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Arrhenius equations; A2 Chemistry coursework

  1. Apr 5, 2005 #1
    Hello everyone,
    I'm in my last year of A-level chemistry here in the UK, and am currently writing up my chemistry coursework. Its a project I was given on the kinetics of the reactions between magnesium ribbon and several different acids (I wanted to explore some much more exciting fields, but sadly couldn't!)

    Anyway, I need some help on something...

    i'm trying to use the 'arrhenius equation' to ascertain the activation energy/rate of reaction for a certain acid/particular molarity at a range of different temperatures, as an extension to the original project. I have the range of data (taken from 'inverted titrations' if you will!) for 5 different temperatures- aka the volume of gas produced every 10 seconds over a two minute period. (for example 8.8cm^2 after 10 seconds, 20.2cm^2 after 20 seconds)

    my problem is I was visiting universities in the last lesson before the holidays started, and we were shown how to use k=Ae (-EA / RT), and how this produces a graph of results... involving logarithms and... things. ahem. My maths isn't outstanding sadly, and I've hit a huge dead end with this. I also can't ask any staff as it's due in the day we return- about a weeks time.

    So, heres my desparate plea. If anyone knows what I'm talking about and feels particulary kind, could you please either show me what to do, or point me in the right direction? If this is in the wrong forum or section of the site i apologise, I haven't been active here very long at all!

    Thankyou in advance to any help :smile:

    (p.s. does anyone know any websites that tell you the percentage error of simple scientific equipment such as burettes/ conical flasks/ measuring cylinders etc? or am i wishing on a star? heh)

    <edited for a spelling mistake, theres probably more, its late!>
     
  2. jcsd
  3. Apr 5, 2005 #2

    Gokul43201

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    Fear not; you've come to the right place. If this were math homework, I would not be doing the following for you :

    [tex]k = Ae^{-E_a/RT} [/tex]

    Take natural logarithms on both sides of the above equation. Remember the following properties of logarithms :

    (i) ln(AB) = lnA + lnB
    (ii)[itex]ln(e^x) = x[/itex]

    Applying these above gives you :
    [tex]ln(k) = ln(A) - \frac {E_a}{RT} [/tex]

    This has the form :

    [tex]y = mx + c [/tex] where y = ln(k), c = ln(A), x = 1/T, m = -Ea/R

    In short, if you plot ln(k) vs. 1/T, the slope (m) will tell you the value of Ea.

    You should be able to find out the accuracy in your glassware by calling the manufacturer or looking in their website.
     
  4. Apr 8, 2005 #3
    heys, thanks for trying to help but I'm still rather confused! (call me stupid, I know I am)
    I'm still endevouring to find the value of Ea by plotting ln(k) vs. 1/T as you so helpfully suggested, but im having trouble with actually finding values to plot!
    To explain further: 2.0mol dm^-3 sulphuric acid was reacted with magnesium at 30, 40, 50, 60 and 70°C. I drew a graph with the reactions plotted on it, and drew tangents to the curve at t=0 to achieve 5 initial rates. (0.98, 1.52, 1.78, 1.99 and 2.25cm^3 s^-1 respectively). Are these the figures I use to try to find the "ln rate?"
    To achieve the ln rate, a formulae in a text book quoted (simply) "ln rate = constant - Ea/R x 1/t", and i still don't understand sadly.Obviously I'm not asking for you or anyone else to do this work for me, just help me understand in laymans terms what to do if possible.

    For the 1/T axis, I take it i convert the temperature into degrees kelvin and divide 1 by it...

    Again, any help would be appreciated more then you realise (and when i said my maths was bad, i meant its GCSE standard. I stupidly didn't take it at A-level!)

    Thankyou in advance!
     
  5. Apr 8, 2005 #4

    Gokul43201

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    Yes, they are - if you want to use ln(rate) instead of ln(k). Either way will work; the only difference will be in the c-intercept of the line. The slope of the stright line will be unaffected.

    Alternatively (and you can do this for brownie points, if you feel like it) you can determine the rate constant (k) from the rate equation and the measured initial rates and concentrations. In this case, the reaction is a first order reaction (I would be surprised if not), so the rate equation would be :

    [tex]\frac{d[A]}{dt}(t=0) = -k[A_0] [/tex] where the LHS = initial rate (slope of curve at t=0), k is the rate constant and Ao is the initial concentration of sulphuric acid. So, from each slope (which will be a negative number as it points downwards), you can determine the corresponding rate constant, k.

    I derived nearly this very same formula for you from the Arrhenius Equation, in my previous post. Go over it again, and in the place of k, simply substitute Ro/Ao, where Ro is the initial rate (the term on the LHS of the above rate equation) . It is simply a matter of taking logarithms and applying the rules for logs.

    Correct.
     
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