CHEM 104 LATEST EXAM MODULE 1-6
PRIORITY EXAM 2024 WINTER-SPRING
QTR
Module 1:
QUIZ 1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following
data tableis obtained:
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
→ 4 NO2 (g) + O2 (g)
• Using the [O2] data from the table, show the calculation of the instantaneous rate
early in thereaction (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.
SELECTED RESPONSE:
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.
QUIZ 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.
SELECTED RESPONSE:
rate = k [A]x [B]y
rate 1 / rate 2 = k [0.50]x [0.50]y / k
[1.00]x [0.50]y2..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]y8..0 = [0.50]y /
[1.00]y
0.125 = 0.5yy = 3
rate = k [A]2 [B]3
,2.0 = k [0.50]2 [0.50]3
k = 64
QUIZ 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 thepaper.
SELECTED RESPONSE:
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
QUIZ 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 SELECTEDRESPONSE.
SELECTED RESPONSE:
, The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous
because ithas relatively large Eact.
QUIZ 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800
mole of COand 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)
SELECTED RESPONSE:
0.309 mole of H2O formed = 0.309 mole of CH4 formed
0.309 mole of H2O formed = 0.800 - 0.309 = 0.491 mole CO
0.309 mole of H2O formed = 3 x 0.309 mole H2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H2
[CO] = 0.491 mole / 8.00 L = 6.1375 x 10-2 M
[H2] = 1.473 mole / 8.00 L = 18.4125 x 10-2 M
[CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M
[H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M
Kc = [3.8625 x 10-2] [3.8625 x 10-2] / [6.1375 x 10-2] [18.4125 x 10-2]3
Kc = 3.89
QUIZ 6
Explain the terms substrate and active site in regard to an enzyme.
SELECTED RESPONSE:
Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction.
The catalysts act only on one type of substance to cause one type of reaction and this is
called a substrate. Active sites are enzymes that are spherical in shape together ith a group of
PRIORITY EXAM 2024 WINTER-SPRING
QTR
Module 1:
QUIZ 1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following
data tableis obtained:
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
→ 4 NO2 (g) + O2 (g)
• Using the [O2] data from the table, show the calculation of the instantaneous rate
early in thereaction (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.
SELECTED RESPONSE:
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.
QUIZ 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.
SELECTED RESPONSE:
rate = k [A]x [B]y
rate 1 / rate 2 = k [0.50]x [0.50]y / k
[1.00]x [0.50]y2..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]y8..0 = [0.50]y /
[1.00]y
0.125 = 0.5yy = 3
rate = k [A]2 [B]3
,2.0 = k [0.50]2 [0.50]3
k = 64
QUIZ 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 thepaper.
SELECTED RESPONSE:
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
QUIZ 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 SELECTEDRESPONSE.
SELECTED RESPONSE:
, The reaction is exothermic since it has a negative heat of reaction and it is nonspontaneous
because ithas relatively large Eact.
QUIZ 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800
mole of COand 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)
SELECTED RESPONSE:
0.309 mole of H2O formed = 0.309 mole of CH4 formed
0.309 mole of H2O formed = 0.800 - 0.309 = 0.491 mole CO
0.309 mole of H2O formed = 3 x 0.309 mole H2 reacted = 2.40 - (3 x 0.309) = 1.473 mole H2
[CO] = 0.491 mole / 8.00 L = 6.1375 x 10-2 M
[H2] = 1.473 mole / 8.00 L = 18.4125 x 10-2 M
[CH4] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M
[H2O] = 0.309 mole / 8.00 L = 3.8625 x 10-2 M
Kc = [3.8625 x 10-2] [3.8625 x 10-2] / [6.1375 x 10-2] [18.4125 x 10-2]3
Kc = 3.89
QUIZ 6
Explain the terms substrate and active site in regard to an enzyme.
SELECTED RESPONSE:
Enzymes are large protein molecules that act as catalysts and increases the rate of a reaction.
The catalysts act only on one type of substance to cause one type of reaction and this is
called a substrate. Active sites are enzymes that are spherical in shape together ith a group of
