Problems 2 & 3 – ANSWERS
2009 A
CH4(g) + 2 Cl2(g)  CH2Cl2(g) + 2 HCl(g)
3.
Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the
equation above.
(a)
A 25.0 g sample of methane is placed in a reaction vessel containing 2.58 mol of Cl2(g).
(i)
(ii)
Identify the limiting reactant when the methane and chlorine gases are combined. Justify your answer
with a calculation.
Calculate the total number of moles of CH2Cl2(g) in the container after the limiting reactant has been
totally consumed.
Initiating most reactions involving chlorine gas involves breaking the Cl-Cl bond, which has a bond energy of 242
kJ mol-1.
(b)
Calculate the amount of energy, in joules, needed to break a single Cl-Cl bond.
(c)
Calculate the longest wavelength of light, in meters, that can supply the energy per photon necessary to break
the Cl-Cl bond.
The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in
the gas phase.
Step 1
Cl2  2 Cl
fast equilibrium
Step 2
CH4 + Cl  CH3 + HCl
slow
Step 3
CH3 + Cl2  CH3Cl + Cl
fast
Step 4
CH3Cl + Cl  CH2Cl2 + H
fast
Step 5
H + Cl  HCl
fast
(d)
In the mechanism, is CH3Cl a catalyst, or is it an intermediate? Justify your answer.
(e)
Identify the order of the reaction with respect to each of the following according to the mechanism. In each
case, justify your answer.
(i) CH4(g)
(ii) Cl2(g)
Answer:
(a)
(i) 25.0 g CH4 
2.58 mol Cl2 
1 mol CH 4 1 mol CH 2Cl2

= 1.56 mol CH2Cl2
16.0 g CH 4
1 CH 4
1 mol CH 2Cl2
= 1.29 mol CH2Cl2
2 mol Cl2
Cl2 is the limiting reactant
(ii) 1.29 mol CH2Cl2 when you run out of chlorine
242000 J
= 4.0210-19 J
23
6.02  10 bonds
(b)
1 Cl-Cl bond 
(c)
hc (6.63  10 34 J  s)(3.0  10 8 m  s1 )


= 4.9510-7 m
E
4.02  10 19 J
(d)
intermediate; it is made in step 3 and consumed in step 4, to be a catalyst it would have to remain unchanged at
the end
(e)
the slowest step (step 2) is the rate determining step
rate of step 1: ratefor = kfor[Cl2] , raterev = krev[Cl]2
at equilibrium: ratefor = raterev, kfor[Cl2] = krev[Cl]2
1
 k for 
[Cl] = 
[Cl2]1/2 = k1[Cl2]1/2

k 
2
rev
rate of step 2: rate = k2[CH4][Cl] ; substituting, rate = k[CH4][Cl2]1/2
therefore, (i) rate is 1st order with respect to CH4 and (ii) ½ order with
respect to Cl2
2006 B
1
CO(g) + 2 O2(g)  CO2(g)
The combustion of carbon monoxide is represented by the equation above.
(a)
Determine the value of the standard enthalpy change, ∆H˚rxn for the combustion of CO(g) at 298 K using the following information.
C(s) +
1
2
O2(g)  CO(g)
∆H˚298 = –110.5 kJ mol-1
C(s) + O2(g)  CO2(g)
(b)
∆H˚298 = –393.5 kJ mol-1
Determine the value of the standard entropy change, ∆S˚rxn, for the combustion of CO(g) at 298 K using the information in the following
table.
Substance
S˚298
(J mol-1 K-1)
CO(g)
197.7
CO2(g)
213.7
O2(g)
205.1
(c)
Determine the standard free energy change, ∆G˚rxn, for the reaction at 298 K. Include units with your answer.
(d)
Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.
(e)
Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.
Answer:
(a) ∆H˚rxn = H f ( prod )  H f (reactants) = (–393.5 kJ mol-1) – (–110.5 kJ mol-1 + 1/2(0)) = –
283.0 kJ
(b) ∆S˚rxn = S prod  Sreactants = (213.7) – (197.7 + 1/2(205.1) J mol-1 K-1 = -86.6 J K-1
(c) ∆G˚ = ∆H˚ – T∆S˚ = (–283.0 kJ) – (298K)(-0.0866 J K-1) = –257.2 kJ
(d) spontaneous; any reaction in which the ∆G˚ < 0 is spontaneous
(e) Keq = e–(∆G/RT) = e–(–257208.1J/(8.31298) = 1.28  1045
2004 B
2 Fe(s) +
3
O (g)  Fe2O3(s)
2 2
∆Hf˚ = -824 kJ mol–1
Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s) is mixed with 11.5 L of O2(g) at 2.66 atm and
298 K.
(a)
Calculate the number of moles of each of the following before the reaction occurs.
(i)
(ii)
Fe(s)
O2(g)
(b)
Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support your answer with calculations.
(c)
Calculate the number of moles of Fe2O3 produced when the reaction proceeds to completion.
(d)
The standard free energy of formation, ∆Gf˚ of Fe2O3 is –740. kJ mol–1 at 298 K.
(i)
Calculate the standard entropy of formation ∆Sf˚ of Fe2O3 at 298 K. Include units with your answer.
(ii)
Which is more responsible for the spontaneity of the formation reaction at 298K, the standard enthalpy or the standard entropy?
The reaction represented below also produces iron(III) oxide. The value of ∆H˚ for the reaction is –280 kJ per mol.
1
2 FeO(s) + O2(g)  Fe2O3(s)
2
(e)
Calculate the standard enthalpy of formation, ∆Hf˚ of FeO(s).
Answer:
1 mol
(a) (i) 75.0 g Fe  55.85 g = 1.34 mol Fe
PV
(ii) PV = nRT, n = RT
(2.66 atm)(11.5 L)
(0.0821
L atm
mol K)
(298 K)
= 1.25 mol O2
3
2
mol O2
(b) Fe; 1.34 mol Fe  2 mol Fe = 1.01 mol O2
excess O2, limiting reagent is Fe
1 mol Fe2O3
(c) 1.34 mol Fe  2 mol Fe = 0.671 mol Fe2O3
(d) (i) ∆Gf˚ = ∆Hf˚ – T∆Sf˚
–740 kJ mol–1 = –824 kJ mol–1 – (298 K)(∆Sf˚)
∆Sf˚ = 0.282 kJ mol–1 K–1
(ii) standard enthalpy; entropy decreases (a non-spontaneous process) so a large change in
enthalpy (exothermic) is need to make this reaction spontaneous
(e) ∆H = ∆Hf(products) – ∆Hf(reactants)
–280 kJ mol–1 = –824 kJ mol–1 – [2(∆Hf˚ FeO) – 1/2(0)]
= -272 kJ mol–1
2005 B
Answer the following questions related to the kinetics of chemical reactions.
I–(aq) + ClO–(aq)
–
OH


IO–(aq) + Cl–(aq)
Iodide ion, I–, is oxidized to hypoiodite ion, IO–, by hypochlorite, ClO–, in basic solution according to the equation above. Three initialrate experiments were conducted; the results shown in the following table.
(a)
[I–]
(mol L–1)
[ClO–]
(mol L–1)
Initial Rate of Formation
of IO– (mol L–1 s–1)
1
0.017
0.015
0.156
2
0.052
0.015
0.476
3
0.016
0.061
0.596
Determine the order of the reaction with respect to each reactant listed below. Show your work.
(i)
(ii)
(b)
Experiment
I–(aq)
ClO–(aq)
For the reaction,
(i)
write the rate law that is consistent with the calculations in part (a);
(ii)
calculate the value of the specific rate constant, k, and specify units.
The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following equation.
2 H2O2(aq)
catalyst
 2 H2O(l) + O2(g)
The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a first-order reaction. Some of the
experimental data are shown in the table below.
(c)
[H2O2]
(mol L–1)
Time (minutes)
1.00
0.0
0.78
5.0
0.61
10.0
During the analysis of the data, the graph below was produced.
(i)
(ii)
Label the vertical axis of the graph
What are the units of the rate constant, k, for the decomposition of H2O2(aq) ?
(iii)
On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H2O2(aq).
Answer:
(a) (i) comparing expt. 1 to expt. 2, while the hypochlorite concentration remains constant, the
0.052
3.06
iodide concentration is essentially tripled { 0.017 =
} and the initial rate is
1
0.476
3.05
essentially tripled { 0.156 = 1
}. This indicates a first order with respect to the
iodide ion.
(ii) comparing expt. 1 to expt. 3, while the iodide concentration remains essentially constant
0.061
4.07
(a 2.7% drop), the hypochlorite concentration is essentially quadrupled { 0.015 = 1
0.596
3.82
} and the initial rate is essentially quadrupled { 0.156 = 1
}. This indicates a first
order with respect to the hypochlorite ion.
OR
(i) from experiments 1 & 2
rate2
k[I–] 2m[ClO–] 2n
=
rate1
k[I–] 1m[ClO–] 1n
0.476
k(0.052)m(0.015)n
=
0.156
k(0.017)m(0.015)n
3.05 =
0.052m
= 3.1m, where m = 1
0.017m
(ii) from experiments 1 & 3
rate3
k[I–] 3m[ClO–] 3n
=
rate1
k[I–] 1m[ClO–] 1n
0.596
k(0.016)m(0.061)n
=
0.156
k(0.017)m(0.015)n
3.82 = (0.94)
0.061n
0.015n
4.06 = 4.1n, where n = 1
(b) (i) rate = k[I–] [ClO–]
rate
0.156mol L–1 s–1
–1 –1
(ii) k = –
=
(0.017mol L–1)(0.015mol L–1) = 610 L mol s
[I ] [ClO–]
(c) (i) vertical axis is “ln of [H2O2]”
(ii) units for k are min–1
uncatalyzed
(iii) ln [conc]
Time(minutes)
2004 B
The first-order decomposition of a colored chemical species, X, into colorless products is monitered with a spectrophotometer by measuring
changes in absorbance over time. Species X has a molar absorptivity constant of 5.00103 cm–1M–1 and the pathlength of the cuvetee containing
the reaction mixture is 1.00 cm. The data from the experiment are given in the table below.
[X]
(M)
Absorbance
Time
(min)
?
0.600
0.0
4.0010–5
0.200
35.0
3.0010–5
0.150
44.2
1.5010–5
0.075
?
(a)
Calculate the initial concentration of the unknown species.
(b)
Calculate the rate constant for the first order reaction using the values given for concentration and time. Include units with your answers.
(c)
Calculate the minutes it takes for the absorbance to drop from 0.600 to 0.075.
(d)
Calculate the half-life of the reaction. Include units with your answer.
(e)
Experiments were performed to determine the value of the rate constant for this reaction at various temperatures. Data from these
experiments were used to produce the graph below, where T is temperature. This graph can be used to determine Ea, the activation energy.
(i)
Label the vertical axis of the graph
(ii)
Explain how to calculate the activation energy from this graph.
Answer:
(a) A = abc; 0.600 = (5000 cm–1M–1)(1.00 cm)(c)
c = 1.2010–4 M
(b) ln[X] t – ln[X] 0 = –kt
ln(4.0010–5) – ln(1.2010–4) = –k(35 min)
k = 0.0314 min–1
(c) ln[X] t – ln[X] 0 = –kt
ln[1.5010–5] – ln[1.2010–4] = –0.0314 min–1t
t = 66.2 min.
0.693
0.693
(d) t1/2 = k = 0.0314 = 22.1 min
(e) (i)
(ii)
–Ea
R = slope of the line, multiply the slope by –R to obtain Ea
2003 B
5 Br–(aq) + BrO3–(aq) + 6 H+(aq)  3 Br2(l) + 3 H2O(l)
In a study of the kinetics of the reaction represented above, the following data were obtained at 298 K.
(a)
Experim
ent
Initial
[Br–] (mol
L-1)
Initial
[BrO3–]
(mol L-1)
Initial
[H+] (mol
L-1)
Rate of
Disappearance of
BrO3– (mol L-1 s-1)
1
0.00100
0.00500
0.100
2.5010-4
2
0.00200
0.00500
0.100
5.0010-4
3
0.00100
0.00750
0.100
3.7510-4
4
0.00100
0.01500
0.200
3.0010-3
From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning.
(i)
Br–
(ii)
BrO3–
(iii)
H+
(b)
Write the rate law for the overall reaction.
(c)
Determine the value of the specific rate constant for the reaction at 298 K. Include the correct units.
(d)
Calculate the value of the standard cell potential, E˚, for the reaction using the information in the table below.
Half-reaction
(e)
E˚ (V)
Br2(l) + 2e-  2 Br–(aq)
+1.065
BrO3–(aq) + 6 H+(aq) + 5e- 1/2 Br2(l) + 3 H2O(l)
+1.52
Determine the total number of electrons transferred in the overall reaction.
2003 B
A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g of O2(g)
(a)
Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.
(b)
The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the following.
(c)
(i)
The mole fraction of N2(g) in the cylinder.
(ii)
The partial pressure, in atm, of N2(g) in the cylinder.
If the cylinder develops a pinhole-sized leak and some of the gaseous mixture escapes, would the ratio
N2 (g )
O2 ( g )
in the cylinder increase,
decrease, or remain the same? Justify your answer.
A different rigid 5.00 L cylinder contains 0.176 mol of NO(g) at 298 K. A 0.176 mol sample of O2(g) is added to the cylinder, where a reaction
occurs to produce NO2(g).
(d)
Write the balanced equation for the reaction.
(e) Calculate the total pressure, in atm, in the cylinder at 298 K after the reaction is complete.
Answer:
(a) 24.5 g N2 
1mol
= 0.875 mol N2
28.0 g
28.0 g O2 
1mol
= 0.875 mol O2
32.0 g
L•atm
1.75mol  0.0821 mol•K
 298K 
nRT
P=
=
5.00L
V
= 8.56 atm
(b) (i)
(ii)
0.875 mol N 2
= 0.500 mole fraction N2
1.75 mol mix
P1 P2
P T (8.56atm)(280K)
 ; P2  1 2 
T1 T2
T1
298K
= 8.05 atm  mole fraction = 8.05 atm  0.500
= 4.02 atm N2
(c) decrease; since N2 molecules are lighter than O2 they have a higher velocity and will escape
more frequently (Graham’s Law), decreasing the amount of N2 relative to O2
(d) 2 NO + O2  2 NO2
(e) all 0.176 mol of NO will react to produce 0.176 mol of NO2, only 1/2 of that amount of O2
will react, leaving 0.088 mol of O2, therefore, 0.176 + 0.088 = 0.264 mol of gas is in the
container.
L•atm
 0.264 mol  0.0821 mol•K
 298 K 
nRT
P=
=
5.00L
V
= 1.29 atm
2002 B
Answer parts (a) through (e) below, which relate to reactions involving silver ion, Ag +.
The reaction between silver ion and solid zinc is represented by the following equation.
2 Ag+(aq) + Zn(s)  Zn2+(aq) + 2 Ag(s)
(a)
(b)
A 1.50 g sample of Zn is combined with 250. mL of 0.110 M AgNO3 at 25˚C.
(i)
Identify the limiting reactant. Show calculations to support your answer.
(ii)
On the basis of the limiting reactant that you identified in part (i), determine the value of [Zn2+] after the reaction is complete.
Assume that volume change is negligible.
Determine the value of the standard potential, E˚, for a galvanic cell based on the reaction between AgNO3(aq) and solid Zn at 25˚C.
Another galvanic cell is based on the reaction between Ag+(aq) and Cu(s), represented by the equation below. At 25˚C, the standard potential, E˚,
for the cell is 0.46 V.
2 Ag+(aq) + Cu(s)  Cu2+(aq) + 2 Ag(s)
(c)
Determine the value of the standard free-energy change, ∆G˚, for the reaction between Ag+(aq) and Cu(s) at 25˚C.
(d)
The cell is constructed so that [Cu2+] is 0.045 M and [Ag+] is 0.010 M. Calculate the value of the potential, E, for the cell.
(e) Under the conditions specified in part (d), is the reaction in the cell spontaneous? Justify
your answer.
Answer:
(a) (i) AgNO3 solution
1 mol Zn
2 mol Ag+
1000 mL
1.50 g Zn  65.39 g Zn  1 mol Zn 
= 417 mL of silver nitrate solution
0.110 mol Ag+
required to completely react the zinc, therefore, AgNO3 is the limiting reagent.
(ii) 250 mL AgNO3 
0.110 mol AgNO3
1 mol Ag+
1 mol Zn


= 0.01375 mol Zn
1000 mL
1 mol AgNO3
2 mol Ag+
0.01375 mol Zn
= 0.0550 M [Zn2+]
0.250 L
(b) 2 Ag+ + 2e-  2 Ag
E˚ = +0.80 v
Zn – 2e-  Zn2+
E˚ = +0.76 v
+1.56 v
(c) ∆G˚ = -nFE˚ = -(2)(96500)(0.46 v) = -89000 J
0.0592
[Cu2+]
0.0592
0.045
(d) Ecell = E˚ - 2
log
= 0.46 - 2
log
= 0.38 v
+
2
[Ag ]
0.0102
(e) yes; any reaction is spontaneous with a positive voltage
2001 B
2 NO(g) + O2(g)  2 NO2(g)
H°= -114.1 kJ, S°= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of photochemical smog.
(a)
Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).
(b)
For the reaction at 25C, the value of the standard free-energy change, G, is -70.4 kJ.
(c)
(i)
Calculate the value of the equilibrium constant, Keq, for the reaction at 25C.
(ii)
Indicate whether the value of G would become more negative, less negative, or remain unchanged as the temperature is
increased. Justify your answer.
Use the data in the table below to calculate the value of the standard molar entropy, S, for O2(g) at 25C.
Standard Molar Entropy, S (J K-1 mol-1)
(d)
NO(g)
210.8
NO2(g)
240.1
Use the data in the table below to calculate the bond energy, in kJ mol -1, of the nitrogen-oxygen bond in NO2 . Assume that the bonds in
the NO2 molecule are equivalent (i.e., they have the same energy).
Bond Energy (kJ
mol-1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in NO2
?
Answer:
1 mol NO 114.1 kJ
(a) 73.1 g  30.007 g  2 mol NO = 139 kJ
(b) (i) Keq = e–G/RT = e–(–70400/(8.31)(298)) = 2.221012
(ii) less negative; G = H – TS; as temperature increases, –TS becomes a larger
positive value causing an increase in G (less negative).
(c) S = S(products) – S(reactants)
-146.5 = [(2)(240.1)] – [(210.8)(2)+ Soxygen] J/K
Soxygen = +205.1 J/K
(d) 2 NO(g) + O2(g)  2 NO2(g) + 114.1 kJ
H = enthalpy of bonds broken – enthalpy of bonds formed
-114.1 = [(607)(2) + 495] - 2X
X = 912 kJ / 2 N=O bonds
456 kJ = bond energy for N=O bond
2001 B
Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.
(a)
The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of
acetylsalicylic acid in the tablet.
(b)
The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields
1.200 g of water and 3.72 L of dry carbon dioxide, measured at 750. mm Hg and 25C. Calculate the mass, in g, of each element in the
3.000 g sample.
(c)
A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using
88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid.
(d)
A 2.00×10–3 mole sample of pure acetylsalicylic acid was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The
equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table
below, determine
(i)
the value of the acid dissociation constant, Ka, for acetylsalicylic acid and
(ii)
the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added (assume that volumes are additive).
Volume of 0.100M
NaOH Added (mL)
pH
0.00
2.22
5.00
2.97
10.00
3.44
15.00
3.92
20.00
8.13
25.00
?
Answer:
0.325 g
(a) 2.00 g  100% = 16.3%
(1.0079)(2) g H
+ 16 g H2O) = 0.134 g H
(1.0079)(2)
750
760 atm (3.72 L)
P•V
n = R•T = (0.0821L•atm·mol-1•K-1)(298 K) = 0.150 mol CO2
(b) 1.200 g H2O 
12.0 g C
0.150 mol CO2  1 mol CO = 1.801 g C
2
3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O
0.102 mol
(c) 0.08843 L 
= 0.00902 mol base
1L
1 mol base = 1 mol acid
1.625 g ASA
0.00902 mol = 180 g/mol
(d) (i) HAsa  Asa– + H+
2.00  10-3 mole
= 0.133 M
0.015 L
pH = –log[H+]; 2.22 = –log[H+]
[H+] = M = [Asa–]
[HAsa] = 0.133 M – 6.03  10-3 M = 0.127 M
[H+][Asa–] (6.03  10-3)2
K = [HAsa] =
= 2.85 10-4
0.127
OR
when the solution is half-neutralized, pH = pKa
at 10.00 mL, pH = 3.44; K = 10–pH
= 10–3.44 = 3.6310-4
(ii) 0.025 L  0.100 mol/L = 2.50  10-3 mol OH2.50  10-3 mol OH- - 2.00  10-3 mol neutralized = 5.0  10-4 mol OH- remaining in (25 +
15 mL) of solution; [OH-] = 5.010-4 mol/0.040 L = 0.0125 M
pH = 14 – pOH = 14 + log[OH-] = 14 – 1.9 = 12.1