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April 18, 2014

Homework Help: Chemistry

Posted by Hannah on Monday, February 20, 2012 at 3:10pm.

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-ls-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

I chose rate=k(XO)^2 (O2)^2

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