<< Chapter < Page | Chapter >> Page > |
The integrated rate law for our second-order reactions has the form of the equation of a straight line:
A plot of versus t for a second-order reaction is a straight line with a slope of k and an intercept of If the plot is not a straight line, then the reaction is not second order.
Test the data given to show whether the dimerization of C 4 H 6 is a first- or a second-order reaction.
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 versus t . The values needed for these plots follow.
Time (s) | 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 versus t is linear, indicating that the reaction is second order.
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 |
Yes. The plot of vs. t is linear:
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 ( 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.
Notification Switch
Would you like to follow the 'Chemistry' conversation and receive update notifications?