Chapter 11 Source Answers and Examples

For overall order n, [k]=M¹⁻ⁿs⁻¹. Thus inverse-second units imply n=1. Units alone do not identify which species appears or its individual exponent. If the law is specifically −d[A]/dt=k[A], it is first order in A.

1. Reciprocals are approximately0.1000,0.1100,0.1200,…,0.1700 M⁻¹, linear with time.
2. The supported model is −d[A]/dt=k[A]²; equality of A and B is maintained by the stated1:1 stoichiometry. k≈(0.1700−0.1000)/35=0.00200 L mol⁻¹ s⁻¹.
3. t½=1/(k[A]₀)=1/(0.00200×10.0)=50.0 s. M means molarity; lowercase m would mean molality. These high concentrations are formal model data.

From the start, t=(1/[A]−1/[A]₀)/k=(1/0.230−1/10.0)/0.00200=2123.913 s≈2.12×10³ s. This is an extrapolation far beyond the35 s data window; it needs the explicitly stated constant-law assumption.

A total-order-two constant does not establish the individual powers. The solid’s amount is not a dissolved molarity; dissolution, accessible surface area, liquid volume changes and the rate-controlling process are unspecified. A defined empirical rate law and a model for the reactive solid are needed to obtain reaction extent after300 s. Stoichiometry gives an upper bound, not the time course:0.100 mol acid is less than10/84.01=0.1190 mol bicarbonate, so no more than0.100 mol CO₂ could form in the stated stoichiometric model. At1 atm/298.15 K this ideal dry-gas upper bound is nRT/P=2.45 L; it is not the gas volume at5 min. The concentrations are formal source quantities, not practical instructions.

The ln([A]/1 M)-versus-time graph is linear with slope−0.0230 s⁻¹. It supports first-order disappearance with k=0.0230 s⁻¹. The concentration and reciprocal plots are curved. A straight line drawn through a curved plot does not make it linear; the replacement figures omit the original misleading fitted lines and correct the vertical units.

1. T=318.15 K and Eₐ=422000 J mol⁻¹. k=4.31exp[−422000/(8.314×318.15)]=2.22326×10⁻⁶⁹ s⁻¹.
2. t½=ln2/k=3.11771×10⁶⁸ s. The first-order half-life is independent of[A]₀.
3. This enormous value follows from the supplied formal model; extrapolating one Arrhenius law over such a timescale has no established physical validity. Do not silently change422 to42.2.

1. Threshold mass=5.00×10.0=50.0 mg=0.0500 g; n_u,target=0.0500/58.44=0.000855578 mol.
2. t=ln(4.00/0.000855578)/0.320=26.4063 s≈26.4 s.
3. The result belongs only to this explicitly supplied remaining-population model. No universal first-order law for real NaCl dissolution is asserted.

1. k₈₅=0.230 s⁻¹ and k₃₅=0.0500 s⁻¹ from the negative slopes. T values are358.15 and308.15 K.
2. Eₐ=Rln(k₈₅/k₃₅)/(1/T₃₅−1/T₈₅)=28005.1 J mol⁻¹≈28.0 kJ mol⁻¹.
3. A=k₈₅exp(Eₐ/RT₈₅)=2794.50 s⁻¹≈2.79×10³ s⁻¹. Two points supply a formal fit, not proof that one mechanism holds over other temperatures.

−d[AB]/dt=r₁+r₂−r₃=0.00325[AB][CD]+0.221[AB]−0.322[A⁺][B⁻]. A negative value means net AB formation. The terms are supplied kinetic models; the schematic formulas do not establish an elementary chemical mechanism.

X₁ disappears with k₁, so t½(X₁)=ln2/0.411=1.68649 s. Step2 has the longest isolated half-life, ln2/0.121=5.72849 s, because its first-order constant is smallest among directly comparable first-order laws. Actual fluxes are k₁[X₁], k₂[X₂] and k₃[X₃]; populations determine them. Identifying the smallest k does not replace the required concentration/mechanism analysis for a general bottleneck.

1. Equality persists under the specified1:1 reaction, so t=(1/3.00−1/10.0)/k.
2. k₃₅=0.0324exp[−24000/(8.314×308.15)]=2.76788×10⁻⁶ L mol⁻¹ s⁻¹; k₂₀=1.71386×10⁻⁶ L mol⁻¹ s⁻¹.
3. t₃₅=84300.3 s≈23.4 h; t₂₀=136145.2 s≈37.8 h. The warmer model reaches the target in1/1.615≈0.619 of the time, about14.4 h sooner. Its rate constant is1.615 times larger.

k′=k[B]=0.734 s⁻¹. A is pseudo-first-order, so t½=ln2/0.734=0.944342 s≈0.944 s. The individual powers and maintained B concentration are explicitly supplied.

1. The known combined constant is0.0009562 s⁻¹, which predicts[A](210)=20exp(−0.0009562×210)=16.3615 M, far above0.596 M. Additional depletion or a model/data problem is present.
2. First and last30 s average total disappearance rates are(20−14.6)/30=0.1800 and(1.72−0.596)/30=0.0374667 M s⁻¹. Subtracting the known sinks at interval mean concentrations gives approximate extra rates0.16345774 and0.03635939 M s⁻¹, which are not constant.
3. Dividing these residual rates by their corresponding mean concentrations17.3 and1.158 M gives0.00945 and0.0314 s⁻¹; they are not one common first-order constant either. The interval approximation and rounding limit precision, but the discrepancy is large.
4. No unique pure zero-, first- or second-order extra law is established by these data. Obtain initial-rate measurements with controlled concentrations, test competing models and report uncertainty. A finite table does not uniquely identify an arbitrary mechanism.

Boiling occurs when the solution vapor pressure equals the external pressure. Pressure, composition effects and an applicable boiling condition are needed; reaction k does not provide them. If100.0 °C is only a stipulated target with a constant observed temperature rise, t=(100.0−35.0)/5.00=13.0 min. This conditional threshold time is not an established boiling time for the concentrated reacting solution. If a heating power rather than a temperature-rise rate were given, heat capacity and reaction enthalpy would also enter.

The effective constant would be k′=k[O₂]fixed in yr⁻¹. Conversion24% leaves fraction0.76, so t=−ln0.76/(3.2421[O₂]fixed) yr with oxygen concentration in M. A numerical time needs that concentration or an independently supplied effective k′. Real corrosion also involves surfaces and changing conditions; no universal homogeneous iron law follows from these data.

−d[A]/dt=r₁+2r₂=0.23511[A]²+0.662[A]. Product formation rates are d[AB]/dt=r₁ and d[A₂]/dt=r₂. The factor2 belongs to A consumption in the second reaction, even though its supplied empirical exponent is1. A numerical time and partition require[A]₀, initial product amounts and assurance that the laws remain valid while B is available. No unique numeric answer follows from the original data. If those were supplied, one would integrate this mixed-order law rather than use a single standard half-life formula.

A constant second-step rate needs a stated zero-order regime, units and sufficient intermediate AB and other reactants. At initial AB=0, a step consuming AB cannot maintain a positive rate indefinitely. The source does not specify its initial AB/DE populations or zero-order range, so no unique physical bottleneck time follows. Conditionally, comparing the first-step rate with a stated capacity0.441 M s⁻¹ gives0.753[A]²=0.441 and[A]=0.765283 M. Calling this a true step-rate crossover still requires the missing conditions.

1. The first stage consumes3.00 M A and R in3.00/0.320=9.375 s, leaving[A]=2.00 M.
2. The second stage takes ln(2.00/1.00)/0.02421=28.6306 s.
3. Total=38.0056 s≈38.0 s. A catalyst participates and is regenerated by the catalytic cycle; a species consumed stoichiometrically is not a catalyst. These two empirical regimes are explicitly stipulated.

k′=kp=1.05 h⁻¹. t=ln(1.00/0.200)/1.05=1.53280 h≈1.53 h. The original question omitted the partial pressure and a defined remaining-material rate law. C₆H₁₀O₅ is a cellulose repeat-unit formula, not the complete chemical composition of wood.

1. C=1/(0.082057×298.15)=0.0408742 M. Each reactant concentration is0.0400C.
2. r=0.321(0.0400C)² and d[AB]/dt=2r because each event produces2 AB.
3. Target[AB]=0.0100C. t=0.0100C/[2×0.321(0.0400C)²]=238.1746 s≈238 s. The fixed-density reservoirs and no-product-removal assumptions are model inputs, not inferred from a real room.