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The integrated rate law for our second-order reactions has the form of the equation of a straight line:

1 [ A ] = k t + 1 [ A ] 0 y = m x + b

A plot of 1 [ A ] versus t for a second-order reaction is a straight line with a slope of k and an intercept of 1 [ A ] 0 . If the plot is not a straight line, then the reaction is not second order.

Determination of reaction order by graphing

Test the data given to show whether the dimerization of C 4 H 6 is a first- or a second-order reaction.

Solution

Trial Time (s) [C 4 H 6 ] ( M )
1 0 1.00 × 10 −2
2 1600 5.04 × 10 −3
3 3200 3.37 × 10 −3
4 4800 2.53 × 10 −3
5 6200 2.08 × 10 −3

In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C 4 H 6 ] versus t and compare it with a plot of 1 [ C 4 H 6 ] versus t . The values needed for these plots follow.

Time (s) 1 [ C 4 H 6 ] ( M −1 ) ln[C 4 H 6 ]
0 100 −4.605
1600 198 −5.289
3200 296 −5.692
4800 395 −5.978
6200 481 −6.175

The plots are shown in [link] . As you can see, the plot of ln[C 4 H 6 ] versus t is not linear, therefore the reaction is not first order. The plot of 1 [ C 4 H 6 ] versus t is linear, indicating that the reaction is second order.

Two graphs are shown, each with the label “Time ( s )” on the x-axis. The graph on the left is labeled, “l n [ C subscript 4 H subscript 6 ],” on the y-axis. The graph on the right is labeled “1 divided by [ C subscript 4 H subscript 6 ],” on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).
These two graphs show first- and second-order plots for the dimerization of C 4 H 6 . Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.

Check your learning

Does the following data fit a second-order rate law?

Trial Time (s) [ A ] ( M )
1 5 0.952
2 10 0.625
3 15 0.465
4 20 0.370
5 25 0.308
6 35 0.230

Answer:

Yes. The plot of 1 [ A ] vs. t is linear:

A graph, with the title “1 divided by [ A ] vs. Time” is shown, with the label, “Time ( s ),” on the x-axis. The label “1 divided by [ A ]” appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).
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Zero-order reactions

For zero-order reactions, the differential rate law is:

Rate = k [ A ] 0 = k

A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.

The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:

[ A ] = k t + [ A ] 0 y = m x + b

A plot of [ A ] versus t for a zero-order reaction is a straight line with a slope of −k and an intercept of [ A ] 0 . [link] shows a plot of [NH 3 ] versus t for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO 2 ). The decomposition of NH 3 on hot tungsten is zero order; the plot is a straight line. The decomposition of NH 3 on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:

slope = k = 1.3110 −6 mol/L/s
A graph is shown with the label, “Time ( s ),” on the x-axis and, “[ N H subscript 3 ] M,” on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled “Decomposition on W.” A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled “Decomposition on S i O subscript 2.”
The decomposition of NH 3 on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO 2 ) surface, the reaction is first order.

The half-life of a reaction

The half-life of a reaction ( t 1/2 ) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide ( [link] ) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H 2 O 2 decreases from 1.000 M to 0.500 M . During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M ; during the third half-life, it decreases from 0.250 M to 0.125 M . The concentration of H 2 O 2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.

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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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