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University of Alberta

Chemistry

CHEM102

Sai Yiu

Winter

Description

page 1
Chemistry 102
Practice Final Examination
Dr. Sai Yiu
Answer all questions.
This exam counts 45% of your course grade
The last two pages are data sheets for calculations
If you need to use back pages, please indicate clearly by writing on the
front page
Please note that whenever you do your calculations, make sure to:
1. Show all your steps one at a time (i.e. showing how the calculation was
done)
2. Give the correct significant figures
3. Give a suitable unit for the answer
Question Maximum Total
Number marks
1 12
Please note that:
The questions in this paper were taken from a
collection ofprevious exams in the U of A and 2 8
different university colleges which I taught. The
goalof this paper is to let you know the format of3 11
my exams so that you can be well prepared for the
upcoming final exam. 4 6
5 10
Please be aware that each exam is unique and is
tailored for a specific class and so the level of 6 7
difficulty of the exam may be different from that
of yours.
7 9
You also need to know that writing an exam at
home is very different from writing it in an 8 14
examination hall.
9 13
Therefore, do not complain that the practice
exam is easier than the real exam. 10 10
Total 100 page 2
1 At 125 C, K =p0.25 for the reaction:
2NaHCO (s)3 Na 2O (s3 + CO (g)2+ H O (g)2
o
A 1.00-L flask containing 10.0 g NaHCO is ev3cuated and heated to 125 C until
equilibrium is established.
(i) Calculate the total pressure of the equilibrium mixture. [6 marks]
pCO2 pH2O = K p 0.25
Because equal number of moles for CO and H O, their partial pressure should be equal.
2 2
(p CO2) = 0.25
pCO2 = √ (0.25) = 0.50 atm
Total pressure = 0.50 x 2 = 1.0 atm
(ii) Calculate the masses of NaHCO and 3a CO pre2ent3at equilibrium. [6 marks]
The number of mole of CO formed can be found from the ideal gas equation:
2
P V
n =
RT
0.50 atm x 1.00 L
n =
-1 -1
0.08206 L atm mol K x (273 + 125) K
= 0.015 mol
From the equation, number of mole of NaHCO requir3d = 2 x 0.015 mol = 0.030 mol
10.0 g
Number of mole of NaHCO used 3
(22.99 + 1.01 + 12.01 + 16.00 x 3) g / mol
= 0.119 mol
Therefore, number of mol of NaHCO in ex3ess = 0.119 mol - 0.030 mol = 0.089 mol
Mass of NaHCO at 3quilibrium
= 0.089 mol x (22.99 + 1.01 + 12.01 + 16.00 x 3) g / mol = 7.5 g page 3
Number of mole of Na CO is the same as CO = 0.015 mol
2 3 2
Mass of Na CO2at e3uilibrium
= 0.015 mol x (22.99 x 2 + 12.01 + 16.00 x 3) g / mol = 1.6 g
2 a) The decomposition of NO (g) o2curs by the following bimolecular
elementary reaction:
2NO (g2 2NO g) + O (g)2
The rate constant at 273 K is 2.3 x 10 -12L mol s and the activation energy is 111 kJ /
mol. How long will it take for the concentration of NO (g) to 2ecrease from an initial
partial pressure of 2.5 atm to 1.5 atm at 500.K? Assume ideal gas behaviour.
[8 marks]
We need to find rate constant k 500at 500 K by using the Arrhenius equation.
-1
111 kJ mol
ln k 273= ln A -
-3 -1 -1
8.314 x 10 kJ mol K x 273 K
-1
111 kJ mol
ln k 500= ln A -
-3 -1 -1
8.314 x 10 kJ mol K x 500 K
Combining gives:
-1
k500 111 kJ mol 1 1
ln = -
-3 -1 -1
k273 8.314 x 10 kJ mol K 273 K 500K
= 22.2
k
500 22.2 9 9
= e = 4 x 10 (4.38 x 10 )
2.3 x 10 -12 L mol s1 -1
k500= 4 x 10 x 2.3 x 10 -12 L mol s1 -1
-1 -1
= 0.01 L mol s
Need to convert pressure to concentration [NO ] (n /2V) by using the ideal gas equation: page 4
n P 2.5 atm
[NO 2 o = =
V RT RT
After some time,
n P 1.5 atm
[NO 2 = = =
V RT RT
nd
For a 2 order reaction, the equation is:
1 1
= + kt
[A] [A]o
Substituting [A] by [N2 ] gives:
1 1
RT - = k t
1.5 atm 2.5 atm
RT 1 1
t = -
k 1.5 atm 2.5 atm
-1 -1
0.08206 L atm mol K x 500 K 1 1
t = -
0.01 L mol s1-1 1.5 atm 2.5 atm
3
= 1 x 10 s (1094 s) page 5
b) The following mechanism for the synthesis of nitrosyl bromide (NOBr) was proposed:
NO (g) + Br (2) k 1 NOBr (g2 fast
NOBr (g2 k -1 NO (g) + Br (2) fast
NOBr (g2 + NO (g) k2 2NOBr (g) slow
(i) What is the equation for the formation of nitrosyl bromide? [1 mark]
2NO (g) + Br (g2 2NOBr(g)
(ii) Derive a rate law for this reaction. [4 marks]
Rate of reaction for step 1 = k [1O][Br ] 2
Rate of reaction for step -1 = k- 1NOBr ] 2
Rate of reaction for step 2 = k [2O] [NOBr ] 2
Rate of reaction = k [2O] [NOBr ] (ra2e-determining step)
Since an equilibrium is established for step 1 and step -1, their rates are equal.
k 1NO][Br ] 2 k [N-1r ] 2
k1[NO][Br ] 2
[NOBr ] 2
k-1
2
k 1 [2O] [Br ] 2
Rate of reaction =
k-1 page 6
3 The following is a cell diagram ofa certain Voltaic cell:
2+ 2+
Fe (s) │Fe (aq) (0.68 M) ║Co (aq) (0.15 M) │Co (s)
(i) Calculate the standard cell potential for the Voltaic cell. [2 marks]
Fe (aq) + 2e Fe (s) E = - 0.440 V
Co (aq) + 2e Co (s) E = - 0.277 V
E cell= E ocathode Eoanode
o
E cell= - 0.277 V – ( - 0.440 V) = 0.163 V
(ii) Write a balanced equation to show the overall cell reaction. Make sure that you
show your steps. [2 marks]
2+
Since Fe / Fe is the anode, therefore the equation should be reversed:
2+ o
Fe (s) Fe (aq) + 2e E = + 0.440 V
Since Co / Co is the cathode, therefore the equation is:
2+ o
Co (aq) + 2e Co (s) E = - 0.277 V
Overall equation:
Co (aq) + Fe (s) Co (s) + Fe (aq) E ocell + 0.440 V - 0.277 V = 0.163 V
(iii) What is the cell potential of the Voltaic cell as indicated by the cell diagram?
[3 marks]
The Nernst equation should be used.
2+
0 0.0592 [Fe ]
Ecell= E cell - log
2 [Co ]+
0.0592 0.68
= 0.163 V - log
2 0.15
= 0.144 V page 7
(b) The following ionic equation is a redoxreaction.
Ce (aq) + Fe (aq)+ Ce (aq) + Fe (aq)
Predict whether the above reaction is spontaneous under standard conditions. Briefly explain your
reasoning. [4 marks]
4+ 3+ o
Ce (aq) + e Ce (aq) E = + 1.70 V
Fe (aq) + e Fe (aq) E = + 0.77 V
Inorder to get the equation above, the Ce / Ce equation needs to be reversed:
3+ 4+ o
Ce (aq) Ce (aq) + e E = - 1.70 V
3+ 2+
Keep the Fe / Fe equation:
Fe (aq) + e Fe (aq) E = + 0.77 V
Overall equation:
3+ 3+ 4+ 2+ o
Ce (aq) + Fe (aq) Ce (aq) + Fe (aq) E cell + 0.77 V - 1.70 V = - 0.93 V
Since the cell potential is negative, the reaction is therefore not spontaneous.
4 (i) Explain the difference between a galvanic cell and an electrolytic cell. [2 marks]
A Galvanic cell generates electric current bya spontaneous chemical reaction
An electrolytic cell brings about a non-spontaneous chemical reaction by using
electric curren.
(ii) In an electrolytic cell a current of23.0 Apasses through a ZnSO sol4tion. How
long does it take to deposit 85.5 g of Zn? [4 marks]
Zn (aq) + 2e Zn(s)
Number of mole of Zndeposited =
85.5 g
= 1.31 mol
65.38 g / mol
From the equation, the number of mole of electronrequired page 8
= 2 x1.31 mol = 2.62 mol
Iftime is t the charge passing throughthe cell = 23.0 C s xt s
The number of mole of electron is
23.0 xt C
96485 C / mol
23.0 xt C
= 2.62 mol
96485 C / mol
2.62 mol x96485 C / mol
t =

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