rate expression reaction

Discussion in 'MCAT Study Question Q&A' started by FiLLiBEANO, May 7, 2008.

  1. YogiDoc23

    YogiDoc23 Flight Surgeon

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    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

    +pissed+ 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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  3. eeyoreDO

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    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 :confused:
     
  4. Zerconia2921

    Zerconia2921 Bring your A-game!

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    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
     

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