DAWSON COLLEGE DEPARTMENT OF CHEMISTRY & CHEMICAL TECHNOLOGY FINAL EXAMINATION CHEMISTRY 202-NYB-05 May 21, 2010 9:30 – 12:30 Print your Name: __________________________________________ Student Number: ____________________________ INSTRUCTORS: Please circle the name of your instructor: J. Ali I. Dionne M. Haniff D. Baril M. Di Stefano S. Holden O. Behar N. Duxin / Y-S. Uh S. Mutic INSTRUCTIONS: This exam set consists of 16 questions. Please ensure that your copy of this examination is complete. Answer all questions in the space provided. 1. Calculators may not be shared. Programmable calculators are not permitted. 2. No books or extra paper are permitted. 3. In order to obtain full credit, you must show the method used to solve all problems involving calculations and express your answers to the correct number of significant figures. 4. Your attention is drawn to the College policy on cheating. 5. A Periodic Table is provided. (last page). 6. If a mathematical equation is used to solve a problem, the equation should be clearly written. 7. Write your answer in the appropriate space when required. USEFUL DATA: Avogadro’s Number N A = 6.022 x 10 23 mol –1 Gas Constant = 0.08206 L⋅atm⋅K –1 ⋅mol –1 R = 8.314 L⋅kPa⋅K –1 ⋅mol –1 = 8.314 J⋅K –1 ⋅mol –1 1 atm = 101.3 kPa = 760 mmHg = 760 torr 1 J = 1 kg⋅m 2 ⋅s –2 101.3 J = 1 L⋅atm MARK DISTRIBUTION 1. / 8 2. / 5 3. / 3 4. / 6 5. / 5 6. / 8 7. / 6 8. / 6 9. / 9 10. / 7 11. / 6 12. / 7 13. / 7 14. / 6 15. / 6 16. / 5 TOTAL /100
Transcript
DAWSON COLLEGE DEPARTMENT OF CHEMISTRY & CHEMICAL TECHNOLOGY
FINAL EXAMINATION CHEMISTRY 202-NYB-05 May 21, 2010 9:30 – 12:30
Print your Name: __________________________________________
Student Number: ____________________________
INSTRUCTORS: Please circle the name of your instructor:
J. Ali I. Dionne M. Haniff
D. Baril M. Di Stefano S. Holden
O. Behar N. Duxin / Y-S. Uh S. Mutic
INSTRUCTIONS:
This exam set consists of 16 questions. Please ensure that your copy of this examination is complete.
Answer all questions in the space provided.
1. Calculators may not be shared. Programmable calculators are not permitted.
2. No books or extra paper are permitted.
3. In order to obtain full credit, you must show the method used to solve all problems involving calculations and express your answers to the correct number of significant figures.
4. Your attention is drawn to the College policy on cheating.
5. A Periodic Table is provided. (last page).
6. If a mathematical equation is used to solve a problem, the equation should be clearly written.
7. Write your answer in the appropriate space when required.
USEFUL DATA: Avogadro’s Number NA = 6.022 x 1023 mol–1 Gas Constant = 0.08206 L⋅atm⋅K–1⋅mol–1 R = 8.314 L⋅kPa⋅K–1⋅mol–1 = 8.314 J⋅K–1⋅mol–1 1 atm = 101.3 kPa = 760 mmHg = 760 torr 1 J = 1 kg⋅m2⋅s–2 101.3 J = 1 L⋅atm
MARK DISTRIBUTION
1. / 8
2. / 5
3. / 3
4. / 6
5. / 5
6. / 8
7. / 6
8. / 6
9. / 9
10. / 7
11. / 6
12. / 7
13. / 7
14. / 6
15. / 6
16. / 5
TOTAL /100
- 2 -
Question 1
Ethanol is the common alcohol with molecular formula C2H5OH. An alcohol-water solution is prepared by dissolving 10.00 cm3 of ethanol, with density dethanol = 0.789 g/cm3, in a sufficient volume of water to produce 100.00 cm3 of solution. Density of solution is dsoln = 0.932 g/cm3. For a given solution calculate the following for ethanol: a. the mass percent
Ans. Mass%: _____________
b. the molarity
Ans. molarity: _____________
c. the molality
Ans. molality: _____________
d. the mole fraction.
Ans. Mole fraction: _____________
(2 marks)
(2 marks)
(2 marks)
(2 marks)
For 100.00 cm3 of solution: 100.00 cm3 x 0.932
g
cm3 = 93.2 g solution
In this solution 10.00 cm3 ethanol are present: 10.00 cm3 x 0.789
g
cm3 = 7.89 g ethanol
Mass percent =
7.89 g93.2 g
x 100% = 8.46% ethanol
For 1.00 L of solution: 1.00 L x 0.932
gmL
x 1000 mLL
= 932 g solution
In this solution 8.46% is ethanol then: 932 g solution x
7.80 g ethanol100. g solution
= 78.8g ethanol
Molarity = mol soluteL solution
=
78.8 g x 1 mol ethanol46.07 g ethanol1.00 L
= 1.71 M ethanol
Mole of ethanol for 1.00L solution = 78.8 g x
1 mol46.07 g
= 1.71 mol
Number of kilogram of solvent in one liter of solution: 932 g solution - 78.8 g ethanol = 853 g solvent
Molality =
mol solutekg solvent
=
1.71 mol
853 g x 1 kg
1000 g solvent
= 2.00 molal ethanol
Mole of ethanol for 1.00 L solution = 78.8 g x
1 mol46.07 g
= 1.71 mol
Mole of solvent for 1.00 L solution = 853 g solvent
1 mol18.015 g
= 47.3 mol
Mole fraction =
mol solutetotal mol solution
=
1.71 mol1.71 mol + 47.3 mol
= 3.49x10-2 ethanol
- 3 -
Question 2
Toluene, C7H8 is a component of gasoline (octane, C8H18). It is present in gasoline as an octane booster at concentrations between 3 to 5% by mass (25% in racing cars gasoline). Consider a solution of octane with 20.% by mass of toluene at 20°C a. Calculate the total vapor pressure of this solution
Data: P°octane = 10.5 mm Hg at 20°C, Tb = 126°C P°toluene 22 mm Hg at 20°C, Tb = 111°C
ans. total vapor pressure:_____________
b. Calculate the mole ratio of toluene to octane in the vapor phase above the solution
ans. mole ratio: toluene/octane:_____________
c. If the actual vapor pressure measured is 15.2 mm Hg, will the boiling point of this solution be higher
or lower than the one expected from Raoult’s law? Explain.
(3 marks)
(1 mark)
(1 mark)
for 100 g of solution you have : (keep one extra sig.fig. for calculations)
100. g solution x 20 % toluene = 20. g toluene 100. g solution - 20. g toluene = 80. g octane
Total vapor pressure = Ptolueneo x χtoluene + Poctane
o x χoctane =
= (22 mm Hg x 0.237) + (10.5 mm Hg x 0.763) = 13 mm Hg
Molar ratio =
ntoluene Vap.
noctane Vap. Since PV = nRT then P ∝ n therefore:
ntoluene Vap.
noctane Vap. =
PtoluenePoctane
=
22 mm Hg x 0.23710.5 mm Hg x 0.763
=
5.21 mm Hg 8.01 mm Hg
= 0.65
The actual vapor pressure is higher than the ideal one (calculated) therefore, the boiling point will be
lower than the one expected. The intermolecular forces between toluene - octane are weaker than the
one of the pure compounds.
- 4 -
Question 3
A 0.461 g sample of cumene, a non-volatile non-ionic compound, is dissolved in 10.0 g cyclohexane (C6H12), producing a solution that freezes at -1.25°C. Cyclohexane has a normal freezing point of 6.50°C and a freezing point depression constant of 20.2°C/m. What is the molar mass of cumene?
Ans. Mol. mass cumene: _____________
(3 marks)
Colligative properties involving the freezing point depression of a non ionic solute:
ΔTf = Kf x m where ΔTf = 6.50°C - (-1.25°C) = 7.25°C
The concentration of cumene can be found:
ΔTf Kf
= m =
7.75°C
20.2°C.m-1 = 0.384 m
The number of mole of cumene is:
molality =
mole solutekg solvent
Mole solute = molality x kg solvent = 0.384 m x 10.0 g x
1 kg1000 g
= 3.84x10-3 mol
Finally, the molar mass of cumene =
mass cumenemole cumene
=
0.461 g0.00384 mol
= 120.
gmol
(or 1.20x102
gmol
)
- 5 -
Question 4
Hydrofluoric acid, (HF) is a weak acid that can be used in the fluoridation of water. An aqueous solution of 0.100 M HF has an osmotic pressure of 2.64 atm at 25°C. a. Calculate the van’t Hoff factor for HF at this concentration
Ans. van’t Hoff factor:_____________
b. Does it differ from the maximum van’t Hoff factor expected for a monoprotic acid? If so, explain why. c. What is the percent ionization of HF at this concentration?
Ans. % ionization:_____________
(2 marks)
(2 mark)
(2 marks)
Colligative properties involving the osmotic pressure of an ionic compound in solution: ∏ = i M R T
i =
Π M R T
=
2.64 atm
(0.100 molL
)(0.0821L.atmK.mol
)(25 + 273)K = 1.08
Yes it differs from the expected van't Hoff factor of 2 for a monoprotic acid.
Therefore. HF is a week acid since its dissociation is partial because of ion pairing at this concentration
corresponding to an incomplete dissociation of the acid (weak acid).
HF à H+ + F-
C-x x x
i =
mole of particle in solutionmole solute added
=
2x + (c - x)mole solute added
=
x + CC
Since I = 1.08 and C = 0.100 M à 1.08 =
x + 0.1000.100
then x = 0.008
% ionization =
xC
x 100% =
0.008 M0.100 M
x 100% = 8%
- 6 -
Question 5
Iodide ion is oxidized in acidic solution to triiodide ion I3� by hydrogen peroxide.
H2O2(aq) + 3 I�(aq) + 2 H+(aq) à I3
�(aq) + 2H2O(l)
A series of four experiments was run at different concentrations, and the initial rates of I3
a. From the table above, obtain the reaction orders with respect to each of the following species: H2O2, I�
, H+.
Ans. Reaction order: H2O2: ______ I�: ______ H+
: ______ b. Find the rate constant with its units.
Ans. rate constant: _____________________
(3 marks)
(2 marks)
The reaction rate is given by: rate = k [H2O2]α [I-]β [H+]γ
For α :
Exp.2Exp.1
=
2.30x10-6
1.15x10-6 =
k(0.020)α (0.010)β(0.00050)γ
k(0.010)α(0.010)β(0.00050)γ à 2 = 2α then α = 1
For β :
Exp.3Exp.1
=
2.30x10-6
1.15x10-6 =
k(0.010)α (0.020)β(0.00050)γ
k(0.010)α(0.010)β(0.00050)γ à 2 = 2β then β = 1
For γ :
Exp.4Exp.1
=
1.15x10-6
1.15x10-6 =
k(0.010)α (0.010)β(0.00100) γ
k(0.010)α(0.010)β(0.00050)γ à 1 = 2γ then γ = 0
The reaction orders are: H2O2: ___1___ I_: ___1___ H+: ___0___
The reaction rate is given by: rate = k [H2O2] [I-]
Using Exp.4 : 1.16x10-6 M.s-1 = k (0.010 M) (0.010 M)
k =
1.16x10-6 M.s-1
(1.0x10-2 M)2 = 1.2x10-2 M-1s-1
- 7 -
Question 6
The reaction below was monitored as a function of time at a temperature of 400 K:
2NOCℓ (g) → 2NO(g) + Cℓ2(g) A plot of 1/[NOCℓ] against time yielded a straight line with slope of 6.7x10-4 L·mol-1·s-1. a. Write the rate law for the reaction. b. What is the half-life for the reaction if the initial concentration of NOCℓ is 0.20 M?
Ans. half-life: _____________________ c. If the initial concentration of NOCℓ is 0.35 M, what is the concentration of NOCℓ after 5.0 min?
Ans. [NOCℓ] after 5.0 min: _____________________
d. If the initial concentration of NOCℓ is 0.35 M, How long will it take for the concentration to drop to 20%
a. Consider the potential energy profiles for three different chemical reactions.
Indicate which reaction is the slowest one. Explain your choice
b. Consider the potential energy profiles for a chemical reaction.
Mechanism 1
2A + B C C D
Mechanism 2
2A C
B + C D Mechanism 3
A C A + B + C 2D
Circle the proposed mechanism that is consistent with the reaction profile shown and explain your choice.
c. Beside concentration and pressure, give two parameters you can change that could affect the reaction
rate of a chemical reaction: i. ii.
(2 marks)
2A + B D
(2 marks)
(2 marks)
Reaction 1 is the slowest one since its activation energy is the highest .
The curve is consistent with mechanism 2 since the slow step is the first one (highest Ea).
The rate law will be rate = k [A]2
Temperature Catalyst or enzyme
- 9 -
Question 8
At elevated temperature (997°C) limestone dissociates according to the equation
CaCO3(s) CaO(s) + CO2(g) ΔH = +42.5 kJ a. If 50.0 g CaCO3 (100.1 g/mol) is placed in an evacuated 4.00 L container and heated up to 997°C,
how many grams of CaCO3 will decompose if the pressure at equilibrium is 392 kPa? b. If the volume of the container is expanded to 10.0 L at 997°C, what will be the CO2 pressure at
equilibrium? c. Calculate Kc for this reaction at 997°C d. Predict the effect of each of the following changes will have on the equilibrium position.
equilibrium position shift
change to the left no change to the right
i. CO2 is added
ii. CaCO3 is added
iii. Pressure is increased (adding N2 gas, volume unchanged)
iv. The temperature is increased
(2 marks)
(1 mark)
(1 mark)
(2 marks)
Mole of CO2 produced = PV = nRT
n = RTPV =
273)K(977 xmol K.
L.kPa 8.31
L 4.00 x kPa 392
+= 0.149 mole
Since it is a 1:1 ratio between CO2 and CaCO3 then:
Mass of CaCO3 consumed = 0.149 mol CO2 x 2
3CO 1
CaCO 1 x 100.1 g.mol-1 = 14.9 g
K = [CO2]. The concentration of CO2 remains the same whatever the volume For a gas, concentration = pressure. Therefore, pressure does not change = 392 kPa or 3.87 atm.
K = [CO2] since the other reactants are all solids.
In part A, the number of mole of CO2 for a container of 4.00 L was 0.149 mole.
K =
0.149 mol4.00 L
= 0.0373 also K = RTKp =
RTP = where P is also obtained from PV = nRT
- 10 -
Question 9
Consider the following set of data:
Formula Ka (at 25°C)
[Al(H2O)6]3+ 1.4×10-5 HNO2 4.0×10-4
HF 7.2×10-4
a. What is the strongest acid in the table? _______HF____________ b. With the help of the table, arrange the following in order of most basic to least basic: H2O, NO2
�, [Al(H2O)5OH]2+
Most basic ____[Al(H2O)5OH]2+ ___ > ___ NO2�___ > ____ H2O ____ Least basic
c. What is the value of Kb for F�
at 25°C ?
Ans. Kb : ___________
d. Write the chemical reaction represented by the Kb for F�
in water and place the species involved in the appropriate place
__H2O_____ + __F- __ __ HF__ + ___OH-__ Acid Base Conjugate Conjugate acid base
e. At 40°C, Kw = 2.9x10-14. What is the neutral pH of water at this temperature?
Ans. ___________
(1 mark)
(2 marks)
(2 marks)
(2 marks)
(2 marks)
Kb = 4-
-14
7.2x1010 = 1.4x10-11
H2O OH- + H+ then Kw = [H+][OH-]
at neutral pH: [H+] = [OH-] then kw = [H+] = 2.9x10-14 = 1.7x10-7
pH = -log(1.7x10-7) = 6.77
- 11 -
Question 10
a. A solution of the basic oxide CaO is prepared by adding water to 0.28 g CaO to make 0.50 L of solution.
i. Write the equations for the reactions that occur when CaO is dissolved in water ii. Assuming that ion-pairing is non-existent, what is the expected pH of this solution?
ans. pH:____________
b. For which of the following salts will the solubility depend on pH? pH sensitive pH independent
i. KCℓO4 ii. Pb(OH)2 iii. AgF iv. Ba(NO3)2
c. For each of the following salts dissolved in water, predict whether the aqueous solution will be acidic,
neutral or basic. acid neutral basic
i. RbOH ii. NaIO iii. NH4OH iv. LiCℓO3
(1 mark)
(2 marks)
(2 marks)
(2 marks)
1) CaO(s) + H2O(l) à Ca(OH)2(aq) 2) Ca(OH)2(aq) à Ca2+(aq) + 2OH—(aq)
or 1) CaO(s) à Ca2+(aq) + O2— (aq) 2) O2—(aq) + H2O(l) à 2OH— (aq)
Mole of OH— in solution: 0.28 g CaO x
1 mol56.08 g
x
Ca(OH)2CaO
x
2 OH-
Ca(OH)2 = 0.010 mol OH—
Concentration of OH— in solution:
0.010 mol OH-
0.50 L = 0.020 M
Finally, the pH will be: pOH = -log(0.020 M) = 1.70 pH = 14 - pOH = 12.30
- 12 -
Question 11
a. Consider 0.500 L of a buffer that consists of 1.50 M KCℓO (Ka HCℓO = 3.5x10-8) and 0.50 M HCℓO. What will be the pH of this buffer after the addition of 250 mL of 1.0 M HNO3?
ans. pH:____________
b. Which of the following mixtures would result in a buffer solution when 100 mL of each of the two solutions are mixed together?
buffer not a buffer
i. 0.1 M KOH and 0.2 M NH3
ii. 0.2 M HCℓ and 0.2 M NH3
iii. 0.2 M HNO3 and 0.4 M NaNO3
iv. 0.1 M HNO3 and 0.2 M NaF
(4 marks)
(2 marks)
The system is the following: HClO(aq) H+(aq) + ClO—(aq) (note H+ = H3O
+)
Mol of HClO = 0.500 L x 0.50
molL
= 0.25 mol
Mol of ClO— = 0.500 L x 1.50
molL
= 0.750 mol
Mol of H+ added = 0.250 L x 1.0
molL
= 0.25 mol
First we have a reaction between a strong acid and a week base: H+ + ClO— HClO Initial 0.25 0.75 0.25 Reaction -0.25 -0.25 +0.25 After reaction 0 0.50 0.50 After, the system reach equilibrium: HClO H+ + ClO— I 0.50 0 0.50 C -x +x +x E 0.50-x +x 0.50+x Since we have a buffer, x << 0.50 AND the number of mole for the acid and the conjugated base are the same, then pH = pKa pH = -log(3.5x10-8) = 7.46
- 13 -
Question 12
Consider the following titration curve of trimethylamine (C3H9N) a weak base with 0.100 M HCℓ at 23°C. Initial solution: 50.0 mL of C3H9N, 4.00x10-2 M a. Draw on the graph the shape of the titration curve if this base had a smaller Kb value.
b. Which letter (A to F) on the graph corresponds to each of the following?
letter
The equivalence point D
The point of half-neutralization B
The point corresponding to the pKa of C3H9NH+ B c. When 15.0 mL of 0.100 M HCℓ is added, the pH of the solution is 9.255. Calculate Kb of
trimethylamine.
ans.
Kb:__
_____
_____
Let's call trimethylamine TMA an its conjugated acid TMAH+ Mole of TMA = 4.00x10-2 M x 0.0500 L = 2.00x10-3 mol Mole of acid added = 1.00x10-1 M x 0.0150 L = 1.50x10-3 mol The HCl reacts to completion (strong acid) with TMA according to:
Then, the system will reach equilibrium from those initial conditions: TMA + H2O TMAH+ + OH—
0.50x10-3 -x mole 1.50x10-3 + x mole x mole In a buffer regime, x is small compared to [base] or [conjugated acid] and can be neglected. In this case ONLY the ratio of mole of TMAH+ / TMA is the same as the one of concentrations. Therefore:
Kb =
[TMAH+][OH −][TMA]
=
(1.50x10-3)(10−(14−9.255) )
0.50x10-3 = 5.4x10-5
smaller Kb value
- 14 -
(1 mark)
(3 marks)
(3 marks)
- 15 -
Question 13
a. A saturated aqueous solution of Mg(OH)2 has a pH of 10.08, what is the Ksp of Mg(OH)2?
ans. Ksp:____________
b. The Ksp of cobalt(III) hydroxide is 2.5x10-43. Calculate the solubility of Co(OH)3 in water in mol/L
ans. solubility (mol/L):____________
c. Does a precipitate form when 25 mL of 0.10 M lithium nitrate LiNO3, is mixed with 35 mL of 0.75 M
sodium carbonate Na2CO3? (Ksp Li2CO3 = 8.15x10-4) Show your work.
ans: yes no
(2 marks)
(2 marks)
(3 marks)
Mg(OH)2(s) Mg2+(aq) + 2 OH—(aq) Ksp = [Mg2+][OH—]2 I Solid 0 0 C -x +x +2x E solid x +2x Ksp = (x)(2x)2 = 4x3
We can use the pH to know [OH—] that is equal to 2x pOH = 14 - 10.08 = 3.92, [OH—] = 10-3.92 = 1.2x10-4 M. since 2x = 1.2x10-4 M. then x = 6.0x10-5. finally: Ksp = 4 (6.0x10-5)3
Ksp = 8.6x10-13
According to the Ksp value, the concentration of OH— generated by the salt will be small compared to the one of natural ionization of water [OH—] = 10-7 M therefore, common ion effect.
Co(OH)3(s) Co3+(aq) + 3OH—(aq) Ksp = [Co3+][OH—]3
I Solid 0 10-7 C -x +x +3x E solid x 3x+10-7 if x<< 10-7 then Ksp = (x)( 10-7)3
And the solubility (or x) is x =
Ksp
10-21 =
2.5x10-43
10-21 = 2.5x10-22 M (check : x = 2.5x10-22 << 10-7 )
A precipitation will occur if Q > Ksp Li2CO3(s) 2Li+(aq) + CO3
—(aq) Ksp = [Li+]2[NO3—]
Li+ concentration =
0.025 L x 0.10 M(0.025 + 0.035)L
= 0.042 M
CO32- concentration =
0.035 L x 0.75 M(0.025 + 0.035)L
= 0.44 M
Q = [Li+]2[NO3
—] = (0.042)2(0.44) = 7.8x10-4. Since Q < Ksp then: No precipitation.
- 16 -
Question 14
a. A system is made of a cylinder of gas with a piston. When 4.0 kJ of heat is transferred from the surroundings to the system, the gas in the piston expands from 12 L to 27 L and performs work on the surroundings. If the system gains 201 J of internal energy from this process, against what constant external pressure, in atmospheres, is the piston working?
Ans. pressure (atm): ____________
b. Bromine is a liquid at room temperature. Calculate the freezing point of bromine if its heat of fusion is + 5.79 kJ⋅mol-1 and its entropy of fusion is 21.8 J⋅K-1⋅mol-1.
Ans. Tf bromine: ____________
(3 marks)
(3 marks)
∆H = ∆E + P∆V (first law: conservation of the energy) ∆H - ∆E = P∆V
∆H - ∆E = [+4000 J - (+ 201 J) ] x
1 L.atm101.3 J
= 37.50 L.atm
37.5 L.atm = P∆V and ∆V = Vfinal - Vinitial
P =
37.5 L.atm27L - 12L
= 2.5 atm
∆G = ∆H - T∆S (second law: spontaneity of a system) The freezing point occurs when all the heat added to a system is converted to entropy. At this point, the system is highly reversible (equilibrium condition, therefore ∆G = 0)
The equation becomes: 0 = ∆H - T∆S and
ΔHΔS
= T
T =
ΔHΔS
=
5.79x103 J.mol
21.8 J.K-1 . mol = 266K or -7ºC
- 17 -
Question 15
a. Circle the substance in each of the following pairs that would have the greater entropy. i. H2O (ℓ, 1 mol, 75˚C, 1 atm) or H2O (g, 1 mol, 75˚C, 1 atm) ii. Fe (s, 50.0 g, 5˚C, 1 atm) or Fe (s, 0.80 mol, 5˚C, 1 atm) iii. Br2 (ℓ, 1 mol, 8˚C, 1 atm) or Br2 (s, 1 mol, –8˚C, 1 atm) iv. SO2 (g, 0.312 mol, 32.5˚C, 0.110 atm) or SO2 (g, 0.284 mol, 22.3˚C, 15 atm)
b. Methyl isothiocyanate, CH3—N═C═S, is a highly irritating pesticide. It can be prepared by reacting
carbon disulfide with methylamine. Given the thermodynamic data at 25˚C below, calculate the standard molar entropy of methyl isothiocyanate.
In the laboratory experiment 4, you want to determine the activation energy of the following reaction:
2I¯(aq) + S2O82¯(aq) I2 (aq) + 2S2O4
2¯
Where the reaction rate is: Rate = - [ΔI¯]/2Δt and the rate law for this reaction is: Rate = k [I¯][S2O82¯]
By recording the reaction rate of several experiments at different temperatures, the following graph based on the linear form of the Arrhenius equation is obtained.
Arrhenius plot for the determination of the activation energy for the reaction of iodide with peroxydisulfate
From this graph, calculate the activation energy (with units) for this reaction.
Ans. Ea : _________________
(5 marks)
We will be using the linear form of the Arrhenius equation: Ln k = -
EaR
1T
+ constant
The graph of Ln k vs.
1T
gives a slope = —
EaR
Slope =
Y2 − Y1X2 − X1
=
(-4.80) − (-5.60)
(3.2x10−3 - 3.35x10−3)K-1 =
0.80
-1.5x10−4K-1 = — 5.3x103 K
Since: slope = —
EaR
then Ea = — (8.31
JK.mol
) x — 5.3x103 K = 44x103
Jmol
or 44
kJmol
.
- 19 -
1A Periodic Table of the Elements 8A
1 2
1 H He 1.008 2A 3A 4A 5A 6A 7A 4.003
3 4 5 6 7 8 9 10 2 Li Be B C N O F Ne 6.941 9.012 10.81 12.01 14.01 16.00 19.00 20.18
11 12 13 14 15 16 17 18 3 Na Mg Al Si P S Cl Ar 22.99 24.31 3B 4B 5B 6B 7B 8B 9B 10B 1B 2B 26.98 28.09 30.97 32.07 35.45 39.95
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.87 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.47 87.62 88.91 91.22 92.91 95.94 98.00 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 6 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 137.3 138.9 178.5 181.0 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 209.0 210.0 222.0
