<< Chapter < Page Chapter >> Page >

If the first step in a mechanism is rate determining as in this case, it is easy to find the rate law forthe overall expression from the mechanism. If the second step or later steps are rate determining, determining the rate law isslightly more involved. The process for finding the rate law in such a case is illustrated in [link] .

Review and discussion questions

  1. When C 60 O 3 in toluene solution decomposes, O 2 is released leaving C 60 O 3 in solution.

    (a) Based on the data in [link] and [link] , plot the concentration of C 60 O as a function of time.

    (b) How would you define the rate of the reaction in terms of the slope of the graph from above ? How is the rate of appearance of C 60 O related to the rate of disappearance ofC 60 O 3 ? Based on this, plot the rate of appearance ofC 60 O as a function of time.

  2. The reaction 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) was found in this study to have rate law given byRate k[N 2 O 5 ]with k = 0.070 s -1 .

    (a) How is the rate of appearance of NO 2 related to the rate of disappearance of N 2 O 5 ? Which rate is larger?

    (b) Based on the rate law and rate constant, sketch a plot of [N 2 O 5 ] ,[NO 2 ],and [O 2 ] versus time all on the same graph.

  3. For which of the reactions listed in [link] can you be certain that the reaction does not occur as a single step collision? Explainyour reasoning.
  4. Consider two decomposition reactions for two hypothetical materials, A and B. The decomposition of A isfound to be first order, and the decomposition of B is found to be second order.

    (a)Assuming that the two reactions have the same rate constant at the same temperature, sketch [ A ] and [ B ] versus time on the same graph for the same initial conditions, i.e. [ A ] 0 [ B ] 0 .

    (b)Compare the half-lives of the two reactions. Under what conditions will the half-life of B be less than the half-life of A?Under what conditions will the half-life of B be greater than the half-life of A?

  5. A graph of the logarithm of the equilibrium constant for a reaction versus 1 T is linear but can have either a negative slope or a positive slope, depending on the reaction, as was observed here . However, the graph of the logarithm of the rate constant for a reaction versus 1 T has a negative slope for essentially every reaction. Using equilibrium arguments, explain why the graph for the rate constantmust have a negative slope.
  6. Using [link] and the data in [link] , determine the activation energy for the reaction H 2 (g)+I 2 (g) → 2HI(g).
  7. We found that the rate law for the reaction H 2 (g)+I 2 (g) → 2HI(g) is Rate = k[H 2 ][I 2 ] . Therefore, thereaction is second order overall but first order in H 2 . Imagine that we start with [H 2 ] 0 =[I 2 ] 0 and we measure [H 2 ] versus time. Will a graph of ln([H 2 ])versus time be linear or will a graph of 1 [ H 2 ] versus time be linear? Explain your reasoning.
  8. As a rough estimate, chemists often assume a “rule of thumb” that the rate of any reaction will double when the temperature is increased by 10°C.

    (a) What does this suggest about the activation energies of reactions?

    (b) Using Equation (18), calculate the activation energy of a reaction whose rate doubles when the temperature is raised from 25°C to 35°C.

    (c) Does this rule of thumb estimate depend on the temperature range? To find out, calculate the factor by which the rate constant increases when the temperature is raised from 100°C to 110°C, assuming the same activation energy you found in part(b). Does the rate double in this case?

  9. Consider a very simple hypothetical reaction A + B ↔ 2Cwhich comes to equilibrium.

    (a) At equilibrium, what must be the relationship between the rate of the forward reaction, A + B → 2Cand the reverse reaction 2C → A + B?

    (b) Assume that both the forward and reverse reactions are elementary processes occurring by a single collision. What is therate law for the forward reaction? What is the rate law for the reverse reaction?

    (c) Using the previous results from here and here , show that the equilibrium constant for this reaction can be calculated from K c k f k r , where k f is the rate constant for the forward reaction and k r is the rate constant for the reverse reaction.

  10. Consider a very simple hypothetical reaction A + B C + D . By examining [link] , provide and explain the relationship between the activation energy in the forwarddirection, E a , f , and in the reverse direction, E a , r . Does this relationship depend on whether the reaction is endothermic( [link] ) or exothermic ( [link] )? Explain.
  11. For the reaction H 2 (g) + I 2 (g) → 2HI(g), the rate law is Rate = k [H 2 ][I 2 ]. Although thissuggests that the reaction is a one-step elementary process, there is evidence that the reaction occurs in two steps, and the secondstep is the rate determining step:

    Step 1

    I 2 2 I

    Step 2

    H 2 + 2 I 2 H I
    Where Step 1 is fast and Step2 is slow.

    If the both the forward and reverse reactions in Step 1 are much faster than Step2 , explain why Step 1 can be considered to be at equilibrium.

    What is the rate law for the rate determining step?

    Since the rate law above depends on the concentration of an intermediateI, we need to find that intermediate. Calculate[I] from Step 1 , assuming that Step1 is at equilibrium.

    Substitute [I]from above into the rate law found previously to find the overall rate law for the reaction. Is this result consistent with the experimentalobservation?

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Concept development studies in chemistry 2012. OpenStax CNX. Aug 16, 2012 Download for free at http://legacy.cnx.org/content/col11444/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Concept development studies in chemistry 2012' conversation and receive update notifications?

Ask