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Reaction Rate and Activation Energy

  1. Feb 4, 2006 #1
    1) Methanol can be produced by the following reaction:
    CO (g) + 2H2 (g) --> CH30H (g)

    How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol?
    -delta[H2]/delta t = ?

    a. + delta [H2]/delta t
    b. + 1delta[CH3OH]/2delta t
    c. + 2 delta [CH3OH]/delta t
    d. none of the above

    Is it d because the rates are supposed to be determined experimentally?


    2) The first order isomerization of cyclopropane -> propene has a rate constant of 0.00026 1/sec at 435 Celsius with an activation energy of 264 kJ/mol. What is the temperature at which the rate constant is 0.00447 1/sec? Calculate your answer in kelvins.

    k2 = 0.00447 1/s
    k1 = 0.00026 1/s
    T1 = 435 C = 708.15 K
    E_a = 264 kJ/mol = 2640 J/mol

    ln (k2/k1) = (-E_a/R)*[(1-T2) - (1/T1)]????

    1/T2 = [-ln (k2/k1)/(-E_a/R)] + (1/T1)
    1/T2 = -ln(0.00447/.00026)/[-2640 J/mol/(8.314)] + (1/708.15 K)
    =0.0096133567 + (1/708.15 K) = 0.011025

    1/0.011025 = 90.69 K = 90.7 K???

    Is the correct final answer and method?

    Thank you.
     
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 4, 2006 #2
    number 1:
    no, it has to do with the mole ratios.

    number 2:
    the equation should be:
    [tex] ln(\frac{K_1}{K_2}) = \frac {E_a}{R} (\frac{1}{T_2}- \frac{1}{T_1}) [/tex]
     
    Last edited: Feb 4, 2006
  4. Feb 5, 2006 #3
    1. So the rates are determined stoichiometrically? (I thought this was wrong.) The answer then would be b.??

    2. ln (k2/k1) = (-E_a/R)*[(1-T2) - (1/T1)] I got this equation in my text. Why is the ratio of k's switched? (The - on the E_a/R will cancel out in solving for T2.)


    Thanks.
     
    Last edited: Feb 5, 2006
  5. Feb 5, 2006 #4
    1. no, hydrogen gas disappears at twice the rate

    2. I'm sorry, I didn't notice the negative sign that you put in front of [itex] E_a [/itex]
     
    Last edited: Feb 5, 2006
  6. Feb 5, 2006 #5

    GCT

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    I'm pretty sure that

    [tex] - \frac{d[H_2]}{2dt}= \frac{d[CH3OH]}{dt} [/tex]

    opposite signs just in case the latex isn't showing up
     
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