0.5 Reaction rates  (Page 10/10)

Based on the data in and , plot the concentration of $C_{60}O$ 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 $C_{60}O$ related to the rate of disappearance of $C_{60}{O}_3$ ? Based on this, plot the rate of appearance of $C_{60}O$ as a function of time.

How is the rate of appearance of $N{O}_2$ related to the rate of disappearance of $N_2{O}_5$ ? Which rate is larger?

Based on the rate law and rate constant, sketch a plot of $\left[N_2{O}_5\right]$ , $\left[N{O}_2\right]$ , and $\left[O_2\right]$ versus time all on the same graph.

Assuming that the two reactions have the same rate constant at the same temperature, sketch $\left[A\right]$ and $\left[B\right]$ versus time on the same graph for the same initial conditions, i.e. ${\left[A\right]}_0={\left[B\right]}_0$ .

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?

What does this suggest about the activation energies of reactions?

Using , 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 constantincreases 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?

At equilibrium, what must be the relationship between the rate of the forward reaction, $A+B\to 2C$ and the reverse reaction $2C\to A+B$ ?

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?

Using the previous results from here and here , show that the equilibrium constant for this reaction can be calculated from $K_c=\frac{k_f}{k_r}$ , where $k_f$ is the rate constant for the forward reaction and $k_r$ is the rate constant for the reverse reaction.

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 $\left[I\right]$ from Step 1 , assuming that Step1 is at equilibrium.

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