9.4 Integrated rate laws  (Page 3/7)

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

A plot of $\frac{1}{\left[A\right]}$ versus t for a second-order reaction is a straight line with a slope of k and an intercept of $\frac{1}{{\left[A\right]}_0}.$ 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 $\times$ 10 −2
2	1600	5.04 $\times$ 10 −3
3	3200	3.37 $\times$ 10 −3
4	4800	2.53 $\times$ 10 −3
5	6200	2.08 $\times$ 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 $\frac{\text{1}}{\left[{\text{C}}_4{\text{H}}_6\right]}$ versus t . The values needed for these plots follow.

Time (s)	$\frac{1}{\left[{\text{C}}_4{\text{H}}_6\right]}(M^{\mathrm{-1}})$	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 $\frac{1}{\left[{\text{C}}_4{\text{H}}_6\right]}$ versus t is linear, indicating that the reaction is second order.

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 $\frac{1}{\left[A\right]}$ vs. t is linear:

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Zero-order reactions

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

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 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:

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.