# Chemistry

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For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below.

run # [XO] [O2] rate,mol L-l s-1
1 0.010 0.010 2.5
2 0.010 0.020 5.0
3 0.030 0.020 45.0

The rate law is therefore

a. rate = k[XO]2 [O2]
b. rate = k[XO][O2]2
c. rate = k[XO][O2]
d. rate = k[XO]2 [O2] 2
e. rate = k[XO]2 / [O2] 2

• Chemistry -

And your problem is what? You find two sets of data in which XO is the same but O2 is different. Compare the change in concn with rate and determine order from that. Then pick another set where XO is not the same but O2 is the came (in concn), and go through the same procedure.
For example, XO is the same in runs 1 & 2. O2 changes by a factor of 2. The rate changes by a factor of 2 also; therefore the order is 1 with respect to O2 since 2^1 = 2.
Do XO the same way.

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