Focus Exercise 7.21

[Stella Nguyen 1J]

Hi everyone!

I don't really remember Dr. Lavelle discussing the plots for situations from part (e) to (h), so I was wondering if you guys could explain the conceptual meaning and relationships behind these plots and how we would know if they are linear or not?

7.21: Which of the following plots will be linear? (a) [A] against time for a reaction that is first order in A; (b) [A] against time for a reaction that is zeroth order in A; (c) ln [A] against time for a reaction that is first order in A; (d) 1/[A] against time for a reaction that is second order in A; (e) kr against temperature; (f) initial rate against [A] for a reaction that is first order in A; (g) half-life against [A] for a reaction that is zeroth order in A; (h) half-life against [A] for a reaction that is second order in A.

Thank you so much!

[Anil Chaganti 3L]

Dr. Lavelle went over the plots of the integrated rate laws of first, second, and zero order of reactions. To emphasize, first order reactions are linear when you plot ln[A] vs t (negative slope=-k). This means that when you plot [A] vs t, you will obtain an exponential graph. Second order reactions are linear when you plot 1/[A] vs t (positive slope=k) and [A] vs t is not necessarily exponential. Zero order reactions have a linear plot at [A] vs t (negative slope=-k). For the relationships between half lives and the graphs, remember that as concentration goes up, the half life of a first order reaction is not affected as the half life is independent of concentration (t1/2=.693/k). For second order graphs, again we cannot deduce that it is linear since t1/2=1/k[A]0. Plotting some values show that this plot appears to detour from a constant linear plot. However, for zero order graphs, as concentration increases, the half life increases (more concentration=longer time), so this plot is linear. For part f) that is referring to the initial rates table we did in 7A, so I would take a look at those problems as well.

[Nico Medina]

I was wondering this too, thanks for the help!