Section 16.1 Collision Theory
2.Experimental Factors Influencing the Rate of Reaction
(a)Yes, greater concentration of reactants increases the rate.
(b)Yes, greater reaction temperature increases the rate.
(c)Yes, by definition, a catalyst increases the reaction rate.
(d)Occasionally, a few reactions are influenced by UV light.
4.Ineffective Collision Geometry
6.No, the catalyst speeds up the rate at which ammonia is produced, but does not increase the amount of ammonia produced.
8.The ultraviolet rays in sunlight speed up the reaction between methane gas and chlorine gas by initiating the production of chlorine atom Òfree radicals.Ó
That is, ultraviolet light dissociates a Cl2 molecule to give two chlorine free radicals (2 Cl¥). The Cl¥ is unstable and very reactive. The Cl¥ attacks a CH4 molecule to give HCl and a methyl free radical (CH3¥). In turn, the CH3¥ attacks a Cl2 molecule to give CH3Cl and regenerates a chlorine free radical.
The steps for the overall reaction are as follows:
Cl2(g) + UV 2 Cl¥(g)
Cl¥(g) + CH4(g) HCl(g) + CH3¥(g)
CH3¥(g) + Cl2(g) CH3Cl + Cl¥(g)
Section 16.2 Energy Profiles of Chemical Reactions
10.2 O3(g) 3 O2(g) + heat
12.H2(g) + Cl2(g) 2 HCl(g) + heat
14.The heat of reaction (H) is independent from the effects of a catalyst.
16.Since the reaction is exothermic, less energy is required for the reaction to proceed in the forward direction toward products. Thus, the Eact is greater in the reverse direction than in the forward direction.
Section 16.3 The Chemical Equilibrium Concept
18.A reversible reaction is said to be at equilibrium if the forward rate of reaction (ratef) is equal to the reverse rate of reaction (rater). In other words, chemical equilibrium occurs when: ratef = rater.
20.(a) False. In the general equilibrium expression, the value of Keq is dependent on temperature.
(b) True. In general, the substances appearing in the Keq expression are in the same physical state. A substance participating in gaseous equilibria reactions in a different physical state (such as solid or liquid) is assigned a concentration of [1], and is incorporated into the Keq value.
Section 16.4 General Equilibrium Constant, Keq
22.Equilibrium ReactionEquilibrium Constant Expression
(a)3 A 2 CKeq =
(b)A + B 2 CKeq =
(c)3 A + 5 B C + 4 D Keq =
24.Since the amount of any liquid present has no effect on the equilibrium of a gaseous state reaction, any substance in the liquid state will not appear in the equilibrium expression.
26.Equilibrium ReactionEquilibrium Constant Expression
(a)H2(g) + Br2(g) 2 HBr(g) Keq =
(b)4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)Keq =
(c)CO(g) + 2 H2(g) CH3OH(l) Keq =
28.N2(g) + 3 H2(g) 2 NH3(g)
[N2] = 0.400 M[H2] = 1.20 M[NH3] = 0.195 M
Keq = = = 0.0550
Section 16.5 Gaseous State Equilibria Shifts
30.N2(g) + 3 H2(g) 2 NH3(g) + heat
(a)If the temperature increases, the heat shifts the equilibrium to the left. Thus, the concentration of NH3decreases.
(b)If the pressure increases, the pressure shifts the equilibrium to the right, which has fewer molecules. Thus, the concentration of NH3increases.
32.CH4(g) + H2O(g) + heat CO(g) + 3 H2(g)
(a)increase [CH4]shift right(b)increase [CO]shift left
(c)decrease [H2O]shift left(d)decrease [H2]shift right
(e)add gaseous xenonno shift(f)decrease temp.shift left
(g)decrease pressureshift right(h)increase pressureshift left
34.2 C2H4(g) + 2 O3(g) 4 CH2O(g) + O2(g) + heat
(a)increase [C2H4]shift right(b)increase [CH2O]shift left
(c)decrease [O3]shift left(d)decrease [O2]shift right
(e)increase temp.shift left(f)decrease pressureshift right
(g)decrease volumeshift left(h)add a metal catalystno shift
Section 16.6 Ionization Constant,Ki
36.Equilibrium ReactionEquilibrium Constant Expression
(a)NH2OH(aq) + H2O(l) NH3OH+(aq) + OHÐ(aq)
Ki =
(b)C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + OHÐ(aq)
Ki =
(c)(CH3)2NH(aq) + H2O(l) (CH3)2NH+(aq) + OHÐ(aq)
Ki =
38.NH4OH(aq) NH4+(aq) + OHÐ(aq)
[NH4+] = [OHÐ] = 2.1 10Ð3M[NH4OH] = 0.245 M
Ki = = = 1.8 10Ð5
40.N2H4(aq) + H2O(l) N2H5+(aq) + OHÐ(aq)
[H+] = 10ÐpH = 10Ð11.00 = 1.0 10Ð11M
[OHÐ] = = = 1.0 10Ð3 = 0.0010 M
[N2H5+] = [OHÐ] = 0.0010 M[N2H4] = 0.139 M
Ki = = = 7.2 10Ð6
Section 16.7 Weak AcidÐBase Equilibria Shifts
42.HNO2(aq) H+(aq) + NO2Ð(aq)
(a)decrease [HNO2]shift left(b)decrease [H+]shift right
(c)increase [HNO2]shift right(d)increase [NO2Ð]shift left
(e)add solid KNO2shift left(f)add solid KClno shift
(g)add solid KOHshift right(h)increase pHshift right
44.NH4OH(aq) NH4+(aq) + OHÐ(aq)
(a)increase [NH4+]shift left(b)decrease [OHÐ]shift right
(c)increase [NH4OH]shift right(d)decrease pHshift right
(e)add gaseous NH3shift right(f)add solid KClno shift
(g)add solid KOHshift left(h)add solid NH4Clshift left
Section 16.8 Solubility Product Equilibrium Constant, Ksp
46.ReactionSolubility Product Expression
(a)Cu2CO3(s) 2 Cu+(aq) + CO32Ð(aq)Ksp = [Cu+]2 [CO32Ð]
(b)ZnCO3(s) Zn2+(aq) + CO32Ð(aq)Ksp = [Zn2+] [CO32Ð]
(c)Al2(CO3)3(s) 2 Al3+(aq) + 3 CO32Ð(aq)Ksp = [Al3+]2 [CO32Ð]3
48.MgF2(s) Mg2+(aq) + 2 FÐ(aq)
Since two moles of FÐ are produced for every mole of Mg2+,
then, [Mg2+] = [FÐ] = (2.3 10Ð3) = 1.2 10Ð3
Ksp = [Mg2+] [FÐ]2 = (1.2 10Ð3) (2.3 10Ð3)2 = 6.3 10Ð9
50.Fe(OH)3(s) Fe3+(aq) + 3 OHÐ(aq)
Since three moles of OHÐ are produced for every mole of Fe3+,
then, [Fe3+] = [OHÐ] = (2.2 10Ð10) = 7.3 10Ð11
Ksp = [Fe3+] [OHÐ]3 = (7.3 10Ð11) (2.2 10Ð10)3 = 7.810Ð40
52.MnCO3(s) Mn2+(aq) + CO32Ð(aq)
[Mn2+] [CO32Ð] = Ksp = 1.8 10Ð11
Since one mole of CO32Ð is produced for every mole of Mn2+,
then, [Mn2+]2 = 1.8 10Ð11
thus, [Mn2+] = 4.2 10Ð6
Mn(OH)2(s) Mn2+(aq) + 2 OHÐ(aq)
[Mn2+] [OH Ð]2 = Ksp = 4.6 10Ð14
Since two moles of OHÐ is produced for every mole of Mn2+,
then, [Mn2+] [2 Mn2+]2 = 4.6 10Ð14
thus, 4 [Mn2+]3 = 4.6 10Ð14
and, [Mn2+]3 = 1.2 10Ð14
[Mn2+] = 2.3 10Ð5
The manganese ion concentration in a saturated solution of Mn(OH)2 is about five times greater than in a saturated solution of MnCO3.
Section 16.9 Solubility Equilibria Shifts
54.CdS(s) Cd2+(aq) + S2Ð(aq)
(a)increase [Cd2+]shift left(b)increase [S2Ð]shift left
(c)decrease [Cd2+]shift right(d)decrease [S2Ð]shift right
(e)add solid Cd(NO3)2shift left(f)add solid NaNO3no shift
(g)add solid CdSno shift(h)add H+shift right
56.SrCO3(s) Sr2+(aq) + CO32Ð(aq)
(a)increase [Sr2+]shift left(b)increase [CO32Ð]shift left
(c)decrease [Sr2+]shift right(d)decrease [CO32Ð]shift right
(e)add solid SrCO3no shift(f)add solid Sr(NO3)2shift left
(g)add solid KNO3no shift(h)decrease pHshift right
General Exercises
58.A solar calculator represents a dynamic process as it is constantly discharging and charging simultaneously. The solar calculator represents a reversible process as it discharges during operation and charges during exposure to light.
60.With regard to concentration, a reversible reaction at equilibrium has the following characteristic: the concentrations of each reactant and product are constant although a small amount of reactants is being converted to products and a small amount of products is being converted to reactants.
62.N2O4(g) 2 NO2(g)
PN2O4 = 0.0014 atmPNO2 = 0.092 atm
Kp = = = 6.0 atm
64.H2O(l) H+(aq) + OHÐ(aq)
(a)add gaseous HClshift left(b)add solid NaOHshift left
(c)add liquid H2SO4shift left(d)add solid NaFshift right
66.AgI(s) Ag+(aq) + IÐ(aq)
Aqueous NaCl dissociates to give Na+(aq) and ClÐ(aq). The ClÐ(aq) can react with Ag +(aq) to form AgCl(s). This decreases the Ag+(aq) concentration which shifts the equilibrium to the right, thus allowing more dissociation of AgI(s).
In summary, AgI is more soluble in aqueous NaCl than in water, owing to the following reaction:
AgI(s) + ClÐ(aq) AgCl(s) + IÐ(aq)
68.Zn(OH)2(s) Zn2+(aq) + 2 OHÐ(aq)
pH = 8.44
[ H+ ] = 10Ð8.44 = 100.56 10Ð9
[ H+ ] = 3.6 10Ð9M
[ OHÐ ] = = = 2.8 10Ð6
Since two moles of OHÐ are produced for every mole of Zn2+,
then, [Zn2+ ] = [OH Ð] = (2.8 10Ð6) = 1.4 10Ð6
Therefore,
Ksp = [Zn2+] [OH Ð]2 = (1.4 10Ð6) (2.8 10Ð6)2 = 1.110Ð17
Chemical Equilibrium 1