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

Question

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

Answer 1

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.