a. Use a procedure similar to the one just to derive an

Chapter 13, Problem 112E

(choose chapter or problem)

The first-order integrated rate law for a reaction \(\mathrm{A} \ \longrightarrow\) products is derived from the rate law using calculus:

        \(\text { Rate }=k[\mathrm{A}]\) (first-order rate law)

        \(\text { Rate }=-\frac{d[\mathrm{A}]}{d t}\)

        \(\frac{d[\mathrm{~A}]}{d t}=-k[\mathrm{A}]\)

The equation just given is a first-order, separable differential equation that can be solved by separating the variables and integrating:

         \(\frac{d[\mathrm{A}]}{[\mathrm{A}]}=-k d t\)

         \(\int_{|\mathrm{A}|_{0}}^{|\mathrm{A}|} \frac{d[\mathrm{~A}]}{[\mathrm{A}]}=-\int_{0}^{t} k d t\)

In the integral just given, \([\mathrm{A}]_{0}\) is the initial concentration of A. We then evaluate the integral:

         

           \(\ln [\mathrm{A}]-\ln [\mathrm{A}]_{0}=-k t\)

           \(\ln [\mathrm{A}]=-k t+\ln [\mathrm{A}]_{0}\)  (integrated rate law)

a. Use a procedure similar to the one just to derive an integrated rate law for a reaction \(\mathrm{A} \ \longrightarrow\) products, which is one-half order in the concentration of A (that is, \(\text{ Rate }=k[\mathrm{A}]^{1/2}\)).

b. Use the result from part a to derive an expression for the half-life of a one-half-order reaction.