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