chemistry

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C2H4O(g) ¨ CH4(g) + CO(g) Find the rate law of C2H4O?

The following kinetic data were observed for the reaction at 688 K.


Initial
Concentration
of Ethylene Oxide

Initial Rate
Exp. 1: 0.00272 M 5.57 10-7 M/s
Exp. 2: 0.00544 M 1.11 10-6 M/s

  • chemistry -

    Here is how you do these.
    rate = k(A)x where A in this case is concn ethylene oxide.

    1.11E-6.....k*(0.00544)x
    -------- = ----------------
    5.57E-7.....k*(0.00272)x

    k cancels
    1.99 = (2)x
    You may look at this and see that x must be 1 for the equality to hold OR you may solve it mathematically by taking the log of both sides.
    log 1.99 = x*log 2
    0.298 = 0.301x
    x = 0.298/0.301 = 0.993 which I would round to 1. To determine k for the reaction, take EITHER trial run, plug in the values and evaluate k.
    rate = k(A)1
    Take run 2.
    1.11E-6 = k(0.00544)1
    and solve for k. The other run should give about the same k value.

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