trial # [A]initial initial initial rate product formation 1 0.120 0.060 6.50x10^-4 2 0.120 0.120 1.30x10^-3 3 0.060 0.240 6.50x10^-4 Based on the results listed in the table below, what is the rate expression for this reaction? Rate = k[A] Rate = k[A]2 -->Rate = k[A]2 Rate = k[A]22 i was able to calculate the exponent for B as 1, but i was struggling to figure out A's exponent. Can you help a brother out?

1 0.120 0.060 6.50x10^-4 2 0.120 0.120 1.30x10^-3 3 0.060 0.240 6.50x10^-4 Sweet, let us compare 1 and 2. You observed that by doubling the concentration of , the rate double, right? If we double it again, the rate should double again to 2.60x10^-3. That would give you an equation that looks like this. 0.120 0.240 2.60x10^-3 Now we have two equations that have 0.240 as concentrations for . Problem solved!!! We observe that doubling the concentration of [A] would make the rate increase by 2.6 divided by 4 means 0.65... So it is second order in [A] Thats the right answer... right?? I hope my kinetics aren't gone to hell

I like working with whole numbers the sif numbers are okay too but I change up the numbers so that they work for me. My table would be like the following [A] Rate Trial 1 2 1 1 Trial 2 2 2 2 Trial 3 1 4 1 I just look for common dividers like for the first column you got .120 and .06 so i divided that entire colume by .120. Same thing with the other columns. So now looking at my tabel I can attack the question quickly. Alright with respect to B (hold the concentration of A constant) : 2^y = 2 therefore B is 1st order. Alright the trick for A is looking for patterns. Look at trial 2 as you see A was held constant but concentration of B was increase by 2 and the rate increased by 2. If we double B again and HOLD A CONSTANT we get 4 and 4. This is the key. So with respect to A : Double trial 2 again you get Trial 3'. [A] Rate Trial 3' 2 4 4 Trial 3 and 3' used to find A : 2^x=4 so A is second order. Your rate law will be k[A]^2 ^1 I hope this made sense to you. Good luck