Asked by Hannah

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. Is this correct?

Answers

Answered by DrBob222
No. It is 2nd order with respect to XO but third order over all (which should tell you the exponent for O2.
Answered by Hannah
Would it be k[XO]^2[O2] because second order is squared?
Answered by DrBob222.
Yes.
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