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    I know this question involves the Arrhenius equation. Basically i want to work out the activation energy. We were given the results and everything, and i keep seeming to get too low a number.


    I have attached the results table. If anyone can give me a worked solution i would be extremely grateful.
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  1. File Type: doc Table of results.doc (27.5 KB, 178 views)
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    Will rep (reputation) anyone who can help me.
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    Ill start you off.

    k = Ae^-Ea/RT

    Ea = activation energy
    R = 8.314 J/mol/K
    T = absolute temperature in Kelvins
    A =constant (this constant has little or no effect to temperature, therefore we can call it 1)
    K = rate coefficient
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    deep breath...

    Rate = k[A]^x[B]^y

    if [A]^x[B]^y is kept constant for two reactions then

    Rate1/Rate2 = k1/k2

    as k = Ae^-Ea/RT

    then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2)

    A cancels out

    Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2)

    from expts 1 and 2

    Rate1/Rate2 = 0.49934

    thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2)

    take logs(e)

    ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1)

    ln 0.49934 x 8.314 = Ea [ (T1 - T2)/T1T2]

    (-0.69447 x 8.314) x T1T2/(T1-T2) = Ea

    (-5.7738 x 101099) /(-10) = Ea

    Ea = 58372 J

    Ea = 58.372 kJ

    better check my sums!!
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    (Original post by charco)
    deep breath...

    Rate = k[A]^x[B]^y

    if [A]^x[B]^y is kept constant for two reactions then

    Rate1/Rate2 = k1/k2

    as k = Ae^-Ea/RT

    then Rate1/Rate2 = (Ae^-Ea/RT1)/(Ae^-Ea/RT2)

    A cancels out

    Rate1/Rate2 = (e^-Ea/RT1)/(e^-Ea/RT2)

    from expts 1 and 2

    Rate1/Rate2 = 0.49934

    thus 0.49934 = (e^-Ea/RT1)/(e^-Ea/RT2)

    take logs(e)

    ln 0.49934 = Ea/RT2 - Ea/RT1 = Ea/R x (1/T2-1/T1)

    ln 0.49934 x 8.314 = Ea [ (T1 - T2)/T1T2]

    (-0.69447 x 8.314) x T1T2/(T1-T2) = Ea

    (-5.7738 x 101099) /(-10) = Ea

    Ea = 58372 J

    Ea = 58.372 kJ

    better check my sums!!
    Wow, i really appreciate that. I am going to have to decipher it lol.
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    r u sure thts how u do it. coz we r goiing to do it sum other way
    log(10) = constant + Ea/2.3RT

    can sum 1 work out my Ea aswell PLEASE

    --------------

    Please
    Attached Files
  2. File Type: doc Activation Energy Graph for the Arrhenius Equation.doc (57.0 KB, 208 views)
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    ok, for nandoz...

    take the gradient of your graph.

    gradient = y/x

    gradient = approx. -1250

    gradient = -Ea/R R = 8.31

    -1250 = -Ea/8.31

    -Ea = -1250 x 8.31 = -10400 in joules per mole, I think

    Ea = +10.4 kJ/mol May be wrong
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    (Original post by Nandoz)
    r u sure thts how u do it. coz we r goiing to do it sum other way
    log(10) = constant + Ea/2.3RT

    can sum 1 work out my Ea aswell PLEASE

    --------------

    Please

    there's more than one way to skin a cat (with apologies to the animal rights people out there - no animals were actuallly harmed while making this statement)
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    props for that


    ...do i do the same with this 1?
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  3. File Type: zip Book1.zip (2.8 KB, 125 views)
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    `I guess so, but why isn't the graph the other way up!? :confused:
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    which 1?
    the 1st 1 was worng so i did it again in exel
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    so wot do i do with this...?
    log(10) = constant + Ea/2.3RT
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    this equation is derived from the Arrhenius equation:

    k= Ae^-Ea/RT

    take logs (either natural logs or base 10)

    natural ogs give you:

    lnk = lnA - Ea/RT

    base 10 logs give:

    log(10)k = log(10)A -Ea/2.303RT

    in either case you can plot graphs as they both have the form y = mx + c ( a straight line graph)

    OK s for your equation, log(10)k = log(10)A -Ea/2.303RT

    so, rearranging gives:

    log(10)k = -Ea/2.303RT + log(10)A

    (similar to y = mx + c)

    a plot of log(10) k (the y values) against -1/2.303RT (the x values) gives the Ea as the gradient of the line and log(10) A as the intercept with the y axis.

    your graph is probably upside down because you have your axis the wrong way round
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    have u checked it?
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    have I checked what?
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    My version of the nasty equation is much easier I think :cool:
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    thanks i got it
 
 
 
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