Posted by Nagaa on Saturday, January 22, 2011 at 1:24am.
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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