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12 October, 22:23

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36*10-5 s-1k=3.36*10-5 s-1 at 720 KK. If the initial cyclopropane concentration is 0.0445 MM, what will the cyclopropane concentration be after 235.0 min?

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Answers (2)
  1. 12 October, 23:05
    0
    The concentration would be 0.0302 M.

    Explanation:

    The reaction follows a first-order in cyclopropane.

    Rate = kC = change in concentration/time

    Let the concentration of cyclopropane after 235 minutes be y

    k is rate constant = 3.36*10^-5 s^-1

    Initial concentration = 0.0445 M

    Change in concentration = (0.0445 - y)

    Time = 235 minutes = 235 * 60 = 14,100 s

    3.36*10^-5y = (0.0445 - y) / 14,100

    3.36*10^-5y * 14,100 = 0.0445 - y

    0.47376y = 0.0445 - y

    0.47376y + y = 0.0445

    1.47376y = 0.0445

    y = 0.0445/1.47376 = 0.0302 M
  2. 13 October, 00:54
    0
    r = -dA/dt

    = k[A]

    Where,

    k = rate constant

    = 3.36*10^-5 s^-1

    Initial concentration, Ao = 0.0445 M

    Change in concentration, dA = (0.0445 - y), y is the concentration at time, t (At)

    Time, t = 235 min

    = 235 * 60

    = 14,100 s

    3.36*10^-5 * y = (0.0445 - y) / 14,100

    3.36*10^-5y * 14,100 = 0.0445 - y

    0.4738 * y = 0.0445 - y

    0.4738y + y = 0.0445

    1.4738y = 0.0445

    y = 0.0445/1.4738

    = 0.0302 M
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