AP Unit 5 Solutions

1. A disappears at (0.0500 −0.0380)/40.0 =3.00×10⁻⁴ M s⁻¹.
2. r is half that, 1.50×10⁻⁴ M s⁻¹; B forms at 3r =4.50×10⁻⁴ M s⁻¹.
3. The finite interval gives an average; the initial tangent slope could differ if the rate changes during the interval.

1. For the solid comparison, hold solid identity/mass, solution composition/volume, temperature and mixing comparable; vary particle size. A suitable signal is gas volume formed per time if a known gas product is formed, or calibrated concentration of a specified reacting species versus time. More exposed sites permit more solid–solution contacts per time. Warming only one trial also changes collision energies and the reactive fraction, confounding the surface-area test.
2. The concentration pair changes only initial [A]. The measured rate doubles across these two concentrations, consistent with more reactant encounters in this range. This pair alone does not establish a universal rate law.
3. The catalyst pair changes catalyst presence while holding [A] and other conditions fixed. Its fourfold initial-rate increase supports a catalytic rate effect under these conditions. A catalyst provides a different route or more effective reaction events; the rates do not show a change in final yield or uniquely identify a mechanism.
4. The light pair changes illumination at fixed temperature. Its threefold rate increase supports light as an environmental factor affecting this reaction. Light can enable a photochemical route, but these data alone do not establish the molecular mechanism. The matched temperature excludes a simple warming explanation.
5. The universal doubling claim is unsupported. Measure additional concentrations under otherwise matched conditions and test a rate law for this reaction and regime. A zero-order or saturated regime can respond differently, and other reactions have different concentration dependences.

1. Trials 1→2: doubling A doubles rate, so order in A is one. Trials 1→3: doubling B quadruples rate, so order in B is two.
2. r =k[A][B]², overall order three; k =0.00120/(0.100×0.100²) =1.20 M⁻² s⁻¹.
3. Predicted r =1.20×0.150×0.300² =0.0162 M s⁻¹.

1. Absorbance can be converted to colored-reactant concentration using fixed path length and calibration; its early slope tracks disappearance.
2. With both concentrations doubled, fourfold rate supports a combined concentration factor, not second order in either separately. First order in each also gives four.
3. Hold the other reactant fixed while changing the colored reactant, then reverse that controlled comparison; use initial slopes and replicates at fixed temperature.

1. A is zero order and B first order in the tested range.
2. r =k[B], overall first order, so k has units s⁻¹.
3. The equation is overall, not declared elementary. Its coefficients count net consumption, while the data identify concentration sensitivity.

1. Equal 40.0 s intervals halve concentration, supporting first order; ln([A]/1 M) versus t is linear.
2. Slope =−ln2/40.0 ≈−0.0173 s⁻¹, so k =0.0173 s⁻¹.
3. At 200 s, five half-lives have passed: [A] =0.160/32 =0.00500 M.

1. Reciprocals are approximately 10.00, 15.00, 20.00 M⁻¹; the second-order model fits.
2. k =(20.00 −10.00)/100.0 =0.1000 M⁻¹ s⁻¹.
3. To reach 0.02500 M, 1/[A] =40.00 M⁻¹; t =(40.00 −10.00)/0.1000 =300.0 s.

1. Equal absolute loss per equal time supports zero order; k =0.0120/30.0 =4.00×10⁻⁴ M s⁻¹.
2. The model reaches zero at 0.0600/k =150 s.
3. A negative concentration beyond depletion is impossible; the zero-order regime cannot continue without reactant.

1. 24.0/8.00 =3.00 half-lives; undecayed mass =64.0(½)³ =8.00 μg.
2. Radioactive decay changes nuclei. A chemical catalyst changes electronic bonding pathways and does not ordinarily change this nuclear decay constant.

1. The declared elementary first step is bimolecular in NO₂, so its event rate is k[NO₂]².
2. The other equation is overall; it can have a multistep mechanism. A simultaneous three-particle event is rare and cannot be inferred from the net coefficients.

1. Draw particle energy on the horizontal axis and probability density on the vertical axis. Both curves have equal normalized total area; the hotter curve has a lower, broader peak shifted toward higher energy.
2. Mark one fixed Eₐ for the unchanged pathway; shade the tails to its right. The hotter curve has a larger tail area even though its peak height is lower. Fractions are areas, not peak heights.
3. More collisions can reach the transition-state region at higher temperature; proper reactive-site orientation is still necessary, and encounter frequency also generally rises.

The left panel shows the requested temperature comparison. The right panel is a separate fixed-temperature catalyst comparison, reinforcing that temperature and pathway thresholds are different changes.

1. The reacting sites may face each other in one collision but not the other.
2. Only the suitable approach allows the necessary bond rearrangement; high energy alone is insufficient.

1. Draw energy (kJ mol⁻¹) vertically and reaction coordinate horizontally. Begin at 10, rise to a transition-state maximum at 75, and end at products at 40. The coordinate is not time.
2. Draw the forward barrier arrow from 10 to75 and the reverse arrow from40 to75: Eₐ,f =65 and Eₐ,r =35 kJ mol⁻¹. Draw the endpoint arrow10→40: overall change +30 kJ mol⁻¹.
3. A–B and C–D are rearranged into A–C and B–D. Breaking initial bonds costs energy; forming new bonds releases energy. In this hypothetical endothermic rearrangement, the net formation release is smaller than the input cost. Actual breaking/formation can overlap along the pathway.
4. Warming increases the energetic collision fraction for this same barrier and normally increases k and rate; it does not lower the drawn barrier merely by increasing temperature.

1. NO is consumed first and regenerated, so it is catalyst. NO₂ is formed then consumed, so it is intermediate.
2. Cancel both: O₃ + O → 2O₂.
3. Detection is consistent with the proposed intermediate and supports the route; it does not prove the sequence uniquely or establish the slow step.

1. The slow first step predicts r =k[A]², with B absent while the later step remains fast.
2. Both observed changes match this law, so the proposal is consistent with those data, but not uniquely proved.
3. Tripling A gives a factor 3² =9 at the same temperature and sufficient B.

1. Rapid equilibrium gives k₁[NO][Br₂] ≈k₋₁[NOBr₂]. Thus [NOBr₂] ≈(k₁/k₋₁)[NO][Br₂].
2. Slow-step law r =k₂[NOBr₂][NO]. Substitution gives r ≈k_eff[NO]²[Br₂].
3. Factor =2²×½ =2. Fast reversible exchange must maintain near-equilibrium before appreciable intermediate removal.

1. M1 predicts r =k[A][B].
2. For M2, [I] ≈(k₁/k₋₁)[A]; substituting into k₂[I][B] predicts r =k_eff[A][B].
3. The same orders support both; other supplied evidence, such as intermediate evidence, would be needed to distinguish them.

1. Draw three maxima at35,60,20 separated by intermediate minima10 and−25; reactants begin at0 and products end at−40 kJ mol⁻¹. Label energy and reaction coordinate.
2. Draw local arrows0→35,10→60 and−25→20. Forward barriers are35,50 and45 kJ mol⁻¹. Draw the overall arrow0→−40, giving ΔE =−40 kJ mol⁻¹.
3. I1 and I2 are intermediate valleys; TS1–TS3 are transition-state peaks. Step2 has the largest local barrier, suggesting slower passage under comparable kinetic conditions; actual rate control also depends on concentrations, reversibility and prefactors.

1. E is a regenerated catalyst; ES and EP are formed/consumed bound intermediates. Adding the enzyme steps gives S → P.
2. Binding may improve orientation and provide lower barriers, increasing productive events. The same reactant/product endpoints mean unchanged ΔH. Total enzyme amount is conserved in the ideal cycle, but free E can be redistributed into ES and EP.
3. In the second model the slow, rate-determining step consumes C when it combines with A to form CA. The subsequent fast step regenerates C when CA gives C + P.
4. Adding C + A → CA and CA → C + P cancels C and CA, yielding A → P. C is the catalyst and CA is the intermediate.
5. The conserved total is [C]_total = [C] + [CA] at constant volume. Each CA contains one catalyst unit. Converting free C to bound CA lowers [C] by the same amount that [CA] increases; regeneration reverses this distribution.

1. Acid catalysis creates a protonated intermediate and a sequence of proton-transfer/bond-rearrangement steps; later loss of H⁺ restores it.
2. Surface catalysis creates adsorbed intermediates at sites; reaction and product desorption restore accessible sites.
3. Both supply alternative pathways, often with smaller barriers or favorable orientation, but the same initial/final species retain the same enthalpy difference.