Questions & explanations
1. For a reaction with Ea = 50 kJ/mol, the fraction of molecules with energy ≥ Ea at 300 K is 2.0×10⁻⁹. At 320 K, the fraction becomes 6.9×10⁻⁹. For another reaction with Ea = 100 kJ/mol at 300 K, the fraction is 4.0×10⁻¹⁸. What is the fraction at 320 K for the second reaction?
- 4.8×10⁻¹⁷
- 1.4×10⁻¹⁷
- 2.0×10⁻¹⁸
- 8.0×10⁻¹⁸
Answer: 4.8×10⁻¹⁷
The fraction is e^(-Ea/RT). For Ea=100 kJ/mol, at 300 K: Ea/RT = 100000/(8.314×300) ≈ 40.1, fraction = e^(-40.1) ≈ 4.0×10⁻¹⁸. At 320 K: Ea/RT = 100000/(8.314×320) ≈ 37.6, fraction = e^(-37.6) ≈ 4.8×10⁻¹⁷. The ratio of fractions at 320 K to 300 K is e^((Ea/R)(1/300 - 1/320)) = e^(100000/8.314 * (0.003333 - 0.003125)) = e^(12028 * 0.000208) = e^(2.5) ≈ 12.2, so fraction = 4.0×10⁻¹⁸ × 12.2 ≈ 4.9×10⁻¹⁷, closest to 4.8×10⁻¹⁷.
2. For a reaction, the rate constant at 298 K is 2.0×10⁻⁵ s⁻¹ and at 308 K is 4.0×10⁻⁵ s⁻¹. What is the activation energy in kJ/mol? (R = 8.314 J K⁻¹ mol⁻¹)
- 53.6 kJ/mol
- 51.2 kJ/mol
- 55.0 kJ/mol
- 49.8 kJ/mol
Answer: 53.6 kJ/mol
Use the Arrhenius two-point form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Here k₂/k₁ = 2, so ln2 = 0.693. 1/T₁ - 1/T₂ = (308-298)/(298×308) = 10/91784 = 1.089×10⁻⁴ K⁻¹. Then Ea = (0.693 × 8.314) / (1.089×10⁻⁴) = 5.76 / 1.089×10⁻⁴ = 5.29×10⁴ J/mol = 52.9 kJ/mol. The closest option is 53.6 kJ/mol (due to rounding).
3. For a first-order reaction, k = 2.0×10⁻³ s⁻¹ at 300 K and Ea = 55 kJ/mol. What is k at 320 K? (R = 8.314 J mol⁻¹ K⁻¹)
- 4.0×10⁻³ s⁻¹
- 8.0×10⁻³ s⁻¹
- 1.0×10⁻² s⁻¹
- 2.0×10⁻³ s⁻¹
Answer: 8.0×10⁻³ s⁻¹
Using Arrhenius two-point form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Convert Ea to J/mol: 55000 J/mol. Compute 1/300 - 1/320 = 2.083×10⁻⁴ K⁻¹. Then (Ea/R) = 55000/8.314 ≈ 6615. So ln(k₂/k₁) = 6615 × 2.083×10⁻⁴ ≈ 1.378. Thus k₂/k₁ = e^1.378 ≈ 3.97 ≈ 4. Hence k₂ = 4 × 2.0×10⁻³ = 8.0×10⁻³ s⁻¹.
4. For a reaction with activation energy 50 kJ/mol, by what factor does the rate constant increase when temperature rises from 300 K to 310 K? (R = 8.314 J/mol·K)
- About 1.9 times
- About 2.0 times
- About 1.5 times
- About 1.2 times
Answer: About 1.9 times
Using the Arrhenius equation, the ratio k2/k1 = e^{(Ea/R)(1/T1 - 1/T2)}. Here Ea = 50000 J/mol, R = 8.314, T1 = 300 K, T2 = 310 K. Compute 1/300 - 1/310 = 0.0001075, then (Ea/R)*0.0001075 = (50000/8.314)*0.0001075 ≈ 0.646. So k2/k1 = e^0.646 ≈ 1.91, about 1.9 times.
5. For the reaction A + B → products, initial rates at 300 K are: [A]₀=0.10 M, [B]₀=0.10 M, rate=2.0×10⁻³ M/s; [A]₀=0.20 M, [B]₀=0.10 M, rate=4.0×10⁻³ M/s; [A]₀=0.10 M, [B]₀=0.20 M, rate=8.0×10⁻³ M/s; [A]₀=0.20 M, [B]₀=0.20 M, rate=1.6×10⁻² M/s. What is the rate constant k?
- 0.20 M⁻¹ s⁻¹
- 0.20 M⁻² s⁻¹
- 2.0 M⁻¹ s⁻¹
- 2.0 M⁻² s⁻¹
Answer: 2.0 M⁻² s⁻¹
From runs 1 and 2, [A] doubles, [B] constant, rate doubles → order in A = 1. From runs 1 and 3, [B] doubles, [A] constant, rate quadruples → order in B = 2. Rate law: rate = k[A][B]². Using run 1: 2.0×10⁻³ = k(0.10)(0.10)² = k(0.001) → k = 2.0 M⁻² s⁻¹.
6. A reaction has activation energy 100 kJ/mol. At 300 K, its rate constant is 1.0×10⁻⁴ s⁻¹. What is the rate constant at 310 K? (R = 8.314 J K⁻¹ mol⁻¹)
- 2.00×10⁻⁴ s⁻¹
- 3.64×10⁻⁴ s⁻¹
- 1.91×10⁻⁴ s⁻¹
- 4.95×10⁻⁴ s⁻¹
Answer: 3.64×10⁻⁴ s⁻¹
Use ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Ea = 100×10³ J/mol, R = 8.314, 1/T₁ - 1/T₂ = 1/300 - 1/310 = 10/(300×310) = 1.075×10⁻⁴ K⁻¹. So ln(k₂/k₁) = (100000/8.314)×1.075×10⁻⁴ = 12027×1.075×10⁻⁴ = 1.293. Thus k₂/k₁ = e^1.293 = 3.64, so k₂ = 3.64×10⁻⁴ s⁻¹.
7. For the first-order gas-phase decomposition SO2Cl2(g) → SO2(g) + Cl2(g), the initial pressure of pure SO2Cl2 is 0.500 atm. At t = 200 s, the total pressure is 0.675 atm. What is the rate constant k?
- 4.15 × 10⁻³ s⁻¹
- 1.15 × 10⁻³ s⁻¹
- 3.15 × 10⁻³ s⁻¹
- 2.15 × 10⁻³ s⁻¹
Answer: 2.15 × 10⁻³ s⁻¹
For first-order reaction, k = (2.303/t) log(P0/(2P0 - P_total)). Here P0 = 0.500 atm, P_total = 0.675 atm, so P_SO2Cl2 = 2×0.500 - 0.675 = 0.325 atm. Then k = (2.303/200) log(0.500/0.325) = 0.011515 × log(1.538) ≈ 0.011515 × 0.187 = 2.15×10⁻³ s⁻¹.
8. Which experimental technique is most suitable to follow the rate of inversion of cane sugar (sucrose → glucose + fructose)?
- Titration of the acid produced
- Conductivity measurement
- Polarimetry
- Pressure measurement
Answer: Polarimetry
Inversion of cane sugar involves a change in optical rotation because sucrose is dextrorotatory and the product mixture is levorotatory. Polarimetry directly measures the change in optical activity, making it the most suitable technique.
9. For a reaction, the half-life at different initial concentrations is: [R]₀ = 0.1 M, t₁/₂ = 200 s; [R]₀ = 0.2 M, t₁/₂ = 100 s; [R]₀ = 0.4 M, t₁/₂ = 50 s. What is the order of the reaction?
- Zero order
- First order
- Second order
- Third order
Answer: Second order
For a second-order reaction, t₁/₂ ∝ 1/[R]₀. Here, when [R]₀ doubles from 0.1 M to 0.2 M, t₁/₂ halves from 200 s to 100 s. When [R]₀ quadruples to 0.4 M, t₁/₂ becomes one-fourth (50 s). This inverse proportionality indicates second order.
10. Which of the following statements about order and molecularity of a reaction is correct?
- Order is always a small positive integer.
- Order is determined experimentally from the rate law.
- Molecularity can be fractional.
- Molecularity applies to both elementary and complex reactions.
Answer: Order is determined experimentally from the rate law.
Order is the sum of powers of concentration terms in the experimentally determined rate law, so it is an experimental quantity. Molecularity is the number of reacting species in an elementary step and is always a small positive integer.
11. In the acidic hydrolysis of ethyl acetate, the volume of NaOH required to neutralize the acetic acid produced at t=0, t=30 min, and t=∞ are 5.0 mL, 9.5 mL, and 25.0 mL respectively. What is the observed rate constant (in min⁻¹)?
- 4.25 × 10⁻³
- 1.70 × 10⁻²
- 8.49 × 10⁻³
- 2.83 × 10⁻³
Answer: 8.49 × 10⁻³
For pseudo first-order hydrolysis, k = (2.303/t) log((V∞ - V0)/(V∞ - Vt)). Substituting V0=5.0, Vt=9.5, V∞=25.0, t=30 min gives k = (2.303/30) log(20.0/15.5) = 0.07677 × log(1.2903) = 0.07677 × 0.1106 = 8.49×10⁻³ min⁻¹.
12. For a reaction R → P, doubling [R]₀ doubles the initial rate, half-life is independent of [R]₀, and log[R] vs t is linear. What is the order of the reaction?
- Fractional order
- Zero order
- Second order
- First order
Answer: First order
Doubling [R]₀ doubles rate → rate ∝ [R]₀¹, so order = 1. Half-life independent of [R]₀ is characteristic of first-order. Linear log[R] vs t plot confirms first-order kinetics. All three methods converge to first order.