Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?