Page 158 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

142 Chapter 9
But tl = 348 s at 32 "C, when [A] = 85% [Ao].

+ kl = (1/348)(1n[AoJ/0.85 [Ao]) = (1/348)(1n 1.1764706) =
0.000467 s- '.
Therefore, from step 6 (t)
Ink2 = 1.051124 + lnkl = 1.051124 + ln0.000467 = -6.6180573
Hence, k2 = e-6.6180573 = 0.001 336023.
7. What are the units? k2 = 0.001336023 s-l
8. tl/2 = (In 2)/(0.001336023) = 518.8 s, as k = (In 2)/t1/2, for a
first -order reaction.
Answer: tllz = 518.8 s

SUMMARY OF THE TWO CHAPTERS ON
CHEMICAL KINETICS
There now follow the five sets of equations which should be remem-
bered for kinetics type problems at this level. In the following two
sections, a multiple-choice test and three longer questions, involving
the working methods covered in Chapters 8 and 9, respectively, are
given:

1. Zero-order [A] = -kt + [&I forA+ Products
Y =mx+ c
Half-Life tl/2 = [&l/(2k)
2. First-order ln[A] = -kt + ln[&] for A + Products
Y =mx+ c
Half-Life tlj2 = (ln2)/k
3. Second-order 1/[A] = kt + l/[AO] for A + Products
Y =mx+ c
Half-Life tl12 = l/(k[Aol)
4. Arrhenius Ink = -E,,/(RT) + In A
Y =mx + c
5. In (kllk2) = (~act/m(Tl - T2)/(Tl T2)I

MULTIPLE-CHOICE TEST
1. The reaction A + B -+ Products obeys the rate law:
--- d[A1 - k[A]-'[BI2
dt