Chem 104 Module 1 to 6 Exam answers Portage learning 2022
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Module 1:
Question 1
In the reaction of gaseous N2
O5
to yield NO2
gas and O2
gas as shown below, the following data table
is obtained:
→ 4 NO2 (g) + O2 (g)
- Using the [O2
] data from the table, show the calculation of the instantaneous rate early in the
reaction (0 secs to 300 sec). - Using the [O2
] data from the table, show the calculation of the instantaneous rate late in the reaction
(2400 secs to 3000 secs). - Explain the relative values of the early instantaneous rate and the late instantaneous rate.
Your Answer:
2 N2
O5 (g)
Data Table #2
Time (sec) [N2
O5
] [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 - rate = (0.014 – 0) / (300 – 0) = 4.67 x 10-5 mol/Ls
- rate = (0.090 – 0.083) / (3000 – 2400) = 1.167 x 10-5 mol/Ls
- 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
- Determine the reaction order with respect to [A].
- Determine the reaction order with respect to [B].
- Write the rate law in the form rate = k [A]n
[B]m (filling in the correct exponents). - 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 / 8.0 = [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 / 64.0 = [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 t
1/2
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
H2
O and corresponding amounts of CO, H2
, and CH4
.
CO (g) + 3 H2 (g) CH4 (g) + H2
O (g)
Your Answer:
0.309 mole of H2
O formed = 0.309 mole of CH4
formed