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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] .
(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.
(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.
(a)Assuming that the two reactions have the same rate constant
at the same temperature, sketch
and
versus time on the same graph for the same initial conditions,
(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?
(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?
(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 , where is the rate constant for the forward reaction and is the rate constant for the reverse reaction.
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:
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?
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