Exam (elaborations)

CHEM 104 MODULE 1 –MODULE 6 Exam (2025/2026)(Portage Learning)

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CHEM 104

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CHEM 104 MODULE 1 –MODULE 6 Exam (2025/2026)(Portage Learning) CHEM 104 MODULE 1 –MODULE 6 Exam (2025/2026)(Portage Learning) CHEM 104 MODULE 1 –MODULE 6 Exam (2025/2026)(Portage Learning)

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chem 104 module 1 –module 6 exam portage learning

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Chem 104 Module 1 to 6 Exam

Portage learning 2022

Module 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) 2 N2O5 (g)

Data Table #2

Time (sec) [N2O5] [O2]

0 0.300 M 0

300 0.272 M 0.014 M

600 0.224 M 0.038 M

900 0.204 M 0.048 M

1200 0.186 M 0.057 M

1800 0.156 M 0.072 M

2400 0.134 M 0.083 M

3000 0.120 M 0.090 M

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).

,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
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0
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 t 0.693 = 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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Exam (elaborations)

CHEM 104 MODULE 1 –MODULE 6 Exam (2025/2026)(Portage Learning)

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