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To determine the activation energy

  1. Mar 9, 2005 #1

    I am having serious problem at plotting the graph for the reaction between Bromide ion and Bromate(V) ion in Acid Solution. The Arrhenius euqation is:
    [tex]k= Ae^{-E_a/RT}[/tex] Taking logarithms on both sides, the equation becomes: [tex]ln k = ln A + E_a/R (1/T) [/tex]

    So generally, a graph of ln k versus 1/T is plotted. However, I don't need to find k in my experiment, so it is suggested that a graph of ln (1/t) versus 1/T is plotted, where t is the time required for the complete disappearing of the red colour.

    of course, if two rate constant values are compared, like: [tex]ln (k1/k2) = -E_a/R (1/T1 - 1/T2) [/tex], the ln(ka/k2) can be replaced by ln(rate1/rate2), since rate is proporational to k....

    Can anybody kindly enough to explain to me why a ln (1/t) versus 1/T is plotted? *confusing*

  2. jcsd
  3. Mar 12, 2005 #2


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    Irrespective of the order of the reaction, you will see that kt(@x=some fixed value) equals either a constant or a function of that fixed point in the reaction co-ordinate. So, it is always posiible to write
    [tex]\frac {1}{t(x=something)} = k*(x=something)^{n-1} [/tex]

    [tex]So,~ln(1/t) = ln(k) + constant [/tex]
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