Kinda hard chem question? :/ please? i keep getting it wrong?

NetherCraft 0

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1 Answer

  • K I figured it out

    Consider this reaction data

    A –> products

    T(K). . . k(s^-1)

    225 . . . 0.374

    625 . . . 0.648

    If you were going to graphically determine the activation energy of this reaction, what points would you plot?

    . . . . . . . . . . . . . . X. . . . . . . . . . . . . . . . . . . .Y . . . . . .

    Point 1 = 1/225 = 0.00444* . . . ln(0.374) = –0.9835

    Point 2 = 1/625 = 0.0016** . . . .ln(0.648) = –0.4339

    Determine the rise, run, and slop of the line formed by these points

    Rise y2 – y1

    –0.4339 – (–0.9835) = 0.5496****

    Run x2 –x1

    0.0016 – 0.00444 = –0.00284

    Slope = rise / run

    0.5496 / –0.00284 = –193.52

    What is the activation energy of this reaction (use arrhenius equation)

    ln(K2/K1) = Ea/R (1/T1 – 1/T2)

    ln(0.648 / 0.374) = Ea/8.314 (1/225 – 1/625)

    0.5496 = Ea/8.314 (0.00284)

    Ea = (0.5496 x 8.314) / 0.00284 = 1608.9 J/mol

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