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M5: Exam - Requires Respondus LockDown
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- Points 100
- Questions 10
- Time Limit 120 Minutes
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Attempt Time Score
LATEST Attempt 1 119 minutes 69.5 out of 100
Score for this quiz: 69.5 out of 100
Submitted Feb 4 at 11:17pm
This attempt took 119 minutes.
Question 1
10 / 10 pts
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Scientific Calculator (Links to an external site.)
Periodic Table
Equation Table
In the reaction of gaseous N 2
O
5
to yield NO 2
gas and O 2
gas as shown below the
following data table is obtained:
2 N
2
O
5
(g) → 4 NO 2
(g) + O 2
(g)
Data Table #
Time (sec) [N 2
O
5
] [O
2
]
0 0.200 M 0
300 0.182 M 0.009 M
600 0.166 M 0.017 M
900 0.152 M 0.024 M
1200 0.140 M 0.030 M
1800 0.122 M 0.039 M
2400 0.112 M 0.044 M
3000 0.108 M 0.046 M
Complete the following three problems:
- Using the [O 2
] data from the table show the calculation of the instantaneous rate early
in the reaction (0 secs to 300 sec).
- Using the [O 2
] 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:
.009-0/300-0 = .009/300 = .00003 = 3.0 x 10
- Determine the reaction order with respect to [B].
- Determine the reaction order with respect to [C].
- Write the rate law in the form rate = k [B]
n
[C]
m
(filling in the correct exponents).
- Show the calculation of the rate constant, k.
Your Answer:
rate = k [b]
x
[c]
y
rate = k [c]
x
[b]
y
rate = k [B]
n
[C]
m
rate = 1.8. = k[.50]
n
[.50]
m
3.6 k [1.00]
n
[.50]
m
rate = .5 =.
n
n=
rate = 3.6 = k[1.00]
n
[.50]
m
7.2 k [1.00]
n
[1.00]
m
rate = .5 =.
m
m=
rate =k [B]
1
[C]
1
1.8 = k [.50]
1
[.50]
1
1.8= k.
1.8/.25 = k
7.2 = k
1. 1st order with respect to [B] : 3.6/1.8 =k (1.00)
n
m
/ k (0.50)
n
m
: n = 1
2. 1st order with respect to [C] : 7.2/3.6 =k (1.00)
n
m
/ k (1.00)
n
m
: m = 1
3. rate = k [B]
1
[C]
1
4. k = 1.8 / (0.50)
1
1
Question 3
3 / 10 pts
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Periodic Table
Equation Table
ln [A] - ln [A] 0
= - k t 0.693 = k t 1/
An ancient sample of cloth was found to contain 21.4 %
14
C content as compared to a
Equation Table
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:
exothermic because of the negative sign, large E is nonspontaneous, require input of
significant energy to cause reaction.
Large E act
= nonspontaneous
ΔH- = exothermic
Question 5
6 / 10 pts
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Scientific Calculator (Links to an external site.)
Periodic Table
Equation Table
Show the calculation of K c
for the following reaction if an initial reaction mixture of
0.700 mole of CO and 2.10 mole of H 2
in a 7.00 liter container forms an equilibrium
mixture containing 0.266 mole of H 2
O and corresponding amounts of CO, H 2
, and CH 4
Your Answer:
equilibrium:
H
2
O = .266 mole
CH
4 =
.266 mole
CO = .700 - .266 = .434 mole
H
2
= 3 - 3x.266 = 2.202 mole
mole/ liter :
H 2 O = .266 / 7.00 L = .038 M
CH
4 =
.266 /7.00 L = .038 M
CO = .434 mole / 7.00 L= .062 M
H
2
= 2.202 mole / 7.00 L = .3145714286 M
Kc = [CH 4
] [H
2
O]
[CO] [H
2
]
3
Kc = [.038] [.038]
[.062] [.3145714286]
3
Kc = [.001444]
[.0019299654]
Kc =.
Kc = 7.249921893 x 10
Your Answer:
K
c
K
c =
products / reactants
Since K c
is a small number, that means there is a high concentration of reactant.
This K c
≈ 1 indicates that the equilibrium mixture will contain significant amounts of
both products and reactants.
Question 8
8 / 10 pts
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Scientific Calculator (Links to an external site.)
Periodic Table
Equation Table
The equilibrium reaction below has the following equilibrium mixture concentrations:
H
2
O = [0.0380], CH
4
= [0.0380], CO = [0.0620] and H 2
= [0.186] with K c
If the concentration of CO at equilibrium is increased to [1.00], how and for what reason
will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and
use this to confirm your answer.
Your Answer:
Equilibrium:
H
2
O = [0.0380], CH
4
= [0.0380], CO = [0.0620] and H 2
= [0.186] with K c
so when CO [1.00], Q c
= [.0380][.0380]/[1.00][.186]
3
Q
c
= .000009291932064 or 9.291932064 x 10
K
c
CO is apart of the reactant, adding CO concentration will shift reaction in the forward
direction.
Q
c
<K
c
reaction will proceed to the right in the direction of the products
K
c
= 3.62 when H 2
O = [0.0380], CH
4
= [0.0380], CO = [0.0620] and H 2
= [0.186 M]
When CO = [1.00],
Q = [0.0380] [0.0380] = 0.
[1.00] [0.186]
3
The reaction must shift briefly in the forward direction to decrease the [CO] to come
back to equilibrium. This is in agreement with Q c
< K
c
which also predicts the reaction will
proceed to the right.
Q = [0.0380] [0.0380] = 0.224 [1.00] [0.186]
Question 9
2.5 / 10 pts
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Scientific Calculator (Links to an external site.)
Periodic Table
Equation Table
The equilibrium reaction below has the K c
= 3.93. If the volume of the system at
equilibrium is increased from 3.00 liters to 6.00 liters, how and for what reason will the
equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this
to confirm your answer.
Your Answer:
K
c
when volume increase from 3.00 to 6.
Q
c
= [CH
4
/2 ] [H
2
O/2] = K
c
[CO/2] [H
2
/2]
3
The delta H symbol is positive which determines that this is endothermic.
The temperature increase from 25 to 100. When there is an increase on endothermic,
there is a forward shift in the reaction causing a higher concentration of products and
lower concentration of reactants, therefore an increase in the K c,
becasue K c =
products
over reactants. K c
at equilibrium equal 1, mathematically in fractions if the numberator is
greater than the denominator the results will be always great than 1.
If the temperature is increased on this reaction at equilibrium which has a +ΔH 0
, the
reaction will briefly shift in the direction that uses up some of the added heat
[+ΔH
0
indicates an endothermic (heat absorbing) forward reaction] so this reaction will
shift in the forward direction. Simultaneously, this forward shift will increase the
concentration of CO and H 2
(in the numerator of K c
) and decrease the concentration of
CH
4
and H 2
O (in the denominator of K c
) which will increase the value of K c
Quiz Score: 69.5 out of 100
