Chemistry 2710 Spring 2000 Test 1 Solutions

1. 1. If we look at the first two experiments, we see that the rate doubles when the ozone concentration doubles. Comparing the first and third experiments, the rate triples when the ozone concentration triples. The order with respect to ozone is therefore one. Similarly, if we compare the third and fourth experiments or the third and fifth, we see a simple proportionality between the rate and nitrogen monoxide concentration. It follows that the rate law is

      

      The rate constant can be calculated from any one of the experiments, for instance the first one:

      

   2. It could be elementary. It involves only two reactants. One can imagine the transfer of an oxygen atom from one molecule to another occuring in one step. Moreover, the rate law is consistent with this being an elementary reaction.
2. 1. B is an intermediate. It should not show up in the overall reaction. To eliminate it, we add twice the second reaction to the first. The result is

      

   2. With the help of the principle of detailed balance, we know that at equilibrium, each reaction will be in equilibrium, i.e. the forward and reverse rates will be equal. This leads to the equations

      

      If we square the second equation and rearrange, we get

      

      If we now eliminate in the first equation using this one, we obtain

      

      or

      

      Therefore

      

      Note that this phenomenological equilibrium constant, unlike the thermodynamic equilibrium constant, has units.

3. 1. The integrated rate laws for and 1 are

      

      We therefore need to compare plots of and vs t. If one plot is clearly linear while the others isn't, that will tell us the order of reaction. First however, we need to calculate a from b. Since the production of each B requires 2 A's and the reaction is irreversible, the long-time value of b is half the initial amount of A, i.e. . Moreover, . The data can therefore be transformed to

      

      The graphs are as follows:

      

      

      The second plot ( vs t) is clearly linear while the other isn't. The order of the reaction with respect to A is therefore one. The rate constant (the negative of the slope of the plot) is or .

   2. It's a first-order reaction, but the reaction involves two equivalents of A. This must therefore be a complex reaction with a rate-determining step in which one molecule of A is somehow activated for reaction.
4. 1. 
   2. Clearly,

      

      The time derivative of c(t) is

      

      The right-hand side of the rate equation is

      

      so the differential equation is satisfied.

   3. Rearrange the solution of the differential equation to isolate t: