4.36 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

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    Payal Gupta | Contributor-Level 10

    5 months ago

    4.36 We know, The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,

    Ln k = ln A- (Ea/RT)

    Thus, log k = log A - (Ea/2.303RT). eqn 1

    The given equation is log k = 14.34 – 1.25 * 104K/T. eqn 2

    Comparing 2 equations, Ea/2.303R = 1.25 * 104K

    Ea = 1.25 * 104K * 2.303 * 8.314

    Ea = 239339.3 J mol-1 (approximately) Ea = 239.34 kJ mol-1

    Also, when t1/2 = 256 minutes,

    k = 0.693 / t1/2

    = 0.693 / 256

    = 2.707 * 10-3 min-1 k = 4.51 * 10-5s–1

    Substitute k = 4.51 * 10-5s–1 in eqn 2,

    log 4.51 * 10-5 s–1 = 14.34 – 1.25 * 104K/T

    log (0.654-5) = 14.34– 1.25 * 104K/T = 1.25 * 104/ [ 14.34- log (0.654-5)] T = 668.9K or T =

    ...more

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