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March 29, 2015

Homework Help: chemistry

Posted by steve on Friday, February 17, 2012 at 9:18pm.

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-l
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

No idea, help

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