Posted by Nagaa on Saturday, January 22, 2011 at 1:24am.
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 5.57 107 M/s
Exp. 2: 0.00544 M 1.11 106 M/s

chemistry  DrBob222, Saturday, January 22, 2011 at 11:51am
Here is how you do these.
rate = k(A)^{x} where A in this case is concn ethylene oxide.
1.11E6.....k*(0.00544)^{x}
 = 
5.57E7.....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.
1.11E6 = k(0.00544)^{1}
and solve for k. The other run should give about the same k value.