Chapter 13 Source Answers and Examples
Chapter problems → · Original reading →
Reviewed solution to Problem 2
Let [B]=x and [A]=0.100−x. 12.4=x/(0.100−x), so x=0.0925373 M. All concentrations are nonnegative and their total remains 0.100 M. The original copper/HCl hydrogen-generation premise is superseded.
Reviewed solution to Problem 3
t=(1/0.200−1/0.400)/0.0422=59.2417 s. This is a concentration/pressure threshold in a specified kinetic model, rather than time to equilibrium.
Reviewed solution to Problem 6
t=ln(0.100/0.0100)/32.0=0.0719558 min=4.31735 s. No equilibrium-time or tanker-purity prediction is inferred.
Reviewed solution to Problem 8
x/(0.600−x)=7.231 gives x=0.527105 M. n(B)=0.800x=0.421684 mol. This complete aqueous model avoids the original missing gas-space volume and mixed-phase convention.
Reviewed solution to Problem 13
t½=ln 2/42.1=0.0164643 days=23.7086 min. Equilibrium composition and approach time require separate thermodynamic and reversible kinetic data.
Reviewed solution to Problem 14
Let the forward extent be ξ mol. √13.21=2ξ/(1−ξ), so ξ=0.645047 mol. Thus n(A)=n(B)=1.00−ξ=0.354953 mol≈0.355 mol each, and n(C)=2ξ=1.290094 mol≈1.29 mol. The total gas amount remains 2.00 mol because Δn=0. P=nRT/V=8.96555 atm≈8.97 atm (three significant figures). This is a complete hypothetical gas-equilibrium task rather than assumed sulfuric-acid vapor chemistry.
Reviewed solution to Problem 15
ΔHsystem=(0.250)(−481.1)=−120.275 kJ≈−120 kJ at three significant figures. It releases about 120 kJ under constant-pressure conditions with only pressure–volume work. The reaction extent is supplied rather than inferred from the original unbalanced equation.
Reviewed solution to Problem 16
t=(1/0.0500−1/0.100)/4.20=2.38095 h. The balanced chemical reaction and reverse kinetics of the original lead-oxide task are not inferred.
Reviewed solution to Problem 17
One mole of Zn forms one mole H₂, so 0.0100 mol H₂ forms. This explicit reviewed metal/nonoxidizing-acid model replaces the original nitric-acid hydrogen premise.
Reviewed solution to Problem 18
[A]=[A]₀e⁻ᵏᵗ gives 0.143639 M at 10.0 min and 0.0891816 M at 120 min. Heating NaCl solution does not establish the source’s proposed chlorate/hydrogen reaction; this is a stated kinetic model.
Reviewed solution to Problem 19
Consumed OH⁻=0.495 mol. The ratio is 3 mol Cl₂ per 6 mol OH⁻, giving 0.2475 mol Cl₂≈0.248 mol. The original unspecified equilibrium and gas-volume conditions are superseded.
Reviewed solution to Problem 20
[A]/[A]₀=0.100, so t=ln(10)/0.3211=7.17093 days. This operational threshold differs from exact equilibrium time.
Reviewed solution to Problem 21
Let x be the pressure extent. √4.19=2x/(1−x), giving x=0.505801 atm. Thus p(H₂)=p(Cl₂)=0.494199 atm and p(HCl)=1.011601 atm. No nitric-acid hydrogen generation or mixed aqueous/gas calculation is assumed.
Worked examples and reviewed revisions
Example Problem 5.1.1
Worked example
Example Problem 5.1.1 — AP edition reviewed replacement.
Reviewed expression task: use Ca²⁺(aq)+CO₃²⁻(aq)+2 H⁺(aq)+SO₄²⁻(aq) ⇌ CaSO₄(s)+CO₂(g)+H₂O(l). Write a dimensionless activity-based equilibrium expression when the specified pure phases are present.
The net equation conserves atoms and charge. If calcium chloride supplies Ca²⁺, its formula is CaCl₂. Na⁺ and Cl⁻ spectators cancel; do not introduce CaCl₃.
K=a(CO₂)/[a(Ca²⁺)a(CO₃²⁻)a(H⁺)²a(SO₄²⁻)]. Each activity is referred to its standard state; the gas activity can be represented as p(CO₂)/p°.
Pure CaSO₄(s) and pure liquid water have unit activity in this model. CO₂ remains in the numerator, and the coefficient 2 on H⁺ gives exponent 2.
Example Problem 5.2.1
Worked example
Example Problem 5.2.1 — AP edition reviewed replacement.
Reviewed source-data model: 3 A+2 B⇌C+4 D, Kc=43, initially [A]=0.960 M, [B]=0.560 M, [C]=0.0760 M and [D]=0.0110 M. Determine equilibrium composition in the ideal concentration model. The numerical solver is retained enrichment.
Let x be the forward extent per litre. [A]=0.960−3x; [B]=0.560−2x; [C]=0.0760+x; [D]=0.0110+4x. All four must be nonnegative.
Qinitial=(0.0760)(0.0110)⁴/[(0.960)³(0.560)²]≈4.01×10⁻⁹, below Kc=43, so the net change is forward.
Solve (0.0760+x)(0.0110+4x)⁴/[(0.960−3x)³(0.560−2x)²]=43 on −0.00275<x<0.280. The physical numerical root is x=0.1881492235 M.
[A]=0.395552330 M; [B]=0.183701553 M; [C]=0.264149223 M; [D]=0.763596894 M. Substitution returns Q=43.0000. Round only at the end.
Example Problem 5.2.2
Worked example
Example Problem 5.2.2 — AP edition reviewed replacement.
Reviewed irreversible first-order threshold task: −d[A]/dt=k[A], k=0.0112 day⁻¹, [A]₀=0.960 M. Find the time to [A]=0.396 M. This is a separate kinetic task; an equilibrium constant does not supply an exact approach time.
Integrate the stated disappearance law: [A]=[A]₀e⁻ᵏᵗ, or ln([A]₀/[A])=kt.
t=ln(0.960/0.396)/0.0112=79.0642 days≈79.1 days.
The specified threshold is reached at this finite time. Exact reversible equilibrium requires a reversible rate model and is generally approached asymptotically.
Example Problem 5.2.3
Worked example
Example Problem 5.2.3 — AP edition reviewed replacement.
Reviewed two-part task. (A) For A+B⇌C, Kc=0.023 and initial [A]=0.362 M, [B]=0.691 M, [C]=0.457 M, find equilibrium concentrations in an ideal concentration model. (B) Separately, an irreversible disappearance model −d[C]/dt=0.320[C] min⁻¹ starts at 0.457 M. Find time to 0.0360 M. The separate rate task does not calculate exact equilibrium time.
Reviewed ideal composition exercise: A+B⇌C, Kc=0.023, with initial concentrations 0.362 M, 0.691 M and 0.457 M. Define x as the forward extent per litre: [A]=0.362−x, [B]=0.691−x and [C]=0.457+x.
Kc=(0.457+x)/[(0.362−x)(0.691−x)]=0.023. The initial Q=0.457/(0.362×0.691)≈1.827>K, so x must be negative: the net reaction proceeds toward A and B.
The correctly expanded product is 0.250142−1.053x+x². Rearranging gives 0.023x²−1.024219x−0.451246734=0.
The physical root is x=−0.436301683 M. The other root is approximately 44.9676 M and gives negative reactant concentrations, so reject it.
The equilibrium concentrations are [A]=0.798301683 M, [B]=1.127301683 M and [C]=0.0206983171 M. Substitution gives Q=0.0230000, and each concentration is nonnegative.
This ICE calculation establishes equilibrium composition only. An irreversible rate constant cannot identify an exact finite time to reversible equilibrium. A separate reviewed kinetic threshold exercise must supply its own disappearance law and target concentration.
For part B only: t=ln(0.457/0.0360)/0.320=7.94114 min≈7.94 min. Neither the equilibrium constant nor the ICE table supplies a reversible kinetic time.