\frac{1}{9.2 \times 10^{-3}s^{-1}}[/latex]

t  =  1.7 x 10² s

Check Your Learning
Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:

¹³¹I   →   ¹³¹Xe  +  e-

All radioactive decay is first order.  The rate constant is 0.138 days⁻¹.  How many days will it take for 90% of the ¹³¹I in a 0.500 M solution of this substance to decay to ¹³¹Xe?

Answer:

16.7 days

We can use integrated rate laws with experimental data of concentration and time over the course of a reaction to graphically determine the order and rate constant of a reaction.

The integrated rate law can be rearranged to a standard linear equation format (y = mx + b, where m = slope and b = the y-intercept):

ln[A]_t =  –kt   +   ln[A]₀

   y    =  mx   +      b_

Thus, a plot of ln[A] versus t for a first-order reaction is a straight line with a slope of −k and a y-intercept of ln[A]₀. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.

Example 2

Determination of Reaction Order by Graphing
For the first-order reaction A  →  B, determine the rate constant for the rate of decomposition of A from this data:

[A]	Time (hr)
1.000 M	0
0.500 M	6
0.250 M	12
0.125 M	18
0.0625 M	24

Solution
First, consider the most useful way to use this data. Because it is a first-order reaction, plot the ln[A] vs time:

Trial	Time (hr)	[A] (M)	ln[A]
1	0	1.000	0.0
2	6.00	0.500	−0.693
3	12.00	0.250	−1.386
4	18.00	0.125	−2.079
5	24.00	0.0625	−2.772

The plot of ln[A] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.

The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H₂O₂] versus time where:

slope  =  [latex]\frac{\Delta\;y}{\Delta\;x}[/latex]  =  [latex]\frac{\Delta\;ln[A]}{\Delta\;time}[/latex]

slope  =  [latex]\frac{-1.386\;\;-\;\;(-0.693)}{12.00h\;\;-\;\;6.00h}[/latex]  =  -0.1155 hr^-1

Check Your Learning

Graph the following data to determine whether the reaction, A  →  B  +  C, is first order.

Trial	Time (s)	[A]
1	4.0	0.220
2	8.0	0.144
3	12.0	0.110
4	16.0	0.088
5	20.0	0.074

Answer:

The plot of ln[A] vs. t is not a straight line. The equation is not first order:

Second-Order Reactions

For the second-order reaction:

A   →   Product

the rate is:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]

Because this is a second-order reaction, the rate law is:

rate  =  k[A]²

These two expressions can be combined:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]  =  k[A]²

Calculus can used to integrate this expression to give us a useful equation relating the rate constant, k, with [A]₀, the initial concentration, and [A]_t, the concentration after time, t, has passed.

[latex]\frac{1}{[A]_{t}}[/latex]  =  kt  +  [latex]\frac{1}{[A]_{0}}[/latex]

Example 3

The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas (C₄H₆) with itself produces C₈H₁₂ gas as follows:

2 C₄H₆(g)   →   C₈H₁₂(g)

The reaction is second order with a rate constant equal to 5.76 × 10⁻²M^-1min^-1 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:

We know three variables in this equation: [A]₀ = 0.200 M, k = 5.76 × 10⁻² M^-1min^-1, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:

[latex]\frac{1}{[A]_{t}}[/latex]  =  (5.76 x 10^-2 M^-1min^-1)(10.0 min)  +  [latex]\frac{1}{0.200 M}[/latex]

[latex]\frac{1}{[A]_{t}}[/latex]  = 5.58 M^-1

[A]  =  0.179 M

Therefore 0.179 M of butadiene remain at the end of 10.0 min, compared to the 0.200 M that was originally present.

Check Your Learning
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?

Answer:

0.0195 M

As with a first-order reaction, we can use integrated rate laws with experimental data of concentration and time over the course of a reaction to graphically determine the order and rate constant of a reaction.

The integrated rate law can be rearranged to a standard linear equation format (y = mx + b, where m = slope and b = the y-intercept):

[latex]\frac{1}{[A]_{t}}[/latex]  =   kt   +  [latex]\frac{1}{[A]_{0}}[/latex]

  y   =  mx  +   b_

Thus, a plot of [latex]\frac{1}{[A]}[/latex] versus t for a second-order reaction is a straight line with a slope of +k and a y-intercept of [latex]\frac{1}{[A]_{0}}[/latex]. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not second order in A.

Example 4

Determination of Reaction Order by Graphing

Test the data given to show whether the dimerization of C₄H₆ is a first- or a second-order reaction.

Solution

Time (s)	[C4H6] (M)
0	1.00 × 10−2
1600	5.04 × 10−3
3200	3.37 × 10−3
4800	2.53 × 10−3
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 [latex]\frac{1}{[C_{4}H_{6}]}[/latex] versus t. The data used for these plots in in the table below:

Time (s)	[latex]\frac{1}{[C_{4}H_{6}]}[/latex] M-1	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 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 versus t is linear, indicating that the reaction is second order.

Check Your Learning: Does the following data fit a second-order rate law?

Time (s)	[A] (M)
5	0.952
10	0.625
15	0.465
20	0.370
25	0.308
35	0.230

Answer:

Yes. The plot of vs. t is linear.

Zero-Order Reactions

For the zero-order reaction:

A   →   Product

the rate is:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]

Because this is a zero-order reaction, the rate law is:

rate  =  k[A]⁰  =  k

These two expressions can be combined:

rate  =  [latex]\frac{-\Delta[A]}{\Delta\;t}[/latex]  =  k

 

Calculus can used to integrate this expression to give us a useful equation relating the rate constant, k, with [A]₀, the initial concentration, and [A]_t, the concentration after time, t, has passed.

[A]_t  =  –kt  +  [A]₀

  y  =   mx  +  b_

The integrated rate law for a zero-order reaction has the form of the equation of a straight line.  A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and a y-intercept of [A]₀.

Experimental Technique Using Integrated Rate Laws and the Flooding Method

The integrated rate laws are quite useful when determining the reaction order and rate constants using all the data from a single experimental trial run. However, we’ve only considered integrated rate laws for reactions involving one reactant, whereas many reactions involve two or more reactants—how can we use integrated rate laws to find the reaction order and rate constants for those reactions? From the previous chapter, we can use the “flooding” method, where one reactant is at a much larger concentration than the other, making that reactant appear to have a constant concentration with respect to the concentration change that occurs with the smaller concentration reactant.

Let’s consider the reaction:

A  +  B   →   Products

where

To isolate the effect of reactant A on the reaction rate, we can run the reaction under a large excess of reactant B, i.e., flooding the reaction mixture with B so that A is the limiting reagent, and reactant B concentration effectively remains constant during the reaction because its consumption is so small that the change in concentration becomes negligible. Under this condition, the effectively constant B concentration and the k constant can be consolidated into a new a rate constant, k_obs, where:

k_obs = k[B]ⁿ

By replacing k[B]ⁿ with k_obs in the rate law, the rate law can be written as:

This rate law equation derived from the flooding technique allows us to use the integrated rate laws and experimental data to determine the reaction order with respect to reactant A and k_obs.

After measuring the concentration of A as the reaction progresses, we can produce plots of [A] vs time,  ln[A] vs time, and [latex]\frac{1}{[A]}[/latex] vs time. By identifying which plot results in a linear line of data points, the order of the reaction with respect to A can be determined.  The absolute value of the slope of that linear plot will be equal to k_obs. An outline of the strategy to be employed can be seen below.

In this example, after measuring the concentration of A as the reaction progresses, plots of [A] vs time,  ln[A] vs time, and [latex]\frac{1}{[A]}[/latex] vs time are produced, as seen in Figure 4. The ln[A] vs time plot is linear, so the reaction is first-order with respect to A and the slope of the plot equals –k_obs1 (this experiment will involve a number of trials and so the k_obs has been given a subscript of “1” to keep track of the data).

We can then run another trial with a different concentration of B, 0.2 M, which is still in large excess. The plot of ln[A]_t vs time would will still be linear but will yield a different slope corresponding to -k_obs2. This can be seen in Figure 5.

The ratio of the two k_obs can be used to find the order of B:

[latex]  \frac{k_{obs2}}{k_{obs1}}[/latex]  =  [latex]\frac{k[B]_{2}^n}{k[B]_{1}^n}[/latex]  =  [latex]\left(\begin{array}{c}\frac{[B]_2^n}{[B]_1^n}\end{array}\right)^{n}[/latex]

[latex]\frac{0.12}{0.03}[/latex]  =  [latex]\left(\begin{array}{c}\frac{0.2 M}{0.1 M}\end{array}\right)^{n}[/latex]

These values can be put together to find the complete rate law:

rate  =  k[A][B]²

Glossary

integrated rate laws
a form of each rate law that relates the concentration of reactants and time