When C60O3 in toluene solution decomposes, O2 is released leaving C60O3 in solution.
Based on the data in figure 2 and figure 3, plot the concentration of C60O as a function of time.
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 C60O related to the rate of disappearance of C60O3? Based on this, plot the rate of appearance of C60O as a function of time.
The reaction 2N2O5(g) 4NO2(g)+O2(g) was found in this study to have rate law given by Rate=k[N2O5] with k = 0.070 s-1.
How is the rate of appearance of NO2 related to the rate of disappearance of N2O5? Which rate is larger?
Based on the rate law and rate constant, sketch a plot of [N2O5], [NO2], and [O2] versus time all on the same graph.
For which of the reactions listed in table 4 can you be certain that the reaction does not occur as a single step collision? Explain your reasoning.
Consider two decomposition reactions for two hypothetical materials, A and B. The decomposition of A is found to be first order, and the decomposition of B is found to be second order.
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.
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 graph of the logarithm of the equilibrium constant for a reaction versus 1
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 constant must have a negative slope.
Using equation 18 and the data in table 5, determine the activation energy for the reaction H2 (g) + I2(g) 2 HI(g).
We found that the rate law for the reaction H2(g)+I2(g) 2HI(g) is Rate=k[H2][I2]. Therefore, the reaction is second order overall but first order in H2. Imagine that we start with [H2]0 = [I2]0 and we measure [H2] versus time. Will a graph of ln[H2] versus time be linear or will a graph of 1
versus time be linear? Explain your reasoning.
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.
What does this suggest about the activation energies of reactions?
Using equation 18, calculate the activation energy of a reaction whose rate doubles when the temperature is raised from 25°C to 35°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 above. Does the rate double in this case?
Consider a very simple hypothetical reaction A+B 2C which comes to equilibrium.
At equilibrium, what must be the relationship between the rate of the forward reaction, A+B 2C and the reverse reaction 2C A+B?
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?
Using the previous results from here and here, show that the equilibrium constant for this reaction can be calculated from Kc = kf/kr, where kf is the rate constant for the forward reaction and kr is the rate constant for the reverse reaction.
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, Eα,f, and in the reverse direction, Eα,r. Does this relationship depend on whether the reaction is endothermic or exothermic? Explain.
|Problem 11 ||
For the reaction H2(g) + I2(g) 2HI(g), the rate law is Rate = k[H2][I2]. 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: I2 2I
Step 2: H2+2I 2HI
Where Step 1 is fast and Step 2 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 intermediate I, 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 experimental observation?