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.

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