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AP Chemistry Name
2 ▪ Chemical Kinetics Period Date
2.2 PROBLEMS – INTEGRATED RATE LAW
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 t = ?
[HI]0 = 0.050 M [HI] = 0.025 M
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 = ?
k = = = 0.257 d–1
ln m – ln m0 = –k·t
ln m – ln (10 mg) = –(0.257 d–1)(7.0 d)
ln m = 0.504
m = 1.65 mg
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
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)
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 t = 0.0500 min
[C2H6N2]0 = 2.00 g [C2H6O2] = ?
ln[C2H6N2] – ln[C2H6N2]0 = – k·t
ln[C2H6N2] – ln(2.00 g) = –(40.8)(0.0500)
[C2H6N2] = 0.260 g
(2.00 – 0.26 g C2H6N2) = 1.74 g C2H6N2 used
1.74 g = 0.0300 mol N2
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 k = = = 0.00283 min–1
[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
|Time |[X] |ln [X] |[X]–1 |
|(minutes) |(mol L–1) | |(L mol–1) |
|0 |0.00633 |–5.062 |158 |
|10. |0.00520 |–5.259 |192 |
|20. |0.00427 |–5.456 |234 |
|30. |0.00349 |–5.658 |287 |
|50. |0.00236 |–6.049 |424 |
|70. |0.00160 |–6.438 |625 |
|100. |0.00090 |–7.013 |1,110 |
AP Chemistry 2005B #3
X → 2 Y + 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.
[pic][pic][pic]
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 = 1.24 × 1022 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 = = – = 0.0195 min–1
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
