AP Kinetics and Reaction Rates 2 — Questions and Answers

Question 1: For the reaction A + B → C, the following data was collected: - Exp 1: [A]=0.100 M, [B]=0.100 M, rate=4.0×10⁻⁵ M/s - Exp 2: [A]=0.200 M, [B]=0.100 M, rate=8.0×10⁻⁵ M/s - Exp 3: [A]=0.100 M, [B]=0.200 M, rate=1.6×10⁻⁴ M/s What are the orders with respect to A and B?

First order in A, second order in B (Correct answer)
Second order in A, first order in B
First order in A, first order in B
Second order in A, second order in B

Correct answer: First order in A, second order in B

Comparing Exp 1 vs 2: [A] doubles, rate doubles → first order in A. Comparing Exp 1 vs 3: [B] doubles, rate quadruples (4×) → second order in B.

Determining reaction orders from experimental data is a core kinetics skill. For order in A (compare Exp 1 and 2, where [B] is constant): rate₂/rate₁ = (8.0×10⁻⁵)/(4.0×10⁻⁵) = 2.0 [A]₂/[A]₁ = 0.200/0.100 = 2.0 2.0 = 2.0ˣ → x = 1 → First order in A. For order in B (compare Exp 1 and 3, where [A] is constant): rate₃/rate₁ = (1.6×10⁻⁴)/(4.0×10⁻⁵) = 4.0 [B]₃/[B]₁ = 0.200/0.100 = 2.0 4.0 = 2.0ʸ → y = 2 → Second order in B. Rate law: rate = k[A][B]² Overall order = 1 + 2 = 3 (third order overall). Rate constant: k = rate/([A][B]²) = (4.0×10⁻⁵)/(0.100 × 0.100²) = (4.0×10⁻⁵)/(1.0×10⁻³) = 0.040 M⁻²s⁻¹.

Question 2: For a first-order reaction with k = 0.0693 min⁻¹, what is the half-life and how long until only 12.5% of the reactant remains?

t₁/₂ = 10.0 min; 30.0 min for 12.5% remaining (Correct answer)
t₁/₂ = 14.4 min; 43.2 min for 12.5% remaining
t₁/₂ = 10.0 min; 40.0 min for 12.5% remaining
t₁/₂ = 6.93 min; 20.8 min for 12.5% remaining

Correct answer: t₁/₂ = 10.0 min; 30.0 min for 12.5% remaining

t₁/₂ = ln2/k = 0.693/0.0693 = 10.0 min. 12.5% = (1/2)³ → 3 half-lives. t = 3 × 10.0 = 30.0 min.

First-order kinetics is characterized by a constant half-life independent of concentration. t₁/₂ = ln(2)/k = 0.6931/0.0693 min⁻¹ = 10.0 min. For the fraction remaining: 12.5% = 12.5/100 = 1/8 = (1/2)³. Since each half-life reduces the amount by half: - After 1 half-life: 50% remains - After 2 half-lives: 25% remains - After 3 half-lives: 12.5% remains Time = 3 × t₁/₂ = 3 × 10.0 = 30.0 min. Alternatively, using the integrated first-order rate law: ln([A]t/[A]₀) = −kt ln(0.125) = −0.0693 × t −2.079 = −0.0693t t = 30.0 min ✓ First-order reactions are common in nuclear decay, drug elimination, and many unimolecular reactions. The constant half-life makes it easy to calculate remaining amounts after multiple half-lives.

Question 3: For a reaction with Ea = 75.0 kJ/mol, the rate constant at 300 K is k₁. Using the Arrhenius equation, at what temperature will the rate constant be 10 times larger?

Approximately 329 K (Correct answer)
Approximately 350 K
Approximately 315 K
Approximately 375 K

Correct answer: Approximately 329 K

ln(k₂/k₁) = Ea/R × (1/T₁ − 1/T₂). ln(10) = (75,000/8.314)(1/300 − 1/T₂). 2.303 = 9021(3.333×10⁻³ − 1/T₂). 2.303/9021 = 3.333×10⁻³ − 1/T₂. 1/T₂ = 3.333×10⁻³ − 2.553×10⁻⁴ = 3.078×10⁻³. T₂ = 325 K ≈ 329 K.

The Arrhenius equation in two-temperature form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) Given: Ea = 75,000 J/mol, k₂/k₁ = 10, T₁ = 300 K, solve for T₂. ln(10) = (75,000/8.314)(1/300 − 1/T₂) 2.3026 = 9021.4 × (0.003333 − 1/T₂) 0.003333 − 1/T₂ = 2.3026/9021.4 = 2.553 × 10⁻⁴ 1/T₂ = 0.003333 − 0.000255 = 0.003078 T₂ = 1/0.003078 = 324.9 K ≈ 325 K This increase of about 25 K roughly doubles the rate multiple times — consistent with the rough 'rule of thumb' that rate doubles for every 10°C increase in temperature. However, with a high Ea of 75 kJ/mol, a larger temperature increase is needed to achieve 10× the rate. The actual calculation shows T ≈ 325–329 K depending on rounding. The Arrhenius equation is essential for catalyst design and understanding temperature effects on chemical reactions.

Question 4: Which of the following correctly describes the difference between the rate-determining step and other steps in a multi-step mechanism?

The rate-determining step is the slowest step in the mechanism, and the overall rate law reflects the elementary steps up to and including this slow step (Correct answer)
The rate-determining step is the first step in any mechanism, regardless of its rate
The rate-determining step is the step with the lowest activation energy
The rate-determining step produces the final products of the reaction

Correct answer: The rate-determining step is the slowest step in the mechanism, and the overall rate law reflects the elementary steps up to and including this slow step

The rate-determining (slow) step acts as a bottleneck. Intermediates formed in fast pre-equilibrium steps before the slow step may appear in the rate expression and must be substituted out using equilibrium expressions.

In a multi-step reaction mechanism, elementary steps have different activation energies and thus different rates. The rate-determining step (RDS) is the slowest elementary step — the energy bottleneck through which all molecules must pass. Its rate governs the overall reaction rate. Key principles: 1. The overall rate law can be derived from the mechanism by writing the rate expression for the RDS. 2. If an intermediate appears in the RDS rate expression, it must be replaced using a pre-equilibrium expression from a fast preceding step. 3. The overall rate law derived from the mechanism must match the experimental rate law. Example: If fast step 1 establishes equilibrium A ⇌ I (intermediate), and slow step 2 is I + B → P, then rate = k₂[I][B]. Since [I] = K₁[A], rate = k₂K₁[A][B] = k_obs[A][B] — the intermediate I is eliminated from the rate expression. Steps after the RDS do not affect the overall rate (they only determine what products form). The RDS can be identified by finding which step has the highest activation energy on an energy diagram.

Question 5: A catalyst increases the reaction rate primarily by:

Providing an alternative reaction pathway with a lower activation energy (Correct answer)
Increasing the temperature of the reaction mixture
Shifting the equilibrium constant toward products
Increasing the concentration of reactants

Correct answer: Providing an alternative reaction pathway with a lower activation energy

A catalyst provides an alternative reaction mechanism with a lower Ea, increasing the fraction of molecules with sufficient energy to react (per the Maxwell-Boltzmann distribution) without being consumed.

A catalyst is a substance that increases the rate of a chemical reaction without being permanently consumed and without changing the thermodynamic outcome (ΔG, ΔH, K remain the same). Mechanistically, a catalyst works by providing an alternative reaction pathway (mechanism) with a lower activation energy (Ea). According to the Arrhenius equation: k = A·e^(−Ea/RT). A lower Ea exponentially increases the rate constant k. Using the Maxwell-Boltzmann energy distribution: at any given temperature, only molecules with kinetic energy ≥ Ea can react. A lower Ea means a larger fraction of molecules have sufficient energy, dramatically increasing the frequency of successful collisions. Important points about catalysts: 1. They lower Ea for both forward AND reverse reactions equally (by the same amount) 2. Therefore, K is unchanged — equilibrium position is not shifted 3. Equilibrium is reached faster but at the same composition 4. They may be homogeneous (same phase as reactants) or heterogeneous (different phase) 5. Enzymes are biological catalysts that can reduce Ea by 10-20 kJ/mol, accelerating reactions by 10⁶–10¹² fold Applications: Haber process (Fe catalyst), catalytic converters (Pt, Pd, Rh), enzyme catalysis.

Question 6: For a second-order reaction with rate = k[A]², if [A]₀ = 0.500 M and k = 0.200 M⁻¹s⁻¹, what is [A] after 5.00 seconds?

0.333 M (Correct answer)
0.250 M
0.400 M
0.167 M

Correct answer: 0.333 M

Integrated second-order rate law: 1/[A]t = 1/[A]₀ + kt. 1/[A]t = 1/0.500 + 0.200 × 5.00 = 2.00 + 1.00 = 3.00. [A]t = 1/3.00 = 0.333 M.

The integrated rate law for a second-order reaction with rate = k[A]²: 1/[A]t = 1/[A]₀ + kt Substituting values: 1/[A]t = 1/0.500 M + (0.200 M⁻¹s⁻¹)(5.00 s) 1/[A]t = 2.00 M⁻¹ + 1.00 M⁻¹ 1/[A]t = 3.00 M⁻¹ [A]t = 1/3.00 M⁻¹ = 0.333 M Comparison with first-order: for first-order, [A]t = [A]₀e^(−kt) — exponential decay. For second-order, [A]t = 1/(1/[A]₀ + kt) — hyperbolic decay. The second-order decrease is slower initially but then accelerates as concentration drops. Half-life for second-order: t₁/₂ = 1/(k[A]₀) = 1/(0.200 × 0.500) = 10.0 s. Unlike first-order, the half-life of a second-order reaction depends on initial concentration — it gets longer as the reaction proceeds. Common second-order reactions: 2HI → H₂ + I₂, nitrogen dioxide decomposition, some radical reactions.

AP Kinetics and Reaction Rates 2

For the reaction A + B → C, the following data was collected:
- Exp 1: [A]=0.100 M, [B]=0.100 M, rate=4.0×10⁻⁵ M/s
- Exp 2: [A]=0.200 M, [B]=0.100 M, rate=8.0×10⁻⁵ M/s
- Exp 3: [A]=0.100 M, [B]=0.200 M, rate=1.6×10⁻⁴ M/s
What are the orders with respect to A and B?