To determine the activation energy

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SUMMARY

The discussion centers on the determination of activation energy for the reaction between Bromide ion and Bromate(V) ion in an acid solution. The Arrhenius equation, represented as k = Ae^{-E_a/RT}, is transformed into a linear form for graphing, specifically ln(1/t) versus 1/T, where t is the time for the complete disappearance of red color. This approach is suggested because it allows for the comparison of rate constants without directly calculating k. The relationship ln(k1/k2) = -E_a/R (1/T1 - 1/T2) is also highlighted, emphasizing the proportionality of rate to k.

PREREQUISITES
  • Understanding of the Arrhenius equation and its components
  • Familiarity with plotting logarithmic graphs
  • Knowledge of reaction kinetics and rate constants
  • Basic principles of chemical reactions in acid solutions
NEXT STEPS
  • Research the Arrhenius equation and its applications in chemical kinetics
  • Learn how to plot and interpret ln(k) versus 1/T graphs
  • Study the relationship between reaction rates and concentration in chemical kinetics
  • Explore the concept of activation energy and its significance in reaction mechanisms
USEFUL FOR

Chemistry students, researchers in chemical kinetics, and professionals analyzing reaction mechanisms will benefit from this discussion.

Clari
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hi!

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*

Thanks!
 
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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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