Chemistry 102 Practice Final Examination
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 of previous exams in the U of A and different university colleges which I taught. The goal of this paper is to let you know the format of my exams so that you can be well prepared for the upcoming final exam.
2
8
3
11
4
6
Please be aware that each exam is unique and is tailored for a specific class and so the level of difficulty of the exam may be different from that of yours.
5
10
6
7
7
9
8
14
9
13
10
10
Total
100
You also need to know that writing an exam at home is very different from writing it in an examination hall. Therefore, do not complain that the practice exam is easier than the real exam.
page 2
1
At 125oC, Kp = 0.25 for the reaction: 2NaHCO3 (s)
Na2CO3 (s) + CO2 (g) + H2O (g)
A 1.00-L flask containing 10.0 g NaHCO3 is evacuated and heated to 125oC until equilibrium is established. (i)
Calculate the total pressure of the equilibrium mixture.
[6 marks]
pCO2 pH2O = Kp = 0.25 Because equal number of moles for CO2 and H2O, their partial pressure should be equal. (pCO2)2 = 0.25 pCO2 = √ (0.25) = 0.50 atm Total pressure = 0.50 x 2 = 1.0 atm (ii)
Calculate the masses of NaHCO3 and Na2CO3 present at equilibrium. [6 marks] The number of mole of CO2 formed can be found from the ideal gas equation: PV n= RT 0.50 atm x 1.00 L n= 0.08206 L atm mol-1 K-1 x (273 + 125) K = 0.015 mol
From the equation, number of mole of NaHCO3 required = 2 x 0.015 mol = 0.030 mol 10.0 g Number of mole of NaHCO3 used = (22.99 + 1.01 + 12.01 + 16.00 x 3) g / mol = 0.119 mol Therefore, number of mol of NaHCO3 in excess = 0.119 mol - 0.030 mol = 0.089 mol Mass of NaHCO3 at equilibrium = 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 Na2CO3 is the same as CO2 = 0.015 mol Mass of Na2CO3 at equilibrium = 0.015 mol x (22.99 x 2 + 12.01 + 16.00 x 3) g / mol = 1.6 g
2
a) The decomposition of NO2 (g) occurs by the following bimolecular elementary reaction: 2NO2(g)
2NO g) + O2 (g)
The rate constant at 273 K is 2.3 x 10-12 L mol-1 s-1 and the activation energy is 111 kJ / mol. How long will it take for the concentration of NO2 (g) to decrease 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 k500 at 500 K by using the Arrhenius equation. 111 kJ mol-1 ln k273 = ln A -
8.314 x 10-3 kJ mol-1 K-1 x 273 K 111 kJ mol-1
ln k500 = ln A -
8.314 x 10-3 kJ mol-1 K-1 x 500 K
Combining gives: 111 kJ mol-1
k500 ln
1
= k273
1 -
8.314 x 10-3 kJ mol-1 K-1
273 K
500K
= 22.2 k500
= e 22.2 = 4 x 109 (4.38 x 109)
2.3 x 10-12 L mol-1 s-1 k500 = 4 x 109 x 2.3 x 10-12 L mol-1 s-1 = 0.01 L mol-1 s-1
Need to convert pressure to concentration [NO2] (n / V) by using the ideal gas equation:
page 4
n [NO2]o =
P
2.5 atm
=
=
V RT After some time, n [NO2] =
RT
P
1.5 atm
= V
= RT
RT
For a 2nd order reaction, the equation is: 1
1 =
[A]
+ kt [A]o
Substituting [A] by [NO2] gives: 1 RT
1
=kt 1.5 atm 2.5 atm
RT
1
t=
1 -
k
1.5 atm
2.5 atm
0.08206 L atm mol-1 K-1 x 500 K 1 t=
1 -
0.01 L mol-1 s-1 = 1 x 103 s (1094 s)
1.5 atm
2.5 atm
page 5
b)
The following mechanism for the synthesis of nitrosyl bromide (NOBr) was proposed:
NO (g) + Br2 (g) k1 NOBr2 (g) k-1 NOBr2 (g) + NO (g) k2
NOBr2 (g) NO (g) + Br2 (g) 2NOBr (g)
fast fast slow
(i) What is the equation for the formation of nitrosyl bromide? 2NO (g) + Br2 (g)
[1 mark]
2NOBr (g)
(ii) Derive a rate law for this reaction. Rate of reaction for step 1 = k1 [NO][Br2]
[4 marks]
Rate of reaction for step -1 = k-1 [NOBr2] Rate of reaction for step 2 = k2 [NO] [NOBr2] Rate of reaction = k2 [NO] [NOBr2] (rate-determining step) Since an equilibrium is established for step 1 and step -1, their rates are equal. k1 [NO][Br2] = k-1 [NOBr2] k1 [NO][Br2] [NOBr2] = k-1 k1 k2 [NO]2 [Br2] Rate of reaction = k-1
page 6
3
The following is a cell diagram of a certain Voltaic cell: Fe (s) │Fe2+ (aq) (0.68 M) ║Co2+ (aq) (0.15 M) │Co (s)
(i)
Calculate the standard cell potential for the Voltaic cell. [2 marks] 2+ o Fe (aq) + 2e Fe (s) E = - 0.440 V Co2+ (aq) + 2e
Eo = - 0.277 V
Co (s)
Eocell = Eocathode – Eoanode Eocell = - 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]
Since Fe2+ / Fe is the anode, therefore the equation should be reversed: Fe2+ (aq) + 2e
Fe (s)
Eo = + 0.440 V
Since Co2+ / Co is the cathode, therefore the equation is: Co2+ (aq) + 2e
Eo = - 0.277 V
Co (s)
Overall equation: Co2+ (aq) + Fe (s)
(iii)
Co (s) + Fe2+ (aq) Eo cell = + 0.440 V - 0.277 V = 0.163 V
What is the cell potential of the Voltaic cell as indicated by the cell diagram? [3 marks] The Nernst equation should be used. [Fe2+ ]
0.0592 Ecell = E
0
cell
-
log [Co2+]
2 0.0592 = 0.163 V -
log 2
= 0.144 V
0.68 0.15
page 7
(b) The following ionic equation is a redox reaction. Ce3+ (aq) + Fe3+ (aq)
Ce4+ (aq) + Fe2+ (aq)
Predict whether the above reaction is spontaneous under standard conditions. Briefly explain your reasoning. [4 marks] Ce4+ (aq) + e
Ce3+ (aq)
Eo = + 1.70 V
Fe3+ (aq) + e
Fe2+ (aq)
Eo = + 0.77 V
In order to get the equation above, the Ce3+ / Ce4+ equation needs to be reversed: Ce3+ (aq)
Ce4+ (aq) + e
Eo = - 1.70 V
Keep the Fe3+ / Fe2+ equation: Fe3+ (aq) + e
Fe2+ (aq)
Eo = + 0.77 V
Overall equation: Ce3+ (aq) + Fe3+ (aq)
Ce4+ (aq) + Fe2+ (aq) Eo 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 by a spontaneous chemical reaction An electrolytic cell brings about a non-spontaneous chemical reaction by using electric current.
(ii)
In an electrolytic cell a current of 23.0 A passes through a ZnSO4 solution. How long does it take to deposit 85.5 g of Zn? [4 marks] Zn2+ (aq) + 2e
Zn (s)
Number of mole of Zn deposited = 85.5 g = 1.31 mol 65.38 g / mol From the equation, the number of mole of electron required
page 8
= 2 x 1.31 mol = 2.62 mol If time is t the charge passing through the cell = 23.0 C s-1 x t s The number of mole of electron is 23.0 x t C 96485 C / mol 23.0 x t C = 2.62 mol 96485 C / mol 2.62 mol x 96485 C / mol t= 23.0 C s-1 4
t = 1.09 x 10 s 5
10.00g of ammonium chloride is added to 100.0 mL of 0.750 M sodium hydroxide solution. Calculate the pH of the resulting solution. [10 marks] 10.00 g Number of mole of ammonium chloride =
= 0.1869 mole (14.01 + 1.008 x 4 + 35.45) g / mol
Because 100.0 mL = 0.1000 L (Volume of solution)
The equation for the reaction is: NH4+ (aq) + OH- (aq)
NH3 (aq) + H2O (l)
Number of mole of OH- added = 0.750 mol / L x 0.1000 L = 0.0750 mole Thus, An ICF table is needed:
I C F(final)
NH4+ 0.1869 0.1869 - 0.0750 0.1119
OH0.0750 - 0.0750 0
NH3 0 + 0.0750 0.0750
page 9
Since both conjugate acid base are present, a buffer solution is formed and we can use the Henderson-Hasselbalch equation to find pH [NH3] pH = pKa + log
[NH4+]
1.00 x 10-14 Ka =
= 5.6 x 10-10 1.8 x 10-5 0.0750 / 0.1000
pH = -log 5.6 x 10-10 + log 0.1119 / 0.1000 = 9.08
6
(i) State the Hess’ Law
[1 mark]
The enthalpy change for a chemical reaction is the same whether it takes place in one step or a series of steps
(ii) Methanol, CH3OH, is being promoted as a clean fuel and is used as an additive in many gasolines. When methanol burns in a gasoline engine the following reaction occurs: [6 marks] CH3OH (l) + 3/2 O2 (g) CO2 (g) + 2 H2O (l) Given the following information, determine the standard enthalpy change, H°, for the above reaction. Show all your work. C (graphite) + 2 H2 (g) + 1/2 O2 (g) CH3OH (l)
H1°= - 238.7 kJ
C (graphite) + O2 (g) CO2 (g)
H2°= - 395.5 kJ
H2 (g) + 1/2 O2 (g) H2O (l)
H3°= - 285.8 kJ
The question applies the Hess’ Law for finding H° Equation 1 is reversed: CH3OH (l) C (graphite) + 2 H2 (g) + 1/2 O2 (g)
H1°= + 238.7 kJ (Equation 4)
Multiply equation 3 by 2: 2H2 (g) + O2 (g) 2 H2O (l)
H3°= 2 x (- 285.8 kJ) (Equation 5)
page 10
Combine equations 2, 4 and 5 gives: C (graphite) + O2 (g) CO2 (g)
H2°= - 395.5 kJ (Equation 2)
CH3OH (l) C (graphite) + 2 H2 (g) + 1/2 O2 (g)
H1°= + 238.7 kJ (Equation 4)
2H2 (g) + O2 (g) 2 H2O (l)
H3°= 2 x (- 285.8 kJ) (Equation 5)
CH3OH (l) + 3/2 O2 (g) CO2 (g) + 2 H2O (l) H°= - 395.5 kJ + 238.7 kJ + 2 x (- 285.8 kJ) = - 728.4 kJ 7
a) Most thermodynamic functions are relative (i.e. no absolute value can be given). With the Third Law of Thermodynamics, we are able to determine the absolute value of a certain thermodynamic function. What is it? [1 mark] Entropy The equation represents an equilibrium at 470o C
b)
H2(g) + I2(g)
2HI(g)
The partial pressures of the components was found to be: H2(g) = 0.472 atm. I2(g) = 0.011 atm. HI(g) = 0.517 atm. (i) Find the equilibrium constant Kp for the equilibrium.
[4 marks]
P HI2 Kp= P H2 PI2 (0.517) 2 = 0.011 x 0.472 = 51 (51.48) (ii)
Find the standard free energy, ∆Go for the reaction.
[4 marks]
∆Go = - RT lnK = - 8.314 J K-1 mol-1 x 743 K ln 51 = - 24300 J mol-1 = - 24.3 kJ mol-1
page 11
8
The following equation shows the formation of ammonia in the Haber Process: N2 (g) + 3H2 (g)
(i)
2NH3 (g)
Calculate ΔHo and ΔSo for the formation of ammonia. From the data sheet, ΔHfo (NH3) = - 46.11 kJ / mol-1
[4 marks]
ΔHo = 2 mol x - 46.11 kJ / mol-1 = - 92.22 kJ ΔSo = 2 mol of NH3 x 192.5 J mol-1 K-1 - 1 mol of N2 x 191.6 J mol-1 K-1 - 3 mol of H2 x 130.7 J mol-1 K-1 = - 198.7 J K-1 = - 0.1987 kJ K-1 (ii)
Calculate the temperature when Kp = 1.00
[6 marks]
ΔGo = - RT ln Kp When Kp = 1.00, ln Kp = 0 and so ΔGo = 0 Applying ΔGo = ΔHo - TΔSo WhenΔGo = 0, ΔHo = TΔSo ΔHo - 92.22 kJ T= = = 464.1 K ΔSo - 0.1987 kJ K-1
(iii)
The production of ammonia in industry is usually carried out at 400. oC. Calculate Kp at 400.oC. [4 marks] ΔGo = ΔHo - TΔSo = - 92.22 kJ – (273 + 400. K) (- 0.1987 kJ K-1) = 41.50 kJ ΔGo = - RT ln Kp = 41.50 kJ 41.50 kJ ln Kp =
= - 7.42 -3
-1
-1
- 8.314 x 10 kJ mol K x (273 + 400. K)
Kp = e
- 7.42
= 6.0 x 10-4
page 12
9
(i) Barium carbonate (BaCO3) is a sparingly soluble solid. Write an equation for the dissolution (dissolving) of barium carbonate in water. [2 marks] BaCO3 (s) Ba2+ (aq) + CO32- (aq) (ii)
Write the solubility product (Ksp) expression for barium carbonate. [1 mark] Ksp = [Ba2+] [CO32-]
(iii) Given that the solubility product of barium carbonate is 7.0 x 10-9 mol2 L-2 at 25oC. What maximum mass of barium carbonate can be dissolved in 500 ml of water at 25 oC? [8 marks] [Ba2+] [CO32-] = 7.0 x 10-9 [Ba2+ ]
BaCO3 Initial Excess Change -x Equilibrium Excess - x Therefore, [Ba2+ ] = x
0 +x x
[CO32-] 0 +x x
[CO32-] = x
Ksp = x2 = 7.0 x 10-9 x = √ (7.0 x 10-9) = 8.4 x 10-5 M Mass of barium carbonate = 8.4 x 10-5 mol / L x (137.33 + 12.01 + 3 x 16.00) g / mol = 0.017 g /L = 0.017 / 2 g / 500 mL = 8.3 x 10-3 g / 500 mL
(iv)
Do you expect the solubility of barium carbonate increase or decrease in a solution of sodium carbonate? Briefly explain. [2 marks]
BaCO3 (s) Ba2+ (aq) + CO32- (aq) In the presence of sodium carbonate, the equilibrium will shift to the left according to the Le Chatelier principle and so the solubility of BaCO3 is expected to decrease.
page 13
10
(i)
Suggest systematic names for the following two coordination compounds [4 marks] [Pt(en)2Br2](NO3)2 Dibromobis(ethylenediamine)platinum(IV) nitrate
K2[Mn(H2O)(CN)5] Potassium aquapentacyanomanganate(III)
(ii) Draw the three-dimensional structures of all possible stereoisomers for the coordination complex [Co(en)2Br2]+ [6 marks] en
en
Br
Br
Br Co
Co Br
en
Br
en
en
End of Paper
13
Co Br
en
page 14
Data Sheet
1 F = 96485 C / mol electron
[A] = - k t + [A]o
PV = nRT
ln [A] = - kt + ln [A]o 1 1 = + kt [A] [A]o
R = 0.08206 L atm mol-1 K-1 R = 8.314 J mol-1 K-1 0.0oC = 273.15 K
[A]o t½ = 2k
∆U = q + w
ln 2
∆H = qp = ∆U + P∆V
t½ = k 1 t½ = k [A]o k = Ae-Ea/RT k2 ln
Ea
1
R
T1
= k1
- ΔHo
1 -
ln K = T2
14
RT
ΔSo + R
Q (charge) = I t
page 15
For thermodynamic properties of substances at 298.15 K, please refer to the University of Alberta Introductory Chemistry Information Sheet* in your WebCT
15
page 16
In case if don’t find the necessary data, please refer to your text book or from the University of Alberta Introductory Chemistry Information Sheet
16
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 of previous exams in the U of A and different university colleges which I taught. The goal of this paper is to let you know the format of my exams so that you can be well prepared for the upcoming final exam.
2
8
3
11
4
6
Please be aware that each exam is unique and is tailored for a specific class and so the level of difficulty of the exam may be different from that of yours.
5
10
6
7
7
9
8
14
9
13
10
10
Total
100
You also need to know that writing an exam at home is very different from writing it in an examination hall. Therefore, do not complain that the practice exam is easier than the real exam.
page 2
1
At 125oC, Kp = 0.25 for the reaction: 2NaHCO3 (s)
Na2CO3 (s) + CO2 (g) + H2O (g)
A 1.00-L flask containing 10.0 g NaHCO3 is evacuated and heated to 125oC until equilibrium is established. (i)
Calculate the total pressure of the equilibrium mixture.
[6 marks]
pCO2 pH2O = Kp = 0.25 Because equal number of moles for CO2 and H2O, their partial pressure should be equal. (pCO2)2 = 0.25 pCO2 = √ (0.25) = 0.50 atm Total pressure = 0.50 x 2 = 1.0 atm (ii)
Calculate the masses of NaHCO3 and Na2CO3 present at equilibrium. [6 marks] The number of mole of CO2 formed can be found from the ideal gas equation: PV n= RT 0.50 atm x 1.00 L n= 0.08206 L atm mol-1 K-1 x (273 + 125) K = 0.015 mol
From the equation, number of mole of NaHCO3 required = 2 x 0.015 mol = 0.030 mol 10.0 g Number of mole of NaHCO3 used = (22.99 + 1.01 + 12.01 + 16.00 x 3) g / mol = 0.119 mol Therefore, number of mol of NaHCO3 in excess = 0.119 mol - 0.030 mol = 0.089 mol Mass of NaHCO3 at equilibrium = 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 Na2CO3 is the same as CO2 = 0.015 mol Mass of Na2CO3 at equilibrium = 0.015 mol x (22.99 x 2 + 12.01 + 16.00 x 3) g / mol = 1.6 g
2
a) The decomposition of NO2 (g) occurs by the following bimolecular elementary reaction: 2NO2(g)
2NO g) + O2 (g)
The rate constant at 273 K is 2.3 x 10-12 L mol-1 s-1 and the activation energy is 111 kJ / mol. How long will it take for the concentration of NO2 (g) to decrease 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 k500 at 500 K by using the Arrhenius equation. 111 kJ mol-1 ln k273 = ln A -
8.314 x 10-3 kJ mol-1 K-1 x 273 K 111 kJ mol-1
ln k500 = ln A -
8.314 x 10-3 kJ mol-1 K-1 x 500 K
Combining gives: 111 kJ mol-1
k500 ln
1
= k273
1 -
8.314 x 10-3 kJ mol-1 K-1
273 K
500K
= 22.2 k500
= e 22.2 = 4 x 109 (4.38 x 109)
2.3 x 10-12 L mol-1 s-1 k500 = 4 x 109 x 2.3 x 10-12 L mol-1 s-1 = 0.01 L mol-1 s-1
Need to convert pressure to concentration [NO2] (n / V) by using the ideal gas equation:
page 4
n [NO2]o =
P
2.5 atm
=
=
V RT After some time, n [NO2] =
RT
P
1.5 atm
= V
= RT
RT
For a 2nd order reaction, the equation is: 1
1 =
[A]
+ kt [A]o
Substituting [A] by [NO2] gives: 1 RT
1
=kt 1.5 atm 2.5 atm
RT
1
t=
1 -
k
1.5 atm
2.5 atm
0.08206 L atm mol-1 K-1 x 500 K 1 t=
1 -
0.01 L mol-1 s-1 = 1 x 103 s (1094 s)
1.5 atm
2.5 atm
page 5
b)
The following mechanism for the synthesis of nitrosyl bromide (NOBr) was proposed:
NO (g) + Br2 (g) k1 NOBr2 (g) k-1 NOBr2 (g) + NO (g) k2
NOBr2 (g) NO (g) + Br2 (g) 2NOBr (g)
fast fast slow
(i) What is the equation for the formation of nitrosyl bromide? 2NO (g) + Br2 (g)
[1 mark]
2NOBr (g)
(ii) Derive a rate law for this reaction. Rate of reaction for step 1 = k1 [NO][Br2]
[4 marks]
Rate of reaction for step -1 = k-1 [NOBr2] Rate of reaction for step 2 = k2 [NO] [NOBr2] Rate of reaction = k2 [NO] [NOBr2] (rate-determining step) Since an equilibrium is established for step 1 and step -1, their rates are equal. k1 [NO][Br2] = k-1 [NOBr2] k1 [NO][Br2] [NOBr2] = k-1 k1 k2 [NO]2 [Br2] Rate of reaction = k-1
page 6
3
The following is a cell diagram of a certain Voltaic cell: Fe (s) │Fe2+ (aq) (0.68 M) ║Co2+ (aq) (0.15 M) │Co (s)
(i)
Calculate the standard cell potential for the Voltaic cell. [2 marks] 2+ o Fe (aq) + 2e Fe (s) E = - 0.440 V Co2+ (aq) + 2e
Eo = - 0.277 V
Co (s)
Eocell = Eocathode – Eoanode Eocell = - 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]
Since Fe2+ / Fe is the anode, therefore the equation should be reversed: Fe2+ (aq) + 2e
Fe (s)
Eo = + 0.440 V
Since Co2+ / Co is the cathode, therefore the equation is: Co2+ (aq) + 2e
Eo = - 0.277 V
Co (s)
Overall equation: Co2+ (aq) + Fe (s)
(iii)
Co (s) + Fe2+ (aq) Eo cell = + 0.440 V - 0.277 V = 0.163 V
What is the cell potential of the Voltaic cell as indicated by the cell diagram? [3 marks] The Nernst equation should be used. [Fe2+ ]
0.0592 Ecell = E
0
cell
-
log [Co2+]
2 0.0592 = 0.163 V -
log 2
= 0.144 V
0.68 0.15
page 7
(b) The following ionic equation is a redox reaction. Ce3+ (aq) + Fe3+ (aq)
Ce4+ (aq) + Fe2+ (aq)
Predict whether the above reaction is spontaneous under standard conditions. Briefly explain your reasoning. [4 marks] Ce4+ (aq) + e
Ce3+ (aq)
Eo = + 1.70 V
Fe3+ (aq) + e
Fe2+ (aq)
Eo = + 0.77 V
In order to get the equation above, the Ce3+ / Ce4+ equation needs to be reversed: Ce3+ (aq)
Ce4+ (aq) + e
Eo = - 1.70 V
Keep the Fe3+ / Fe2+ equation: Fe3+ (aq) + e
Fe2+ (aq)
Eo = + 0.77 V
Overall equation: Ce3+ (aq) + Fe3+ (aq)
Ce4+ (aq) + Fe2+ (aq) Eo 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 by a spontaneous chemical reaction An electrolytic cell brings about a non-spontaneous chemical reaction by using electric current.
(ii)
In an electrolytic cell a current of 23.0 A passes through a ZnSO4 solution. How long does it take to deposit 85.5 g of Zn? [4 marks] Zn2+ (aq) + 2e
Zn (s)
Number of mole of Zn deposited = 85.5 g = 1.31 mol 65.38 g / mol From the equation, the number of mole of electron required
page 8
= 2 x 1.31 mol = 2.62 mol If time is t the charge passing through the cell = 23.0 C s-1 x t s The number of mole of electron is 23.0 x t C 96485 C / mol 23.0 x t C = 2.62 mol 96485 C / mol 2.62 mol x 96485 C / mol t= 23.0 C s-1 4
t = 1.09 x 10 s 5
10.00g of ammonium chloride is added to 100.0 mL of 0.750 M sodium hydroxide solution. Calculate the pH of the resulting solution. [10 marks] 10.00 g Number of mole of ammonium chloride =
= 0.1869 mole (14.01 + 1.008 x 4 + 35.45) g / mol
Because 100.0 mL = 0.1000 L (Volume of solution)
The equation for the reaction is: NH4+ (aq) + OH- (aq)
NH3 (aq) + H2O (l)
Number of mole of OH- added = 0.750 mol / L x 0.1000 L = 0.0750 mole Thus, An ICF table is needed:
I C F(final)
NH4+ 0.1869 0.1869 - 0.0750 0.1119
OH0.0750 - 0.0750 0
NH3 0 + 0.0750 0.0750
page 9
Since both conjugate acid base are present, a buffer solution is formed and we can use the Henderson-Hasselbalch equation to find pH [NH3] pH = pKa + log
[NH4+]
1.00 x 10-14 Ka =
= 5.6 x 10-10 1.8 x 10-5 0.0750 / 0.1000
pH = -log 5.6 x 10-10 + log 0.1119 / 0.1000 = 9.08
6
(i) State the Hess’ Law
[1 mark]
The enthalpy change for a chemical reaction is the same whether it takes place in one step or a series of steps
(ii) Methanol, CH3OH, is being promoted as a clean fuel and is used as an additive in many gasolines. When methanol burns in a gasoline engine the following reaction occurs: [6 marks] CH3OH (l) + 3/2 O2 (g) CO2 (g) + 2 H2O (l) Given the following information, determine the standard enthalpy change, H°, for the above reaction. Show all your work. C (graphite) + 2 H2 (g) + 1/2 O2 (g) CH3OH (l)
H1°= - 238.7 kJ
C (graphite) + O2 (g) CO2 (g)
H2°= - 395.5 kJ
H2 (g) + 1/2 O2 (g) H2O (l)
H3°= - 285.8 kJ
The question applies the Hess’ Law for finding H° Equation 1 is reversed: CH3OH (l) C (graphite) + 2 H2 (g) + 1/2 O2 (g)
H1°= + 238.7 kJ (Equation 4)
Multiply equation 3 by 2: 2H2 (g) + O2 (g) 2 H2O (l)
H3°= 2 x (- 285.8 kJ) (Equation 5)
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Combine equations 2, 4 and 5 gives: C (graphite) + O2 (g) CO2 (g)
H2°= - 395.5 kJ (Equation 2)
CH3OH (l) C (graphite) + 2 H2 (g) + 1/2 O2 (g)
H1°= + 238.7 kJ (Equation 4)
2H2 (g) + O2 (g) 2 H2O (l)
H3°= 2 x (- 285.8 kJ) (Equation 5)
CH3OH (l) + 3/2 O2 (g) CO2 (g) + 2 H2O (l) H°= - 395.5 kJ + 238.7 kJ + 2 x (- 285.8 kJ) = - 728.4 kJ 7
a) Most thermodynamic functions are relative (i.e. no absolute value can be given). With the Third Law of Thermodynamics, we are able to determine the absolute value of a certain thermodynamic function. What is it? [1 mark] Entropy The equation represents an equilibrium at 470o C
b)
H2(g) + I2(g)
2HI(g)
The partial pressures of the components was found to be: H2(g) = 0.472 atm. I2(g) = 0.011 atm. HI(g) = 0.517 atm. (i) Find the equilibrium constant Kp for the equilibrium.
[4 marks]
P HI2 Kp= P H2 PI2 (0.517) 2 = 0.011 x 0.472 = 51 (51.48) (ii)
Find the standard free energy, ∆Go for the reaction.
[4 marks]
∆Go = - RT lnK = - 8.314 J K-1 mol-1 x 743 K ln 51 = - 24300 J mol-1 = - 24.3 kJ mol-1
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The following equation shows the formation of ammonia in the Haber Process: N2 (g) + 3H2 (g)
(i)
2NH3 (g)
Calculate ΔHo and ΔSo for the formation of ammonia. From the data sheet, ΔHfo (NH3) = - 46.11 kJ / mol-1
[4 marks]
ΔHo = 2 mol x - 46.11 kJ / mol-1 = - 92.22 kJ ΔSo = 2 mol of NH3 x 192.5 J mol-1 K-1 - 1 mol of N2 x 191.6 J mol-1 K-1 - 3 mol of H2 x 130.7 J mol-1 K-1 = - 198.7 J K-1 = - 0.1987 kJ K-1 (ii)
Calculate the temperature when Kp = 1.00
[6 marks]
ΔGo = - RT ln Kp When Kp = 1.00, ln Kp = 0 and so ΔGo = 0 Applying ΔGo = ΔHo - TΔSo WhenΔGo = 0, ΔHo = TΔSo ΔHo - 92.22 kJ T= = = 464.1 K ΔSo - 0.1987 kJ K-1
(iii)
The production of ammonia in industry is usually carried out at 400. oC. Calculate Kp at 400.oC. [4 marks] ΔGo = ΔHo - TΔSo = - 92.22 kJ – (273 + 400. K) (- 0.1987 kJ K-1) = 41.50 kJ ΔGo = - RT ln Kp = 41.50 kJ 41.50 kJ ln Kp =
= - 7.42 -3
-1
-1
- 8.314 x 10 kJ mol K x (273 + 400. K)
Kp = e
- 7.42
= 6.0 x 10-4
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(i) Barium carbonate (BaCO3) is a sparingly soluble solid. Write an equation for the dissolution (dissolving) of barium carbonate in water. [2 marks] BaCO3 (s) Ba2+ (aq) + CO32- (aq) (ii)
Write the solubility product (Ksp) expression for barium carbonate. [1 mark] Ksp = [Ba2+] [CO32-]
(iii) Given that the solubility product of barium carbonate is 7.0 x 10-9 mol2 L-2 at 25oC. What maximum mass of barium carbonate can be dissolved in 500 ml of water at 25 oC? [8 marks] [Ba2+] [CO32-] = 7.0 x 10-9 [Ba2+ ]
BaCO3 Initial Excess Change -x Equilibrium Excess - x Therefore, [Ba2+ ] = x
0 +x x
[CO32-] 0 +x x
[CO32-] = x
Ksp = x2 = 7.0 x 10-9 x = √ (7.0 x 10-9) = 8.4 x 10-5 M Mass of barium carbonate = 8.4 x 10-5 mol / L x (137.33 + 12.01 + 3 x 16.00) g / mol = 0.017 g /L = 0.017 / 2 g / 500 mL = 8.3 x 10-3 g / 500 mL
(iv)
Do you expect the solubility of barium carbonate increase or decrease in a solution of sodium carbonate? Briefly explain. [2 marks]
BaCO3 (s) Ba2+ (aq) + CO32- (aq) In the presence of sodium carbonate, the equilibrium will shift to the left according to the Le Chatelier principle and so the solubility of BaCO3 is expected to decrease.
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(i)
Suggest systematic names for the following two coordination compounds [4 marks] [Pt(en)2Br2](NO3)2 Dibromobis(ethylenediamine)platinum(IV) nitrate
K2[Mn(H2O)(CN)5] Potassium aquapentacyanomanganate(III)
(ii) Draw the three-dimensional structures of all possible stereoisomers for the coordination complex [Co(en)2Br2]+ [6 marks] en
en
Br
Br
Br Co
Co Br
en
Br
en
en
End of Paper
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Co Br
en
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Data Sheet
1 F = 96485 C / mol electron
[A] = - k t + [A]o
PV = nRT
ln [A] = - kt + ln [A]o 1 1 = + kt [A] [A]o
R = 0.08206 L atm mol-1 K-1 R = 8.314 J mol-1 K-1 0.0oC = 273.15 K
[A]o t½ = 2k
∆U = q + w
ln 2
∆H = qp = ∆U + P∆V
t½ = k 1 t½ = k [A]o k = Ae-Ea/RT k2 ln
Ea
1
R
T1
= k1
- ΔHo
1 -
ln K = T2
14
RT
ΔSo + R
Q (charge) = I t
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For thermodynamic properties of substances at 298.15 K, please refer to the University of Alberta Introductory Chemistry Information Sheet* in your WebCT
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In case if don’t find the necessary data, please refer to your text book or from the University of Alberta Introductory Chemistry Information Sheet
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