Question

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?

Answer 1

No. It is 2nd order with respect to XO but third order over all (which should tell you the exponent for O2.

Answer 2

Would it be k[XO]^2[O2] because second order is squared?

Answer 3

Yes.