©Prep101 Chem203 Final Booklet Gases Problem 1. An engine cylinder has a volume of 600 mL at 1.00 atm and 7.00°C. When ignition occurs, the volume increases to 1200 mL and the pressure rises to 2.00 atm. What is the temperature of the cylinder at that time? PV
PV  nRT  nR 
T
2.00  atm   1200  mL 
PV
PV
PV
1 1
 2 2  T2  T1  2 2   273  7.00  C    K  
 1120  K  T1
T2
PV
1.00  atm   600  mL 
1 1
T2  1120  K   273  847  C 
Solutions at www.prep101.com/solutions 1 of 69
©Prep101 Chem203 Final Booklet Practice Questions Problem 2. The increasing vapor pressure caused by heating a liquid is due to: a) increased intermolecular interactions b) increasing potential energy of molecules c) increasing kinetic energy of molecules. d) decreasing surface tension. e) weaker solvation. Answer: C Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure. Problem 3. The density of phosphorous vapor at 310°C and 775 mmHg is 2.64 g/L. What is the molar mass of the phosphorous? The atomic mass of P is 31.0 g/mol. What is the molecular formula of the phosphorous vapor under these conditions? 775  mmHg 
 1  atm   1.02  atm 
1  atm   760  mmHg  ; P  atm   775  mmHg   P  atm  
760  mmHg 
T  K   273  310  C   583  K 
P
RT
 MrPx  d Px 
RT
P
0.08206  L  atm  mol 1  K 1   583  K 
 123.8  g / mol  MrPx  2.64  g / L  
1.02  atm 
d Px  MrPx 
x
MrPx
MrP

123.8  g / mol 
 3.99  4  P4
31.0  g / mol 
Solutions at www.prep101.com/solutions 2 of 69
©Prep101 Chem203 Final Booklet Problem 4. The empirical formula of a substance is CH . When 0.9720 g of the liquid is evaporated completely, the gas occupies 559 mL at 110 °C and 735 Torr. Determine the molar mass and the molecular formula of the gas. 735  mmHg 
1  atm   760  torr  ; P  atm   735  torr   P  atm  
 1  atm   0.967  atm 
760  mmHg 
T  K   273  110  C   383  K 
M CH 2   mCH 2  
x
x
MrCH 2 
x
x
MrCH 2 

0.08206  L  atm  mol 1  K 1   383  K 
RT
 0.9720  g  
 56.5  g / mol 
PV
0.967  atm   0.559  L 
56.5  g / mol 
 4.04  4  C4 H 8
14.0  g / mol 
Problem 5. How many liters of O , measured at 0.0°C and 1.00 atm are required to oxidize 1.50 g of C H O ? C6 H12O6  6O2  6CO2  6 H 2O RT
PV  nRT  VO2  nO2 
P
mC6 H12O6
nO2
6
  nO2  6  nC6 H12O6  6 

nC6 H12O6 1
MrC6 H12O6
 VO2  6 
mC6 H12O6
MrC6 H12O6
0.08206  atm  L  K 1  mol 1   273  K 
1.50  g 
RT

 6

 1.12  L 
P
180.0  g / mol 
1.00  atm 
Solutions at www.prep101.com/solutions 3 of 69
©Prep101 Chem203 Final Booklet Problem 6. When 225 mg of an organic compound were vaporized at 30.0 °C and a pressure of 103.2 kPa, it occupied a volume of 70.4 mL. Calculated the molar mass of the compound. Solution: T K   273  30.0 C   303 .0 K 




8.3145 J  mol 1  K 1  303 .0 K
RT
M X  mX 
 0.225 g 
 79.5 g /mol
PV
101 .3 kPa  70.4  10 3 L

 


Problem 7. A sample contains 0.25 atm of oxygen gas. To this was added 440 Hg of nitrogen gas. What is the total pressure, assuming the volume and temperature was held constant? a) 1200 mmHg b) 440.25 mmHg c) 630 mmHg d) 760 mmHg e) none of the above Answer: C 760mmHg
= 190 mmHg O2 Solution: 0.25atmO 2 
1atm
Total pressure = 190mmHg O2 + 440mmHg N2 = 630mmHg Solutions at www.prep101.com/solutions 4 of 69
©Prep101 Chem203 Final Booklet Problem 8. How many moles of helium gas occupy a volume of 22.4 L at 30.0 C and 1.00 atm? a) 0.11 b) 0.90 c) 1.00 d) 1.11 e) 1.90 Answer: B Solution: PV = nRT (1.00atm)(22.4L) = n(0.0821atmL/molK)(303K) n = 0.90 moles helium gas Problem 9. A 30 L vessel contains 10 g O and 10 g N at 100  C . The partial pressure (kPa) of O is? a) 8.7 b) 32 c) 48 d) 61 e) 65 Answer: B Solution: 10 g
nO2 
 0.3125
32
nRT 0.3125  8.314  373
P

 32.3kPa
V
30
Problem 10. Which gas will have the highest density at 25 C and 100 kPa pressure? Solutions at www.prep101.com/solutions 5 of 69
©Prep101 Chem203 Final Booklet a F b C H c H S
d NO
e SiH Answer: A Solution: To determine which gas has the highest density, consider the equation g
molarmass
pressure kPa
mol
Density
R gasconstant
temperature K
If measured at the same pressure and temperature, the gas that will have the highest density will have the largest molar mass. The molar masses of the gases are as follows: F2, 38g/mol; C2H6, 30.08g/mol; H2S, 34.086g/mol; NO, 30.01g/mol and SiH4, 32.13g/mol. Therefore, F2 will have the highest density because it has the largest molar mass. Problem 11. An engine cylinder has a volume of 600 mL at 1.00 atm and 7.00C. When ignition occurs, the volume increases to 1200 mL and the pressure rises to 2.00 atm. What is the temperature of the cylinder at that time? a) 103 C b) 203 C c) 280 C d) 847 C e) 923 C Answer: D PV PV
Solution: PV = nRT 1 1  2 2 T1
T2
(1.00atm)(600mL) (2.00atm)(1200mL)

280 K
T2
T2 = 847C Problem 12. 0.5 g of water is placed inside a sealed flask whose volume is 0.5 L. If the temp is 70 ˚C and the vapor pressure of water at 70 ˚C is 233.7 mmHg, what mass of water is in the liquid phase? Solutions at www.prep101.com/solutions 6 of 69
©Prep101 Chem203 Final Booklet Solution: We need to determine what mass is in the gas phase, and the rest must be in the liquid phase. We make the assumption that the liquid takes up a negligible volume of the container. The pressure of the water vapour will be 233.7mmHg P=1 atm / 760mmHg * 233.7 mmHg = 0.3075 atm Now calculate the moles of gas: PV
0.3075atm ∗ 0.5L
n
0.0055
RT
0.08206Latmmol 1K 1 ∗ 273.15 70 K
mass of gaseous water is: g
0.0055mol 0.098g 18
mol
So the mass of liquid water is 0.5‐0.098 = 0.402 g. Solutions at www.prep101.com/solutions 7 of 69
©Prep101 Chem203 Final Booklet Thermodynamics Problem 13. In a coffee cup calorimeter, 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed to yield the following reaction: Cl
→ AgCl Ag
The two solutions were initially at 22.60 ˚C, and the final temperature is 23.40 ˚C. Calculate the heat that accompanies this reaction in of AgCl formed. Solution: Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05 moles Cl Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced (you can prove this to yourself using an ICE table) Heat lost by reaction = heat gained by solution Heat gain = m  s  T J
100.0gH Ox4.18
1
23.40°C– 22.60°C g°C
= 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl Therefore, the heat produced per mol of AgCl = 330 J
1 kJ

 6.60 kJ mol 1 0.050 mol AgCl 1000 J
Problem 14. When a bomb calorimeter is heated electrically for exactly 6 minutes using a 4.00 kilowatt heater (1 watt = 1 J s‐1), its temperature increases from 25.0 C to 50.3 C. When 0.100 mol C H OH is burned completely in oxygen in the same calorimeter, the temperature increases from 25.0C to 27.4C. What is the constant volume heat of combustion of C H OH ? Solution C H OH
3 O 2 CO
3H O Answer: ‐1370 kJ mol‐1 Heat capacity of calorimeter = C q=(360 s)  (4000 J/s) = 1440 kJ q = C ∆T = C (50.3 ‐ 25.0) rearrange and solve  C = 56.92 kJC‐1 (56.92 kJ C 1 )(27.4  25.0 C)
Q rxn  
 1370 kJ / mol
(0.100 mol)
Solutions at www.prep101.com/solutions 8 of 69
©Prep101 Chem203 Final Booklet Problem 15. A gas absorbs 1.0 J of heat and performs 15.2 J of work. What is the change in the internal energy of the gas? Solution: ‐14.2 J Absorbing heat is an endothermic process (positive) Work done by the system on the surroundings is negative by convention, therefore E = q + w = 1.0 J ‐ 15.2 J= ‐14.2 J Problem 16. A balloon filled with 39.1 mol helium has a volume of 876 L at 0.0 °C and 1.00 atm pressure. The temperature of the balloon is 38.0 °C as it expands to a volume of 998 L, the pressure remaining constant. Calculate q, w, and ΔE for the helium in the balloon. The molar heat capacity for helium gas is 20.8
°
Solution: q n c
T 1 kJ
= 39.1mol
20.8
38.0– 0.0 °C x
= +30.9 kJ °
1000 J
Pressure‐Volume work can be calculated by: 101.3 J
1 kJ
x = ‐12.4 kJ w = ‐PΔV = ‐1.00 atm x (998 – 876 L) = ‐122 L atm x L atm
1000 J
So, the total internal energy, ΔE q w
30.9kJ
12.4kJ
18.5kJ Solutions at www.prep101.com/solutions 9 of 69
©Prep101 Chem203 Final Booklet Problem 17. Given the following data: → 3O ΔH
427kJ
2O
O
→ 2O ΔH 495kJ
NO
O
→ NO
O ΔH 199kJ Calculate ΔH for the reaction: NO
O → NO Solution: O
→ NO
O
ΔH
199 kJ
NO
3
O
→ O ΔH
½ 427kJ 2
O → ½O ΔH ½ 495kJ NO
O → NO ΔH 233kJ
Problem 18. Using equations below, find H  at 25  C for the oxidation of C H OH : C H OH
3O
→ 3H O
2CO C H
3O
→ 2CO
2H O ΔH ‐1411kJ
3H
½O
→ C H OH ΔH ‐278kJ
C
H O → C H OH ΔH ‐44kJ
C H
Solution: 1367 kJ C H OH → C H
H O ΔH
44 kJ 3O
→ 2CO
2H O ΔH
1411kJ C H
3O
→ 3H O
2CO ΔH 1367kJ
C H OH
Solutions at www.prep101.com/solutions 10 of 69
©Prep101 Chem203 Final Booklet Standard Enthalpies of Formation The standard enthalpy of formation H ) is the enthalpy change for the formation of one mole of the substance in the standard state from the reference forms of the elements in their standard states. Standard state is 1 atm pressure at 25 oC (298 °K). The standard enthalpy of formation of a pure element in its reference form is 0 J/mol. Standard Enthalpies of Reaction The enthalpy change for the overall reaction is the sum of the standard enthalpy changes of the individual steps. H nH Products ‐nH Reactants Where n is moles of products or reactions taken from the stoichiometric coefficients in the reaction. Problem 19. Based on the following the data at 25°C, and assuming ΔH° and ΔS° are independent of T, : J
kJ
Compound S°
ΔH
mol K
mol
NO 90.29 210.4 NO 233.82 240.2 Calculate ΔH° at 25°C for 2NO
O
→ 2NO Answer: ΔH°
2 33.82
2 90.29
0
112.94
Solutions at www.prep101.com/solutions 11 of 69
©Prep101 Chem203 Final Booklet Problem 20. Energy can also be drawn from the combustion of simple alcohols such as methanol and ethanol. Determine the enthalpy of reaction for the combustion of ethanol using bond energies: 3O
→ 2CO
3H O Solution: CH CH OH
The energy lost from breaking bonds and energy gained in bond formation must be determined. Bonds Broken Bonds Formed 5 C‐H Bonds (413 kJ mol‐1) 4 C=O bonds (CO2) (799 kJ mol‐1) 1 C‐C Bond (347 kJ mol‐1) 6 O‐H bonds (467 kJ mol‐1) 1 C‐O Bond (358 kJ mol‐1) 1 O‐H Bond (467 kJ mol‐1) 3 O=O Bonds (498 kJ mol‐1) Total = 4731 kJ Total = 5998 kJ H = BE(Broken)  BE(Formed) H = 4731 kJ  5998 kJ = ‐1267 kJ Solutions at www.prep101.com/solutions 12 of 69
©Prep101 Chem203 Final Booklet Problem 21. Which of the following pairs has the greater entropy for each system? a) Water at 105 oC or water at 15 oC. Cl b) NaCl orNa
c) Pure solutions of H2O and MeOH or an H2O/MeOH mixture d) The reactants or products for the reaction H
½O
→ H O e) A mole of F or Cl at the same temperature. Answer: a) Water at 105 oC b) Na
Cl c) H2O/MeOH mixture d) Reactants e) F Solutions at www.prep101.com/solutions 13 of 69
©Prep101 Chem203 Final Booklet Problem 22. Calculate S and H for the reaction from entropies and enthalpies of formation. Is the reaction spontaneous at 298 K? Without calculating G, indicate whether this reaction will be spontaneous at higher temperatures, or lower temperatures? At what temperature does this reaction become spontaneous? CH
½Cl
→ CH Cl
½H Enthalpies and entropies of formation are: Cl CH Cl H CH ΔH
S
kJ
mol
J
molK
‐74.85 0 90 0 186.2 223 245.2 130.6 
Solution: S reaction
  coeff pS  ( products )   coeff p S  ( reactants ) 



= S
+ ½ SH2
‐ ( SCH4
+ ½  S Cl2
) = (245.2 J mol‐1K‐1) + ½(130.6 J mol‐1K‐1) – [(186.2 J mol‐1K‐1) + ½ (223 J mol‐1K‐1)] = +12.8 J mol‐1 K‐1 
  coeff p H  ( products )   coeff p H  ( reactants ) H reaction

CH3Cl

= H f CH3Cl + ½ H f H2 ‐ ( H f CH4 + ½ H f Cl2 ) 



= (+90 kJ mol‐1) + ½(0 kJ mol‐1) – [(‐74.85 kJ mol‐1) + ½(0 kJ mol‐1)] = +164.8 kJ mol‐1 At 298 K, ΔGo = ΔHo ‐TΔSo = 164.8 kJ – (298K)(12.8 J mol‐1 K) = 161 kJ mol‐1 so non‐spontaneaous Since both ΔHo and ΔSo are positive, the reaction will be spontaneous at high temperatures. To determine at which temperature the reaction becomes spontaneous, find when ΔG < 0 Solve for ΔG = 0 = ΔHo ‐TΔSo 16480
 o
J mol 1
T = =
= 12875 °K S o 12.8 J K 1 mol 1
Solutions at www.prep101.com/solutions 14 of 69
©Prep101 Chem203 Final Booklet 0
Problem 23. The following thermodynamic data are valid at 25 C: J
kJ
Compound S
ΔH
mol K
mol
C H ‐84.7 229.5 H 130.6 For the reaction: H
→C H C H
ΔH =‐137.0 kJ and ΔG =‐101.1 kJ at 25 0C. Calculate ΔH (C H ) at 25 0C. Solution: ΔH
ΔH C H
ΔH C H
ΔH H
137.0
84.7 ΔH C H
0
kJ
ΔH C H
52.3
mol
Calculate ΔS0 for the reaction at 25 0C. Solution: ΔG0= ΔH0‐TΔS0 101.1
137.0 298
120.5J
ΔS
K
ΔS Calculate S0 (C H ) at 25 0C. Solution: 0
0
S 0   S prod
  Sreact
120.5  229.5  S 0 (C2 H 4 ( g ))  130.6 S 0 (C2 H 4 ( g ))  219.4 JK 1
Give a brief rationalization in physical terms for the algebraic sign of ΔS0 calculated in part (ii) above. Solutions at www.prep101.com/solutions 15 of 69
©Prep101 Chem203 Final Booklet Solution: The product side contains fewer gas molecules than the reactant side→ more ordered. ΔS0 is negative Solutions at www.prep101.com/solutions 16 of 69
©Prep101 Chem203 Final Booklet Problem 24. The manufacture of sulfuric acid involves the oxidation process 1
O
→ SO SO
2
carried out at high temperature. The following data are for 300 K:  G 0f (kJ mol‐1 ) S  (JK‐1 mol‐1)
SO
O
SO
‐300.4 ‐370.3 248.5 205.0 256.1 Calculate ΔG0, ΔS0, ΔH0 and ΔE0 for this reaction at 300K. Answer: ΔG0=‐69.9kJ, ΔS0=‐94.9kJ, ΔH0=98.4kJ, ΔE0=‐97.1kJ Solution: G 0  G 0f ( SO3( g ) )  G 0f ( SO2( g ) )  370.3  300.4  69.9kJ
S 0  S  ( SO3( g ) )  S  ( SO2( g ) )  S  (O2( g ) )  256.1  248.5 
1
 205.0  94.9 JK 1
2
G 0  H 0  T S 0
94.9  300K
 98.4kJ
1000
1
H 0  E 0  PV  E 0  ng RT where ng  
2
8.314 1
E 0  H 0  ng RT  98.4kJ 
  300kJ  97.1kJ
1000 2
H 0  G 0  T S 0  69.9kJ 
Solutions at www.prep101.com/solutions 17 of 69
©Prep101 Problem 25. Calculate K at 298 K for the following reactions: NO
½O
⇌ NO
Br
⇌ 2HBr
2HCl
Given:  G of NO O NO HCl (kJ mol‐1) 86.60 0 51 ‐95.30 Chem203 Final Booklet Cl
HBr
‐53.5 Cl
0 Br
0 Solution: (a) Greaction 


coeff G
p

f ( products)
 coeffpGf
(reactants)
= (51 kJ mol‐1) – [(½(0 kJ mol‐1) + 86.60 kJ mol‐1) = ‐35.6 kJ mol‐1 G    RT ln Keq ln K
∆G
RT
35,600J
8.314Jmol L
298K
ln K eq  14.4 K eq  1 .8  10 6 (b) Same method. Answer: K eq  4 . 5  10 14 Solutions at www.prep101.com/solutions 18 of 69
©Prep101 Problem 26. Keq for 2SO
Chem203 Final Booklet O
⇌ 2SO
24
at 298 K is 7.0 x 10 What is the value of Keq for the same reaction at T = 273 K, and what is the Gibbs free energy change, when the reaction is brought down to this temperature and sealed containers are filled with 0.500 mol of SO , 0.0100 mol of O , and 0.100 mol of SO ? kJ
ΔH
198.4
mol
The Keq under standard conditions is given, and the Gibbs free energy is needed under non‐standard conditions, not at equilibrium. This problem can be solved in three steps: Step 1: Solve for Keq at 273 K: K
H   1 1 
ln 2  
   K1
R  T2 T1 
K 273K  K 298 K e

H   1
1 



R  273 K 298 K 
K 273K  7.0  10 e
K 273 K  1.1  10 28 24

( 198.4 kJmol 1 )  1
1 



8.314 JKmol 1  273 K 298 K 
o
Step 2: Solve for  G273
K : o
 G273 K   RT ln K eq o
 G273
K = ‐(8.314 J mol‐1 K‐1) (298 K) ln(1.1 x 1028) = ‐146 kJ mol‐1 Step 3: Solve for  G273K with given concentrations (0.100) 2
[ SO3 ]2
=
= 4.0 [ SO2 ]2 [O2 ] (0.500) 2 (0.0100)
o
o
 G 273
K   G 273 K  RT ln Q Q = o
1
 G 273
 (8.314 J mol 1 K 1 )(298 K ) ln 4.0 K   146 kJ mol
o
 G273
K = ‐143 kJ mol‐1 Problem 27. Exposure of metallic iron to moist air results in the formation of rust, one component of which is iron(III) oxide, Fe O . Metallic iron can be recovered from iron(III) oxide by heating it in the presence of solid graphitic carbon: 2Fe O
3C → 4Fe
CO For the substances involved in this reaction we have the following data at 25°C: Solutions at www.prep101.com/solutions 19 of 69
©Prep101 Substance Fe O
C ‐ graphite Fe CO
kJ
mol
‐824.2 ΔH
Chem203 Final Booklet J
S
mol K
87.4 5.7 27.3 ‐393.5 213.7 Assuming ideal behavior, (a) Calculate ΔH° for the above reaction at 25°C. Answer: ΔH° = 467.9kJ H 0   n p H 0f ( prod )   n r H 0f (react ) = (4x0+3(‐393.5))‐(2(‐824.2)+3(0)) = +467.9kJ (b) Calculate ΔS° for the above reaction at 25°C. Answer: ΔS°= 558.4 JK‐1 ΔS°=  n p S 0 ( prod )   n r S 0 ( react ) = (4(27.3)+3(213.7))‐(2(87.4)+3(5.7)) = 558.4 JK‐1 (c) Account for the algebraic sign of ΔS° for the above reaction. Answer: Entropy (disorder) increases (ie. ΔS° is positive) because 5 moles of solid are connected to 4 moles of solid and 3 moles of gas. (d) Calculate ΔG° for the above reaction at 25 °C. Solutions at www.prep101.com/solutions 20 of 69
©Prep101 Answer: 301.5 kJ Solution: ΔG°= ΔH°‐ TΔS°=467.9‐298x558.4/1000=301.5kJ Chem203 Final Booklet (e) Calculate the equilibrium constant, K, for the above reaction at 25°C. Answer: K=1.4x10‐53  G 0
301,500
ln K 

 121 .7
8.314  298
RT
K  1.4  10  53
(f) Determine the temperature above which this reaction would become spontaneous under standard state conditions. Answer: 838K Solution: ie. We require ΔG°≤ 0 ie ΔG°=0 when T=
H 0 467,900J

 838K S 0 558.4JK 1
Solutions at www.prep101.com/solutions 21 of 69
©Prep101 Chem203 Final Booklet Kinetics Problem 28. For the following reaction: S O
→I
2SO
2I
initial rates were determined as described in the table below: mol
mol
I
S O
L
L
Initial Rate 0.080 0.040 1.25 x 10‐5 0.040 0.040 6.25 x 10‐6 0.080 0.020 6.25 x 10‐6 0.032 0.040 5.00 x 10‐6 0.060 0.030 7.00 x 10‐6 Calculate the rate law and the rate constant of the reaction. Solution: v1 k [ I  ]1m [ S2O82 ]1n

v2 k [ I  ]2m [ S2O82 ]n2
1.25  105 k (0.080) m (0.040) n

 2  2m  m  1
6
m
n
6.25  10
k (0.040) (0.040)
v1 1.25  105 k (0.080) m (0.040) n


 2  2n  n  1
6
m
n
v3 6.25  10
k (0.080) (0.020)
rate  k [ I  ]1[ S2O82 ]1
k
rate
[ I ][ S2O82 ]
k1 
1.25  105
 3.91  103 M 1 s 1
(0.080)  (0.040)

Solutions at www.prep101.com/solutions 22 of 69
©Prep101 Problem 29. For the reaction H O
CH Cl
Chem203 Final Booklet ⇌ CH OH
HCl
the following data was obtained: Initial Values [CH Cl ] / M [H O ] / M Experiment 1 0.0100 0.0300 2 0.0200 0.0300 3 0.0085 0.0450 What is the rate law for this equation? Solution: Rate
k H O CH Cl x
4
Rate1 k (0.0100)
3.6  10


x
Rate2 k (0.0200)
1.44  104
0.5x = .25, therefore x = 2 and the reaction is second‐order in H O Rate / Ms‐1 3.60 10 1.44 10 3.90 10 Rate1 k (0.0100) x 3.6  104


Rate2 k (0.0200) x 1.44  104
5.54(2/3)y=3.69 (2/3)y = 2/3, therefore y = 1 and the reaction is first‐order in CH Cl Rate k H O CH Cl Problem 30. Derive the equation of the half‐life of a first order reaction.  [ A]0 
ln[ A]t1/2  ln[ A]0  kt1/2 , ln 
 kt1/2
 [ A]t 
1/2 

[ A]0
for t1/2 [A]t1/2 
2
 [ A]0 
 [ A]0 
ln 
 ln 
  ln 2  kt1/2
 [ A]t 
[
A
]
2
0


1/2


ln  2 
 t1/2 
k
Solutions at www.prep101.com/solutions 23 of 69
©Prep101 Chem203 Final Booklet Problem 31. The compound N O decomposes according to a first‐order rate law. If 20.0% of a sample of N O decomposes in 50.0 s, what is the half‐life of N O ? Solution: If 20% decomposes, then 80% of the sample remains. ln[ A]  ln[ A]0  kt rearrange 0.8[ A]o 
ln
 k(50s)
 [ A]o 
k  4.463 103 sec1 t1/ 2 
ln 2
 155 sec
k
Problem 32. In experiments investigating the thermal decomposition of azomethane in a flask of fixed volume, CH NNCH
→C H
N The following data were obtained: Initial [CH NNCH ] Exp. Initial Rate 1 5.13102 M 2.8106 M/s 2 2.05101 M 1.1105 M/s Determine the rate law and the value of the rate constant. In Exp. 1 above, how long would it take for 90% of the original azomethane to decompose? In Exp. 1 above, what would be the rate of decomposition of CH NNCH once 90% of the original azomethane had decomposed? Solutions at www.prep101.com/solutions 24 of 69
©Prep101 Solution: (a) v1 k [CH 3 NNCH 3 ]1n

v2 k [CH 3 NNCH 3 ]n2
2.8  106 (5.13  102 ) n

 0.25  0.25n  n  1
1.1  105 (2.05  101 ) n
Therefore the reaction is the first order. Rate = k [CH NNCH
], therefore k = Chem203 Final Booklet 2.8 106 M / s
= 5.46 x 10‐5 s‐1 5.13 102 M
(b) ln[A] = ln[A]o – kt ln[0.15.13102] = ln[5.13102] – (5.4610‐5)t ‐5.27 = ‐2.97 ‐ (5.4610‐5)t (5.4610‐5)t = 2.30 t = 4.21104 s = 11.7 hours (c) [A]10% = 0.15.13102 = 5.1210‐3M Rate = k[A] = (5.4610‐5 s‐1)( 5.1210‐3 M) = 2.8010‐7 M/s Solutions at www.prep101.com/solutions 25 of 69
©Prep101 Chem203 Final Booklet Problem 33. Consider the bromination of acetone: CH COCH
Br → CH COCH Br H
Br Determine the value of the rate constant from the following data, assuming the above step is elementary: Experiment [CH COCH ] / M [Br ] / M [H ] / M Initial rate / Ms‐1 1 0.01 0.01 0.01 4.0010‐5 2 0.02 0.01 0.01 8.0010‐5 3 0.02 0.02 0.01 8.0010‐5 4 0.01 0.02 0.01 4.0010‐5 Solution: Note – the bromination of acetone isn’t in reality a one‐step mechanism but for this exercise you can only solve by assuming it is and that it follows the equation given. We use the method of initial rates to find the order in each reagent. First we use experiments 1 and 2 to find the order in CH COCH : CH COCH
Br
CH COCH
Br
8.00E 5
0.02 0.01
4.00E 5
0.01 0.01
0.02
8.00E 5
,2 2 ,
1 4.00E 5
0.01
*To solve for the exponents we can either rationalize the possible values or take the log of both sides and solve. Then we use experiments 2 and 3 to find the order in Br CH COCH
Br
CH COCH
Br
0.02 0.02
8.00E 5
0.02 0.01
8.00E 5
8.00E 5
0.02
,1 2 ,
0 8.00E 5
0.01
From this we get CH COCH
Br Then, we use any one of the trials with complete data to find the rate constant CH COCH
Br , 8.00E 5
0.01 0.02 M
4.00E 5 s
M
4.00E 5
0.01 ,
4.00E 3s s
0.01M
Notice that with the data as given, the concentration of H+ is not necessary to be used. It was provided as extraneous data. Solutions at www.prep101.com/solutions 26 of 69
©Prep101 Chem203 Final Booklet Problem 34. The decomposition of compound A follows a second‐order rate law. Draw a decomposition plot with a linear slope. Label the axes, and slope. Solution: Because this reaction is second order 1
1

 kt if 1/[A] was plotted vs. t then the y‐
[ A] [ A]o
intercept would be 1/[A]0 and the slope would be k 1/[A] slope = +k 1/[A]0 t Problem 35. The reaction 2H O → 2H O
has the following mechanism. O H O
I  → H O IO 
H O
IO → H O O
I What is the catalyst in the reaction? Solution: I is the catalyst as it is used up in the first reaction and is produced in the second. It does not show up in the overall reaction. Solutions at www.prep101.com/solutions 27 of 69
©Prep101 Chem203 Final Booklet Problem 36. Answer this question based on the following 3 step reaction S O ↔ SO slow SO
½O ↔ SO fast H O ↔ H SO fast SO
Assume all three steps are exothermic. Draw and label a potential energy diagram (include, intermediates, Ea, reactants, products, ∆H, etc.) How will the potential energy diagram change if step 1 is catalyzed? Solutions at www.prep101.com/solutions 28 of 69
©Prep101 Problem 37. Given that the activation energy for the reaction A
change for the reaction is 41
Chem203 Final Booklet B ⇌ C D, is 66 and the enthalpy , what is the activation energy for the reverse reaction, C
D ⇌ A
B? Solution: Since Hrxn > 0, then Hproducts > Hreactants Thus, Ea rev = 66 – 41 = 25 kJ/mol (draw an energy diagram to see this better) Solutions at www.prep101.com/solutions 29 of 69
©Prep101 Problem 38. The reaction: NO
Chem203 Final Booklet O → NO
O has reaction rate constants of 1.21010 at 25 oC, and 3.551010 at 96 C. Determine the activation energy for this reaction. Solution: Let T1 = 96 C = 369 K, and T2 = 25 C = 298 K. Using the Arrhenius Equation  3.55 x 1010 s 1 
Ea
1
 1
ln 


10  1 
1
1 
8.314 JK mol  369 K 298 K
 1.2 x 10 s 
 

and solve for Ea, Ea = 12.8 kJ/mol Problem 39. The rate of a certain reaction is 3.0 times faster at 50 oC than at 40 oC. Which of the following is the best estimate of the activation energy for the reaction?  k2
 k1
Solution: from Arrhenius equation ln 

E  1
1
   a 
  R  T 2 T1 

since vk, then k  v  E 1 1 
ln 2  ln 2   a   
R T2 T1 
k1  v1 
Ea  1
1 
ln(3)  


 Ea  77433.3 J / mol  77.4 kJ / mol
8.314 325 313
Solutions at www.prep101.com/solutions 30 of 69
©Prep101 Chem203 Final Booklet Problem 40. The reaction 2A → A is an elementary process. It takes 35.2 minutes for the concentration of A to decrease from 0.20 M to 0.10 M. What is the rate constant for the reaction? Solution: Since we have an elementary process, then rate of the reaction is rate  k[A]2 , This reaction is second order therefore 1
1

 kt [ A] [ A]o
1
1

 k(35.2 min) 0.1M 0.2 M
5 M‐1 = k(35.2 min) k = 1.42 x 10‐1 M‐1min‐1 Solutions at www.prep101.com/solutions 31 of 69
©Prep101 Chem203 Final Booklet Problem 41. Nitrogen oxide converts ozone into molecular oxygen, as follows: O
NO → O
NO The elementary steps for the reaction are given below. What is the overall rate law? Mechanism I k1
 O  NO3
O3  NO 
(slow)
k2
 2 O2
O  O3 
(fast) k3
 2 NO2
NO3  NO 
(fast)
The steps are elementary, and the first step is the rate determining step. So the overall rate law only depends on the first step: rate = k[O3][NO] Solutions at www.prep101.com/solutions 32 of 69
©Prep101 Problem 42. The rate law for a reaction is rate
this rate law? i) A B  E
E +B  C +D
iii) Chem203 Final Booklet B . Which of the following mechanisms supports k A
ii) A B  E
(fast)
E  AC  D
(slow)
(fast)
(slow) A A E
(slow)
E  B C  D
(fast)
Solution: Slow step rate k E B For the fast equilibrium, rate
Rearrange to: [E] 
k
E k1
kk
[A][B], therefore, rate  1 2 [ A][B]2 This is inconsistent with the rate law. k1
k 1
(ii): slow step rate
[E] 
k A B
k E A and fast equilibrium rate
k A B
k
E k1
kk
[A][B], therefore, rate 1 2 [A]2[B] This is consistent with the rate law k1
k1
(iii) The first step is the rate determining step. The rate law would be rate k A This is inconsistent with the proposed rate law. Problem 43. The reaction of NO with O2 to form NO2 contributes to the formation of urban smog. The following mechanism has been proposed: NO NO ⇌ N O fastequilibrium O → 2NO slow N O
What is the rate law derived from this mechanism? Solution: N2O2 is an intermediate, therefore cannot be present in the rate law. Rate is found from the slower of the two reactions: Rate = k2[N2O2][O2], where k‐1[N2O2] = k1[NO]2 Therefore, rate
k1k2
[NO]2[O2 ]  k [NO]2[O2 ] k1
Solutions at www.prep101.com/solutions 33 of 69
©Prep101 Chem203 Final Booklet Equilibrium Problem 44. Draw equilibrium constant expressions for the following reactions: 5O
⇌ 3CO
4H O 1.C H
H O ⇌ CH COO
Na 2.CH COONa
Solution: CO2 3eq
1. K eq 
C3 H 8 1eq O2 5eq

2. Keq  CH3COO eq  Na eq Problem 45. Given the reactions: 1 H
Br
2 H
Cl
, withK
3.5 10 at1495K,
, withK
9.5 10 at1495K Give the equilibrium constants for the following reactions at 1495 K. a) HBr ⇌ ½H
½Br ½Cl
½Br
⇌ HCl
HBr b) H
Solution: (a) HBr(g) ⇌½ H2(g) + ½ Br2(g) From H2(g) + Br2(g) ⇌2 HBr(g), K1 = 3.5 x 104, you must reverse the reaction: 2 HBr(g) ⇌H2(g) + Br2(g), K‐1 = 1/K1 = 2.9 x 10‐5, and then half the reaction: HBr(g) ⇌½ H2(g) + ½ Br2(g), K = K 1 = 5.3 x 10‐3 (b) H2(g) + ½ Cl2(g) + ½ Br2(g) ⇌HCl(g) + HBr(g) This reaction is the sum of half of each of the reactions above. H2(g) + Br2(g) ⇌2 HBr(g), with K1 = 3.5 x 104 ½ H2(g) + ½ Br2(g) ⇌HBr(g), with K3 = K 1 = 187 H2(g) + Cl2(g) ⇌2 HCl(g), with K2 = 9.5 x 103 ½ H2(g) + ½ Cl2(g) ⇌HCl(g), with K 4 = K 2 = 97 H2(g) + ½ Cl2(g) + ½ Br2(g) ⇌ HCl(g) + HBr(g), K = K3 x K4 = 1.8 x 104 ⇌ 2HBr
⇌ 2HCl
Solutions at www.prep101.com/solutions 34 of 69
©Prep101 Problem 46. For the following reaction: 5O
⇌PO
,K
P
In which direction must the reaction proceed to reach equilibrium if: (a) [P ], [O ], and [P O
] = 1.0 M ] = 5.0 M (b) [P ] = 8.0 M, [O ] = 0.70 M, and [P O
Solution:  P4O10 eq
The expression for K 
 3  P4 eq O2 5eq
Chem203 Final Booklet 3  P4O10 init
 P4 init O2 5init
 P4O10 init
In (b) Q 
 P4 init O2 5init
In (a) Q 


1M
1 M  1 M 
5
 1  3 , the reaction proceeds to the right. 5.0 M
8.0 M   0.70 M 
5
 3.7  3 , the reaction proceeds to the left. Problem 47. Consider the reaction: A
B
⇌ 2AB K 800K 0.5 If 1.00 mol of A , 2.00 mol of B , and 2.00 mol of AB , are placed in a 250 mL vessel at 800 K, what is the composition of the equilibrium mixture? Solutions at www.prep101.com/solutions 35 of 69
©Prep101 Chem203 Final Booklet Solution: To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q. [A
] = and Q
1.00 mol
= 4.00 M, [AB ] = [B
0.250 L
2
AB init


 A2 init  B2 init

] = 2.00 mol
= 8.00 M, 0.250 L
(8.00) 2
 2.0 (4.00) (8.00)
The direction of the reaction needs to be determined. To do this, we compare Q vs K. Since Q = 2.0 > K = 0.5, the reaction proceeds to the left. To determine what final concentrations will be from initial concentrations, a handy tool – the Initial, Change, Equilibrium (ICE) table can be used. All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A2 and B2 will be formed and AB will be consumed. A
B
Concentration (M) 2 AB Initial 8.00 M 4.00 M 8.00 M Change ‐2x +x +x Equilibrium 8.00 – 2x 4.00 + x 8.00 + x As the reaction proceeds, x moles of A and B are formed as 2x moles of AB are consumed. Make sure to consider stoichiometric coefficients appropriately. The equilibrium values are simply the sum of the initial + change concentrations. Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described):  AB 2eq
(8.00  2 x ) 2
K 

 0.5  A2 eq  B2 eq (4.00  x ) (8.00  x )
so, (8.00  2 x )2  0.5(4.00  x )(8.00  x )
64  32 x  4 x 2  0.5(32  12 x  x 2 )
64  32 x  4 x 2  16  6 x  0.5 x 2
3.5 x 2  38 x  48  0
To solve for x, the quadratic formula must be used, x
 b  b 2  4 ac 38  (  38) 2  4(3.5)(48)

 1.46 or 9.39, 2a
2(3.5)
Since 2 x 9.39 = 18.78 > 8.00 (which would leave the equilibrium concentration of AB at equilibrium negative), then the x = 1.46 Solutions at www.prep101.com/solutions 36 of 69
©Prep101 Chem203 Final Booklet So, final concentrations are: [A2] = 4.00 + 1.46 = 5.46 M [B2] = 8.00 + 1.46 = 9.46 M [AB] = 8.00 – 2(1.46) = 5.08 M To check, substitute these concentrations into the Equilibrium constant expression,  AB 2eq
(5.08) 2
K 

 0.5 , matches up.  A2 eq  B2 eq (5.46) (9.46)
Solutions at www.prep101.com/solutions 37 of 69
©Prep101 Chem203 Final Booklet Practice Questions Problem 48. What is the equilibrium constant expression for the reaction below? 4H
4NO
H O ⇌ 3H SnO
4NO 3Sn
[ NO ]4
Answer: K c 
[ H  ]4 [ NO3 ]4
Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts. Problem 49. Give an expression for Kc for the reaction ⇌ CaO
CO ? CaCO
Solution: K
CO
Remember: Solids and liquids are not included in the expression. Solutions at www.prep101.com/solutions 38 of 69
©Prep101 Chem203 Final Booklet Problem 50. Given the equilibrium constant for equation 1, what is the equilibrium constant for equation 2? 1
3
⇌ N
H K
0.157
Equation1:NH
2
2
3H
⇌ 2NH K
?
Equation2:N
Solution: Equation 2 is equal the double and reverse of equation 1 therefore 1
K
K
40.6 0.157
Problem 51. Given: P
6Cl
⇌ 4PCl EquilibriumConstant K What is the equilibrium constant for the following reaction? 2PCl
⇌ 3Cl
½P Answer: /
EquilibriumConstant K
Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2. Solutions at www.prep101.com/solutions 39 of 69
©Prep101 Chem203 Final Booklet Problem 52. At 25 ˚C Kc for the reaction of carbon monoxide with oxygen to produce carbon dioxide is 3.3x1091. What is Kp for this reaction under the same conditions? Answer: So the reaction equations is: O
⇌ 2CO K
3.3 10 2CO
The relationship between K and K is: K RT ∆ In this case there are 3 moles of gas in the reactants and 2 moles of gas in the products, K
so ∆n = ‐1 So solving for K : Latm
0.0821
K RT ∆ 3.3 10
1.35 10 K
298K
molK
Problem 53. The equilibrium constant for the reaction below is K
concentrations were N
C H
HCN
1.00
2.5
10 at 25C. If the initial , then what is [HCN] at equilibrium? N
C H
⇌ 2HCN
Solution: N
i: 1.00 c: +x e: 1.00+ x  Q=1>K so then rxn goes to the left (1.00  2 x)2
(1.00  2 x)
 Kc 
 Kc 2
(1.00  x)
(1.00  x)
x
1  Kc
2  Kc
1.00 +x 1.00 + x ⇌
2HCN
1.00 ‐2x 1.00‐2x  0.488 Solutions at www.prep101.com/solutions C H
40 of 69
©Prep101 Problem 54. Phosgene, COCl
Chem203 Final Booklet , is a toxic gas that dissociates according to the chemical equation: COCl
⇌ CO
Cl
For the reaction above, K
8.05 10 at 673 K. Exactly 1.00 mol COCl is placed in an empty 25.0‐L flask. The flask is capped and then warmed to 673 K. What is the equilibrium concentration of CO at this temperature? Solution: COCl
⇌ CO
Cl
initial: 0.04 0 0 change: ‐x +x +x equil: 0.04‐x x x x2
Kc = 0.04  x
x2 + Kcx – 0.04Kc = 0  Kc  Kc 2  (4)(0.04)( Kc)
x
 5.29  103 2
Problem 55. The equilibrium constant for the gas‐phase reaction CO H O ⇌ CO
H 5.10 at 700 K. One mole each of CO, H O, CO , and H is placed in a 10.0‐L flask and the temperature is K
is raised to 700 K. What is the concentration of H when the reaction reaches equilibrium? Solution: Kc 
[CO2 ][ H 2 ]
 5.10 [CO][ H2O]
CO I C E H O ⇌ 0.1M ‐x 0.1‐x 0.1 ‐x 0.1‐x CO 0.1 +x 0.1+x H 0.1 +x 0.1+x Kc 
(0.1  x )2 0.01  0.2 x  x 2

 5.10
(0.1  x )2 0.01  0.2 x  x 2
0.01  0.2 x  x 2  0.051  1.02 x  x 2
0  0.041  1.22 x
1.22 x  0.041
x  0.034 M
Therefore at equilibrium [H2] = 0.1 + x = 0.1 + 0.034M = 0.134M Solutions at www.prep101.com/solutions 41 of 69
©Prep101 Chem203 Final Booklet Problem 56. For the following equilibrium, which is endothermic in the forward direction, are the following statements true or false? CO CH CHO ⇌ CH
a) A decrease in temperature will increase the pressure in the system. b) Addition of methane will increase the heat absorbed by the system. c) An increase in the volume of the system will increase the total moles of CO at equilibrium. d) At constant total pressure, an increase in temperature will increase the amount of CH CHO at equilibrium. e) As temperature increases, the equilibrium constant for the reaction decreases. Solution: Only C is true An increase in volume will shift the equilibrium to the side with more moles of gas (to the right) thus causing an increase in the total moles of CO at equilibrium. B is false because adding CH will shift to system to the left, creating CH CHO and heat (not absorbing heat). Problem 57. The following reaction has reached a state of dynamic equilibrium, in the presence of a catalyst. 2SO
O
⇌ 2SO ,
∆H
198kJ Which direction will the equilibrium shift with the following changes? a) increasing the temperature b) removing the catalyst c) decreasing the volume of the reaction vessel d) adding some more O Solution: a) shift to the left b) no effect c) shift to the right d) shift to the right Solutions at www.prep101.com/solutions 42 of 69
©Prep101 Chem203 Final Booklet Problem 58. Consider the following reaction: ⇌ Ni
4CO Ni CO
A mixture of Ni CO
and CO , each with a concentration of 0.800M, and an excess of Ni were confined in a 1.0 L container at 300K. In which direction will the reaction shift to reach equilibrium? Solution: The reaction will shift left forming Ni CO
to reach equilibrium Problem 59. The equilibrium constant for the reaction Cl 2NOCl ⇌ 2NO
is 0.51 at a certain temperature. A mixture of NOCl , NO and Cl with concentrations 1.3, 1.2, and 0.60 M, respectively, was introduced into a container at this temperature. Which of the following are true? a) The concentration of Cl increases until equilibrium is reached. b) [NOCl ] = [NO ] = [Cl ] at equilibrium c) We are at equilibrium and thus no net change takes place. d) The concentration of NOCl increases until equilibrium is reached. Answer: C Solution: Q
.
.
.
0.51
Solutions at www.prep101.com/solutions K, therefore we are already at equilibrium. 43 of 69
©Prep101 Chem203 Final Booklet Acids and Bases Problem 60. In a laboratory experiment, students measured the pH of samples of rainwater and household ammonia. Determine (a) [H O ] in the rainwater, with pH measured at 4.35; (b) [OH ‐ ] in the ammonia, with pH measured at 11.28. Solution: (a) By definition, pH = ‐log[H3O+], or log[H3O+] = ‐pH = ‐4.35 [H3O+] = 10‐4.35 = 4.5  10‐5 M (b) First, determine pOH pOH = 14.00 – pH = 14.00 – 11.28 = 2.72 Now, use the definition pOH = ‐log[OH‐], log[OH‐] = ‐pOH = ‐2.72 [OH‐] = 10‐2.72 = 1.9  10‐3 M Problem 61. Calculate the pH of a 0.20 M solution of HNO . Solution: HNO dissociates via HNO
H O⇌H O
NO Since HNO is a strong acid, it dissociates completely, producing 0.5 M H O . Taking the negative log, we get a pH of 0.70 (2 decimal places since there are two significant figures in 0.20 M). Weak Acids and Weak Bases Identification All compounds that donate protons, but do so only partially, are weak acids. The extent of dissociation of a weak acid is described by its K . Weak acids are all acids outside of the six strong acids listed earlier. All compounds that accept protons, but do so only partially, are weak bases. The extent of dissociation of a weak base is described by its K . Weak bases include all the bases not included in the list of strong bases. Solutions at www.prep101.com/solutions 44 of 69
©Prep101 Chem203 Final Booklet Problem 62. A chemist dissolves 0.020 moles of sodium methanoate (NaHCOO), a weak base, in 200 mL of pure water, and measures its pH. What value should she get? Solution: Sodium methanoate dissolves and then dissociates following: ⇌ Na
HCOO
NaHCOO
HCOO
H O ⇌ HCOOH
OH 1 mol HCOO 
1 mol NaCOOH
 0.1M HCOO  0.200 L
0.020 mol NaCOOH 
[HCOO‐]initial = Using an initial, change, equilibrium table: HCOOH
HCOO
Initial 0.1 M 0 M Change ‐x +x Equilibrium 0.1 M – x x [ HCOOH ][OH  ]
x2

 5.6  1011 The expression for K b 
[ HCOO  ]
0.1  x
Assuming that x << 0.1, OH
0 M +x x x2
 5.6 1011 0.1
and x = [OH‐] = (5.6x10‐12)1/2 = 2.4 x 10‐6 M pOH=‐log10(2.4x10‐6 M) = 5.63 and pH =14.00 ‐ 5.63 = 8.37 Solutions at www.prep101.com/solutions 45 of 69
©Prep101 Chem203 Final Booklet : Problem 63. Calculate the pH of a 1 M solution of H SO
The acid‐base dissociations for H SO
are as follows: H SO
H O ⇌H O
HSO
K 1, goestocompletion HSO
H O ⇌ H O
SO
K
1.2 10 Solution: For a strong acid, the moles of H O produced from the first dissociation are equal to the concentration of the acid, H SO
1
H O
1
so From the second (weak) acid dissociation, the number of moles of H O can be determined using an ICE table. HSO
H O SO
Initial 1 1 0 Change ‐x +x +x Equilibrium (1 – x) (1 + x) x SO
H O
1
K
1.2 10
1
HSO
Because products already exist, we don’t expect a lot of additional H O to form and so x < 1, ∴ we can approximate 1.2
10
∴
1.2
10 From this we get H O
1
1 0.012 1.012 pH
log 1.012
0.005181 *If the H O
1 then pH=0, if it is below one then pH can in fact be a negative value as it is in this case.
*For most polyprotic weak acids it is not necessary to even calculate the second ionization – the added amount of ions in solution will be insignificant. Periodic Trends and Effects of Structure on Acid‐Base Strength. Strong Acids Strong Bases (any 1st or 2nd column/group hydroxides) HCl LiOH
HBr NaOH
HI KOH
HClO RbOH
HNO CsOH
H SO Ca OH The molecular structure of the acid or base is usually a good indicator of strength. For the hydroxide bases in the previous table, their strength originates from the fact that there is a large electronegativity (ability to accept and hold negative charges) difference between the cations (metal ions) and the anions (hydroxide ions). This facilitates the ability of the base to dissociate into free metal and hydroxide (OH‐) ions, making a basic solution (in accordance with the Arrhenius theory). Solutions at www.prep101.com/solutions 46 of 69
©Prep101 Chem203 Final Booklet Ionic Charge and Number of Protons Problem 64. Will a 0.1 M solution of AlF be acidic or basic? Solution: AlF dissociates in water following: AlF
H O ⇌ Al
Al
H O ⇌ Al OH
H O
H O ⇌ HF
F
3F K 1.0
OH 10 1  1014
Kw
 1.4  1011 = 4
K a(HF) 7  10
Since Ka > Kb, AlF will be acidic. Kb = Problem 65. Calculate the pH of a 1.0 M solution of HF and CaF knowing that K 7.0E 4. Assume that CaF dissociates completely in water. Solution: HF
H O ⇌F
H O
This is a buffer. If both HF and F are present, their interconversion will be negligible (e.g. the concentrations will stay stable). Our initial concentration of HF is 1.0M. Our initial concentration of F is based off: H O ⇌ Ca
2F CaF
Considering CaF dissolves completely: 2F
1.0MCaF
2.0MF 1CaF
PREFERRABLY, we just use the Henderson Hassel‐Bach equation (as given on the formula sheet): Base
log
pH pK
Acid
F
pH
log 7.0E 4
log
HF
2
3.1549 0.30103 3.456 3.46 pH 3.1549 log
1
Alternatively, we can use an ICE table: HF F
H O ⇌ H O Initial 1.0 2.0 0 Change ‐x +x +x Equilibrium (1.0 – x) (2.0 + x) x K
7.0E
4
.
, .
assume x << 1.0, 7 7.0E
4
Then H O
Solutions at www.prep101.com/solutions 2.0
,
1.0
3.5
3.5
4 ,
4 3.46 47 of 69
©Prep101 Chem203 Final Booklet Problem 66. Determine if a solution prepared by mixing 5.0 mL of concentrated HCl (12 M) with 75 mL of 1.0 M acetate is a buffer solution. Solution: The question asks if this is a buffer solution. A buffer solution contains both a weak acid and its conjugate base as major species. Thus, to answer the question, we must calculate the concentrations of acetate anions and acetic acid in the solution. Obtaining these concentrations requires quantitative calculations. The species in the two solutions that are mixed together are H2O and the following: HCl:H O andCl‐ Sodiumacetate:CH CO andNa Because the hydronium ion is a strong acid and acetate is a weak base, mixing the solutions results in near‐
quantitative reaction between these two ions: 1
H O
CH CO ⇌ H O CH CO HK
5.6 10 K
We calculate the initial amounts of the species from the volumes and concentrations, using n = MV: n (H O ) = (12 M)(5.0 mL) = 60 mmol n (CH CO ) = (1.0 M)(75 mL) = 75 mmol When the solutions are mixed, H3O+ reacts quantitatively with CH CO , consuming all the hydronium ions and generating acetic acid molecules, so after mixing we have the following amounts: n (CH CO ) = 75 mmol  60 mmol = 15 mmol n (CH CO H) = 60 mmol We see that both the weak acid and its conjugate base are major species in the mixture, so this is indeed a buffer solution. Solutions at www.prep101.com/solutions 48 of 69
©Prep101 Chem203 Final Booklet Problem 67. Buffer solutions with a pH of about 10 are prepared using sodium carbonate (Na CO ) and sodium hydrogen carbonate (NaHCO ). What is the pH of a solution prepared by dissolving 10.0 g each of these two salts in enough water to make 0.250 L of solution? Solution: We are asked to calculate the pH of a buffer solution. . Both compounds are salts that dissolve in water, the major species are H O , Na , HCO
A buffer solution must contain a weak acid and its conjugate base as major species; HCO
is the weak acid and CO
is the conjugate base: H O
HCO
⇌ CO
H O This proton transfer reaction involves the second acidic hydrogen atom of carbonic acid, so the appropriate equilibrium constant is Ka2 whose pK is found to be pKa2 = 10.33. Because this is a buffer solution, we apply the Henderson‐Hasselbach equation: Base
log
pH pK
Acid
Get the initial concentrations from the masses of the salts and the volume of the solution: 10.0 g 
m
HCO
= = 0.476 M 
 Mr V 84.01 g /mol  0.250 L 
CO
= 10.0 g 
m
= 0.377 M 
 Mr V 106.0 g /mol  0.250 L 
Now we substitute the appropriate values into the buffer equation and evaluate:  0.377 M 
pH pK
log
= 10.33 + log 
 = 10.33 + (0.10) = 10.23  0.476 M 
The pH is close to the pKa of the conjugate acid‐base pair, so this is a reasonable result. Solutions at www.prep101.com/solutions 49 of 69
©Prep101 Chem203 Final Booklet Problem 68. By how much does the pH of the buffer solution of example in the previous section change on the addition of 3.50 mL of 6.0 M HCl? Solution: The first four steps of the seven‐step strategy are identical to the ones in previous example. In this example, addition of a strong acid or base modifies the concentrations that go into the buffer equation. We need to determine the new concentrations (step 5) and then apply the buffer equation (step 6). In dealing with changes in amounts of acid and base, it is often convenient to work with moles rather than molarities. The units cancel in the concentration term of the buffer equation, so the ratio of concentrations can be expressed as a ratio of moles as well as a ratio of molarities:  A   mol A / V  mol 
A


 HA  molHA / V  molHA
We do this problem using moles. First, determine the amounts present in the solution before addition of the HCl: 10.0 g
= 1.19101 mol molHA = 84.01 g/mol
10.0 g
= 9.43102 mol molA = 106.0 g/mol
Next, calculate the amount of hydronium ions added: mol H O = MV = (6.0 mol/L)(3.5 mL)(103 L/mL) = 2.1102 mol 3
The hydronium ions react completely with conjugate base anions, increasing the amount of HCO
reducing the amount of CO
molHA = 1.19101 mol + 2.1102 = 1.40101 mol molA = 9.43102  2.1102 = 7.3102 mol Finally, substitute these new amounts into the buffer equation to compute the new pH: mol A
 0.073 mol 
pH = pKa + log = 10.33 + log 
 = 10.33  0.28 = 10.05 mol HA
 0.140 mol 
Note that the pH changes be a small amount (0.18), a reasonable outcome for a buffer solution. Solutions at www.prep101.com/solutions and 50 of 69
©Prep101 Buffer Capacity and Buffer Range Chem203 Final Booklet Practice Questions Problem 69. Which one of the following mixtures are buffers? i. Equal volumes of 0.1 M NaNO and 0.1 M HF ii. Equal volumes of 0.1 M NaF and 0.1 M HF iii. Equal volumes of 0.1 M KCH CO and 0.1 M NaOH Solution: Equal volumes of 0.1 M NaF and 0.1 M HF Adding an acid and the salt of its conjugate base can form a buffer, this is the case in (b). Problem 70. Which 2 of the following solutions are buffers? 1. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M NaOH 2. 100 mL of 0.1 M acetic acid mixed with 25 mL of 0.5 M NaOH 3. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M sodium acetate 4. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.2 M NaOH 5. 50 mL of 0.1 M sodium acetate mixed with 25 mL of 0.05 M NaOH Solution: 1 and 3. A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid, so all of the acetic acid is consumed so it is not a buffer. 4 is not a buffer solution since all of the weak acid is consumed with strong base, 5 is not a buffer because sodium acetate is a weak base, so you have a weak base with a strong base in solution. Solutions at www.prep101.com/solutions 51 of 69
©Prep101 Chem203 Final Booklet Problem 71. Exactly 0.600 mol NaHSO and 0.400 mol Na SO are dissolved in water to a final volume of 1.00 L. What is the pH of this solution? (KaHSO
1.20 10 ) Solution: B This is a buffer solution, therefore: 
 [ A ]eq 
 0.400 
2
pH  pKa  log 
   log(1.20  10 )  log 
  1.74  0.600 
 [ HA]eq 
Problem 72. How many moles of sodium acetate, CH COONa, must be added to 1.0 L of 0.100 M acetic acid, CH COOH (pKa = 4.75), to give the solution a pH of 4.75? Solution: This is a buffer solution, therefore:  [ A ]eq 
 [CH 3COONa ] 
pH  pKa  log 
 = 4.75  log 
 =4.75 0.100


 [ HA]eq 
 [CH 3COONa ] 
 [CH 3COONa ] 
Therefore log 
 =0 and 
 =1 0.100
0.100




[CH3COONa] = 0.10M # moles CH3COONa = 0.10 moles Solutions at www.prep101.com/solutions 52 of 69
©Prep101 Problem 73. For the acid‐base indicator HIn, Ka
conjugate base for this system? Solution: Base
Acid
In
log 1 10
log
HIn
In
8 6 log
HIn
In
HIn
,
10 ,
HIn
In
pH
8
2
log
In
HIn
1
Chem203 Final Booklet 10 . At pH = 8.0, what is the ratio of acid to 
pK
log
10
1
100
Problem 74. Consider a solution of 2.0 M HCN and 1.0 M NaCN. Ka for HCN is 6.2
of this solution? Solution: pH = 8.91 CN 
pH = pKa + log
HCN
1. 0
= 9.20 + log
2.0
= 8.91 Solutions at www.prep101.com/solutions 10 . What is the pH 53 of 69
©Prep101 Chem203 Final Booklet 
Problem 75. Ka for HA is 1.800 10 . One liter of a solution contains 0.500 moles each of HA and NaA. If 0.0100 moles of HCl is added to it, what would be the pH of the resulting solution assuming that its volume remains constant? a) 0.300 b) 2.000 c) 3.727 d) 3.745 e) 3.762 Solution: C This is a buffer system as there are equal amounts of conjugate acid and base, but the addition of H+ can change the pH slightly. The original pH of the buffer is approximately equal to pKa. When H+ ions are added from the strong acid HCl, A‐ is converted into HA. Therefore the addition of 0.01 moles H+ produces 0.01 moles HA and consumes 0.01 moles A‐. nHA= 0.5 + 0.01 = 0.51 mol nA‐ = 0.5 – 0.01 = 0.49 mol [H+]=Ka x nHA/nA‐ = (1.8 x 10‐4) (0.51/0.49) = 1.873 x 10‐4 pH = ‐log (1.873 x 10‐4) = 3.727 Problem 76. Which of the following will have the smallest pH change upon the addition of 0.01 mole of strong acid? a) 50 mL of 0.10 M pH = 7.2 carbonate buffer. b) 100 mL of 0.050 M pH = 7.2 carbonate buffer. c) 75 mL of 0.067 M pH = 7.2 carbonate buffer. d) 250 mL of 0.020 M pH = 7.2 carbonate buffer. e) None of the above. Solution: C There are more moles of carbonate buffer in solution c than any of the other solutions Solutions at www.prep101.com/solutions 54 of 69
©Prep101 Chem203 Final Booklet Problem 77. Consider a solution of 2.0 M HCN and 1.0 M NaCN. Ka for HCN is 6.2 10 . Which statement is TRUE? a) The solution is not a buffer because [HCN] is not equal to [CN ]. b) The pH is below 7.00 because the concentration of the acid is greater than that of the base. c) OH
H d) The buffer is more resistant to pH changes from addition of strong acid than of strong base. e) All of the above are false. Solution: E The solution is a buffer with pH above 7. A buffer is resistant to both addition of strong acid and strong base and the concentration of the hydronium ion is not more than the hydroxide ion (pH >7). Problem 78. How much solid NaCN must be dissolved in 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? (K
6.2 10 for HCN.) a) 0.0034 g b) 11 g c) 160 g d) 24 g e) 0.15 g Solution: E CN
H O HCN H O ⇌ I 0.5 0 C 1 10 1 10 1 10 E 0.5 1 10
1 10
Ka = 6.2 x 10‐10 = [(1x10‐7)( Y+1x10‐7)]/ 0.5 Y=0.003M x 1L = 0.003moles Mass = 49.0075g/molNaCN x 0.003 moles = 0.15g Solutions at www.prep101.com/solutions 55 of 69
©Prep101 Chem203 Final Booklet Problem 79. Which of the following solutions is not a buffer? 1. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M NaOH 2. 100 mL of 0.1 M acetic acid mixed with 25 mL of 0.5 M NaOH 3. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M sodium acetate 4. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.2 M NaOH 5. 50 mL of 0.1 M sodium acetate mixed with 25 mL of 0.05 M NaOH a) none of the above b) 5 only c) 1, 2 and 5 d) 3, 4 and 5 e) 2, 4 and 5 Solution: E A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid. Problem 80. Which 2 of the following solutions would be buffer solutions? a) An equimolar solution of nitric acid, HNO , and NaNO b) A solution produced by mixing 0.1mol of NaOH with 0.05mol NaCl c) A solution containing 0.1mol CH COOH and 0.1mol of CH COONa d) A solution produced by mixing 100mL of 0.100M HCl with 400mL of 0.025M NH e) A solution produced by mixing 100mL of 0.100HF with 200mL of 0.02M NaOH Solution: C or E In order to determine which solutions are able to act as buffer solutions, determine what ions will be found in the solution and whether those ions are acidic, basic or spectator ions. If acidic and/or basic ions are found, calculate the amount of ions that are present. A) all ions present (H+, NO3‐, and Na+) are spectator ions. No buffer abilities possible. B) all ions present (Na+, OH‐ and Cl‐) are spectator ions. No buffer abilities possible. C) This is a 1:1 ratio of conjugate acid to its conjugate base. This IS a buffer D) the HCl completely dissociates to form H+ and Cl‐ ions. NH3 can form an equilibrium where NH3 + H+ ↔ NH4+. However, HCl and its dissociated ions are present in 0.01mol amounts—the same as the amount of NH3. Therefore, all the NH3 is “used up” in reacting with the HCl, so no buffering abilities are possible. E) the NaOH completely dissociates to form Na+ and OH‐ ions. The 0.004mol of OH‐ will react with the 0.01mol of HF to form 0.004mol of F‐ and have 0.006mol left of HF since the OH‐ is the limiting reagent. This means we have HA and A‐ both present in a ratio of 3:2. This is a buffer. Solutions at www.prep101.com/solutions 56 of 69
©Prep101 Chem203 Final Booklet Problem 81. A buffer solution that contains 0.40 mol of base B and 0.25 mol of its conjugate acid BH , and has a pH of 8.88. The Kb value of the base is a 8.22 10 b 1.22 10 c 4.74 10 d 4.13 10 e 2.45 10 Solution: C B + H2O ↔ BH+ + OH‐ Calculate from the given pH, the concentration of OH‐ ions that dissociate form from the reaction of the base with water. pOH = 14 – pH = 14 – (8.88) = 5.12 [OH‐] = 10‐pOH = 10‐5.12 = 7.585 x 10‐6M Assume 1 L of solution, therefore the [B] = 0.40mol/L and [BH+] = 0.250mol/L. K = [BH+][OH‐] = (0.250)( 7.585 x 10‐6) = 4.74 x 10‐6 [B] (0.40) Problem 82. A buffer solution is prepared by mixing equal volumes of 0.100M nitric acid, HNO , and 0.300M ammonia, NH . What is the pH of this solution? (K
1.8 10 ) a) 5.28 b) 8.72 c) 11.13 d) 9.56 e) 4.44 Solution: D Because equal volumes of the acid and weak base are being mixed, all concentrations (M) can be treated as moles (mol). HNO3 is a strong acid so it completely dissociates HNO3  H+ + NO3‐ 0.1 0.1 0.1 NH3 will react with the H+ released by the HNO3 to form NH4+. Initially there is 0.3M or 0.3moles of NH3. Upon addition of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leaving 0.2mol NH3 unreacted. Therefore, NH3 + H+ ↔ NH4+ Using the following equation, solve for [H+]. [H+] = Ka x [HA] [A‐] Kw = Ka x Kb so that Ka = Kw = 1.0 x 10‐14 = 5.56 x 10‐10 Kb 1.8 x 10‐5 [H+] = 5.56 x 10‐10 x [0.1] = 2.78 x 10‐10M [0.2] pH = ‐log[H+] = ‐log[2.78 x 10‐10] = 9.56 Solutions at www.prep101.com/solutions 57 of 69
©Prep101 Chem203 Final Booklet 1.41
Problem 83. A buffer solution is prepared by adding 0.10 mol of propionic acid, C H COOH, (K
10 ) and 0.10 mol of sodium propionate, C H COONa, to 1 L of water. If 0.01 mol of HCl is added to the buffer solution, what is the final pH? a) 2.85 b) 4.76 c) 4.85 d) 4.94 e) 3.85 Solution: B Write out the equations that are occurring in the solution described above. Because both solutes are being added to 1L of water, all concentrations (M) can be treated as moles (mol). C2H5COONa is a soluble salt so it completely dissociates C2H5COONa  C2H5COO‐ + Na+ 0.1 0.1 0.1 HCl is a strong acid so it completely dissociates HCl  H+ + Cl‐ 0.01 0.01 0.01 C2H5COO‐ will react will all the H+ to form C2H5COOH. Initially there is 0.1 mole of C2H5COO‐. When 0.01mol of H+ is added, the C2H5COO‐ reacts leaving 0.09mol. There is also 0.1mol of C2H5COOH to start, but when the C2H5COO‐ reacts with the H+, it forms 0.01mol more C2H5COOH so that the total amount of C2H5COOH is 0.11mol. C2H5COOH ↔ C2H5COO‐ + H+ Before HCl is added 0.1 0.1 After HCl is added 0.11 0.09 Calculate [H+] using the following equation. (n = moles) [H+] = Ka x nHA = 1.41 x 10‐5 x (0.11) = 1.72 x 10‐5M nA‐ (0.09) Calculate pH from the [H+]. pH = ‐log[H+] = ‐log[1.72 x 10‐5] = 4.76 Solutions at www.prep101.com/solutions 58 of 69
©Prep101 Chem203 Final Booklet Electrochemistry Problem 84. Consider two other spontaneous reactions: H O 6ClO ⇌ Cr O
6ClO
2H 2Cr
3NO
Cr O
8H ⇌ 3NO
2Cr
4H O
Based only on the fact that these three reactions are spontaneous and no other information, will the following reactions proceed as written (balance the reactions on your own)? Which species is the oxidizing agent, reducing agent? a Cd
ClO ⇌ Cd ClO 2Cr b Cd Cr O ⇌ Cd
Solution: (a) No, Cd cannot oxidize NO (this reaction is spontaneous in the opposite direction), where Cr O could and Cr O cannot oxidize ClO (this reaction is spontaneous in the opposite direction). Therefore Cd cannot oxidize ClO . The reaction must proceed in reverse where Cd is the reducing agent and ClO is the oxidizing agent. (b)Yes, Cd cannot oxidize NO (this reaction is spontaneous in the opposite direction), while Cr O can oxidize NO , therefore Cr O is a stronger oxidizing agent and can oxidize Cd . Cr O is the oxidizing agent, and Cd is the reducing agent. Solutions at www.prep101.com/solutions 59 of 69
©Prep101 Chem203 Final Booklet Practice Questions Problem 85. Find the equilibrium constant at 25 C for Br
Solution: List the oxidation and reduction steps: 
Reduction: Br
2e ⇌ 2Br
1.080 V E cathode
e ⇌ Ag
Oxidation: Ag



E cell
= E cathode ‐ E anode 
E anode
Ag
⇌ 2Ag
2Br
0.799V = 1.080 V – 0.799 V = 0.281 V 0.0257 V
0.0592 V


Ecell

ln K eq
E cell

log K eq
n
n
or 
nEcell
0.0592 V 2(0.281V )
log K eq 
 9.49
0.0592 V
log K eq 
K eq  3.1  10 9
Problem 86. A voltaic cell is constructed as follows: Pb Pb ’ Pb . Pb E
0.0700V What is the K for PbSO under these conditions? Solution: The half‐cell reactions are as follows (note: these are not at standard state or E
Reduction: Pb
2e ⇌ Pb E cathode
? Oxidation: Pb
2e ⇌ Pb
E anode
= 0 V): ? E cell
= E cathode ‐ E anode 0.0592 V
1
= ‐ 0.1546 V log
E cathode =  0.125 
2
0.100 M
E anode  0.125  0.0592 V log 1
= [ Pb2 ]
2
0.0592 V
1 

log
 E cell  0.0700 V =  0.1546    0.125 
2
[ Pb 2 ] 

Pb
4.3 10 Solubility PbSO Pb
Initial Some 0 0 Change ‐s +s +s Equilibrium Some‐s S S K
Pb
SO K
4.3 10 2 1.9 10 Solutions at www.prep101.com/solutions SO 60 of 69
©Prep101 Chem203 Final Booklet Problem 87. What is the oxidation state of all atoms in the following compounds: a CrO b Ca HCO c Fe CO Answers: *It is easiest to do compound c) by stating that CO is a common polyatomic ion, by treating CO as one piece we have: Fe CO , then 2x+3(2‐)=0, followed by 2x‐6=0 and x=3+ is the charge of Fe Cr = +3, O = ‐2 Ca = +2, C = +4, H = +1, O = ‐2 Fe = +3, C = +4, O = ‐2 Problem 88. Consider an electrochemical cell with a cobalt electrode immersed in 1.0 M Co and a lead electrode immersed in 1.0 M Pb . o
Co
2e ⇌ CoE
0.28V
o

Pb
2e ⇌ PbE
0.13V
(a) Calculate E  for this cell Answer: 0.15 V o
Pb
2 e ⇌ Pb E
0.13 V
o

Co ⇌ Co
2e E
0.28V
E
0.28 – 0.13 0.15 (b) Which of the electrodes is the cathode? a) the cobalt electrode b) the lead electrode Answer: B Reduction occurs at the cathode, therefore the lead electrode is the cathode. Solutions at www.prep101.com/solutions 61 of 69
©Prep101 Chem203 Final Booklet Problem 89. Consider the following half‐cell reductions: 4H
3e ⇌ MnO
2H O E 1.68V
MnO


I
2e ⇌ 2I E 0.54V
Zn
2e ⇌ Zn E 0.76V
Place the following species in order from strongest to weakest oxidizing agent: , I
, Zn , MnO
, I  Zn , MnO
Answer: MnO
I
Zn An oxidizing agent gets reduced therefore MnO
is the strongest oxidizing agent as it has the most positive E for reduction. For the following 2 questions refer to the half‐cell reactions, and the galvanic cell shown below which uses inert platinum electrodes and operates at a temperature of 25C. V Salt Bridge
Pt Pt
[Br] = 0.20 M [I] = 10.0 M [Br2] = 0.20 M [I2] = 0.10 M Br
2e ⇌ 2Br  E 1.09V
I
2e ⇌ 2I  E 0.54V
Problem 90. If current is allowed to flow through the above cell at standard state conditions, which species, will be oxidized and which species will be reduced? Solution: 2 e ⇌ 2 Br  E
1.09 V
Br


2I
⇌I
2e E
0.54V 
Br
will be reduced and I will be oxidized. Problem 91. What is the value of E ? Answer: 0.66 V Br
2 e ⇌ 2 Br  E
1.09 V


2I
⇌I
2e E
0.54V 

Br
2I
⇌ 2Br
I
E
0. 55V
Then put the Ecell into the Nerst equation: 0.0592
E
E
log Q n
Solutions at www.prep101.com/solutions 62 of 69
©Prep101 E
E
Chem203 Final Booklet 0.0592
n
log
Br 
I
Br
I
E
0.55
0.0592
2
log
0.2 0.1
0.2 10
0.66V
Problem 92. Under standard conditions, the following redox reactions are observed to occur in aqueous solution. B⇌A B A
A
C ⇌ noreaction D ⇌ 2B D 2B
What is the order of reactivity (most easily oxidized to least easily oxidized) of species A, B, C, and D? Answer: D>B>A>C Solution: Reducing agent undergoes oxidation (most easily oxidized = strongest reducing agent) From A
B⇌A B B can oxidize to B+,  B is a stronger reducing agent than A. From A
C ⇌ noreaction C cannot oxidize to C+  A is a stronger reducing agent than C. From 2B
D ⇌ 2B D D can oxidize to D  D is a stronger reducing agent than B. The decreasing order of reactivity (most easily oxidized to least easily oxidized) is: D>B>A>C, where D is the strongest reducing agent. Problem 93. Which of the following half‐reactions is a reasonable anode reaction for a cell using the reaction below? 2Hg
2Br
⇌ Hg Br 2e ⇌ 2Hg a 2Hg
b 2Hg
2Br
⇌ Hg Br
2e 2e ⇌ 2Hg
2Br c Hg Br
d 2Hg
2Br
⇌ Hg Br
2e e) 2Hg
2Br
2e ⇌ Hg Br Solutions at www.prep101.com/solutions 63 of 69
©Prep101 Chem203 Final Booklet Answer: B 2Br
⇌ 2Hg Br Given: 2Hg
To Find: anode reaction ANODE allows oxidation to take place, which is the loss of electrons. ‐ Answers A, C and E are incorrect as they are reduction reactions. ‐ Answer D is incorrect as the charges on both sides are NOT balanced. ANODE ∶ 2Hg
2Br
⇌ Hg Br
2e 2e ⇌ 2Hg CATHODE:2Hg
2Hg acts as an intermediate Problem 94. Given the following standard reduction potentials: Sn
2e ⇌ SnE 0.14V
Ag
e ⇌ AgE
0.80V

2e ⇌ ZnE 0.76V
Zn
Cu
2e ⇌ CuE
0.34V
Which of the cations Sn , Ag , Zn could be reduced by Cu metal? Answer: Ag only
Cu ⇌ Cu
2 e E
0.34V So addition of this to the cathode reaction should be a positive number for the reaction to proceed. The only one that this works for is Ag Problem 95. Given the following standard reduction potentials 2HIO
10H
10e ⇌ I
6H O E 1.20V



ClO
6H
6e ⇌ Cl
3H O E 1.45V
The standard cell potential for the reaction 3I
5ClO
3H O ⇌ 6HIO
5Cl is? Solutions at www.prep101.com/solutions 64 of 69
©Prep101 Answer: 0.25 V 3
5
3I
I
6H O
ClO
5ClO
Chem203 Final Booklet ⇌ 2 HIO
10 H
10 e E 1.20V
6H
6e ⇌ Cl
3H O E 1.45V
3H O ⇌ 6HIO
5Cl E 0.25V Problem 96. Given the following data Sn
2e ⇌ Sn E 0.15V
e ⇌ Fe E 0.77V
Fe
What is the E for a cell where the reaction is Sn
2Fe
⇌ Sn
2Fe ? Answer: 0.62 V In the reaction above, Sn is being oxidized to Sn (it lost 2 e ) and Fe is being reduced to Fe (gaining 1e ). E
E
E
. Since all E values are given as E
values, the E value for Fe in the reaction is +0.77 V. The E value for Sn in the reaction must be reversed because Sn is undergoing oxidation, therefore E
0.15V. E
E
0.77V
0.15V
0.62V E
Problem 97. Given the following standard reduction potentials Cd
2e ⇌ Cd E
0.40
Sn
2e ⇌ Sn E
0.14
2e ⇌ Sn E
0.15
Sn
Fe
e ⇌ Fe E
0.77
1
ClO
6H
5e ⇌ Cl
3H O E
2
Solutions at www.prep101.com/solutions 0.146
65 of 69
©Prep101 Chem203 Final Booklet Predict which of the following reactions will occur spontaneously. Cl
6H O 5Cd
ClO
12H a) 5Cd
b) 2Sn
⇌ Sn
Sn c) Sn
2Fe
⇌ Sn
2Fe Answer: A only Calculate the E for all three reactions. In statement I), E = ‐1.86V (Cd is being reduced and Cl is being oxidized). In statement II), E = ‐0.29V (Sn is being oxidized and reduced into Sn and Sn respectively). In statement II) E = 0.91V (Sn is being oxidized and Fe3+ is being reduced). E values that are positive means that the redox reactions will occur spontaneously, while a negative E value means the redox reaction is not spontaneous. Problem 98. Given the following half‐cell reduction potentials Mg
2e ⇌ MgE 2.36V
Cr
3e ⇌ CrE 0.74V
2e ⇌ H E 0.00V
2H
Cu
2e ⇌ CuE 0.34V
I
e ⇌ 2I E 0.54V
Which of the following statements is /are CORRECT? I.
Cu will oxidize Cr toCr II.
H will reduce I to I III.
HCl will oxidize Mg to Mg Answer: I, II, and III Solution: Statement I) is true becauseCu has a more positive E
value than Cr , which means that Cu is a better oxidizing agent than Cr (remember that an oxidizing agent oxidizes other substances and becomes reduced in the process). Problem 99. Consider the reaction 2Ag
Ni ⇌ 2Ag
Ni Ecell 1.06V If E for the Ag reduction half‐reaction is +0.80V, the standard reduction potential for Ni
2e ⇌ Ni at 25C in V, must be…? Solutions at www.prep101.com/solutions 66 of 69
©Prep101 Chem203 Final Booklet Answer: ‐0.26 In the reaction given above, Ag is being reduced to Ag while Ni is being oxidized to Ni . The overall E for the reaction is calculated as E
E
E
The reduction half of the reaction is given as + 0.80V. In order to calculate the E value for the oxidation half, rearrange the E equation to solve for E
. E
E
E
1.06V – 0.80V
0.26V. The E
value is the opposite sign from the E
value, so in order to solve for the E
n value as asked, the sign on the E
value must be reversed. Therefore, E
for Ni
0.26V Problem 100. What is the value of n when the Nernst equation, E E
reaction, 2H
O ⇌ 2H O? Answer: n = 4 2H
O ⇌ 2H O can be broken down into its half‐cell reactions: 2H ⇌ 4H
4e 4H
4e ⇌ 2H O O
Since there is an exchange of four electrons, n = 4 in the Nernst Equation. .
log
, is applied to the Problem 101. Given Pb
If [Co
] is 0.0010 M and [Pb
Co
Pb E
0.15V ] is 0.10 M, calculate E. Solutions at www.prep101.com/solutions ⇌ Co
67 of 69
©Prep101 Answer: 0.21 V Solution: E
Chem203 Final Booklet 0.0257V [Co 2  ]
0.0257V
ln
lnQ = 0.15V – = 0.21V E –
2
n
[ Pb 2  ]
Problem 102. Refer to the galvanic cell below (the contents of each half‐cell are written beneath each compartment): V Pt 0.10 M MnO4 0.20 M Mn2+ 0.010 M H+ Pt 0.40 M Cr3+
0.30 M Cr2O72 0.010 M H+ The standard reduction potentials are as follows: 5e ⇌ Mn
4H OE
MnO 8H


Cr O
14H
5e ⇌ 2Cr
7H OE
(a) When current is allowed to flow, which species is oxidized? Solution: 5 e ⇌ Mn
4H O E
MnO 8H
2Cr
7H O ⇌ Cr O  14H
5e E
Oxidation is a loss of electrons. Cr is being oxidized to Cr O  1.51V
1.33V
1.51 V
1.33V
(b) When current is allowed to flow, which species is reduced? Solutions at www.prep101.com/solutions 68 of 69
©Prep101 Solution: MnO is being reduced to Mn Chem203 Final Booklet (c) What is the value of Ecell ? Solution: 5 e ⇌ Mn
4H O E
1.51 V
MnO 8H


2Cr
7H O ⇌ Cr O
14H
5e E
1.33V
MnO 2Cr
3H O ⇌ Cr O  6H Mn E
0.18V (d) What is the oxidation state of Cr in Cr O  ? Solution: oxid. # = [7(‐2) + 2]/2 = +6 Solutions at www.prep101.com/solutions 69 of 69