South Pasadena • AP Chemistry
Name
2 ▪ Chemical Kinetics
Period
2.2
PROBLEMS
1. The decomposition of hydrogen iodide into
hydrogen and iodine is a second order reaction.
The rate constant k = 0.080 L mol-1s-1. How long
does it take an initial concentration of 0.050 M to
decrease to half this concentration?
k = 0.080 M–1 s–1
[HI]0 = 0.050 M
t=?
[HI] = 0.025 M
1
1
–
= +kt
[HI] [HI]0
1
1
–
= (0.080)t
0.025 0.050
–
Date
INTEGRATED RATE LAW
4. The decomposition of N2O5 in CCl4 is a first-order
reaction. If 256 mg of N2O5 is present initially, and
2.50 mg is present after 4.26 min at 55°C, what is
the value of the rate constant, k? (Ch 15, #27)
[N2O5]0 = 256 mg [N2O5]t = 2.50 mg
k=?
t = 4.26 min
ln [N2O5] – ln [N2O5]0 = –k·t
ln(2.50) – ln(256) = –k(4.26)
t = 250 s
2. The gold-198 isotope has a half-life of 2.7 days. If
you start with 10 mg at the beginning of the week,
how much remains at the end of the week, seven
days later?
t1/2 = 2.7 d
t = 7.0 d
m0 = 10 mg
m=?
ln 2 0.693
k=
=
= 0.257 d–1
t1/2 2.7 d
ln m – ln m0 = –k·t
k = 1.09 min–1
5. Gaseous azomethane, C2H6N2, decomposes in a
first-order reaction when heated.
C2H6N2 (g) → N2 (g) + C2H6 (g)
The rate constant for this reaction at 425°C is 40.8
min–1. If the initial amount of azomethane in the
flask is 2.00 g, how much remains after 0.0500
min? How many moles of N2 are formed in this
time? (Ch 15, #34)
k = 40.8 min–1
[C2H6N2]0 = 2.00 g
t = 0.0500 min
[C2H6O2] = ?
ln m – ln (10 mg) = –(0.257 d–1)(7.0 d)
ln[C2H6N2] – ln[C2H6N2]0 = – k·t
ln m = 0.504
ln[C2H6N2] – ln(2.00 g) = –(40.8)(0.0500)
m = 1.65 mg
1 mol 
[C2H6N2] = 0.260 g
58.09 g = 0.00448 mol
3. The decomposition of SO2Cl2 is a first-order
reaction:
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
The rate constant for the reaction is 2.8 × 10–3
min–1 at 600 K. If the initial concentration of
SO2Cl2 is 1.24 × 10–3 M, how long will it take for
the concentration to drop to 0.31 × 10–3 M?
(Ch 15, #28)
k = 2.8 ×10–3 min–1 t = ?
[SO2Cl2]0 = 1.24 × 10–3 M
[SO2Cl2] = 0.31 × 10–3 M
ln[SO2Cl2]–ln[SO2Cl2]0 = –k·t
ln(0.31×10–3)–ln(1.24×10–3) = –(2.8 × 10–3)(t)
t = 495 min
6. The decomposition of SO2Cl2
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
is first order in SO2Cl2, and the reaction has a half
life of 245 min at 600 K. If you begin with
3.6 × 10–3 mol of SO2Cl2 in a 1.0-L flask, how
long will it take for the quantity of SO2Cl2 to
decrease to 2.0 × 10–4 mol? (Ch 15, #33)
t½ = 245 min
ln 2 0.693
k=
=
= 0.00283 min–1
t½
245
[SO2Cl2]0 = 3.6 × 10–3 M
[SO2Cl2] = 2.0 × 10–4 M
ln[SO2Cl2] – ln[SO2Cl2]0 = – k·t
ln(2.0 × 10–4) – ln(3.6 × 10–3) = –(0.00283)t
t = 1022 min
AP Chemistry 2005B #3
X→2Y+Z
The decomposition of gas X to produce gases Y and Z
is represented by the equation above. In a certain
experiment, the reaction took place in a 5.00 L flask at
428 K. Data from this experiment were used to
produce the information in the table below, which is
plotted in the graphs that follow.
Time
(minutes)
[X]
(mol L–1)
ln [X]
[X]–1
(L mol–1)
0
10.
20.
30.
50.
70.
100.
0.00633
0.00520
0.00427
0.00349
0.00236
0.00160
0.00090
–5.062
–5.259
–5.456
–5.658
–6.049
–6.438
–7.013
158
192
234
287
424
625
1,110
(a) How many moles of X were initially in the flask?
n = V × M = (5.00 L)(0.00633 M) = 0.0317 mol X
(b) How many molecules of Y were produced in the first 20. minutes of the reaction?
nY at 20 min = 5.00 L(0.00427 M) = 0.0214 mol
0.0317 mol X – 0.0214 mol X = 0.0103 mol X
0.0103 mol X 
2 mol Y 6.022 × 1023 molecules Y
22
1 mol Y
1 mol X 
 = 1.24 × 10 molecules Y
(c) What is the order of this reaction with respect to X? Justify your answer.
The reaction is first order with respect to X because the ln[X] vs. time plot yields a linear graph.
(d) Write the rate law for this reaction.
rate = k[X]1
(e) Calculate the specific rate constant for this reaction. Specify units.
k = – slope =
∆ ln[X]
(–7.013) – (–5.062)
=–
= 0.0195 min–1
∆t
(100 – 0)
(f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction.
ln [X] – ln [X]0 = –k∙t
ln [X] – (–5.062) = –(0.0195 min–1)(150. min)
ln [X] = –7.99
[X] = 0.00034 M