AP Chemistry Unit 5: Kinetics — Free Review Games.

Higher temperature means molecules move faster, collide more often, and a greater fraction of collisions have enough energy to overcome the activation energy barrier.

A catalyst provides an alternative reaction mechanism with a lower activation energy. It is not consumed and does not change the equilibrium position.

Reaction rate measures how quickly concentrations change over time, typically expressed in units of mol/(L*s) or M/s.

If \(\text{rate} = k[A]^1\), then doubling \([A]\) doubles the rate. This is the definition of first-order dependence.

Decreasing temperature slows molecular motion and reduces the fraction of molecules with sufficient energy to react, thus decreasing the rate.

\(\text{Rate} = k[NO]^2[O_2]\). Tripling \([NO]\): \(\text{rate} = k(3[NO])^2[O_2] = 9k[NO]^2[O_2]\). The rate increases by a factor of 9.

60 minutes = 3 half-lives. After each half-life, half remains: (1/2)^3 = 1/8 of the original concentration.

Overall order = sum of exponents in the rate law = 1 + 2 = 3 (third order).

The Arrhenius equation shows rate depends exponentially on -Ea/RT. Halving the activation energy has a dramatic exponential effect on the rate constant, far greater than modest changes in temperature or concentration.

The rate-determining step is the slowest step in the mechanism, acting as a bottleneck. The overall rate cannot exceed the rate of this step.

Comparing Expts 1 and 3: doubling \([B]\) does not change rate, so order in B = 0. Comparing Expts 1 and 2: doubling \([A]\) quadruples rate, so order in A = 2. $Rate = k[A]^2$.

The rate is determined by the slow step: \(\text{Rate} = k_1[NO_2][F_2]\). This involves one molecule each of \(NO_2\) and \(F_2\).

For first-order reactions, t1/2 = ln(2)/k = 0.693/0.0693 = 10.0 minutes.

The Arrhenius equation: k = Ae^(-Ea/RT). Taking ln: ln(k) = ln(A) - Ea/(RT). Plotting ln(k) vs 1/T gives slope = -Ea/R.

An intermediate is formed in one elementary step and consumed in a later step. It does not appear in the overall equation. This distinguishes it from a catalyst (regenerated) and a product (remains).

For a zero-order reaction, rate = k, meaning the rate is entirely independent of reactant concentration. Doubling [A] has no effect. This contrasts with first-order reactions (rate doubles) and second-order reactions (rate quadruples) when [A] is doubled.

For a second-order reaction, rate = k[A]², so k = rate/[A]² = (M·s⁻¹)/M² = M⁻¹·s⁻¹. The general formula for units of k in an nth-order reaction is M^(1-n)·s⁻¹. For first-order k has units of s⁻¹; for zero-order k has units of M·s⁻¹.

Collision theory requires that reacting molecules must (1) collide with kinetic energy at least equal to the activation energy, and (2) approach each other with the correct geometric orientation so that bonds can break and form appropriately. Even an energetically sufficient collision fails to produce products if the molecules are misaligned.

The integrated rate law for a zero-order reaction is [A] = [A]₀ - kt, which is the equation of a straight line when [A] is plotted against time. By contrast, a first-order reaction gives a straight line for ln[A] vs. time, and a second-order reaction gives a straight line for 1/[A] vs. time.

Reactant concentration decreases over time, making Δ[A]/Δt negative by definition. The negative sign converts this to a positive rate value. The factor of 1/2 normalizes the rate expression to the stoichiometric coefficient so the same rate value is obtained regardless of which species is monitored.

The defining distinction is phase: a homogeneous catalyst (such as an acid dissolved in aqueous solution) exists in the same phase as the reactants. A heterogeneous catalyst (such as solid platinum used in gas-phase hydrogenation) exists in a different phase. Both types are regenerated after reaction and lower the activation energy for both forward and reverse reactions equally.

According to k = Ae^(-Ea/RT), when the pre-exponential factor A is identical for all reactions, the reaction with the lowest activation energy has the largest rate constant because a greater fraction of molecules in the Maxwell-Boltzmann distribution possesses at least that threshold energy. Reaction 1 (Ea = 20 kJ/mol) has the smallest barrier and therefore the fastest rate at any given temperature.

Starting from the integrated second-order law 1/[A] = kt + 1/[A]₀ and setting [A] = [A]₀/2 yields t½ = 1/(k[A]₀). Unlike first-order reactions where t½ = 0.693/k is constant, the second-order half-life decreases as the reaction proceeds because the effective [A]₀ at each stage decreases.

Each reaction order linearizes a specific graph: zero order gives a straight line for [A] vs. time; first order gives a straight line for ln[A] vs. time (from the integrated law ln[A] = -kt + ln[A]₀); second order gives a straight line for 1/[A] vs. time. The linear graph II confirms first-order kinetics. Graphs I and III are curved, ruling out zero-order and second-order.

The overall rate is linked to each species by its stoichiometric coefficient: rate = -(1/3)(Δ[H₂]/Δt) = (1/2)(Δ[NH₃]/Δt). Solving for the rate of NH₃ formation: Δ[NH₃]/Δt = (2/3) × 0.060 M/s = 0.040 M/s. A common error is to equate the rate of H₂ consumption directly to the rate of NH₃ formation, ignoring the 3:2 stoichiometric ratio.

The Arrhenius equation k = Ae^(-Ea/RT) shows k increases as T increases because raising temperature shifts the Maxwell-Boltzmann energy distribution so a larger fraction of molecules exceeds Ea. Choice D is a common approximation that only holds roughly for reactions with Ea near 50-60 kJ/mol around room temperature; the actual factor depends on both Ea and the temperature range.

Comparing Exp 1 and Exp 2: [X] doubles while [Y] is constant and the rate doubles, indicating first order in X. Comparing Exp 1 and Exp 3: [Y] doubles while [X] is constant and the rate is unchanged, indicating zero order in Y. The rate law is rate = k[X]. A critical misconception is assuming order must equal the stoichiometric coefficient; here Y appears in the equation but does not affect the rate.

For elementary reactions, the rate law follows directly from molecularity because the step describes an actual molecular event. A bimolecular step requires one A molecule and one B molecule to collide simultaneously; collision frequency is proportional to [A][B]. This reasoning applies only to elementary steps, not to overall balanced equations, where rate laws must always be determined experimentally.

A catalyst lowers the activation energy equally for both forward and reverse reactions, leaving their ratio and therefore K unchanged. K is a thermodynamic quantity determined by ΔG°, which kinetics cannot alter. At equilibrium, a catalyst has no effect on concentrations. Spontaneity (ΔG°) is also unaffected — a catalyst cannot drive a thermodynamically unfavorable reaction forward.

When [B] >> [A], [B] stays essentially constant throughout the reaction (pseudo-first-order conditions). The rate simplifies to rate = k[B][A] = k_obs[A], where k_obs = k[B]. Measuring k_obs under several large [B] values and plotting k_obs vs. [B] gives a straight line with slope equal to the true second-order rate constant k, allowing accurate determination of k without needing to vary two concentrations simultaneously.

The rate-determining step (Step 2) gives rate = k₂[B][C]. However, B is an intermediate and must be eliminated from the rate law. Using the fast equilibrium from Step 1: K_eq = [B]/[A], so [B] = K_eq[A]. Substituting: rate = k₂K_eq[A][C] = k[A][C]. Choice B is incorrect because intermediates cannot appear in the final experimentally testable rate law — they must be expressed in terms of reactants.

From the Arrhenius equation, d(ln k)/dT = Ea/RT², meaning the temperature sensitivity of ln k is proportional to Ea. A larger activation energy produces a steeper slope in the ln k vs. 1/T plot. When temperature rises from 300 to 400 K, the ratio k₄₀₀/k₃₀₀ is larger for Reaction Q (higher Ea) than for Reaction P. The pre-exponential factor A cancels in the ratio and does not affect temperature sensitivity.

Since [A]⁰ = 1 contributes no units, the effective rate expression is rate = k[B]², so k = rate/[B]² = (mol·L⁻¹·s⁻¹)/(mol·L⁻¹)² = (mol·L⁻¹·s⁻¹)·(L²·mol⁻²) = L·mol⁻¹·s⁻¹. This matches the standard units for any second-order rate constant. A common error is choosing s⁻¹ (first-order units), forgetting that the overall order is 0 + 2 = 2, not 1.

The slow step gives rate = k₂[A₂][B]. A₂ is a reaction intermediate (produced in Step 1, consumed in Step 2) and cannot appear in the final rate law. Using the fast equilibrium: K = [A₂]/[A]², so [A₂] = K[A]². Substituting: rate = k₂K[A]²[B] = k[A]²[B]. Choice C is incorrect because intermediates must always be eliminated by expressing them in terms of actual reactants.

Using the two-temperature Arrhenius form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Here ln(8.0×10⁻³/2.0×10⁻³) = ln 4 ≈ 1.39. The temperature term: 1/300 - 1/320 = (320 - 300)/(300 × 320) = 20/96000 ≈ 2.08×10⁻⁴ K⁻¹. Therefore Ea = (8.314 × 1.39)/(2.08×10⁻⁴) ≈ 11.56/2.08×10⁻⁴ ≈ 55,600 J/mol ≈ 55 kJ/mol. Choice A results from incorrectly using ΔT = 20 K in the denominator instead of the proper 1/T difference.

Using the integrated second-order rate law: 1/[A] = kt + 1/[A]₀ = (0.050)(100) + 1/(0.40) = 5.0 + 2.5 = 7.5 L·mol⁻¹. Therefore [A] = 1/7.5 ≈ 0.13 M. Choice A (0.20 M) corresponds to t = 50 s, which is the first half-life (t½ = 1/(k[A]₀) = 50 s). Choice D (0.10 M) is the error of treating the half-life as constant and applying first-order logic: 0.40 × (1/2)² = 0.10 M, which ignores that the second-order half-life grows as the reaction proceeds.

If Step 2 is rate-determining: rate = k₂[NOCl₂][NO]. NOCl₂ is an intermediate; using the pre-equilibrium from Step 1: K = [NOCl₂]/([NO][Cl₂]), so [NOCl₂] = K[NO][Cl₂]. Substituting: rate = k₂K[NO]²[Cl₂] = k[NO]²[Cl₂], which matches the observed third-order rate law. If Step 1 were rate-determining, the predicted rate law would be rate = k[NO][Cl₂] (second-order), which contradicts the experimental data.

First determine the rate law: Exp 1 vs. Exp 2 — [A] doubles, rate doubles → first order in A. Exp 1 vs. Exp 3 — [B] triples (0.10 to 0.30 M), rate triples → first order in B. Rate law: rate = k[A][B]. Solving for k using Exp 1: k = rate/([A][B]) = (1.2×10⁻⁴ M/s)/((0.10 M)(0.10 M)) = 0.012 L·mol⁻¹·s⁻¹. Note that despite the stoichiometric coefficient of 2 for B, the reaction is first order in B — reaction orders are always determined experimentally, not from coefficients.

The rate-determining step is identified by the largest activation energy relative to the immediately preceding minimum. Step 1: Ea = 80 - 0 = 80 kJ. Step 2: Ea = 60 - 30 = 30 kJ. Step 1 is rate-determining. The overall reaction is exothermic (products at -40 kJ, below reactants). The reverse overall activation energy is from products (-40 kJ) to the highest transition state (80 kJ): 80 - (-40) = 120 kJ, not 40 kJ. Choice B confuses absolute energy level with activation energy relative to the preceding minimum.

For first-order reactions, k = 0.693/t½. At T₁ = 298 K: k₁ = 0.693/200. At T₂ = 338 K: k₂ = 0.693/50. The ratio k₂/k₁ = 200/50 = 4, so ln(k₂/k₁) = ln 4 ≈ 1.39. Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): the temperature term is 1/298 - 1/338 = (338 - 298)/(298 × 338) = 40/100724 ≈ 3.97×10⁻⁴ K⁻¹. Therefore Ea = (8.314 × 1.39)/(3.97×10⁻⁴) ≈ 11.56/3.97×10⁻⁴ ≈ 29,100 J/mol ≈ 29 kJ/mol.

The rate constant k depends on temperature only (for a given reaction), as described by the Arrhenius equation: k = Ae^(-Ea/RT). Increasing temperature increases k. Concentrations of reactants appear in the rate law but do not change k — k is a constant at a fixed temperature regardless of concentration. Choice C is a common misconception that conflates the rate law with the rate constant.

For a first-order reaction, t½ = ln(2)/k. Since t½ depends only on k, and k depends only on temperature, the half-life is independent of initial concentration. This is the defining feature that distinguishes first-order from second-order kinetics. Choice B applies to second-order reactions, where t½ = 1/(k[A]₀), which increases as [A]₀ decreases over time.

For a zero-order reaction, rate = k, meaning k must have the same units as the rate: mol·L⁻¹·s⁻¹. The units of k always depend on the overall reaction order. Choice A (s⁻¹) applies to first-order rate constants, and Choice B (L·mol⁻¹·s⁻¹) applies to second-order rate constants. Choice D applies to third-order rate constants.

A catalyst works by offering an alternative reaction pathway with a lower activation energy, which means a greater fraction of molecular collisions have sufficient energy to produce products. A catalyst does not change temperature or concentrations, and critically, it does not alter the position of equilibrium — a catalyst affects kinetics (how fast equilibrium is reached) but not thermodynamics (ΔG or the final equilibrium constant).

The integrated rate law for a zero-order reaction is [A] = [A]₀ − kt, which is the equation of a straight line when [A] is plotted against time (slope = −k, y-intercept = [A]₀). Choice A (ln[A] vs t) is linear for first-order reactions. Choice B (1/[A] vs t) is linear for second-order reactions. Recognizing which integrated rate law produces a linear graph is a key AP Chemistry skill.

The rate-determining step is the slowest step in a mechanism — it acts as a bottleneck and governs the overall reaction rate. Its elementary rate law directly gives the overall observed rate law. The rate-determining step is not necessarily the first step (Choice A) and typically has the highest activation energy, not the lowest (Choice B). The final product may form in a fast step after the rate-determining step.

The overall rate is defined as rate = −(1/2)d[H₂]/dt = −d[O₂]/dt. Given that d[H₂]/dt = 0.040 mol·L⁻¹·s⁻¹, the overall rate = 0.040/2 = 0.020 mol·L⁻¹·s⁻¹, which equals the rate of disappearance of O₂. H₂ is consumed twice as fast as O₂ due to the 2:1 stoichiometric ratio. Choice C incorrectly assumes a 1:1 correspondence between H₂ and O₂ consumption rates.

A lower activation energy means a greater fraction of colliding molecules possess enough energy to overcome the energy barrier and form products. Per the Arrhenius equation k = Ae^(−Ea/RT), decreasing Ea exponentially increases k and therefore the reaction rate. Choice A confuses activation energy (energy input needed) with reaction enthalpy (energy released). Choice D is incorrect because both Ea and T appear in the Arrhenius equation, so Ea directly affects rate.

This rate law is third order overall (second order in A, first order in B). For units to balance: mol·L⁻¹·s⁻¹ = k × (mol·L⁻¹)² × (mol·L⁻¹) = k × mol³·L⁻³. Solving: k = mol·L⁻¹·s⁻¹ ÷ mol³·L⁻³ = L²·mol⁻²·s⁻¹. Choice A (s⁻¹) is for first-order reactions, and Choice B (L·mol⁻¹·s⁻¹) is for second-order reactions. The units of k are determined by the overall reaction order.

Using the integrated first-order rate law: ln[A] = ln[A]₀ − kt. Substituting: ln[A] = ln(0.500) − (0.0150)(75.0) = −0.693 − 1.125 = −1.818. Therefore [A] = e^(−1.818) ≈ 0.162 mol/L. The half-life is ln(2)/0.0150 ≈ 46.2 s, so 75 s is slightly more than 1.5 half-lives, making the answer between 0.500/2 = 0.250 and 0.500/4 = 0.125 — consistent with 0.162. Choice A incorrectly applies a linear (zero-order) approach.

The overall rate expression is: rate = −(1/3)d[H₂]/dt = (1/2)d[NH₃]/dt. Since d[H₂]/dt = 0.60 mol·L⁻¹·s⁻¹, the overall rate = 0.60/3 = 0.20 mol·L⁻¹·s⁻¹. Then d[NH₃]/dt = 2 × 0.20 = 0.40 mol·L⁻¹·s⁻¹. Each species is divided (or multiplied) by its stoichiometric coefficient. Choice C (0.60) incorrectly assumes H₂ and NH₃ are consumed and produced at the same rate, ignoring stoichiometry.

If rate = k[A]^n, then tripling [A] multiplies the rate by 3^n. Since 3^n = 9 = 3², the order is n = 2 (second order). For zero order, tripling [A] would have no effect on the rate. For first order, tripling [A] would triple the rate (factor of 3). For third order, the rate would increase by a factor of 27. The rate factor equals the concentration factor raised to the power of the reaction order.

The linearized Arrhenius equation is ln(k) = ln(A) − Ea/(RT), so a plot of ln(k) versus 1/T has slope = −Ea/R. Therefore Ea = −(slope) × R = −(−8500 K) × 8.314 J·mol⁻¹·K⁻¹ = 70,669 J/mol ≈ 70.7 kJ/mol. Choice A incorrectly uses the slope value directly in kJ without multiplying by R. Choice C results from an arithmetic error in unit conversion, and Choice D from forgetting to convert J to kJ.

The overall rate is determined by the slowest (rate-determining) step. Step 1 is slow, so rate = k₁[A][B]. C is a reaction intermediate — it is produced in Step 1 and consumed in Step 2. Intermediates must not appear in the final rate law, making Choice C incorrect. Choice B would require two molecules of B in the slow step, but Step 1 involves only one B. The fast second step does not influence the overall rate.

For a second-order reaction, t½ = 1/(k[A]₀) = 1/(0.100 × 0.200) = 1/0.0200 = 50.0 s. This is a key distinction from first-order reactions: the second-order half-life depends on the initial concentration and increases as the reaction proceeds. Choice A might result from mistakenly applying ln(2)/k (the first-order formula), though the units of k would not match. Choice D would result from dividing 1/k only without including [A]₀.

The defining distinction between catalyst types is phase: a heterogeneous catalyst is in a different phase from the reactants (e.g., solid platinum catalyzing a gas-phase reaction), while a homogeneous catalyst is in the same phase (e.g., an acid dissolved in an aqueous reaction mixture). Both types lower activation energy and are regenerated after the reaction (Choice A and C are wrong). Choice D overgeneralizes — heterogeneous catalysts are also used in liquid-phase reactions.

A linear plot of 1/[A] versus time is the graphical signature of second-order kinetics. The integrated second-order rate law is 1/[A] = 1/[A]₀ + kt, giving a straight line with positive slope equal to k. By comparison, a linear [A] vs time plot indicates zero order (slope = −k), and a linear ln[A] vs time plot indicates first order (slope = −k). Identifying reaction order from these three diagnostic plots is a fundamental AP Chemistry skill.

The rate is set by the slow step: rate = k₂[AB][A]. Since AB is an intermediate, it must be eliminated using the fast equilibrium expression: K₁ = [AB]/([A][B]), so [AB] = K₁[A][B]. Substituting: rate = k₂K₁[A][B][A] = k_obs[A]²[B], where k_obs = k₂K₁. Choice C is not an acceptable final rate law because it contains the intermediate AB. Choice A would require the slow step to involve only one molecule of A, but Step 2 requires two A atoms total (one from AB plus one free A).

Using the two-temperature Arrhenius form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Here ln(3) = 1.099, and 1/290 − 1/310 = (310 − 290)/(290 × 310) = 20/89,900 = 2.225 × 10⁻⁴ K⁻¹. Solving: Ea = (1.099 × 8.314)/(2.225 × 10⁻⁴) ≈ 41,000 J/mol ≈ 41.1 kJ/mol. A common error is computing 1/T₂ − 1/T₁ instead of 1/T₁ − 1/T₂, which yields a negative value. Another error is substituting ln(k₂/k₁) = 3 instead of ln(3) ≈ 1.099.

The ratio of rate constants is k_cat/k_uncat = e^(−Ea_cat/RT) / e^(−Ea_uncat/RT) = e^((Ea_uncat − Ea_cat)/RT). The exponent is (80,000 − 50,000)/(8.314 × 298) = 30,000/2478 ≈ 12.11. Therefore k_cat/k_uncat = e^12.11 ≈ 1.8 × 10⁵. This dramatic increase illustrates why even a modest reduction in Ea produces an enormous rate enhancement — because the Boltzmann factor depends exponentially on Ea. Choice A would result from using the ratio of Ea values (80/50) rather than applying the Arrhenius equation properly.

The concentration halves every 100 s regardless of the starting concentration (0.800 → 0.400 → 0.200 → 0.100 M), indicating a constant half-life — the hallmark of first-order kinetics. The rate constant is k = ln(2)/t½ = 0.693/100 s = 6.93 × 10⁻³ s⁻¹. Choice B (second order) is wrong because second-order half-lives increase each cycle as concentration decreases. Choice D (zero order) would show a constant decrease in concentration per unit time (e.g., 0.004 M/s), not a constant ratio.

F is produced in Step 1 and consumed in Step 2 without appearing in the overall balanced equation, making it a reaction intermediate. The rate law comes from the slow step: rate = k₁[NO₂][F₂]. F₂ is a reactant, not an intermediate (Choice A). NO₂F is a product, not an intermediate (Choice C). Choice D would require two molecules of NO₂ in the slow step, but Step 1 involves only one NO₂. The fast step does not contribute to the rate law.

For the reverse reaction, the starting point is the products at −20 kJ/mol relative to the forward reactants. The reverse reaction must cross the highest point on the energy profile, which is the second transition state at +50 kJ/mol. The activation energy for the reverse reaction equals the energy of the highest transition state minus the energy of the products: 50 − (−20) = 70 kJ/mol. Equivalently, Ea_reverse = Ea_forward(overall) + |ΔH|: the overall forward Ea is 50 kJ/mol and ΔH = −20 kJ/mol, so Ea_reverse = 50 + 20 = 70 kJ/mol.

The data shows a constant half-life of 10.0 min (0.100 → 0.0500 → 0.0250 → 0.0125 M each 10 min), which is the signature of first-order kinetics, not second order. Choice B reverses the truth: it is first-order reactions that have constant half-lives, while second-order half-lives increase each successive cycle. Choice C confuses zero-order (linear [A] vs t) with second-order. Choice D is incorrect — second-order reactions can involve a single reactant with rate = k[A]², as in many decomposition reactions.

Because tripling [A] still increases the rate by a factor of 9 (= 3²) with the catalyst present, the reaction remains second order — catalysts alter the reaction pathway but do not change the rate law or reaction order. However, the lower Ea increases k via the Arrhenius equation (k = Ae^(−Ea/RT)). Choice C is incorrect: a catalyst lowers Ea but does not change the pre-exponential factor A, which reflects molecular collision geometry and frequency. Choice D is incorrect: catalysts do not alter the thermodynamics (ΔH or ΔG) of a reaction — only the kinetics.

The instantaneous rate equals the magnitude of the slope of the tangent line to the concentration-versus-time curve at a specific point. Choice B describes average rate, which uses total concentration change over a total time interval rather than the value at a single instant.

For a zero-order reaction, rate = k[A]^0 = k, so the rate equals the rate constant and is completely independent of reactant concentration. Choices A and B describe first-order behavior, and choice D describes second-order behavior.

For a second-order reaction, rate = k[A]², so k = rate/[A]² = (mol·L⁻¹·s⁻¹)/(mol·L⁻¹)² = L·mol⁻¹·s⁻¹. Choice B (s⁻¹) applies to first-order, choice C (mol·L⁻¹·s⁻¹) applies to zero-order, and choice D applies to third-order reactions.

A catalyst provides an alternative mechanism with a lower activation energy barrier, allowing a greater fraction of collisions to be successful. Catalysts do not change temperature, concentration, or the thermodynamic equilibrium constant — they only affect how quickly equilibrium is reached.

For a first-order reaction, t₁/₂ = 0.693/k. Because k is constant at a given temperature, the half-life is constant regardless of how much reactant remains. This distinguishes first-order from second-order reactions, where t₁/₂ = 1/(k[A]) grows as [A] falls. Choice B is wrong because t₁/₂ = 0.693/k, not 1/k.

Collision theory requires that molecules collide with energy at least equal to the activation energy AND with the proper geometric orientation so reactive regions can interact. Gas phase and catalyst surfaces are not required by the theory. Choice C is wrong because identical kinetic energies are not necessary.

The integrated rate law for a zero-order reaction is [A] = [A]₀ - kt, which is linear in [A] vs. t with slope -k. A first-order reaction gives a straight line only when ln[A] is plotted versus t, and a second-order reaction gives a straight line only when 1/[A] is plotted versus t.

Comparing Trials 1 and 2 (B constant): doubling [A] doubles the rate → first order in A. Comparing Trials 1 and 3 (A constant): doubling [B] quadruples the rate (8.00/2.00 = 4 = 2²) → second order in B. Therefore rate = k[A][B]². Choice C would predict only a 2-fold rate increase when [B] doubles, which contradicts the data.

The second-order integrated rate law is 1/[A] = 1/[A]₀ + kt. Substituting: 1/[A] = 1/0.500 + (0.100)(5.00) = 2.00 + 0.500 = 2.50 M⁻¹, so [A] = 0.400 M. Choice B (0.303 M) results from incorrectly applying the first-order formula: [A] = 0.500 × e^(−0.500) ≈ 0.303 M. Choice C confuses this with a half-life calculation.

Using the two-temperature Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂) = (55000/8.314)(1/298 - 1/308) = 6613 × 1.09 × 10⁻⁴ = 0.720. So k₂/k₁ = e^0.720 = 2.05, and k₂ = 1.50 × 10⁻³ × 2.05 = 3.08 × 10⁻³ s⁻¹. Choice B incorrectly applies the rough rule of thumb that rates triple per 10°C, which is not accurate for all activation energies.

The rate law is determined by the slow (rate-determining) step. Step 1 involves X and Y, giving rate = k[X][Y]. Z is a reaction intermediate and cannot appear in the observed rate law. Choice A is wrong because Y appears only once in the slow step; second-order dependence on Y would require Y to appear twice in that elementary step.

Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(4) = (Ea/8.314)(1/300 - 1/320), so 1.386 = (Ea/8.314)(2.08 × 10⁻⁴). Solving: Ea = (1.386 × 8.314)/(2.08 × 10⁻⁴) = 55,400 J/mol = 55.4 kJ/mol. Choice B is half the correct answer, resulting from using the sum of the two temperatures rather than their reciprocal difference.

When [B] is held constant in large excess, rate = (k[B])[A] = k_obs[A], where k_obs = k[B] is a constant. The reaction appears first order in A alone — called pseudo-first-order kinetics. Choice D is wrong because k_obs must include [B], not [A]. Choice A is wrong because [B] no longer varies and does not appear as a changing factor in the observed kinetics.

A reaction intermediate is produced in one elementary step and consumed in a subsequent step; it does not appear in the overall balanced equation. N₂O₄ is produced in Step 1 and consumed in Step 2, so it is the intermediate. The overall reaction is 2NO₂ + Cl₂ → 2NO₂Cl, so NO₂ and Cl₂ are reactants and NO₂Cl is the product — none of these qualify as intermediates.

If rate = k[A]^n, then tripling [A] multiplies the rate by 3^n = 9, so n = 2. A nine-fold increase is the square of 3, indicating second-order dependence. For first order, tripling [A] would triple the rate. For third order, tripling [A] would give a 27-fold increase. For zero order, the rate would be unaffected.

For a zero-order reaction, [A] = [A]₀ - kt. Setting [A] = [A]₀/2: t₁/₂ = [A]₀/(2k) = 0.800/(2 × 0.0400) = 10.0 s. Choice B (17.3 s) incorrectly applies the first-order formula t₁/₂ = 0.693/k = 0.693/0.0400 = 17.3 s. Unlike first-order reactions, the zero-order half-life is not constant — it decreases as [A]₀ decreases with each successive half-life.

For a zero-order reaction, rate = k[A]^0 = k, so k has the same units as rate: mol·L⁻¹·s⁻¹. Choice A (s⁻¹) applies to first-order reactions. Choice B (L·mol⁻¹·s⁻¹) applies to second-order reactions. Choice C applies to third-order reactions. A quick way to check units is to use k = rate/[A]^n and substitute the reaction order.

Since both share the same pre-exponential factor A, k_P/k_Q = e^((Ea_Q - Ea_P)/RT) = e^(40000/(8.314 × 298)) = e^16.1 ≈ 10⁷. The relationship is exponential — k_P is roughly ten million times larger than k_Q. Choice D incorrectly treats the relationship as linear (ratio ≈ 2) rather than exponential. Choice C is wrong because different Ea values create enormous differences in rate constants even when A is identical.

The slow step gives rate = k₂[A₂][B]. Because A₂ is an intermediate, it must be eliminated using the fast equilibrium: Keq = [A₂]/[A]², so [A₂] = Keq[A]². Substituting: rate = k₂Keq[A]²[B] = k[A]²[B]. Choice B is wrong because intermediates cannot appear in the final rate law — they must be expressed in terms of reactant concentrations.

A catalyst does not alter reaction thermodynamics — it only lowers the kinetic barrier by providing an alternative pathway. The energies of reactants and products are identical in both pathways, so ΔH = -40 kJ/mol in both cases. Choices B and C confuse activation energy with enthalpy change; these are independent quantities. Choice D is wrong because catalysts do not supply energy to the reaction.

Using the two-temperature Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂) = (92000/8.314)(1/500 - 1/600) = 11065 × 3.33 × 10⁻⁴ = 3.69. Therefore k₂/k₁ = e^3.69 ≈ 40.0. Choice A (3.69) is the value of ln(k₂/k₁), not k₂/k₁ itself — a common error of forgetting to exponentiate. Choice D would result from mistakenly squaring the ratio.

The slow step gives rate = k₂[NOBr₂][NO]. Since NOBr₂ is an intermediate, it is eliminated using the fast equilibrium: Keq = [NOBr₂]/([NO][Br₂]), so [NOBr₂] = Keq[NO][Br₂]. Substituting: rate = k₂Keq[NO]²[Br₂] = k[NO]²[Br₂]. Choice C is the rate expression for the slow step before the intermediate is eliminated — intermediates cannot appear in the experimentally observed rate law.

The successive half-lives are 100 s, 200 s, and 400 s — each doubling. A constant half-life indicates first order, but a doubling half-life is the signature of second-order kinetics: t₁/₂ = 1/(k[A]), so as [A] halves, t₁/₂ doubles. Using the first interval: k = 1/(t₁/₂ × [A]₀) = 1/(100 × 1.00) = 0.0100 L·mol⁻¹·s⁻¹. Choice A misidentifies the order as first order despite the non-constant half-lives.

With identical pre-exponential factors, k_A/k_B = e^((Ea_B - Ea_A)/RT) = e^(50000/(8.314 × 298)) = e^20.2 ≈ 5.8 × 10⁸. Even a 50 kJ/mol difference in activation energy produces an approximately billion-fold difference in rate constants at room temperature, illustrating the exponential sensitivity of reaction rates to Ea. Choice A (≈2) incorrectly treats the relationship as linear (100/50 = 2) rather than exponential.

For a second-order reaction: 1/[A] - 1/[A]₀ = kt. Substituting: 1/0.400 - 1/2.00 = 0.250 × t → 2.50 - 0.500 = 0.250t → t = 2.00/0.250 = 8.00 s. Choice B (6.44 s) results from applying the first-order formula: t = ln([A]₀/[A])/k = ln(5)/0.250 = 6.44 s. Choice C (10.0 s) results from omitting the 1/[A]₀ correction term and computing only 1/([A] × k).

The average rate of a reaction is defined as the change in concentration of a reactant or product per unit time. For a reactant, rate = -Δ[reactant]/Δt; for a product, rate = Δ[product]/Δt. Choice B describes total yield, not rate. Choice C combines Arrhenius terms into a ratio that has no standard kinetic definition. Choice D conflates the equilibrium constant K with the rate constant k, which are fundamentally different quantities.

For a reaction second order in A, rate = k[A]². When [A] is tripled, the new rate = k(3[A])² = 9k[A]², so the rate increases by a factor of 9 = 3². Choice A (factor of 3) correctly describes a first-order reaction, not second order. Choice D (factor of 27 = 3³) describes a third-order reaction. Choice B has no basis in standard rate law mathematics and may result from confusing order with a simple multiplier.

For a first-order reaction, t½ = ln(2)/k = 0.693/k, which depends only on the rate constant k and not on the initial concentration. This concentration-independence is a defining characteristic of first-order kinetics. Choice C describes second-order behavior, where t½ = 1/(k[A]₀), making each successive half-life longer. Choices A and D describe dependencies that do not arise in any standard reaction order.

A catalyst works by providing an alternative mechanism with a lower activation energy, allowing a greater fraction of molecular collisions to result in successful reaction. A catalyst is not consumed and does not change the overall thermodynamics of the reaction, so it cannot alter the temperature (Choice A) or the equilibrium constant (Choice D), since ΔG° is unchanged. A catalyst also does not change reactant concentrations over time (Choice B); it only speeds up the approach to equilibrium.

For a zero-order reaction, rate = k[A]⁰ = k, meaning the rate constant has the same units as the rate itself: mol·L⁻¹·s⁻¹ (equivalently M·s⁻¹). Choice A (s⁻¹) corresponds to a first-order rate constant. Choice B (L·mol⁻¹·s⁻¹) corresponds to a second-order rate constant. Choice D (L²·mol⁻²·s⁻¹) corresponds to a third-order rate constant. The general formula for units of k is (mol·L⁻¹)^(1−n)·s⁻¹, where n is the overall reaction order.

A reaction intermediate is formed in one step of a mechanism and then consumed in a later step, so it does not appear in the overall balanced equation. Choice C describes the transition state (activated complex), which is distinct from an intermediate — intermediates occupy energy minima (valleys) on the reaction coordinate diagram, while transition states are at energy maxima (peaks). Choice D describes a catalyst. Choice A is incorrect because intermediates should not appear in the final rate law; if they do appear in an elementary step rate expression, they must be substituted out using equilibrium or steady-state expressions.

In the Arrhenius equation, the exponent is −Ea/RT. For k to be maximized, the magnitude of this negative exponent must be minimized, which occurs when Ea is small and T is large. Choice A is partially correct about temperature but wrong about activation energy — higher Ea makes the exponent more negative, decreasing k. Choice B is partially correct about Ea but wrong about temperature — lower T makes the exponent more negative. Choice D is incorrect on both counts.

The rate-determining step is the slowest step in a mechanism, which corresponds to the elementary step with the highest activation energy barrier. This step acts as a bottleneck — the overall reaction rate cannot exceed the rate of this step, regardless of how fast all other steps proceed. Choice A is incorrect because the first step is not necessarily the slowest. Choice B describes the fastest step (lowest barrier), not the slowest. Choice D describes termolecular or higher-order collisions, which are statistically rare but unrelated to identifying which step is rate-determining.

Comparing Experiments 1 and 2: [A] doubles while [B] is constant, and the rate doubles (from 2.0 to 4.0 × 10⁻³ M/s), so the reaction is first order in A. Comparing Experiments 1 and 3: [B] triples (0.10 to 0.30 M) while [A] is constant, and the rate increases by a factor of (1.8 × 10⁻²)/(2.0 × 10⁻³) = 9 = 3², so the reaction is second order in B. The rate law is therefore rate = k[A][B]². Choice A would predict only a 3-fold rate increase when B triples. Choice C would predict a 4-fold rate increase when A doubles, which is not observed.

For a first-order reaction, the integrated rate law is [A] = [A]₀e^(−kt). Substituting: [A] = 0.500 × e^(−0.0450 × 30.0) = 0.500 × e^(−1.35) = 0.500 × 0.259 = 0.130 M. This can be confirmed by noting that t½ = 0.693/0.0450 = 15.4 s, so 30.0 s is approximately two half-lives: 0.500 × (0.5)² = 0.125 M, which is close to 0.130 M. Choice C (0.250 M) corresponds to exactly one half-life. Choice D results from calculating for a shorter time period, such as using t = 20 s instead of 30.0 s.

For a first-order reaction, the relationship between rate constant and half-life is k = ln(2)/t½ = 0.6931/25.0 min = 0.0277 min⁻¹. Choice A (0.0139 min⁻¹) would correspond to a half-life of 50.0 minutes, arising from the error of dividing ln(2) by twice the given half-life. Choice C (0.0554 min⁻¹) corresponds to a half-life of 12.5 minutes, half the given value. Choice D reverses the formula entirely, computing t½/ln(2) = 36.1 rather than ln(2)/t½, which would give units of minutes rather than min⁻¹.

According to the Boltzmann distribution, the fraction of molecules with kinetic energy exceeding Ea increases exponentially as Ea decreases. Since Reaction X has a lower activation energy (40.0 vs. 80.0 kJ/mol), proportionally more molecules can successfully react, making Reaction X faster. Choice A incorrectly equates activation energy with energy release; Ea is an energy barrier to be overcome, not ΔH. Choice B ignores the critical role of the Ea term in the Arrhenius equation — when Ea differs, equal values of A do not produce equal rate constants. Choice D is self-contradictory: needing less energy to react means more molecules qualify, giving a faster rate.

The integrated rate law for a first-order reaction is ln[A] = ln[A]₀ − kt, which has the linear form y = b + mx, where y = ln[A], slope = −k, and y-intercept = ln[A]₀. A plot of ln[A] versus time is therefore linear for a first-order reaction. Choice A ([A] vs. time) gives a straight line for a zero-order reaction (integrated law: [A] = [A]₀ − kt). Choice C (1/[A] vs. time) gives a straight line for a second-order reaction (integrated law: 1/[A] = 1/[A]₀ + kt). Choice D has no corresponding standard integrated rate law.

Substituting into the rate law: rate = (3.00 × 10⁻²)(0.200)²(0.500) = (3.00 × 10⁻²)(0.0400)(0.500) = (3.00 × 10⁻²)(0.0200) = 6.00 × 10⁻⁴ M/s. Choice A (3.00 × 10⁻³ M/s) results from treating [X] as first order (forgetting to square it): (3.00 × 10⁻²)(0.200)(0.500) = 3.00 × 10⁻³. Choice C results from an arithmetic error in the intermediate steps. Choice D results from squaring the product [X][Y] together rather than squaring only [X] as specified by the rate law.

A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount, since the transition state energy is lowered relative to both reactants and products. Both rate constants therefore increase by the same factor, and their ratio (which equals the equilibrium constant) remains unchanged. Choice A is incorrect because the reverse activation energy, Ea(reverse) = Ea(forward) − ΔH, is also reduced when the transition state is lowered. Choices B and D are incorrect because a catalyst cannot alter the thermodynamic equilibrium — ΔG° and K are determined by the relative stabilities of reactants and products, not by the pathway between them.

For a second-order reaction, the half-life formula is t½ = 1/(k[A]₀) = 1/((0.500 L·mol⁻¹·s⁻¹)(0.400 mol·L⁻¹)) = 1/0.200 s⁻¹ = 5.00 s. This formula differs fundamentally from the first-order half-life (0.693/k), and a common error is mixing them up. Choice A (1.39 s) results from applying the first-order formula 0.693/(k[A]₀), incorrectly inserting [A]₀ into the denominator of the first-order expression. Choice D (13.9 s) comes from using 0.693/k, the first-order formula, with only the rate constant and no initial concentration.

Collision theory requires exactly two conditions for a productive collision: (1) the collision energy must equal or exceed the activation energy (sufficient energy condition), and (2) the reacting molecules must approach in the correct geometric orientation so that the appropriate bonds can interact and rearrange. Choice B is partially relevant (temperature increases average collision energy) but molecular mass alone is not a distinct requirement in collision theory. Choice C is wrong because activation energy of zero would mean every collision is productive, which is not required — only that energy exceeds Ea. Choice D incorrectly ties all reactions to elevated pressure and complete bond dissociation.

The rate is determined by the slow step: rate = k₂[C][B]. Since C is an intermediate, it must be eliminated using the fast equilibrium from Step 1: K₁ = [C]/([A][B]), giving [C] = K₁[A][B]. Substituting: rate = k₂(K₁[A][B])[B] = k_obs[A][B]², where k_obs = k₁k₂/k₋₁. Choice A (first order in B total) fails to account for B appearing in both steps. Choice B uses the intermediate C directly in the final rate law, which is not acceptable since C is not a reactant in the overall equation. Choice D would require A to appear twice before or in the slow step, which is not the case here.

Using the two-temperature Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). First, ln(7.20 × 10⁻²/1.50 × 10⁻³) = ln(48.0) = 3.871. Then, 1/298 − 1/350 = 0.003356 − 0.002857 = 4.99 × 10⁻⁴ K⁻¹. Solving: Ea = (3.871)(8.314 J·mol⁻¹·K⁻¹)/(4.99 × 10⁻⁴ K⁻¹) = 32,180/4.99 × 10⁻⁴ ≈ 64,500 J/mol = 64.5 kJ/mol. Choice A (32.5 kJ/mol) results from omitting the division by the temperature-difference term. Choice D (89.3 kJ/mol) arises from incorrectly computing the temperature factor using (T₂ − T₁) = 52 directly instead of (1/T₁ − 1/T₂).

When the first step is the rate-determining (slow) step, the overall rate equals the rate of that elementary step: rate = k₁[A][B]. The fast second step does not limit or alter the overall reaction rate. Choice A ([A]²[B]) would arise if both A molecules reacted in the slow step, but Step 1 involves only one A. Choice B uses the intermediate AB in the rate law, which is not acceptable for a final rate law since AB is not an initial reactant in the overall equation. Choice D incorrectly combines both rate constants — when there is a single slow step, only its rate constant appears in the overall rate expression.

Choice A: The slow step is X₂ + Y → XY + X, giving rate = k[X₂][Y], which matches the experimental rate law exactly. Choice B: The fast equilibrium gives [X] = K^(1/2)[X₂]^(1/2), so the slow-step rate = k[X][Y] = kK^(1/2)[X₂]^(1/2)[Y], making the order in X₂ equal to 1/2 — inconsistent with the observed first-order dependence on X₂. Choice C: A single step with stoichiometry X₂ + 2Y would give rate = k[X₂][Y]², which is second order in Y — inconsistent. Choice D: [Y₂] = K[Y]², so rate = k[X₂][Y₂] = kK[X₂][Y]², again second order in Y — inconsistent.

For a second-order reaction, the integrated rate law is 1/[A] = 1/[A]₀ + kt. Substituting: 1/[A] = 1/2.50 + (0.180)(4.00) = 0.400 + 0.720 = 1.120 L·mol⁻¹. Therefore [A] = 1/1.120 = 0.893 M. Choice C (1.25 M) represents [A]₀/2, which would require exactly one half-life to elapse. The actual first half-life is t½ = 1/(k[A]₀) = 1/((0.180)(2.50)) = 2.22 s, so after 4.00 s more than one half-life has passed and the concentration must be below 1.25 M. Choices A and D result from arithmetic errors in setting up or solving the second-order integrated equation.

Converting to Kelvin: T₁ = 298 K, T₂ = 308 K. With k₂/k₁ = 2, the Arrhenius equation gives: ln(2) = (Ea/R)(1/T₁ − 1/T₂). Computing the temperature factor: 1/298 − 1/308 = (308 − 298)/(298 × 308) = 10/91,784 = 1.089 × 10⁻⁴ K⁻¹. Solving: Ea = ln(2) × R/(1.089 × 10⁻⁴) = 0.6931 × 8.314/1.089 × 10⁻⁴ = 5.763/1.089 × 10⁻⁴ ≈ 52,900 J/mol = 52.9 kJ/mol. Choice A (21.5 kJ/mol) results from incorrectly using temperatures in Celsius rather than Kelvin in the denominator. Choice D (76.3 kJ/mol) would correspond to a rate tripling (k₂/k₁ = 3) rather than doubling over this same temperature interval.

Choice A: The slow step NO₂ + F₂ → NO₂F + F gives rate = k[NO₂][F₂], which matches the experimental rate law. Choice B: A single concerted step with stoichiometry 2NO₂ + F₂ gives rate = k[NO₂]²[F₂], which is second order in NO₂ — inconsistent with the observed first-order dependence on NO₂. Choice C: From the fast equilibrium, [F] = K^(1/2)[F₂]^(1/2); the slow-step rate = k[NO₂][F] = kK^(1/2)[NO₂][F₂]^(1/2), making the order in F₂ equal to 1/2 — inconsistent. Choice D: [N₂O₄] = K[NO₂]², so rate = k[N₂O₄][F₂] = kK[NO₂]²[F₂], again second order in NO₂ — inconsistent.

Heterogeneous catalysis occurs at the surface of the solid catalyst, where reactant molecules adsorb, react, and then desorb as products. Doubling the surface area while keeping the total catalyst mass constant provides twice as many active adsorption sites, allowing more reactant molecules to participate simultaneously and increasing the overall reaction rate. Choice A is incorrect because activation energy is a property of the catalytic pathway itself, not a function of how much surface is present — a given catalyst either provides a lower-Ea pathway or it does not. Choice C is incorrect because catalysts cannot alter the thermodynamic equilibrium constant K. Choice D is incorrect because k in the Arrhenius equation depends on Ea and temperature, not on catalyst surface area.

For a zeroth-order reaction, rate = k[A]⁰ = k. Since rate has units of mol·L⁻¹·s⁻¹ and [A]⁰ is dimensionless, k must also have units of mol·L⁻¹·s⁻¹. Choice C (s⁻¹) is the unit for a first-order rate constant. Choice A (L·mol⁻¹·s⁻¹) is the unit for a second-order rate constant.

An intermediate is formed in one elementary step and then used up in a subsequent step, so it does not appear in the overall balanced equation. Choice C describes a catalyst, not an intermediate — a catalyst is regenerated at the end while an intermediate is consumed. Choice A describes a reactant, not an intermediate.

Molecularity refers specifically to the number of species (molecules, atoms, or ions) that come together in a single elementary step. Unimolecular steps involve 1 species, bimolecular involve 2, and termolecular involve 3. Molecularity applies only to elementary steps — it is not the same as overall reaction order, which is determined experimentally.

A catalyst works by providing a different mechanism with a lower activation energy barrier, so a greater fraction of collisions have sufficient energy to react. Choice A is incorrect because catalysts do not raise system temperature. Choice D is incorrect because a catalyst speeds up both the forward and reverse reactions equally, so it does not change the equilibrium position or equilibrium constant.

The instantaneous rate is the rate at one specific moment in time, found graphically as the slope of the tangent to the concentration-time curve at that point. Choice A describes the average rate, not the instantaneous rate. Choice D is incorrect — for most reactions, the instantaneous rate decreases over time as reactant concentrations fall.

If the rate is independent of concentration, the reaction is zeroth order: rate = k[A]⁰ = k. The rate equals the rate constant regardless of [A]. Choice A (first-order) would show the rate doubling when [A] doubles. Choice B (second-order) would show the rate quadrupling when [A] doubles. Choice D is not a standard expression for a simple reaction order.

The defining distinction is phase: a homogeneous catalyst is in the same phase as the reactants (for example, both dissolved in aqueous solution), while a heterogeneous catalyst is in a different phase (for example, a solid metal catalyst with gaseous reactants). Choice C is incorrect because all catalysts — homogeneous and heterogeneous alike — are regenerated and not consumed in the net reaction.

When [A] doubles and the rate quadruples (2² = 4), the reaction is second order in A. When [B] doubles and the rate doubles (2¹ = 2), the reaction is first order in B. Therefore the rate law is rate = k[A]²[B] with an overall order of three. Choice A would predict that doubling [A] only doubles the rate, which contradicts the experimental observation of quadrupling.

The integrated rate law for a zeroth-order reaction is [A] = [A]₀ − kt. Substituting: [A] = 0.800 − (0.0200)(30.0) = 0.800 − 0.600 = 0.200 mol/L. Unlike first-order reactions, zeroth-order concentration decreases linearly with time, so concentration drops by the same absolute amount in each equal time interval. Choice B would result from using only half the elapsed time in the calculation.

Because t₁/₂ = [A]₀/(2k), the half-life of a zeroth-order reaction depends on both [A]₀ and k. As the reaction proceeds and the effective starting concentration of each successive half-life decreases, each successive half-life becomes shorter. This contrasts with first-order reactions (constant half-life) and second-order reactions (half-life increases as [A] decreases). Choice D is incorrect because k is a constant at fixed temperature.

For a second-order reaction, t₁/₂ = 1/(k[A]). As the reaction proceeds and [A] decreases, the denominator gets smaller and t₁/₂ increases — each successive half-life takes longer than the previous one. This is the opposite of zeroth-order behavior. Choice A describes a first-order reaction, which uniquely has a constant half-life independent of concentration.

Because the first step reaches equilibrium rapidly, its equilibrium constant expression can be rearranged to express [I] in terms of original reactant concentrations. This substitution yields a final rate law containing only measurable species. Choice D is incorrect because experimentally observed rate laws cannot contain intermediates — their concentrations are not directly controllable or easily measured as independent variables.

Two peaks on a reaction coordinate diagram correspond to two distinct transition states, and the valley between them represents the energy of a reaction intermediate. This is the signature of a two-step mechanism. Choice A would produce only one peak. Choice C is incorrect — a catalyst lowers the heights of existing barriers but does not create two peaks by splitting a reaction into separate competing pathways on the same diagram.

The unified reaction rate is defined as rate = −(1/2)(Δ[H₂O₂]/Δt) = +(1/1)(Δ[O₂]/Δt). Therefore Δ[O₂]/Δt = (1/2) × 0.400 = 0.200 mol·L⁻¹·s⁻¹. For every 2 moles of H₂O₂ consumed, only 1 mole of O₂ is produced, so the rate of O₂ formation is half the rate of H₂O₂ consumption. Choice A mistakenly doubles rather than halves the rate.

Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂) with T₁ = 298 K and T₂ = 308 K: ln(1.80) ≈ 0.588 = (Eₐ/8.314)(1/298 − 1/308) ≈ (Eₐ/8.314)(1.09 × 10⁻⁴ K⁻¹). Solving gives Eₐ ≈ 44,800 J/mol ≈ 45 kJ/mol, squarely in the 40–50 kJ/mol range. Choice C would require a rate ratio far greater than 1.80 over a 10-degree temperature interval.

The integrated first-order rate law is ln[A] = ln[A]₀ − kt. The y-intercept equals ln[A]₀, so ln[A]₀ = −2.303. Therefore [A]₀ = e^(−2.303) = 0.100 mol/L (since ln(0.100) = −2.303). The slope equals −k, giving k = 0.0500 s⁻¹. Choice C confuses the numerical value of the y-intercept with the concentration itself, ignoring the required antilogarithm step.

A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount, increasing both rate constants by the same factor. Because K = k_forward/k_reverse, and both constants increase equally, K remains unchanged. Choice A is incorrect because ΔH is a thermodynamic state function determined solely by reactants and products — a catalyst cannot change it. Choice C is wrong for the same reason: equilibrium position is unaffected by a catalyst.

For rate = k[A][B], dimensional analysis requires: mol·L⁻¹·s⁻¹ = k × (mol·L⁻¹) × (mol·L⁻¹). Solving: k = (mol·L⁻¹·s⁻¹) / (mol²·L⁻²) = L·mol⁻¹·s⁻¹. Choice B (s⁻¹) is the unit for a first-order rate constant. Choice D (L²·mol⁻²·s⁻¹) corresponds to a third-order rate constant. As a general rule, units of k are L^(n−1)·mol^(1−n)·s⁻¹ where n is the overall reaction order.

The slow step is rate-determining, so the rate law is derived from its stoichiometry: rate = k[A][B]. Species C is an intermediate (produced in Step 1, consumed in Step 2). Adding both steps and canceling C gives the overall reaction: A + 2B → D + E. Choice B is incorrect because the fast second step does not contribute to the rate law. Choice C is wrong because intermediates cannot appear in a valid final rate law — they must be expressed in terms of original reactants.

Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂): the ratio k₂/k₁ = (8.00 × 10⁻²)/(2.50 × 10⁻³) = 32, so ln(32) ≈ 3.466. The temperature factor: 1/300 − 1/350 = (350 − 300)/(300 × 350) = 50/105,000 ≈ 4.76 × 10⁻⁴ K⁻¹. Therefore Eₐ = (3.466 × 8.314)/(4.76 × 10⁻⁴) ≈ 60,500 J/mol = 60.5 kJ/mol. Choice A would result from incorrectly computing the ratio as k₂/k₁ = 16 rather than 32.

Comparing Experiments 1 and 2: [A] doubles and the rate doubles → first order in A. Comparing Experiments 1 and 3: [B] doubles and the rate quadruples (2.00 × 10⁻³ to 8.00 × 10⁻³) → second order in B. Rate law: rate = k[A][B]². Solving for k using Experiment 1: k = (2.00 × 10⁻³) / (0.100 × (0.100)²) = (2.00 × 10⁻³) / (1.00 × 10⁻³) = 2.00 L²·mol⁻²·s⁻¹. Choice C reverses the orders of A and B, contradicting the data.

The rate-determining step gives rate = k₂[N₂O₂][O₂]. Since N₂O₂ is an intermediate, the fast-step equilibrium is used: K₁ = [N₂O₂]/[NO]², so [N₂O₂] = K₁[NO]². Substituting: rate = k₂K₁[NO]²[O₂] = k_obs[NO]²[O₂]. Choice B is the preliminary rate expression before eliminating the intermediate — it cannot be directly tested because [N₂O₂] is not an independently controllable concentration. Choice A would require a mechanism where NO and O₂ collide directly in the slow step.

The slope of an Arrhenius plot equals −Eₐ/R, so a steeper (more negative) slope indicates higher Eₐ. The y-intercept equals ln A, so identical y-intercepts mean identical pre-exponential factors A. Choice A incorrectly infers a different A from the slope. Choice D is wrong because if A is identical and Eₐ(P) > Eₐ(Q), then k_P = A·e^(−Eₐ_P/RT) < A·e^(−Eₐ_Q/RT) = k_Q at every finite temperature without exception — a larger negative exponent always yields a smaller value.

The new rate = k(2[A])²(¼[B]) = k · 4[A]² · (1/4)[B] = k[A]²[B] = original rate. Doubling [A] multiplies the rate by 2² = 4, and reducing [B] to one-fourth multiplies the rate by 1/4. These two effects exactly cancel: 4 × (1/4) = 1, so the rate is unchanged. This problem illustrates that simultaneous concentration changes can produce counterintuitive results when the exponents in the rate law differ.

When [B] is in large excess, it remains essentially constant throughout the reaction, so rate = k[B]₀[A] = k'[A], where k' = k[B]₀ = 5.00 × 1.00 = 5.00 s⁻¹. The reaction then behaves as first order in A with this effective rate constant. Choice A would result from multiplying k by [A]₀ instead of [B]₀. This flooding technique is widely used experimentally to isolate and measure individual reaction orders one at a time.

The instantaneous rate is the rate at a specific moment, found by taking the slope of the tangent line to the concentration-time curve at that point. Choice A describes the average rate (delta-concentration divided by delta-time over an interval). Choices C and D are approximations, not true instantaneous values.

For a zero-order reaction, rate = k[A]⁰ = k. The rate equals the rate constant and has no dependence on reactant concentration. Choice A describes first-order behavior, and choice B describes second-order behavior. Choice D is incorrect because the rate remains constant regardless of [A].

The integrated rate law for a zero-order reaction is [A] = [A]₀ − kt, which has the linear form y = b + mx with y = [A] and x = t. A plot of [A] vs time is therefore linear with slope −k. Choice A (ln[A] vs t) is the linear diagnostic for first-order reactions, and choice B (1/[A] vs t) is linear for second-order reactions.

For a first-order reaction, rate = k[A]. Since rate has units of mol·L⁻¹·s⁻¹ and [A] has units of mol·L⁻¹, k must carry units of s⁻¹ to balance the equation. mol·L⁻¹·s⁻¹ corresponds to zero-order, L·mol⁻¹·s⁻¹ to second-order, and L²·mol⁻²·s⁻¹ to third-order rate constants.

Activation energy is the minimum energy threshold that colliding molecules must meet or exceed for a productive collision that converts reactants into products. Choice A describes bond formation energy, which is a component of enthalpy. Choice C describes delta-H (enthalpy change), and choice D describes mean thermal kinetic energy, which is related to temperature but is not activation energy.

A catalyst accelerates a reaction by offering an alternative mechanism with a lower activation energy barrier, allowing a greater fraction of molecular collisions to be successful. Crucially, the catalyst is regenerated at the end of the mechanism and does not change delta-H or the equilibrium position. Raising temperature and increasing concentration also increase rates, but those are not the mechanisms by which a catalyst operates.

The rate-determining step is the slowest elementary step, acting as the bottleneck of the entire reaction — the overall reaction can proceed no faster than this step. It is not necessarily the first step, the most exothermic step, or the step involving the most molecules. Its identity is determined by which step has the highest activation energy relative to the preceding energy level on the reaction coordinate diagram.

Collision theory requires two simultaneous conditions for a productive collision: the collision energy must equal or exceed the activation energy, and the molecules must approach each other with the correct orientation so that reactive sites make proper contact. Even a high-energy collision fails if the molecular geometry is wrong. The other choices describe factors that are not part of collision theory.

For an elementary step, the rate law is written directly from the stoichiometric coefficients because the step occurs exactly as written at the molecular level. One molecule of A and two molecules of B must collide simultaneously, giving rate = k[A][B]². This direct connection between stoichiometry and rate law is valid ONLY for individual elementary steps — it cannot be applied to overall balanced equations.

A catalyst lowers activation energy by the same absolute amount for both the forward and reverse reactions. Delta-H (the energy difference between reactants and products) is a fixed thermodynamic property unaffected by a catalyst. If the forward barrier drops by 25 kJ/mol, the reverse barrier must also drop by 25 kJ/mol to preserve the same delta-H. This is why a catalyst accelerates both directions equally and does not shift the equilibrium position.

The integrated rate law for zero-order is [A] = [A]₀ − kt. Setting [A] = [A]₀/2 at the half-life gives [A]₀/2 = [A]₀ − kt₁/₂, so t₁/₂ = [A]₀/(2k). This contrasts with first-order (choice A, t₁/₂ = 0.693/k, which is constant) and second-order (choice C, t₁/₂ = 1/(k[A]₀), which increases over time). For zero-order, the half-life depends on the initial concentration and decreases as the reaction progresses.

The pre-exponential factor A (also called the frequency factor) encodes both the collision frequency and the steric factor — the fraction of collisions that approach with the correct geometric orientation to be productive. The separate exponential term e^(−Eₐ/RT) accounts for the fraction of collisions with sufficient energy, which is the Boltzmann factor described in choice A. Choices B and C do not correctly describe the physical meaning of A.

When [B] is held in large excess and remains approximately constant, the rate simplifies to rate ≈ k[B][A] = k_obs[A] where k_obs = k[B]. This creates apparent first-order behavior known as a pseudo-first-order condition. The true rate law is still second-order overall; the simplified kinetics arise from the experimental design of flooding one reactant. This technique is widely used to isolate and measure the kinetic contribution of individual reactants.

The Arrhenius equation shows rate constants increase exponentially with temperature through the e^(−Eₐ/RT) term. Even a modest temperature increase significantly shifts the Maxwell-Boltzmann distribution, raising the fraction of molecules with energy at or above Eₐ. Collision frequency increases only as the square root of T (choice B), far too small to account for a doubling of rate. The activation energy itself does not change with temperature (choice A).

For a second-order reaction, t₁/₂ = 1/(k[A]). As [A] decreases over the course of the reaction, the half-life increases — a diagnostic signature of second-order kinetics. First-order reactions (choice A) have a constant half-life independent of concentration (t₁/₂ = 0.693/k). Zero-order half-lives (choice B) actually decrease over time because t₁/₂ = [A]/(2k), so as [A] falls, the half-life shortens.

An elementary step must occur exactly as written at the molecular level. A termolecular step requires three molecules to collide at the same instant and location, with proper orientations, and with sufficient combined energy — a statistically very improbable event compared to bimolecular collisions. As a result, nearly all proposed mechanisms are constructed from unimolecular and bimolecular elementary steps, which are far more likely.

Both intermediates and catalysts appear in the mechanism but not in the net balanced equation. The distinction is directional: an intermediate is first produced as a product of an early step, then consumed as a reactant in a later step. A catalyst is consumed (used up) in an early step but regenerated in a later step. Choice C is incorrect — intermediates do not appear in the final experimentally derived rate law because they are replaced using equilibrium or steady-state expressions.

Adding the two steps and canceling the intermediate I gives the overall reaction: 2A + B → 2P. Since Step 1 is the rate-determining step and involves one A and one B, the rate law is rate = k[A][B], first-order in each reactant. Choice B incorrectly reads the rate law from the overall stoichiometric coefficients, giving k[A]²[B] — a common error. Rate laws must be derived from the mechanism, not the balanced equation. Choice D incorrectly retains the intermediate I in the rate law.

Using the two-temperature Arrhenius equation: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). Here k₂/k₁ = 0.240/0.0300 = 8.00, so ln(8.00) = 2.079. The temperature factor: 1/300 − 1/340 = (340 − 300)/(300 × 340) = 40/102000 = 3.92 × 10⁻⁴ K⁻¹. Therefore Eₐ = (2.079 × 8.314)/(3.92 × 10⁻⁴) = 17.29/3.92 × 10⁻⁴ ≈ 44,100 J/mol = 44.1 kJ/mol. Choice B results from dividing by 2 in error; choice C from doubling the result.

Comparing Experiments 1 and 2: [A] doubles while [B] is constant, and the rate doubles — first-order in A. Comparing Experiments 1 and 3: [B] triples (0.100 to 0.300 mol/L) while [A] is constant, and the rate increases by a factor of 9 (1.08 × 10⁻²/1.20 × 10⁻³). Since 3ⁿ = 9 gives n = 2, the reaction is second-order in B. The rate law is rate = k[A][B]². Choice A reverses the orders, and choice C incorrectly identifies both reactants as first-order.

Using the second-order integrated rate law: 1/[A] − 1/[A]₀ = kt. Substituting: 1/0.100 − 1/0.500 = 10.0 − 2.00 = 8.00 mol⁻¹·L. Setting this equal to kt: 0.200 × t = 8.00, giving t = 40.0 s. Choice A (8.05 s) results from incorrectly applying the first-order integrated law: ln([A]₀/[A]) = kt gives ln(5)/0.200 = 8.05 s. This common error arises when the reaction order is misidentified before selecting the integrated rate law.

The relationship between activation energies and reaction enthalpy is: delta-H = Eₐ(forward) − Eₐ(reverse). Rearranging: Eₐ(reverse) = Eₐ(forward) − delta-H. For the catalyzed case: Eₐ(rev, cat) = 80 − (−50) = 130 kJ/mol. A catalyst does not change delta-H — only the heights of the energy barriers. Choice A (80 kJ/mol) wrongly equates forward and reverse catalyzed barriers. Choice D (170 kJ/mol) is the reverse barrier for the uncatalyzed reaction, not the catalyzed one.

From the slow step: rate = k₂[A*][B]. From the fast equilibrium: Keq = [A*]/[A], so [A*] = Keq[A]. Substituting gives rate = k₂Keq[A][B] = k_obs[A][B], which matches the experimental rate law. The mechanism is fully consistent. Choice A is incorrect — intermediates routinely appear in rate law derivations; they are simply eliminated using equilibrium or steady-state expressions before writing the final rate law. Choice C reverses the logic: the slow step controls the rate, not the fast equilibrium.

Rate laws are determined experimentally and reflect the reaction mechanism, not the overall stoichiometry. When the observed rate law (rate = k[NO₂][F₂]) differs from what stoichiometry would predict (rate = k[NO₂]²[F₂]), it indicates the reaction proceeds through multiple elementary steps rather than one concerted collision. Only for individual elementary steps can stoichiometric coefficients be used directly to write the rate law. Choice B is incorrect — equilibrium constant expressions and kinetic rate laws are fundamentally different quantities.

Setting k_R = k_S: A_R·e^(−Eₐ_R/RT) = A_S·e^(−Eₐ_S/RT). Rearranging: A_R/A_S = e^((Eₐ_R − Eₐ_S)/RT). Substituting: (10¹²/10¹³) = 0.10 = e^((50000 − 80000)/RT) = e^(−30000/RT). Taking the natural log: ln(0.10) = −2.303 = −30000/(8.314 × T). Solving: T = 30000/(8.314 × 2.303) = 30000/19.14 ≈ 1570 K. Choice A (785 K) results from an arithmetic error of dividing the numerator by 2. Choice D (390 K) results from multiplying the denominator by an extra factor of 4.

As a reactant is consumed, its concentration decreases over time, making d[A]/dt inherently negative. Multiplying by −1 yields a positive rate value, which is the universal convention for expressing reaction rates. Choice A is wrong because concentrations are always positive values. Choice C confuses kinetic sign conventions with thermodynamic ones (exothermic/endothermic). Choice D is incorrect because rate constants are always positive.

A catalyst works by offering a different mechanism whose transition state has a lower energy barrier, so a greater fraction of molecular collisions have sufficient energy to be productive. Catalysts do not raise temperature (choice A), change reactant concentrations (choice C), or alter the equilibrium constant K (choice D). Because the catalyst lowers activation energy equally for the forward and reverse reactions, K is unchanged.

For a first-order reaction, t₁/₂ = 0.693/k (where 0.693 = ln 2). Because k is a constant and [A]₀ does not appear in this expression, the half-life is independent of how much reactant remains and stays the same throughout the reaction. Choice A gives the second-order half-life formula t₁/₂ = 1/(k[A]), not the first-order formula. Choices B and D incorrectly include [A]₀, which would make the half-life concentration-dependent.

Collision theory requires that a productive collision have (1) energy at least equal to the activation energy Ea and (2) molecules oriented so that reactive sites interact correctly. Choice B confuses macroscopic conditions with molecular-level requirements. Choice C — molecular mass and collision duration — are not criteria in collision theory. Choice D conflates the energy barrier (which must be overcome, not minimized to zero) with concentration, which affects collision frequency but not per-collision success.

Integrating the first-order differential rate law −d[A]/dt = k[A] yields ln[A]t = ln[A]₀ − kt, which predicts a straight line when ln[A] is plotted against time. Choice A is the zero-order integrated rate law. Choice C is the second-order integrated rate law. Choice D has the wrong sign in the exponent; the correct exponential form is [A]t = [A]₀ × e^(−kt), meaning concentration decays toward zero, not grows without limit.

For a first-order reaction, rate (M·s⁻¹) = k × [A] (M), so k = rate/[A] = (M·s⁻¹)/M = s⁻¹. Choice A (M·s⁻¹) is the unit of reaction rate itself, not the rate constant. Choice B (M⁻¹·s⁻¹) is the correct unit for a second-order rate constant. Choice D (M⁻²·s⁻¹) would correspond to a third-order rate constant.

The activation energy is the energy barrier separating reactants from the transition state; only molecules with at least this much energy can react upon collision. Choice A describes the magnitude of the reaction's exothermicity, not Ea. Choice C describes average kinetic energy, which is related to temperature (3RT/2 per mole for an ideal gas) but is not Ea. Choice D approximately describes the enthalpy of reaction (ΔH), which can be positive or negative, whereas Ea is always positive.

Doubling [A] (first-order in A) multiplies the rate by 2¹ = 2. Doubling [B] (second-order in B) multiplies the rate by 2² = 4. Because both changes apply simultaneously, the overall factor is 2 × 4 = 8. Choice A (factor of 2) would only apply if [A] alone were doubled with B unchanged. Choice B (factor of 4) would result from doubling only [B]. Choice D (factor of 16) would require second-order dependence on both A and B.

The Maxwell-Boltzmann distribution shows that the fraction of molecules with energy greater than or equal to Ea increases exponentially with temperature. Even a 10 K rise can shift many more molecules above the threshold, dramatically increasing productive collision frequency and thus the rate constant k. Choice A is incorrect because Ea is a property of the reaction pathway determined by bond strengths and geometry — temperature does not change it. Choice D confuses kinetics with thermodynamics; K can also vary with temperature via the van't Hoff equation, but that is a separate effect from rate acceleration.

A catalyst lowers the activation energy by the same amount for both the forward and reverse reactions because it provides the same alternative transition state for both directions. As a result, both rates increase equally, the system stays at the same equilibrium composition, and K is unchanged. Choices A and B are wrong because catalysts do not alter thermodynamic quantities like K or the equilibrium position. Choice D is incorrect because the reverse rate also increases; catalysts simply allow equilibrium to be reached more quickly if the system is disturbed.

For an overall second-order reaction, rate (M·s⁻¹) = k × [concentration]² (M²), so k = (M·s⁻¹)/M² = M⁻¹·s⁻¹. Choice A (s⁻¹) is the unit for a first-order rate constant. Choice B (M·s⁻¹) represents the unit of reaction rate itself, not the rate constant. Choice D (M⁻²·s⁻¹) would correspond to a third-order rate constant.

Substituting into the integrated rate law: 1/[A]t = (0.050)(30.0) + 1/(0.200) = 1.50 + 5.00 = 6.50 M⁻¹. Therefore [A]t = 1/6.50 = 0.154 M. Choice A (0.100 M) would result from incorrectly halving [A]₀ as if this were a half-life calculation. Choice C could arise from an algebraic error when computing 1/[A]₀. Choice D would indicate no reaction occurred at all.

The slow (rate-determining) step gives rate = k₂[XY][Z]. However, XY is an intermediate and cannot appear in the final rate law. Using the fast equilibrium: Keq = [XY]/([X][Y]), so [XY] = Keq[X][Y]. Substituting: rate = k₂Keq[X][Y][Z] = k[X][Y][Z]. Choice A is the raw rate expression for the slow step before eliminating the intermediate. Choice C omits Z from the rate law, ignoring its participation in the rate-determining step. Choice D has only one species, which is inconsistent with a bimolecular slow step.

An elementary step describes a single molecular event (such as a bimolecular collision or a unimolecular decomposition). Because it is an actual molecular act, the rate law for that step can be written directly from the stoichiometric coefficients of the species involved — unlike an overall balanced equation, whose rate law must be determined experimentally. Choice A confuses the rate-determining step (slowest, highest activation energy) with the definition of elementary. Choice C describes a unimolecular step, which is one type of elementary step but not the definition of the term. Choice D describes a pre-equilibrium step, which may or may not be elementary.

Adding an inert gas at constant volume and temperature does not change the partial pressures or molar concentrations of A and B. Since the collision frequency between A and B is unaffected, so is the rate. Choice A increases [A], raising the rate directly. Choice B compresses all gases, increasing the concentrations of both A and B and raising the rate. Choice D (catalyst) lowers the activation energy, increasing the rate constant k.

For a first-order reaction, t₁/₂ = 0.693/k — independent of concentration and therefore constant. For a second-order reaction, t₁/₂ = 1/(k[A]), which increases as [A] decreases over time, meaning successive half-lives become progressively longer. Choice A reverses the two behaviors. Choice C is incorrect because the first-order half-life never changes. Choice D is not a general rule; the relative magnitudes depend on the specific values of k and [A]₀.

When [A] doubles from 0.10 to 0.20 M, the rate doubles (from 2.5 × 10⁻³ to 5.0 × 10⁻³ M/s), confirming first-order dependence (rate ∝ [A]¹). The rate constant is k = rate/[A] = (2.5 × 10⁻³ M/s)/(0.10 M) = 2.5 × 10⁻² s⁻¹. Choice B (second order) is wrong because doubling [A] would quadruple the rate for a second-order reaction. Choice C (zero order) is ruled out because the rate clearly changes with concentration. Choice D gives incorrect units for k of a first-order reaction (k must have units of s⁻¹, not M/s).

Rewriting the Arrhenius equation in linear form: ln k = −(Ea/R)(1/T) + ln A. Comparing with y = mx + b, the slope m = −Ea/R and the y-intercept equals ln A. Choice B has the correct magnitude but the wrong sign; the plot has a negative slope because k increases as T increases (i.e., as 1/T decreases). Choice C omits the gas constant R from the denominator. Choice D is the y-intercept of the Arrhenius plot, not the slope.

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂): the ratio k₂/k₁ = 8.00 × 10⁻³ / 2.00 × 10⁻³ = 4, so ln(4) = 1.386. The temperature factor: 1/298 − 1/328 = (328 − 298)/(298 × 328) = 30/97,744 = 3.07 × 10⁻⁴ K⁻¹. Therefore Ea = (1.386 × 8.314) / (3.07 × 10⁻⁴) = 11.52 / 3.07 × 10⁻⁴ = 37,500 J/mol = 37.5 kJ/mol. Choice B (52.3 kJ/mol) would result from using ln(8) instead of ln(4), incorrectly treating the ratio as 8 rather than 4. Choice C underestimates Ea and could arise from an error in computing the temperature difference term.

The slow step gives rate = k₂[O][O₃]. Since O is an intermediate, eliminate it using the fast equilibrium: Keq = [O₂][O]/[O₃], so [O] = Keq[O₃]/[O₂]. Substituting: rate = k₂ × (Keq[O₃]/[O₂]) × [O₃] = k₂Keq[O₃]²/[O₂] = k[O₃]²/[O₂]. Note that O₂ appears in the denominator because it is a product of Step 1 and suppresses the equilibrium concentration of O. Choice C is the raw rate expression for the slow step before the intermediate is eliminated. Choice A ignores the squared concentration of O₃ and the O₂ denominator. Choice D has O₂ in the numerator rather than the denominator, which reverses the equilibrium substitution.

First, find the half-life: t₁/₂ = 0.693/k = 0.693/0.0231 = 30.0 min. Then 120 min represents 120/30 = 4 half-lives, so [A] = [A]₀ × (1/2)⁴ = 0.800 × (1/16) = 0.0500 M. This can be verified using the integrated law: ln[A] = ln(0.800) − (0.0231)(120) = −0.223 − 2.772 = −2.995, giving [A] = e^(−2.995) ≈ 0.0500 M. Choice B (0.100 M) corresponds to only 3 half-lives (90 min). Choice C (0.0250 M) corresponds to 5 half-lives (150 min). Choice D (0.200 M) corresponds to only 2 half-lives (60 min).

The slow step gives rate = k₂[N₂O₂][O₂]. N₂O₂ is an intermediate, so use the equilibrium from Step 1: Keq = [N₂O₂]/[NO]², giving [N₂O₂] = Keq[NO]². Substituting: rate = k₂Keq[NO]²[O₂] = k[NO]²[O₂], which matches the experimental rate law exactly. Choice A is incorrect — intermediates do appear in elementary step rate expressions; they are eliminated algebraically using equilibrium or steady-state conditions. Choice C contains an arithmetic error in the substitution. Choice D is false; many valid mechanisms place the slow step second or later in the sequence.

Going from 0.500 M to 0.125 M requires two sequential half-lives. First half-life: t₁ = 1/(0.400 × 0.500) = 5.00 s, reducing [A] from 0.500 to 0.250 M. Second half-life: t₂ = 1/(0.400 × 0.250) = 10.0 s, reducing [A] from 0.250 to 0.125 M. Total = 5.00 + 10.0 = 15.0 s. This result is confirmed by the integrated rate law: 1/0.125 − 1/0.500 = 8.00 − 2.00 = 6.00 = (0.400)t, giving t = 15.0 s. Choice A gives only the first half-life. Unlike first-order reactions, successive second-order half-lives grow longer as concentration falls.

When v = Vmax/2, the Michaelis-Menten equation gives [S] = Km, so Km = 5.0 × 10⁻³ M. At [S] = 1.0 × 10⁻² M: v = Vmax(0.010)/(0.005 + 0.010) = Vmax(0.010/0.015) = (2/3)Vmax. Choice A is incorrect; doubling [S] to 2Km does not fully saturate the enzyme — full saturation requires [S] >> Km. Choice C is wrong because the rate did increase when [S] was raised. Choice D incorrectly invokes substrate inhibition, which is a distinct kinetic phenomenon not described in this scenario.

Since both reactions share the same A: kP/kQ = exp(−Ea(P)/RT) / exp(−Ea(Q)/RT) = exp((Ea(Q) − Ea(P))/RT) = exp((60,000 − 40,000)/(8.314 × 300)) = exp(20,000/2494) = exp(8.02) ≈ 3030. The 20 kJ/mol difference in activation energy produces a rate ratio of roughly 3000 at 300 K, illustrating the exponential sensitivity of reaction rates to Ea. Choice C (approximately 55) would correspond to a smaller energy gap of about 10 kJ/mol. Choice D (approximately 7.5) drastically underestimates the exponential amplification inherent in the Arrhenius equation.

For a zero-order reaction, the rate law is rate = k[A]⁰ = k. Because the concentration term is raised to the zero power it equals 1, and the rate depends only on the rate constant k. The rate is therefore constant throughout the reaction and does not change as [A] decreases. Choice B describes first-order behavior where rate = k[A], and choice C describes second-order behavior where rate = k[A]².

Collision theory states that two requirements must be met for a productive collision: (1) the collision energy must equal or exceed the activation energy, and (2) the molecules must approach each other with the correct spatial orientation so reactive bonds or atoms are properly aligned. Choice B is wrong because low-energy collisions simply bounce off without forming products. Choices C and D are not requirements stated in collision theory.

The slowest elementary step is called the rate-determining step (RDS) because it acts as a bottleneck: no matter how fast the other steps are, the overall reaction cannot proceed faster than this step allows. This is analogous to a multi-lane highway merging into one lane — traffic moves at the speed of the bottleneck. The fastest step (choice B) has no limiting effect on the overall rate; the first step (choice A) is only the RDS if it also happens to be the slowest.

The integrated rate law for a zero-order reaction is [A] = [A]₀ - kt, which has the linear form y = b + mx. Plotting [A] on the y-axis versus time on the x-axis yields a straight line with slope -k and y-intercept [A]₀. By contrast, a plot of ln[A] versus time is linear for a first-order reaction (choice A), and a plot of 1/[A] versus time is linear for a second-order reaction (choice B). Choice D is not a standard integrated rate law linearization.

Comparing trials 1 and 2: [A] doubles (0.100 to 0.200 M) and the rate increases by a factor of (1.44 × 10⁻²)/(3.60 × 10⁻³) = 4.00. Since 2ⁿ = 4 implies n = 2, the reaction is second-order. This is confirmed by trials 2 and 3: doubling [A] again produces another fourfold rate increase. Choice B (first-order) would require the rate to double — not quadruple — when concentration doubles. Choice D (third-order) would require an eightfold rate increase when concentration doubles.

A catalyst speeds up a reaction by providing a new mechanistic pathway with a lower activation energy (Ea). According to the Boltzmann distribution, a lower Ea means a larger fraction of molecular collisions have sufficient energy to react, dramatically increasing the rate constant. Crucially, a catalyst does not change ΔG° or the equilibrium constant — it speeds up both the forward and reverse reactions equally, so choice A is wrong. It also does not heat the system (choice B) or generate new reactants (choice D).

For a first-order reaction, t₁/₂ = 0.693/k = 0.693/(1.386 × 10⁻² min⁻¹) = 50.0 min. A defining feature of first-order kinetics is that the half-life is constant and independent of concentration — each successive half-life period reduces the remaining concentration by exactly half regardless of how much reactant is present. In contrast, the half-life of a second-order reaction increases as concentration decreases, while the half-life of a zero-order reaction decreases as concentration decreases.

The rate-determining step (slow step 2) gives an initial expression of rate = k₂[XY][Z]. However, XY is a reaction intermediate and cannot appear in the final rate law because intermediates are not measurable reactants. Using the fast equilibrium from step 1: K₁ = [XY]/([X][Y]), so [XY] = K₁[X][Y]. Substituting: rate = k₂K₁[X][Y][Z] = k_obs[X][Y][Z]. This pre-equilibrium approximation is a core technique for deriving rate laws from mechanisms containing intermediates. Choice B is wrong because it retains the intermediate XY rather than expressing it in terms of starting materials.

Using the two-temperature Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Substituting ln(2k₁/k₁) = ln(2) = 0.6931, Ea = 75,000 J/mol, T₁ = 298 K: 0.6931 = (75,000/8.314)(1/298 - 1/T₂) = 9,020 × (1/298 - 1/T₂). Solving: 1/298 - 1/T₂ = 7.68 × 10⁻⁵, so 1/T₂ = 3.356 × 10⁻³ - 7.68 × 10⁻⁵ = 3.279 × 10⁻³, giving T₂ ≈ 305 K. This result highlights that only a 7 K temperature increase is needed to double the rate constant when Ea is moderately large — a consequence of the exponential dependence of k on temperature in the Arrhenius equation.

Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂) with T₁ = 298 K, T₂ = 323 K: ln(6.20 × 10⁻²/1.00 × 10⁻²) = ln(6.20) = 1.825. The temperature factor: 1/298 - 1/323 = 3.356 × 10⁻³ - 3.096 × 10⁻³ = 2.60 × 10⁻⁴ K⁻¹. Therefore Ea = R × ln(k₂/k₁)/(1/T₁ - 1/T₂) = (8.314)(1.825)/(2.60 × 10⁻⁴) = 58,400 J/mol ≈ 58.4 kJ/mol. A common error producing choice D (71.2 kJ/mol) is using Celsius temperatures directly instead of converting to Kelvin — always convert to Kelvin before applying the Arrhenius equation.