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12.4 Integrated rate laws  (Page 2/7)

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ln [ A ] = ( − k ) ( t ) + ln [ A ] 0 y = m x + b

A plot of ln[ A ] versus t for a first-order reaction is a straight line with a slope of − k and an 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 .

Determination of reaction order by graphing

Show that the data in [link] can be represented by a first-order rate law by graphing ln[H 2 O 2 ] versus time. Determine the rate constant for the rate of decomposition of H 2 O 2 from this data.

Solution

The data from [link] with the addition of values of ln[H 2 O 2 ] are given in [link] .

A graph is shown with the label “Time ( h )” on the x-axis and “l n [ H subscript 2 O subscript 2 ]” on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).
The linear relationship between the ln[H 2 O 2 ] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.
Trial Time (h) [H 2 O 2 ] ( M ) ln[H 2 O 2 ]
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[H 2 O 2 ] 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[H 2 O 2 ] versus time where:

slope = change in y change in x = Δ y Δ x = Δln [ H 2 O 2 ] Δ t

In order to determine the slope of the line, we need two values of ln[H 2 O 2 ] at different values of t (one near each end of the line is preferable). For example, the value of ln[H 2 O 2 ] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:

slope = −1.386 − ( −0.693 ) 12.00 h − 6.00 h = −0.693 6.00 h = −1.155 × 10 −2 h −1 k = − slope = − ( −1.155 × 10 −1 h −1 ) = 1.155 × 10 −1 h −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:

A graph, labeled above as “l n [ A ] vs. Time” is shown. The x-axis is labeled, “Time ( s )” and the y-axis is labeled, “l n [ A ].” The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).
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Second-order reactions

The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:

Rate = k [ A ] 2

For these second-order reactions, the integrated rate law is:

1 [ A ] = k t + 1 [ A ] 0

where the terms in the equation have their usual meanings as defined earlier.

The integrated rate law for a second-order reaction

The reaction of butadiene gas (C 4 H 6 ) with itself produces C 8 H 12 gas as follows:

2C 4 H 6 ( g ) ⟶ C 8 H 12 ( g )

The reaction is second order with a rate constant equal to 5.76 × 10 −2 L/mol/min 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:

1 [ A ] = k t + 1 [ A ] 0

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

1 [ A ] = ( 5.76 × 10 −2 L mol −1 min −1 ) ( 10 min ) + 1 0.200 mol −1 1 [ A ] = ( 5.76 × 10 −1 L mol −1 ) + 5.00 L mol −1 1 [ A ] = 5.58 L mol −1 [ A ] = 1.79 × 10 −1 mol L −1

Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L 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:

0.0196 mol/L

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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