Chemistry 2710 Spring 2000 Test 2 Solutions

1. 1. Let . Then

      

      The graph has the following appearance: From the slope and intercept, we find and . Since , we have

      

   2. The Haldane relationship allows us to calculate the equilibrium constant directly:

      

      The agreement between the two values isn't spectacular, but then we weren't told much about the conditions under which the enzyme-catalyzed reaction occurred.

   3. 

2. 1. Oxygen radicals (atoms) can be expected to be quite reactive. Since they appear as intermediates in this reaction, the steady-state approximation should be appropriate.
   2. One oxygen atom occurs in each of the two reactions, so we can simply add them. The result is

      

   3. As argued above, we can use the steady-state approximation for O.

      

      We solve this equation for [O]:

      

      The rate of reaction is

      

   4. At high pressure (high [X]), so that

      

      The reaction is of the second order with respect to . Additionally, the reaction displays an order of -1 with respect to ). In other words, molecular oxygen inhibits the reaction at high pressure.

      On the other hand, at low pressure (low [X]), the rate approaches

      

      and the overall order is 2, with first-order dependencies on both [X] and . Molecular oxygen ceases to exert any significant influence on the rate of reaction, but the rate becomes sensitive to the total pressure.

3. 1. The values are nearly constant, which is compatible with competitive inhibition. It should also be the case that a plot of vs is linear. The data do fit a line rather well and so are compatible with competitive inhibition.
   2. is found by averaging the individual estimates. The result is .

      The plot of vs has an intercept and a slope . It follows that .