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Half life and activation energy

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I have two questions that I just dont even know where to start

    The activation energy for the reaction CH3CO CH3 + CO is 71 kJ/mol. How many times greater is
    the rate constant for this reaction at 170°C than at 150°C?
    A) 0.40 B) 1.1 C) 2.5 D) 4.0 E) 5.0

    A certain reaction A products is second order in A. If this reaction is 10.% complete after 20. s, how
    long would it take for the reaction to be 90.% complete?
    A) 180 s B) 1600 s C) 440 s D) 18,000 s E) 540 s

    2. Relevant equations

    I know the second one is a second order reaction rate which has
  2. jcsd
  3. Dec 16, 2012 #2


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    Staff: Mentor

    Have you heard about Arrhenius equation?
  4. Dec 16, 2012 #3
    I cant see how I can apply the Arrhenius equation if I dont have any activation energies
  5. Dec 16, 2012 #4
    Activation energy is given in the problem statement.

    You will have to derive an equation using the Arrhenius equation to relate the rate constants. Take logarithm on both the sides of Arrhenius equation. Let the rate constant at temperature T1 be k1 and at temperature T2, let the rate constant be k2. You get the two equations:
    Take logarithm on both the sides of the equation and subtract the equations you get.
  6. Dec 16, 2012 #5
  7. Dec 16, 2012 #6
    Yes, that's right if you mean ln(k1/k2)=Ea/R((1/t_2)-(1/t_1)). (Take care of parentheses. :smile:)
  8. Dec 16, 2012 #7
    ok so I did

    ln(Rate 2/rate1)=71/.008314-((1/423)-(1/443))

    got .911452

    took e^.911452 = 2.5 so 2.5 times greater is rate 2 to rate 1
  9. Dec 16, 2012 #8
    I haven't checked the calculations but your result matches with one of the options so I guess it is correct.
  10. Dec 16, 2012 #9
    Do you know how i could tackle the second one ?
  11. Dec 16, 2012 #10
    You do have posted an equation in the main post. Did you try applying it?
  12. Dec 16, 2012 #11
    Would I just be able to just substitute the percentages in as if they were concentrations?
  13. Dec 16, 2012 #12
    You can do that but the problem is you are not given the percentage of concentration left.
    The percentage left after 20 seconds is 90% of the initial concentration. So substitute [A]=0.9[A0], assuming [A0] to be the initial concentration.
  14. Dec 16, 2012 #13
    (1/.9)=1/.1+k(20 seconds)

    I got k to be .005555

    so If I find t i get



    that look like the right path to get to the answer?
  15. Dec 16, 2012 #14
    How you got 0.1? Did you assume the initial concentration [A0] to be 0.1 M? If so, [A] would be equal to 0.09 M, not 0.9 M.
  16. Dec 16, 2012 #15
    oh i did it wrong then i just assumed it was .1 because 10%=.1
    same for 90% being .9
  17. Dec 16, 2012 #16
    What's the 10% of 0.1?
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