Review Assignment #5

Chapter 14 - Chemical Kinetics

5)
 

 
The average rate for each interval was calculated by the following equation:

The negative sign is included because the concentration provided is that of a reactant, that is, it is decreasing as the reaction progresses.

a.
b.
c.

16a.     rate law: rate = k [N₂O₅]

b.

c. when the concentration of N₂O₅ is doubled, the rate doubles since it is a first order process. That is,

20a. rate law:

b. If the concentration of iodide is tripled, the rate will also triple.

c. If the concentration of hydroxide is doubled, the rate will drop by one-half.

22. To find the rate law, you first must determine the order of the reaction with respect to each reactant. A comparison of experiments 1 and 2 will give the order with respect to NH₃ (since [BF₃] is not varied). In these experiments, the ammonia concentration is reduced by one-half and so is the reaction rate falls by one half as a result. Since it is a proportional change, the reaction must be first order in NH₃. Or, more systematically, we can say:

 

Thus m, the order of the reaction with respect to NH₃, is equal to one.

Similarly, a comparison of the rates observed in experiments three and four will yield the order of the reaction in BF₃ since the initial concentration of NH₃ is the same in both cases. By the above procedure,

Thus n, the order of the reaction with respect to BF₃, is equal to one.

So, the rate law is: rate = k [NH₃] [BF₃]. This reaction is second order overall (m + n =2).

The rate constant can be calculated from any one of the experiments. Using the data from experiment 1,

28. In this problem, it is important to realize that pressure can be used in place of concentration in the integrated rate law (because the ratio of pressures of time will be exactly equal to the ratio of concentrations).

The appropriate equation is the first-order integrated rate equation:

 

b) When the pressure is equal to one-tenth the initial pressure, we can set up the ratio,

Rearranging the integrated rate equation, we can say,

So,

Taking the natural log of both sides:

35. To determine the order of this reaction, we need to plot the concentration data two ways:

first order plot: ln [NO₂] vs. t

second order plot: 1/[NO₂] vs. t

The plot that gives the strait line indicates the order of the reaction. The plots are shown below.

 
Since the second order plot is linear, the reaction follows second order kinetics. With this type of plot, the rate constant is equal to the slope. So, the rate constant is 10.2 M^-1 sec^-1.

42. The energy profile for this reaction is similar to what is shown in Figure 14.15, although the relative size of E_a and DE are not the same. The activation energy of the reverse reaction will be 73 kJ.

47. Here's that damn question I messed up in lecture. Let's try it one more time. We can set up a ratio as follows:

a)

 

 

k₆₀ = 2.88 s^-1
 

b)

 

 

k₆₀ = 68.6 s^-1

56. Since each of these are elementary processes the rate laws can be determined from the balanced equations:

a) bimolecular, rate = k [CH₂] [Cl₂]
b) unimolecular, rate = k [C₃H₆]
c) unimolecular, rate = k [SO₃]

74. Comparing experiments 1 and 2:

Thus, n = 2.

Comparing experiments 1 and 4 yields:

So, m = 1. The rate law is then: rate = k [HgCl₂] [C₂O₄^2-]²

b) The rate constant is (using data from experiment 1)

 

c)

 

b. [A]_t = [A]_oe^-kt

[A]_t = (0.500 M)e^-(4.28 ´ 10-4 s-1) (5 ´ 103 s) = 0.059 M
 

The overall reaction is: 2 H₂O₂ (aq) ® 2 H₂O (l) + O₂ (g)

IO^- is the intermediate in this mechanism.