Exam (elaborations)
CHEM 104 MODULE 1 –MODULE 6 Exam (2024/2025)(Portage Learning)
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- CHEM 104
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- Portage Learning
CHEM 104 MODULE 1 –MODULE 6 Exam (2024/2025)(Portage Learning)
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Chem 104 Module 1 to 6 Exam Portage learning 2022Module 1:
Question 1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table is obtained:
→ 4 NO2 (g) + O2 (g)
1.Using the [O2] data from the table, show the calculation of the instantaneous rate early in the
reaction (0 secs to 300 sec).
2.Using the [O2] data from the table, show the calculation of the instantaneous rate late in the reaction
(2400 secs to 3000 secs).2 N2O5 (g)
Data Table #2
Time (sec)[N2O5][O2]
00.300 M 0
3000.272 M 0.014 M
6000.224 M 0.038 M
9000.204 M 0.048 M
12000.186 M 0.057 M
18000.156 M 0.072 M
24000.134 M 0.083 M
30000.120 M 0.090 M 3.Explain the relative values of the early instantaneous rate and the late instantaneous rate.
Your Answer:
1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10-5 mol/Ls
2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls
3. The late instantaneous rate is smaller than the early instantaneous rate. Question 2
The following rate data was obtained for the hypothetical reaction: A + B → X + Y
Experiment # [A][B]rate
10.500.502.0
21.000.508.0
31.001.0064.0
1.Determine the reaction order with respect to [A].
2.Determine the reaction order with respect to [B].
3.Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents).
4.Show the calculation of the rate constant, k.
Your Answer:
rate = k [A]x [B]y
rate 1 / rate 2 = k [0.50]x [0.50]y / k [1.00]x [0.50]y 2..0 = [0.50]x / [1.00]x
0.25 = 0.5x
x = 2
rate 2 / rate 3 = k [1.00]x [0.50]y / k [1.00]x [1.00]y 8..0 = [0.50]y / [1.00]y
0.125 = 0.5y
y = 3
rate = k [A]2 [B]3
2.0 = k [0.50]2 [0.50]3
k = 64
Question 3
ln [A] - ln [A]0 = - k t0.693 = k t1/2
An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the
paper. Your Answer:
0.693 = k t1/2
0.693 = k (5720)
k = 1.21 x 10-4
ln [A] - ln [A]0 = - k t
ln 19.8 - ln 100 = - 1.21 x 10-4 t
t = 13, 384 years
Question 4
Using the potential energy diagram below, state whether the reaction described by the diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your answer.
Your Answer:
The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous because it
has relatively large Eact.
Question 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO
and 2.40 mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of H2O and corresponding amounts of CO, H2, and CH4.
CO (g) + 3 H2 (g)CH4 (g) + H2O (g)
Your Answer:
0.309 mole of H2O formed = 0.309 mole of CH4 formed
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