2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
1/22
In the reaction of 0.200 M gaseous N2O5 to yield NO2 gas and O2 gas as
shown
... [Show More] below:
2 N2O5 (g) → 4 NO2 (g)
+
O2 (g)
the following data table is obtained:
Not yet graded / 0 pts
Question 1
PORTAGE LEARNING
CHEM 104
2022
Module 1: Problem Set
Due No due date Points 0 Questions 18 Time Limit None
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Time (sec)
[N2O5]
[O2]
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
2/22
0
0.200 M
0
300
0.180 M
0.010 M
600
0.019 M
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
3/22
900
0.146 M
0.027 M
1200
0.132 M
0.034 M
1800
0.110 M
0.045 M
2400
0.096 M
0.052 M
3000
0.092 M
0.054 M
(a) Use the [O2] data from the table to calculate the average rate over the measured time interval from 0 to 3000 secs.
(b) Use the [O2] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 300 sec).
(c) Use the [O2] data from the table to calculate the instantaneous rate late in the reaction (2400 secs to 3000 secs).
(d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate.
Your Answer:
.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
4/22
In the reaction of gaseous CH3CHO to yield CH4 gas and CO gas as
shown below:
CH3CHO (g) →
CH4 (g)
+
CO (g)
the following data table is obtained:
Not yet graded / 0 pts
Question 2
(a) The average rate over the measured time interval from 0 to 3000 secs is:
rate = Δ[O2] / Δt = (0.054 - 0) / 3000 - 0 = 1.80 x 10-5 mol/L•s
(b) The instantaneous rate early in the reaction from 0 to 300 secs is:
rate = Δ[O2] / Δt = (0.010 - 0) / 300 - 0 = 3.33 x 10-5 mol/L•s
(c) The instantaneous rate late in the reaction from 2400 to 3000 secs is:
rate = Δ[O2] / Δt = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10-6
mol/L•s
(d) It can be seen that the early instantaneous rate is the largest since the concentrations of reactants is highest during the earliest stages of the reaction and the late instantaneous rate is smallest since the concentrations of reactants is lowest during the late stages of the reaction.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
5/22
Time (sec)
[CH3CHO]
0
0.0500 M
1200
0.0300 M
2000
0.0240 M
6000
0.0120 M
10,000
0.0080 M
15,000
0.0056 M
20,000
0.0043 M
(a) Use the [CH3CHO] data from the table to calculate the average rate over the measured time interval from 0 to 20,000 secs.
(b) Use the [CH3CHO] data from the table to calculate the instantaneous rate early in the reaction (0 secs to 1200 sec).
(c) Use the [CH3CHO] data from the table to calculate the instantaneous rate late in the reaction (15,000 secs to 20,000 secs).
(d) Explain the relative values of the average rate, early instantaneous rate and late instantaneous rate.
Your Answer:
.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
6/22
The following rate data was obtained for the reaction which takes place in
a solution of OH-:
ClO- +
I-
→ IO- + Cl-
Not yet graded / 0 pts
Question 3
(a) The average rate over the measured time interval from 0 to 20,000 secs is:
rate = - Δ[ CH3CHO] / Δt = - (0.0044 - 0.0500) / 20,000 - 0 = 2.30 x
10-6 mol/L•s
(b) The instantaneous rate early in the reaction from 0 to 1200 secs is:
rate = - Δ[ CH3CHO] / Δt = - (0.0300 - 0.0500) / 1200 - 0 = 1.67 x
10-5 mol/L•s
(c) The instantaneous rate late in the reaction from 15,000 to 20,000 secs is:
rate = - Δ[ CH3CHO] / Δt = (0.0043 - 0.0056) / 20,000 - 15,000 =
2.60 x 10-7 mol/L•s
(d) It can be seen that the early instantaneous rate is the largest since the concentration of reactant is highest during the earliest stages of the reaction and the late instantaneous rate is smallest since the concentrations of reactants is lowest during the late stages of the reaction.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
7/22
Experiment # [I-] [ClO-] [OH-] rate
1 0.0030 0.0010 1.00 1.8 x
10-4
2 0.0030 0.0020 1.00 3.6 x
10-4
3 0.0060 0.0020 1.00 7.2 x
10-4
4 0.0030 0.0010 0.50 9.0 x
10-5
Determine the reaction order with respect to (1) ClO-, (2) I- and (3) OH-
Your Answer:
.
Rate = k [ClO-]x [I-]y [OH-]z
If data from experiments 1 and 2 was substituted into the equation we would obtain
1.8 x 10-4 = k [0.0010]x [0.0030]y [1.00]z
3.6 x 10-4 k [0.0020]x [0.0030]y [1.00]z
and if we cancel all common terms we would obtain 1.8 x 10-4 = k [0.0010]x [0.0030]y [1.00]z
3.6 x 10-4 k [0.0020]x [0.0030]y [1.00]z
1.8 x 10-4 = [0.100]x 0.50 = (0.5)x
3.6 x 10-4 [0.200]x
(1) which yields x = 1 as the order of reaction with respect to ClO-
If data from experiments 2 and 3 was substituted into the equation we would obtain:
3.6 x 10-4 = k [0.0020]x [0.0030]y [1.00]z
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
8/22
Following the reaction order that was determined in the previous questions, answer these two parts:
(4) write the rate law and then (5) determine the value of the rate constant, k.
Your Answer:
.
Not yet graded / 0 pts
Question 4
7.2 x 10-4 k [0.0020]x [0.0060]y [1.00]z
and if we cancel all common terms we would obtain 3.6 x 10-4 = k [0.0020]x [0.0030]y [1.00]z
7.2 x 10-4 k [0.0020]x [0.0060]y [1.00]z
3.6 x 10-4 = [0.0030]y 0.5 = (0.5)y
7.2 x 10-4 [0.0060]y
(2) which yields y = 1 as the order of reaction with respect to I-.
If data from experiments 1 and 4 was substituted into the equation we would obtain:
1.8 x 10-4 = k [0.0010]x [0.0030]y [1.00]z
9.0 x 10-5 k [0.0010]x [0.0030]y [0.50]z
and if we cancel all common terms we would obtain 1.8 x 10-4 = k [0.0010]x [0.0030]y [1.00]z
9.0 x 10-5 k [0.0010]x [0.0030]y [0.50]z
1.8 x 10-4 = [1.00]z 2.0 = (2.0)z
9.0 x 10-5 [0.50]z
(3) which yields z = 1 as the order of reaction with respect to OH-.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
9/22
The following rate data was obtained for the reaction:
2 ClO2 + 2 OH- → ClO - +
3
ClO - + H O
2
2
Determine the reaction order with respect to (1) ClO2 and (2) OH-, (3)
write the rate law and then (4) determine the value of the rate constant, k.
Your Answer:
.
Rate = k [ClO2]x [OH-]y
If data from experiments 1 and 2 was substituted into the equation we would obtain
Not yet graded / 0 pts
Question 5
Experiment #
[ClO2]
[OH-]
rate
1
0.060
0.030
2.48 x 10-2
2
0.020
0.030
2.76 x 10-3
3
0.020
0.090
8.28 x 10-3
The overall rate law can now be written as follows: (4) Rate = k [ClO-]1 [I-]1 [OH-]1
(5) and using the data from experiment 1 we can determine the rate constant as follows:
1.8 x 10-4 = k [0.0010] [0.0030] [1.00]
k = 1.8 x 10-4 / [0.0010] [0.0030] [1.00] = 60
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
10/22
2.48 x 10-2 = k [0.060]x [0.030]y
2.76 x 10-3 k [0.020]x [0.030]y
and if we cancel all common terms we would obtain 2.48 x 10-2 = k [0.060]x [0.030]y
2.76 x 10-3 k [0.020]x [0.030]y
2.48 x 10-2 = [0.060]x 9 = (3)x
2.76 x 10-3 [0.020]x
(1) which yields x = 2 as the order of reaction with respect to ClO2
If data from experiments 2 and 3 was substituted into the equation we would obtain
2.76 x 10-3 = k [0.020]x [0.030]y
8.28 x 10-3 k [0.020]x [0.090]y
and if we cancel all common terms we would obtain 2.76 x 10-3 = k [0.020]x [0.030]y
8.28 x 10-3 k [0.020]x [0.090]y
2.76 x 10-3 = [0.030]y 0.333 = (0.333)y
8.28 x 10-3 [0.090]y
(2) which yields y = 1 as the order of reaction with respect to OH-.
(3) the overall rate law can now be written as follows:
Rate = k [ClO2]2 [OH-]1
(4) and using the data from experiment 1 we can determine the rate constant as follows:
2.48 x 10-2 = k [0.060]2 [0.030]
k = 2.48 x 10-2 / [0.060]2 [0.030] = 229.63
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
11/22
t1/2 = 0.693 / 3.47 x 10-3 = 199.7 days
0.693 = (3.47 x 10-3) t1/2
0.693 = k t1/2
10-3
k = - 1.3863 / - 400 = 3.47 x
- 1.3863 = - k (400)
days) 3.2189 - 4.6052 = - k (400)
ln 25 - ln 100 = - k (400
ln [A] - ln [A]0 = - k t
Your Answer:
.
Determine the decay constant (k) and the half-life of a radioactive nucleus
if 75% of the material has decayed in 400 days.
Not yet graded / 0 pts
Question 6
A sample of wood from an ancient tomb was found to contain 15.7 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720
yrs. What is the age of the wood?
Your Answer:
.
Not yet graded / 0 pts
Question 7
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
12/22
Calculate the Kc for the following reaction if an initial reaction mixture of
0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.
CO (g)
+
3 H2 (g)
CH4 (g)
+
H2O (g)
Your Answer:
.
Not yet graded / 0 pts
Question 8
0.693 = k t1/2
= 1.2115 x 10-4
0.693 = k (5720 yrs)
k = 0.693 / 5720
ln [A] - ln [A]0 = - k t
ln 15.7 - ln 100 = - (1.2115 x
10-4) t
2.7537 - 4.6052 = - (1.2115 x 10-4) t
1.8515 / - 1.2115 x 10-4 = 15,283 yrs
t (age) = -
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
13/22
Calculate the Kc for the following reaction if an initial reaction mixture of 0.700 mole of NH3 and 0.910 mole of O2 in a 1.00 liter container forms an equilibrium mixture containing 0.230 mole of N2 and corresponding amounts of NH3, O2, and H2O.
Not yet graded / 0 pts
Question 9
If the equilibrium mixture contains 0.198 mole of H2O then the following amounts of materials must be present in the equilibrium mixture:
H2O = 0.198 mole (as stated)
CH4 = 0.198 mole (1 mole of CH4 forms for every mole of H2O that is formed)
CO = 0.500 - 0.198 mole (1 mole of CO reacts for every mole of H2O that is formed)
H2 = 1.500 - 3 x 0.198 mole (3 mole of H2 reacts for every mole of H2O that is formed)
Change all amounts to moles/L before entering in Kc expression:
H2O = 0.198 mole / 5.00 L = 0.0396 M CH4 = 0.198 mole / 5.00 L = 0.0396 M CO = 0.302 mole / 5.00 L = 0.0604 M
H2 = 0.906 mole / 5.00 L = 0.1812 M
Kc = [CH4] [H2O] = [0.0396] [0.0396] = 4.36
[CO] [H2]3
[0.0604] [0.1812]3
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
14/22
4 NH3 (g)
+ 3 O2 (g)
2 N2 (g)
+ 6 H2O (g)
Your Answer:
.
If the equilibrium mixture contains 0.230 mole of N2 then the following amounts of materials must be present in the equilibrium mixture:
N2 = 0.230 mole of (as stated)
H2O = 0.230 x 6/2 mole (6 mole of H2O forms for every 2 mole of N2 that is formed)
NH3 = 0.700 - 4/2 x 0.230 mole (4 mole of NH3 reacts for every 2 mole of N2 that is formed)
O2 = 0.910 - 3/2 x 0.230 mole (3 mole of O2 reacts for every 2 mole of N2 that is formed)
Change all amounts to moles/L before entering in Kc expression:
N2 = 0.230 mole / 1.00 L = 0.230 M H2O = 0.690 mole / 1.00 L = 0.690 M NH3 = 0.240 mole / 1.00 L = 0.240 M O2 = 0.565 mole / 1.00 L = 0.565 M
Kc = [N2]2 [H2O]6 = [0.230]2 [0.690]6 = 9.54
[NH3]4 [O2]3
[0.240]4 [0.565]3
Not yet graded / 0 pts
Question 10
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
15/22
The very small Kc indicates that this equilibrium mixture will be
composed of mostly reactants.
Your Answer:
.
Kc = 4.6 x 10-31
2 NO (g)
O2 (g)
+
N2 (g)
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure
to explain your answer.
Not yet graded / 0 pts
Question 11
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of predominately reactants, predominately products or significant amounts of both products and reactants. Be sure
to explain your answer.
N2 (g)
+
3 H2 (g)
2 NH3 (g)
Kc = 4.1 x 108
Your Answer:
.
The very large Kc indicates that this equilibrium mixture will be
composed of mostly products.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
16/22
Question 12
Not yet graded / 0 pts
Explain the following:
(a) Explain why reaction rate is affected by a reactant concentration change.
(b) Explain why reaction rate is affected by a temperature change.
(c) Explain why powdered zinc reacts faster than a chunk of
zinc.
(d) Explain how a catalyst works.
(e) Explain why reactions in solution are usually faster than heterogeneous reactions.
(f) Explain why stirring speeds up the rate of heterogeneous reactions.
Your Answer:
.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
17/22
The equilibrium reaction below has the following equilibrium mixture concentrations:
SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00
If the concentration of O2 at equilibrium is decreased to [0.100], 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.
2 SO3 (g)
2 SO2 (g) + O2 (g)
Your Answer:
Not yet graded / 0 pts
Question 13
(a) A change in reactant concentration changes the number of particles colliding which changes the probability of the particles undergoing the reaction.
(b) As the temperature is changes, the kinetic energy of the reacting particles is changed and this changes the probability of particles colliding with sufficient energy to undergo the reaction.
(c) Solid particles only react at their surface and since powdered zinc particles have a greater surface energy than a chunk of zinc the powdered zinc will react faster.
(d) A catalyst speeds up the rate of a chemical reaction by causing formation of an alternate transition state with lower Energy of Activation, thereby allowing more reactant molecules to form the Transition State at lower energy.
(e) The reactants particles can collide with one another much more readily in solution whereas particles in heterogeneous reactions can only collide by way of their surfaces. An increase in collisions causes an increase in reaction rate.
(f) Stirring heterogeneous reactions increases the collisions between particles which speeds up the rate of reaction.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
18/22
The equilibrium reaction below has the following equilibrium mixture concentrations:
SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200] with Kc = 4.00
If the concentration of SO3 at equilibrium is increased to [0.300], 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.
2 SO3 (g)
2 SO2 (g) + O2 (g)
Your Answer:
.
Not yet graded / 0 pts
Question 14
.
Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200]
When O2 = [0.100], Q = [0.400]2 [0.100] = 2.00
[0.0894]2
The reaction must shift briefly in the forward direction to increase the [O2] to come back to equilibrium. This is in agreement with Qc
< Kc which also predicts the reaction will proceed to the right.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
19/22
The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is decreased from 4.00 liters to 2.00 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.
2 SO3 (g)
2 SO2 (g) + O2 (g)
Your Answer:
.
Not yet graded / 0 pts
Question 15
Kc = 4.00 when SO3 = [0.0894], SO2 = [0.400] and O2 = [0.200]
When SO3 = [0.300], Q = [0.400]2 [0.200] = 0.356
[0.300]2
The reaction must shift briefly in the forward direction to decrease the [SO3] to come back to equilibrium.This is in agreement with Qc
< Kc which also predicts the reaction will proceed to the right.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
20/22
The equilibrium reaction below has the Kc = 4.00. If the volume of the system at equilibrium is increased from 3.00 liters to 9.00 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.
2 SO3 (g)
2 SO2 (g) + O2 (g)
Your Answer:
.
Not yet graded / 0 pts
Question 16
When volume decreases from 4.00 to 2.00, the pressure doubles and the concentration of all gases (SO3, SO2 and O2) doubles so:
(at equilibrium) Qc =Kc = [SO2]2 [O2] = 4.0
[SO3]2
(volume halved = pressure doubled = conc doubled) Qc = [2 SO2]2 [2O2] = 2Kc
[2 SO3]2
The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (reverse direction : 3 moles of gas yields 2 moles of gas) to come back to equilibrium. This is in agreement with Qc > Kc: the reaction will proceed to the left (in the direction of the reactants).
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
21/22
The equilibrium reaction below has the Kc = 4.00 at 25oC. If the temperature of the system at equilibrium is increased to 100oC, how and for what reason will the equilibrium shift. Also show and explain how and
why the Kc value will change.
2 SO3 (g)
2 SO2 (g) + O2 (g)
ΔH0 = +
94.1 kJ
Your Answer:
Not yet graded / 0 pts
Question 17
When volume increases from 3.00 to 9.00, the pressure is cut to 1/3 of original and the concentration of all gases (SO3, SO2 and
O2) decreases to 1/3 so:
(at equilibrium) Qc =Kc = [SO2]2 [O2] = 4.0
[SO3]2
(volume tripled = Pi/3 = conc/3) Qc = [1/3 SO2]2 [1/3O2] =
Kc
[1/3 SO3]2
3
The reaction must shift briefly in the direction that increases the pressure by going toward the side with the greater moles of gas (forward direction : 2 moles of gas yields 3 moles of gas) to come back to equilibrium. This is in agreement with Qc < Kc: the reaction
will proceed to the right (in the direction of the products).
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
22/22
The equilibrium reaction below has the Kc = 0.250 at 25oC. If the temperature of the system at equilibrium is decreased to 0oC, how and for what reason will the equilibrium shift. Also show and explain how and
why the Kc value will change.
2 SO2 (g) + O2 (g)
2 SO3 (g)
ΔH0 = -
94.1 kJ
Your Answer:
.
Not yet graded / 0 pts
Question 18
.
If the temperature is increased on this reaction at equilibrium which has a +ΔH0, the reaction will briefly shift in the direction that uses up some of the added heat [+ΔH0 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 SO2 and O2 (in the numerator of Kc) and decrease the concentration of SO3 (in the denominator of Kc) which will
increase the value of Kc.
2022 Module 1: Problem Set: General Chemistry 2 with Lab- Schiren
23/22
Quiz Score: 0 out of 0
If the temperature is decreased on this reaction at equilibrium which has a -ΔH0, the reaction will briefly shift in the direction that produces some heat [-ΔH0 indicates an exothermic (heat producing) forward reaction] so this reaction will shift in the forward direction. Simultaneously, this forward shift will increase the concentration of SO3 (in the numerator of Kc) and decrease the concentration of SO2 and O2 (in the denominator of Kc) which will increase the value of Kc. [Show Less]