0.19 Reaction kinetics  (Page 7/7)

There are a few important notes about the mechanism. First, one product of the reaction is produced in the first step, and the other is produced in the second step. Therefore, the mechanism does lead to the overall reaction, consuming the correct amount of reactant and producing the correct amount of reactant. Second, the first reaction produces a new molecule, NO ₃ , which is neither a reactant nor a product. The second step then consumes that molecule, and NO ₃ therefore does not appear in the overall reaction, Equation (10). As such, NO ₃ is called a “reaction intermediate.” Intermediates play important roles in the rates of many reactions.

If the first step in a mechanism is rate determining as in this case, it is easy to find the rate law for the overall expression from the mechanism. If the second step or later steps are rate determining, determining the rate law is slightly more involved. The process for finding the rate law in such a case is illustrated in Discussion Question 7.

Review and discussion questions

1. For the reactions listed in Table 4 of the previous concept development study, which of these can you be certain that the reaction does not occur as a single step collision? Explain your reasoning.
2. 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 in Concept Development Study 12. 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 constant must have a negative slope.
3. Using Equation (18) and the data in Table 5, determine the activation energy for the reaction H ₂ (g) + I ₂ → 2HI(g) .
4. 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 ºC.
   1. What does this suggest about the activation energies of reactions?
   2. Using Equation (18), calculate the activation energy of a reaction whose rate doubles when the temperature is raised from 25 ºC to 35 ºC.
   3. 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 ºC to 110 ºC, assuming the same activation energy you found in part (b). Does the rate double in this case?
5. Consider a very simple hypothetical reaction which comes to equilibrium.
   1. At equilibrium, what must be the relationship between the rate of the forward reaction, and the reverse reaction ?
   2. Assume that both the forward and reverse reactions are elementary processes occurring by a single collision. What is the rate law for the forward reaction? What is the rate law for the reverse reaction?
   3. Using the results of (a) and (b), show that the equilibrium constant for this reaction can be calculated from , where k f is the rate constant for the forward reaction and k r is the rate constant for the reverse reaction.
6. Consider a very simple hypothetical reaction A + B ↔ C + D . By examining Figure 8, provide and explain the relationship between the activation energy in the forward direction, E _a,f and in the reverse direction, E _a,r . Does this relationship depend on whether the reaction is endothermic (Figure 8a) or exothermic (Figure 8b)? Explain.
7. For the reaction H ₂ (g) + I ₂ (g) → 2HI(g) , the rate law is Rate = k[H ₂ ][I ₂ ] . Although this suggests that the reaction is a one-step elementary process, there is evidence that the reaction occurs in two steps, and the second step is the rate determining step: Step 1: I ₂ ↔ 2I (fast) Step 2: H ₂ + 2I → 2HI (slow)
   1. If both the forward and reverse reactions in step 1 are much faster than step 2, explain why step 1 can be considered to be at equilibrium.
   2. What is the rate law for the rate determining step?
   3. Since the rate law in (b) depends on the concentration of an intermediate I, we need to find that intermediate. Calculate [I] from Step 1, assuming that Step 1 is at equilibrium.
   4. Substitute [I] from part (c) into the rate law in part (b) to find the overall rate law for the reaction. Is this result consistent with the experimental observation?