rate expression reaction

[DrKendrickLamar]

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?

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[eeyoreDO]

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

 

[Zerconia2921]

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?

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