Chem Kinetics

What is chemical kinetics?

Chemical kinetics is the area of chemistry that concerns reaction rates – the speed at which chemical reactions occur. It also studies the steps (mechanism) by which a reaction takes place.

What is the difference between spontaneity (thermodynamics) and reaction rate (kinetics)?

Spontaneity tells whether a reaction has an inherent tendency to occur (e.g., H₂ + O₂ → H₂O is spontaneous). Reaction rate tells how fast it occurs. Many spontaneous reactions are very slow (e.g., diamond turning to graphite). Thermodynamics and kinetics are independent.

Define reaction rate in terms of concentration change.

Reaction rate = change in concentration of a reactant or product per unit time.
Rate = Δ[A]/Δt, where [A] is concentration in mol/L and t is time. For a reactant whose concentration decreases, Δ[A] is negative, so we often add a negative sign to make the rate positive.

For the reaction 2NO₂(g) → 2NO(g) + O₂(g), how do you calculate the average rate of decomposition of NO₂ over the first 50 seconds given [NO₂]₀ = 0.0100 M and [NO₂]₅₀ = 0.0079 M?

Average rate = –Δ[NO₂]/Δt = –(0.0079 – 0.0100)/50 = –(–0.0021)/50 = 4.2×10⁻⁵ mol L⁻¹ s⁻¹. The negative sign makes the rate positive.

What is the instantaneous rate of a reaction?

The instantaneous rate is the rate at a particular moment in time. It is equal to the slope of the tangent line to the concentration vs. time curve at that time. For a reactant, instantaneous rate = –(slope).

Why does the average rate of a reaction decrease over time (for most reactions)?

As the reaction proceeds, reactant concentrations decrease. Since rate depends on concentration (rate = k[A]ⁿ), lower concentration leads to a slower rate. The rate is fastest at t=0 (initial rate).

What is the rate law (differential rate law) for a reaction?

A rate law is an equation that shows how the reaction rate depends on the concentrations of reactants. For a reaction aA + bB → products, the general form is Rate = k[A]ⁿ[B]ᵐ. The exponents n and m are orders (determined experimentally, not from coefficients). k is the rate constant.

Why can't the rate law be written directly from the balanced chemical equation?

The balanced equation shows the stoichiometry (overall reactants and products), but the rate law depends on the reaction mechanism (the actual steps). Orders must be determined by experiment; they often do not match the coefficients.

What is the difference between a differential rate law and an integrated rate law?

• Differential rate law (often just "rate law"): expresses rate as a function of concentration (e.g., Rate = k[A]ⁿ).
• Integrated rate law: expresses concentration as a function of time (e.g., ln[A] = –kt + ln[A]₀). The two are mathematically related (by calculus).

For a reaction aA → products with rate = k[A]ⁿ, what is the integrated rate law for n = 0 (zero order)?

[A] = –kt + [A]₀. A plot of [A] vs. t is a straight line with slope = –k. Half‑life: t₁/₂ = [A]₀ / (2k).

For a reaction aA → products with rate = k[A]ⁿ, what is the integrated rate law for n = 1 (first order)?

ln[A] = –kt + ln[A]₀. A plot of ln[A] vs. t is a straight line with slope = –k. Half‑life: t₁/₂ = 0.693 / k (independent of initial concentration).

For a reaction aA → products with rate = k[A]ⁿ, what is the integrated rate law for n = 2 (second order)?

1/[A] = kt + 1/[A]₀. A plot of 1/[A] vs. t is a straight line with slope = k. Half‑life: t₁/₂ = 1 / (k[A]₀) (depends on initial concentration).

What is the half‑life (t₁/₂) of a reaction?

The half‑life is the time required for the concentration of a reactant to decrease to one‑half of its initial concentration. For first‑order reactions, t₁/₂ is constant; for second‑order, it increases as concentration drops; for zero‑order, it decreases.

A first‑order reaction has a half‑life of 100 s. What fraction of the reactant remains after 300 s?

After 1 half‑life (100 s): 1/2 remains. After 2 half‑lives (200 s): 1/4 remains. After 3 half‑lives (300 s): 1/8 remains. Each half‑life multiplies the remaining fraction by 1/2.

A first‑order reaction has a rate constant k = 0.0693 s⁻¹. Calculate its half‑life.

t₁/₂ = 0.693 / k = 0.693 / 0.0693 = 10.0 s.

A second‑order reaction has k = 0.050 L mol⁻¹ s⁻¹ and [A]₀ = 0.10 M. Calculate the first half‑life.

t₁/₂ = 1 / (k[A]₀) = 1 / (0.050 × 0.10) = 1 / (0.005) = 200 s.

For a zero‑order reaction, how does the half‑life change as the reaction proceeds?

t₁/₂ = [A]₀ / (2k). As the reaction proceeds, [A] decreases, so each successive half‑life becomes shorter. (For first order, constant; for second order, each half‑life doubles.)

What is the method of initial rates?

The method of initial rates is an experimental technique to determine the rate law. Several experiments are run with different initial concentrations, and the instantaneous rate at t≈0 (initial rate) is measured for each. By comparing how the initial rate changes when each reactant concentration is changed, the order for each reactant is found.

For the reaction NH₄⁺(aq) + NO₂⁻(aq) → N₂(g) + 2H₂O(l), the following initial rates were obtained. Determine the order with respect to NO₂⁻. Experiment 1: [NH₄⁺]=0.100 M, [NO₂⁻]=0.0050 M, rate=1.35×10⁻⁷; Experiment 2: [NH₄⁺]=0.100 M, [NO₂⁻]=0.010 M, rate=2.70×10⁻⁷.

[NH₄⁺] constant, [NO₂⁻] doubles (0.0050 → 0.010). Rate doubles (1.35→2.70). 2 = (2)ᵐ → m=1. First order in NO₂⁻.

Using the same reaction, Experiment 2: [NH₄⁺]=0.100 M, [NO₂⁻]=0.010 M, rate=2.70×10⁻⁷; Experiment 3: [NH₄⁺]=0.200 M, [NO₂⁻]=0.010 M, rate=5.40×10⁻⁷. Determine the order with respect to NH₄⁺.

[NO₂⁻] constant, [NH₄⁺] doubles (0.100→0.200). Rate doubles (2.70→5.40). 2 = (2)ⁿ → n=1. First order in NH₄⁺. Overall rate = k[NH₄⁺][NO₂⁻].

For the reaction BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O, the following data were obtained: Experiment 1: [BrO₃⁻]=0.10, [Br⁻]=0.10, [H⁺]=0.10, rate=8.0×10⁻⁴; Experiment 2: [BrO₃⁻]=0.20, [Br⁻]=0.10, [H⁺]=0.10, rate=1.6×10⁻³. Find the order with respect to BrO₃⁻.

[Br⁻] and [H⁺] constant, [BrO₃⁻] doubles (0.10→0.20). Rate doubles (8.0→16.0 ×10⁻⁴). 2 = (2)ⁿ → n=1. First order in BrO₃⁻.

Using the same reaction, Experiment 2: [BrO₃⁻]=0.20, [Br⁻]=0.10, [H⁺]=0.10, rate=1.6×10⁻³; Experiment 3: [BrO₃⁻]=0.20, [Br⁻]=0.20, [H⁺]=0.10, rate=3.2×10⁻³. Find the order with respect to Br⁻.

[BrO₃⁻] and [H⁺] constant, [Br⁻] doubles (0.10→0.20). Rate doubles (1.6→3.2 ×10⁻³). 2 = (2)ᵐ → m=1. First order in Br⁻.

Using the same reaction, Experiment 1: [BrO₃⁻]=0.10, [Br⁻]=0.10, [H⁺]=0.10, rate=8.0×10⁻⁴; Experiment 4: [BrO₃⁻]=0.10, [Br⁻]=0.10, [H⁺]=0.20, rate=3.2×10⁻³. Find the order with respect to H⁺.

[BrO₃⁻] and [Br⁻] constant, [H⁺] doubles (0.10→0.20). Rate increases by factor 3.2/0.8 = 4. 4 = (2)ᵖ → p=2. Second order in H⁺. Rate = k[BrO₃⁻][Br⁻][H⁺]².

For the reaction 2NO(g) + Cl₂(g) → 2NOCl(g), the rate law is Rate = k[NO]²[Cl₂]. What is the overall reaction order?

Overall order = sum of individual orders = 2 (for NO) + 1 (for Cl₂) = third order overall.

What are the units of the rate constant k for a first‑order reaction? For a second‑order reaction?

• First order: rate = k[A] → k has units of s⁻¹ (since rate in mol L⁻¹ s⁻¹, [A] in mol L⁻¹).
• Second order: rate = k[A]² → k has units of L mol⁻¹ s⁻¹.
• Zero order: rate = k → k has units of mol L⁻¹ s⁻¹.

The decomposition of N₂O₅ is first order with k = 6.93×10⁻³ s⁻¹ at a certain temperature. If [N₂O₅]₀ = 0.100 M, what is [N₂O₅] after 150 s?

ln[N₂O₅] = –kt + ln[N₂O₅]₀ = –(6.93×10⁻³)(150) + ln(0.100) = –1.0395 – 2.3026 = –3.3421. [N₂O₅] = e^(–3.3421) = 0.0353 M.

A first‑order reaction is 75% complete in 40. minutes. Calculate its half‑life.

75% complete means 25% remains → [A]/[A]₀ = 0.25. ln([A]₀/[A]) = ln(4) = 1.386 = kt = k(40 min). So k = 1.386/40 = 0.03465 min⁻¹. t₁/₂ = 0.693/k = 0.693/0.03465 = 20.0 min.

For a second‑order reaction, a plot of 1/[A] vs. time gives a straight line with slope 0.0614 L mol⁻¹ s⁻¹. What is the rate constant? If [A]₀ = 0.0100 M, what is the half‑life?

Slope = k = 0.0614 L mol⁻¹ s⁻¹. t₁/₂ = 1/(k[A]₀) = 1/(0.0614 × 0.0100) = 1/(6.14×10⁻⁴) = 1630 s.

What is the pseudo‑first‑order method?

When a reaction involves multiple reactants, we can make one reactant (A) have a very low concentration compared to others (B, C). Then [B] and [C] remain nearly constant, and the rate law simplifies to Rate = k'[A]ⁿ, where k' = k[B]₀ᵐ[C]₀ᵖ. This allows us to determine the order with respect to A by measuring [A] vs. time.

What is a reaction mechanism?

A reaction mechanism is the series of elementary steps by which a chemical reaction occurs. Each elementary step describes a single molecular event (bond breaking/forming). The sum of all elementary steps gives the overall balanced equation.

What is an elementary step? What is molecularity?

An elementary step is a reaction that occurs exactly as written at the molecular level. Its molecularity is the number of molecules colliding in that step: unimolecular (1 molecule), bimolecular (2 molecules), termolecular (3 molecules – rare).

How do you write the rate law for an elementary step?

The rate law for an elementary step follows directly from its molecularity:

• Unimolecular: A → products → Rate = k[A]
• Bimolecular: A + A → products → Rate = k[A]²
• Bimolecular: A + B → products → Rate = k[A][B]
• Termolecular: 2A + B → products → Rate = k[A]²[B], etc.

What are the two requirements for a proposed reaction mechanism to be acceptable?

1. The sum of the elementary steps must give the overall balanced equation.
2. The mechanism must agree with the experimentally determined rate law.

What is the rate‑determining step (RDS)?

The rate‑determining step is the slowest step in a multistep reaction mechanism. The overall reaction cannot go faster than this step. The rate law for the overall reaction is the rate law of the RDS (after substituting for any intermediates).

For the mechanism: Step 1 (slow): NO₂ + NO₂ → NO₃ + NO, Step 2 (fast): NO₃ + CO → NO₂ + CO₂. The overall reaction is NO₂ + CO → NO + CO₂. What is the predicted rate law?

The slow step is rate‑determining. Its molecularity is bimolecular (two NO₂). Rate law = k₁[NO₂]². This matches the experimental rate law for this reaction.

For the mechanism: Step 1 (slow): NO₂ + F₂ → NO₂F + F, Step 2 (fast): F + NO₂ → NO₂F. Overall: 2NO₂ + F₂ → 2NO₂F. What is the predicted rate law?

Slow step is bimolecular (NO₂ + F₂). Rate law = k₁[NO₂][F₂]. This is an acceptable mechanism if it matches experiment.

What is an intermediate? How is it different from a catalyst?

An intermediate is a species that is produced in one elementary step and consumed in a later step. It does not appear in the overall equation. A catalyst is also consumed and later regenerated, but a catalyst is added to speed up the reaction; an intermediate is not added separately.

For the ozone decomposition mechanism: Step 1 (fast equilibrium): O₃ ⇌ O₂ + O, Step 2 (slow): O + O₃ → 2O₂. The experimental rate law is Rate = k[O₃]²/[O₂]. Show how this is derived.

Rate = k₂[O][O₃] (slow step). From fast equilibrium: k₁[O₃] = k₋₁[O₂][O] → [O] = (k₁/k₋₁)[O₃]/[O₂]. Substitute: Rate = k₂ (k₁/k₋₁)[O₃]²/[O₂] = k[O₃]²/[O₂].

What is the steady‑state approximation?

The steady‑state approximation assumes that the concentration of any reactive intermediate remains constant (low and unchanging) after an initial brief period. Thus, the rate of formation of the intermediate equals its rate of consumption (d[I]/dt = 0). This allows us to solve for [I] in terms of reactants.

For the mechanism: 2NO ⇌ N₂O₂ (fast), N₂O₂ + H₂ → N₂O + H₂O (slow). Derive the rate law using the steady‑state approximation for N₂O₂.

Rate = k₂[N₂O₂][H₂]. For N₂O₂: formation rate = k₁[NO]², consumption = k₋₁[N₂O₂] + k₂[N₂O₂][H₂]. Set equal: k₁[NO]² = [N₂O₂](k₋₁ + k₂[H₂]) → [N₂O₂] = k₁[NO]²/(k₋₁ + k₂[H₂]). Substitute: Rate = k₂k₁[NO]²[H₂]/(k₋₁ + k₂[H₂]). If k₂[H₂] >> k₋₁, Rate ≈ k₁[NO]²; if k₋₁ >> k₂[H₂], Rate ≈ (k₂k₁/k₋₁)[NO]²[H₂].

What is the collision model of chemical kinetics?

The collision model states that for a reaction to occur, molecules must collide with sufficient energy and proper orientation. The rate depends on collision frequency, fraction of collisions with energy ≥ activation energy (Eₐ), and steric factor (proper orientation).

What is activation energy (Eₐ)?

Activation energy is the minimum kinetic energy that colliding molecules must possess to overcome the energy barrier and form products. It is the energy needed to reach the transition state (activated complex). A higher Eₐ means a slower reaction at a given temperature.

What is the Arrhenius equation? Write its two common forms.

k = A e^(–Eₐ/RT) where k = rate constant, A = frequency factor (includes collision frequency and steric factor), Eₐ = activation energy (J/mol), R = 8.314 J mol⁻¹ K⁻¹, T = Kelvin temperature.
Log form: ln k = –Eₐ/(RT) + ln A or ln(k₂/k₁) = (Eₐ/R)(1/T₁ – 1/T₂).

How can you determine Eₐ experimentally?

Measure the rate constant k at several temperatures. Plot ln k vs. 1/T (Kelvin). The slope of the straight line = –Eₐ/R. Then Eₐ = –R × slope. Alternatively, use two temperatures with the two‑point Arrhenius equation.

The rate constant for a reaction doubles when the temperature is increased from 300 K to 310 K. Estimate the activation energy.

ln(k₂/k₁) = ln(2) = 0.693 = (Eₐ/R)(1/T₁ – 1/T₂) = (Eₐ/8.314)(1/300 – 1/310) = (Eₐ/8.314)(0.003333 – 0.003226) = (Eₐ/8.314)(0.000107). So Eₐ = 0.693 × 8.314 / 0.000107 ≈ 0.693 × 77700 ≈ 53,800 J/mol = 53.8 kJ/mol.

What is the effect of a catalyst on activation energy and reaction rate?

A catalyst provides an alternative reaction pathway with a lower activation energy. This increases the fraction of molecules with sufficient energy (e⁽–Eₐ/RT⁾), so the rate constant k increases dramatically. The catalyst is not consumed and does not change ΔE or the equilibrium constant.

What is the difference between a homogeneous and a heterogeneous catalyst?

• Homogeneous catalyst: exists in the same phase as the reactants (e.g., aqueous acid catalyst in a liquid reaction).
• Heterogeneous catalyst: exists in a different phase, typically a solid catalyst with gaseous or liquid reactants (e.g., iron in the Haber process, platinum in catalytic converters).

How does a heterogeneous catalyst work? Describe the four steps.

1. Adsorption of reactants onto the catalyst surface.
2. Migration of adsorbed reactants on the surface.
3. Reaction among adsorbed species.
4. Desorption of products from the surface.

What is an enzyme? How does it catalyze biochemical reactions?

An enzyme is a biological catalyst, typically a protein. It binds the substrate (reactant) at its active site to form an enzyme‑substrate complex (E·S). This lowers the activation energy. The product is released, and the enzyme is recycled. Enzymes are highly specific and very efficient.

For the reaction 2NO(g) + O₂(g) → 2NO₂(g), the rate law is Rate = k[NO]²[O₂]. A proposed mechanism has a fast equilibrium 2NO ⇌ N₂O₂ followed by slow N₂O₂ + O₂ → 2NO₂. Is this mechanism consistent?

Yes. The slow step gives Rate = k₂[N₂O₂][O₂]. From fast equilibrium, [N₂O₂] = (k₁/k₋₁)[NO]². So Rate = k₂(k₁/k₋₁)[NO]²[O₂] = k[NO]²[O₂], matching experiment.

Why are termolecular elementary steps rare?

The probability of three molecules colliding simultaneously at the same spot with the correct orientation and sufficient energy is extremely low. Most reactions proceed via bimolecular or unimolecular steps.

For the reaction A → products, a plot of ln[A] vs. time is linear with slope –0.030 s⁻¹. What is the order? What is k? What is the half‑life?

Linear ln[A] vs. t → first order. k = –slope = 0.030 s⁻¹. t₁/₂ = 0.693/k = 0.693/0.030 = 23.1 s.

For the reaction A → products, a plot of 1/[A] vs. time is linear with slope 0.040 L mol⁻¹ s⁻¹. What is the order? If [A]₀ = 0.50 M, what is the half‑life?

Linear 1/[A] vs. t → second order. k = slope = 0.040 L mol⁻¹ s⁻¹. t₁/₂ = 1/(k[A]₀) = 1/(0.040 × 0.50) = 1/(0.020) = 50 s.

For the reaction A → products, a plot of [A] vs. time is linear with slope –0.010 mol L⁻¹ s⁻¹. What is the order? If [A]₀ = 0.20 M, what is the half‑life?

Linear [A] vs. t → zero order. k = –slope = 0.010 mol L⁻¹ s⁻¹. t₁/₂ = [A]₀/(2k) = 0.20/(2×0.010) = 0.20/0.020 = 10 s.

What is the frequency factor A in the Arrhenius equation? What two factors does it include?

A (pre‑exponential factor) includes:

1. Collision frequency (z): how often molecules collide.
2. Steric factor (p): the fraction of collisions with proper orientation for reaction.

The activation energy of a reaction is 100 kJ/mol. By what factor does the rate constant increase when the temperature is raised from 300 K to 310 K? (R = 8.314 J/mol·K)

ln(k₂/k₁) = (Eₐ/R)(1/T₁ – 1/T₂) = (100000/8.314)(1/300 – 1/310) = 12026 × (0.003333 – 0.003226) = 12026 × 0.000107 = 1.287. k₂/k₁ = e^(1.287) = 3.62. Rate increases by factor of about 3.6.

For a first‑order reaction, the concentration of reactant drops from 0.100 M to 0.025 M in 60 minutes. What is the rate constant and half‑life?

0.025 is 1/4 of 0.100. That is two half‑lives (1/2 then 1/2 again). So 2 × t₁/₂ = 60 min → t₁/₂ = 30 min. k = 0.693 / t₁/₂ = 0.693/30 = 0.0231 min⁻¹.

The decomposition of hydrogen peroxide is first order with k = 1.0×10⁻³ s⁻¹. How long will it take for 90% of a sample to decompose?

90% decomposed → 10% remains → [A]/[A]₀ = 0.10. ln([A]₀/[A]) = ln(10) = 2.3026 = kt = (1.0×10⁻³)t. t = 2.3026 / 0.001 = 2303 s (about 38.4 min).

What is the relationship between the rate constant k and temperature according to the collision model?

As temperature increases, the fraction of collisions with energy ≥ Eₐ increases exponentially (e^(–Eₐ/RT)). Thus k increases exponentially with T, as described by the Arrhenius equation.

Draw an energy profile for an exothermic reaction. Label: reactants, products, activation energy (Eₐ), and ΔE (enthalpy change).

Reactants at higher energy than products (exothermic). Eₐ is the energy difference from reactants to the top of the peak (transition state). ΔE is the energy difference between products and reactants (negative for exothermic).

How does a catalyst change the energy profile of a reaction?

A catalyst provides a different pathway with a lower activation energy (Eₐ decreased). The energies of reactants and products are unchanged, so ΔE is unchanged. The peak (transition state) is lower.

In the reaction 2HI → H₂ + I₂, the rate law is Rate = k[HI]². Propose a plausible elementary step mechanism.

A single bimolecular elementary step: 2HI → H₂ + I₂. This would give Rate = k[HI]², matching experiment. (Many reactions occur in a single step, but not all.)

What is the difference between an activated complex (transition state) and an intermediate?

The activated complex (transition state) is the highest‑energy configuration along the reaction pathway; it is a temporary, unstable species that cannot be isolated. An intermediate is a local minimum on the energy diagram (a valley between two peaks); it has a finite lifetime and can sometimes be detected.

For the mechanism: Step 1: A + B ⇌ C (fast), Step 2: C + A → D (slow). Derive the rate law.

Rate = k₂[C][A] (slow step). From fast equilibrium: k₁[A][B] = k₋₁[C] → [C] = (k₁/k₋₁)[A][B]. Substitute: Rate = k₂ (k₁/k₋₁)[A][B][A] = k[A]²[B], where k = k₂k₁/k₋₁.

The rate law for the reaction H₂ + Br₂ → 2HBr is Rate = k[H₂][Br₂]¹⁄². Does this suggest a one‑step mechanism? Why?

No. A one‑step mechanism (bimolecular) would give Rate = k[H₂][Br₂] (first order in each). The half‑order indicates a more complex mechanism involving a fast equilibrium (Br₂ ⇌ 2Br) and then a slow step.

What is the role of a catalyst in the Haber process (N₂ + 3H₂ → 2NH₃)?

The Haber process uses a heterogeneous catalyst (iron with promoters). The catalyst adsorbs N₂ and H₂, weakening the strong N≡N and H–H bonds, lowering the activation energy, and allowing the reaction to proceed at a practical rate at moderate temperatures.

In a zero‑order reaction, why does the rate not depend on concentration?

Zero‑order kinetics occur when the reaction rate is limited by something other than reactant concentration, such as a saturated catalyst surface (e.g., decomposition on a metal where all active sites are occupied) or a fixed light intensity in a photochemical reaction. Rate = k (constant).

For a first‑order reaction, how long does it take for 99.9% of the reactant to decompose in terms of half‑lives?

After 1 half‑life: 50% remains; 2: 25%; 3: 12.5%; 4: 6.25%; 5: 3.125%; 6: 1.5625%; 7: 0.78125%; 8: 0.3906%; 9: 0.1953%; 10: 0.09766% remains. 99.9% decomposed means 0.1% remains, which is between 9 and 10 half‑lives (about 9.97 half‑lives). So t ≈ 10 × t₁/₂.

The rate constant for a first‑order reaction is 0.035 s⁻¹. What percentage of the reactant remains after 100 s?

ln([A]/[A]₀) = –kt = –0.035 × 100 = –3.5. [A]/[A]₀ = e^(–3.5) = 0.0302. So 3.0% remains.

For a second‑order reaction with k = 0.050 L mol⁻¹ s⁻¹ and [A]₀ = 0.010 M, calculate the time required for [A] to drop to 0.0025 M.

1/[A] – 1/[A]₀ = kt → 1/0.0025 – 1/0.010 = 400 – 100 = 300 = 0.050 × t → t = 300 / 0.050 = 6000 s.

Why does the rule of thumb "rate doubles for every 10°C increase" not hold for all reactions?

The temperature effect depends on the activation energy. Reactions with higher Eₐ show a larger increase in rate for a given temperature rise. The rule of thumb works for many reactions with Eₐ around 50 kJ/mol near room temperature, but not universally.

What is the difference between absorption and adsorption?

Absorption is the penetration of one substance into the interior of another (e.g., water absorbed by a sponge). Adsorption is the collection of molecules on the surface of a solid (e.g., gas molecules sticking to a catalyst surface).

In the acid‑catalyzed hydration of ethylene (CH₂=CH₂ + H₂O → CH₃CH₂OH), what is the role of H⁺?

H⁺ attacks the double bond, forming a carbocation intermediate (CH₃–CH₂⁺), which then reacts with water. The H⁺ is regenerated in the final step. The catalyst provides a lower‑energy pathway.

The decomposition of N₂O on a hot platinum surface is zero order. Explain why.

The reaction occurs on the platinum surface. At sufficiently high N₂O pressure, the surface becomes saturated with adsorbed N₂O molecules. Further increases in gas‑phase N₂O concentration do not increase the surface coverage, so the rate remains constant (zero order).

What is the integrated rate law for a zero‑order reaction? What plot gives a straight line?

[A] = –kt + [A]₀. A plot of [A] vs. time is linear with slope –k.

What is the integrated rate law for a first‑order reaction? What plot gives a straight line?

ln[A] = –kt + ln[A]₀. A plot of ln[A] vs. time is linear with slope –k.

What is the integrated rate law for a second‑order reaction (one reactant)? What plot gives a straight line?

1/[A] = kt + 1/[A]₀. A plot of 1/[A] vs. time is linear with slope k.

For the reaction 2A → products, if a plot of 1/[A] vs. time is linear, what is the order with respect to A? What is the unit of k?

Second order with respect to A. The units of k: L mol⁻¹ s⁻¹ (or M⁻¹ s⁻¹).

For the reaction A + B → products, the rate law is Rate = k[A]²[B]⁰. How does the rate change if [A] is tripled and [B] is doubled?

Rate ∝ [A]², and [B] does not affect rate (zero order). Tripling [A] increases rate by 3² = 9 times. Doubling [B] has no effect. Overall: 9 times faster.

The half‑life of a first‑order reaction is 30 s. What is the rate constant? How long will it take for 87.5% of the reactant to decompose?

k = 0.693/30 = 0.0231 s⁻¹. 87.5% decomposed means 12.5% remains = 1/8 remaining. That is 3 half‑lives (1/2 → 1/4 → 1/8). So t = 3 × 30 = 90 s.

In the mechanism for the reaction of NO₂ with CO, the rate law is Rate = k[NO₂]². The proposed mechanism has a slow step NO₂ + NO₂ → NO₃ + NO and a fast step NO₃ + CO → NO₂ + CO₂. Why is CO not in the rate law?

CO appears only in the fast step after the rate‑determining step. The overall rate is controlled by the slow step, which does not involve CO. Therefore, the rate law depends only on [NO₂]².

What is a free radical? Give an example of a radical intermediate in a reaction mechanism.

A free radical is a species with an unpaired electron, making it highly reactive. Example: Cl· (chlorine atom) in the photochemical chlorination of methane, or OH· in atmospheric chemistry.

For a reaction that is zero order, what happens to the rate when the concentration of reactant is doubled?

The rate does not change (remains constant = k). Zero order means rate is independent of concentration.

The rate constant for a certain reaction at 300 K is 2.0×10⁻⁴ s⁻¹. At 400 K, it is 1.6×10⁻² s⁻¹. Calculate the activation energy.

ln(k₂/k₁) = ln(1.6×10⁻² / 2.0×10⁻⁴) = ln(80) = 4.382. 1/T₁ = 1/300 = 0.003333, 1/T₂ = 1/400 = 0.002500, difference = 0.000833. Eₐ = R × ln(k₂/k₁) / (1/T₁ – 1/T₂) = 8.314 × 4.382 / 0.000833 = 36.43 / 0.000833 ≈ 43,740 J/mol = 43.7 kJ/mol.

What is the transition state theory?

Transition state theory assumes that reactants form an activated complex (transition state) at the top of the energy barrier. The rate depends on the concentration of this complex and the frequency of its decomposition to products. It provides a more detailed picture than simple collision theory.

In the reaction of ozone with chlorine atoms (Cl + O₃ → ClO + O₂), the activation energy is very low (about 2 kJ/mol). What does that imply about the rate?

A very low activation energy means that at room temperature, almost all collisions are energetic enough to react. The reaction is extremely fast. Such low Eₐ is typical for radical‑radical or radical‑molecule reactions.

The decomposition of ammonium nitrite (NH₄NO₂) in aqueous solution is first order with k = 0.015 s⁻¹. If you start with 0.100 M, what concentration remains after 1 minute?

t = 60 s. ln([A]/0.100) = –0.015 × 60 = –0.90. [A]/0.100 = e^(–0.90) = 0.4066. [A] = 0.100 × 0.4066 = 0.0407 M.

What is the difference between a rate law and the law of mass action?

The rate law (kinetics) is determined experimentally and gives the rate as a function of concentration. The law of mass action (equilibrium) gives the equilibrium constant expression (ratio of product concentrations to reactant concentrations at equilibrium). They are not the same.

For the reaction 2A + B → C, the rate law is Rate = k[A][B]. If [A] is doubled and [B] is halved, how does the rate change?

New rate = k(2[A])(0.5[B]) = k[A][B] × (2×0.5) = k[A][B] × 1. The rate stays the same.

What is a common method to determine the order of a reaction with respect to a specific reactant when the reaction involves multiple reactants?

Use the isolation method (pseudo‑order). Make all other reactants have much higher concentrations than the one being studied. Their concentrations remain nearly constant, so the rate law simplifies to Rate = k'[reactant]ⁿ. Then plot the appropriate function (ln[reactant] vs. t for n=1, 1/[reactant] vs. t for n=2, etc.) to determine n.

The Arrhenius equation can be written in logarithmic form as ln k = –Eₐ/(RT) + ln A. What are the slope and intercept when ln k is plotted against 1/T?

Slope = –Eₐ/R, intercept = ln A.

Why do many reactions go faster in solution than in the gas phase?

In solution, the solvent can assist in breaking bonds or stabilizing transition states, effectively lowering the activation energy. However, diffusion in solution is slower, so the net effect varies. For many reactions, solvent effects can increase or decrease rate.

What is the molecularity of the elementary step: O₃ + O → 2O₂?

Bimolecular (two molecules: O₃ and O collide).

For the elementary step NO₂ + NO₂ → NO₃ + NO, write the rate law.

Rate = k[NO₂]² (since it is bimolecular with two identical molecules).

For the elementary step Cl + CHCl₃ → HCl + CCl₃, write the rate law.

Rate = k[Cl][CHCl₃] (bimolecular with two different molecules).

The reaction 2O₃ → 3O₂ has the rate law Rate = k[O₃]²/[O₂]. Propose a mechanism with a fast equilibrium first step and a slow second step.

Step 1 (fast equilibrium): O₃ ⇌ O₂ + O. Step 2 (slow): O + O₃ → 2O₂. This gives Rate = k₂[O][O₃] = k₂(k₁/k₋₁)[O₃]²/[O₂] = k[O₃]²/[O₂].

What is a catalyst poison? Give an example.

A catalyst poison is a substance that binds strongly to the catalyst surface, blocking active sites and reducing catalytic activity. Example: Lead (in gasoline) poisons platinum catalysts in catalytic converters, which is why leaded gasoline was phased out.

In enzyme kinetics, the Michaelis‑Menten equation is Rate = Vmax[S]/(K_M + [S]). What does K_M represent?

K_M (Michaelis constant) is the substrate concentration at which the reaction rate is half of Vmax. It is a measure of the enzyme's affinity for the substrate (lower K_M means higher affinity).

At very high substrate concentrations ([S] >> K_M), what is the rate of an enzyme‑catalyzed reaction?

Rate ≈ Vmax (zero order in [S]). The enzyme is saturated with substrate; all active sites are occupied, so adding more substrate does not increase the rate.

For a reaction that is second order in A, what happens to the half‑life if the initial concentration is doubled?

t₁/₂ = 1/(k[A]₀). Doubling [A]₀ halves the half‑life. (Second‑order half‑life is inversely proportional to initial concentration.)