applicationsofkinetics_answers

Applications of Kinetics Problems

Problem 1: Rate Constant Calculation

• Given Data:

  • Rate constant (k1) = 8.25 x 10^5 M^-1 s^-1 at T1 = 302.0°C

  • Activation energy (Ea) = 23 kJ/mol

  • Temperature T2 = 323.0°C

• Conversion of temperatures:

  • T1(K) = 302.0 + 273.15 = 575.15 K

  • T2(K) = 323.0 + 273.15 = 596.15 K

• Formula to find k2:

  • [ \ln k_2 - \ln k_1 = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) ]

• Constants Required:

  • R = 8.314 J/mol*K

  • Convert Ea:

    • 23 kJ/mol = 23000 J/mol

• Calculation:

  • [ \ln k_1 = \ln (8.25 \times 10^5) = 14.01 ]

• Plugging Values into the Formula:

  • [ \ln k_2 - 14.01 = \frac{23000}{8.314} \left( \frac{1}{575.15} - \frac{1}{596.15} \right) ]

  • Calculate ( \frac{1}{T_1} - \frac{1}{T_2} ):

    • Result: 0.1695 K^-1

  • [ \ln k_2 - 14.01 = 0.1695 ]

  • [ \ln k_2 = 14.01 + 0.1695 \approx 14.1795 ]

  • [ k_2 = e^{14.1795} \approx 9.8 \times 10^5 M^{-1} s^{-1} ]

Problem 2: Overall Chemical Reaction and Intermediates

• Given Steps of Mechanism:

  1. Br2 (g) → 2Br (g)

  2. Br (g) + OCl2 (g) → BrOCl (g) + Cl (g)

  3. Br (g) + Cl (g) → BrCl (g)

• Overall Reaction:

  • [ \text{Overall: } Br_2(g) + OCl_2(g) → BrOCl(g) + BrCl(g) ]

• Intermediates present:

  • Br (g), Cl (g)

Problem 3: Observable Rate Law

• Given Mechanism:

  1. Step 1: 2NO (g) → N2O2 (g) (rate constant = k1)

  2. Step 2: N2O2 (g) + O2 (g) → 2NO2 (g) (rate constant = k2)

• Observation:

  • k1 << k2 indicates that Step 1 is slower and thus rate-determining.

• Observable Rate Law:

  • Rate = k [NO]^2

  • This reflects the dependence on the concentration of NO due to the slow initial step.