# chemistry

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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

• chemistry - ,

You can do these by reason or mathematically. Usually the initial ones are easy enough to do by reasoning. Here is how you do it.
Look for a trial (run) in which the concn of 1 material changes and the other material does not. For example, trial 1 has 0.01 and trial 2 has 0.01 for (XO) while trial 1 shows rate of 2.5 for (O2) and trial 2 shows 5.0 for (O2).
The reasoning part is this. Forget (XO) at this point since we did not change the concn (it stayed at 0.01). We changed concn O2 from 0.01 to 0.02 (doubled it) and we changed the rate from 2.5 to 5.0 (doubled it). So doubling concn caused doubling of rate so the order is 1 because 2^what = 2; obviously what is 1.

Now we look to find a trial in which concn of O2 does not change but (XO) does.That's trial 2 and trial 3.
(XO) changes from 0.01 to 0.03 (triples) while rate changes from 5.0 to 45.0 (45.0/5.0 = 9.0 times). Tripling concn XO made rate change by 9 so 3^what = 9 and that's 2 (3 squared = 9). Therefore, the reaction is second order with respect to XO. Thus the rate law expression is
rate = k(XO)2(O2)1 and the 1 as an exponent isn't needed; I put it there to make the point that it was 1. The reaction is third order overall.