Question

A reaction known to be first-order has a half-life of 23.6 min. If the initial concentration of the reactant is 2.50 M, what will the concentration of the reactant be after 2.00 hr?

Answer 1

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1) Determine the rate constant for the reaction:

k = (ln 2) / t_1/2 <--- I'll leave you to figure out how that came to be. Hint: use the integrate form of the first-order rate law

k = (ln 2) / 23.6 min

k = 0.02937 min^-1 <--- keep a few extra digits

1) use the integrated form of the first-order rate law:

ln A = -kt + ln A_o

ln A = - (0.02937 min^-1) (120 min) + ln 2.50

ln A = -3.5244 + 0.91629

ln A = -2.60811

A = 0.07367 M <--- round off more as you see fit

Here's another way:

120 min / 23.6 min = 5.106383 half-lives

(0.5)^5.106383 = 0.02902856 <--- the decimal amount remaining after 5.106383 half-lives

0.02902856 x 2.50 M = the answer