Help Determining the Activation Energy

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    Activation Energy
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SUMMARY

The discussion focuses on determining the activation energy (Ea) for the reaction X + Y --> Z using the Arrhenius equation. The provided rate constants (k) at various temperatures (0°C, 25°C, 40°C, and 60°C) are essential for this calculation. The Arrhenius equation, k=Ae^-(Ea/RT), can be transformed into a linear form, ln(k) = -(Ea/R)(1/T) + ln(A), allowing for the determination of Ea from the slope of the ln(k) vs. 1/T graph. The correct activation energy calculated from the data is 30.3 Joules.

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  • Understanding of the Arrhenius equation and its components
  • Familiarity with first-order reaction kinetics
  • Ability to perform linear regression analysis on experimental data
  • Knowledge of SI units, particularly Kelvin and Joules
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  • Learn how to plot and interpret Arrhenius plots for reaction kinetics
  • Study the derivation and application of the Arrhenius equation in chemical kinetics
  • Explore the concept of rate constants and their significance in reaction mechanisms
  • Investigate the impact of temperature on reaction rates and activation energy
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Chemistry students, researchers in chemical kinetics, and professionals involved in reaction mechanism studies will benefit from this discussion.

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This question comes from a study guide and I can't get it. Any help would be greatly appreciated.


A number of experiments were performed to determine the effect of the temperature on the rate constant, k, for the reaction X + Y --> Z which had been determined to be first order in X. The data are shown below

Temperature °C...k (the rate constant)
0 ...... 0.010
25...... 0.084
40...... 0.129
60...... 0.176


From this data, determine the value of the activation energy for the reaction.
 
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what do the k (s-1) mean? It is not too clear to me

can you write out the rate law for the reaction?
how can you use the arrhenius equation here?
 
I fixed the format a little, sorry about that. This was the exact question given. The answer given is 30.3 Joules. I have tried everything and can't get an answer close to it. I know that we are supposed to use Arrenhuis's equation and something about a slope of a graph. I just can't put everything together.
 
Ahh much better now. The Arrhenius eq. which you probably have seen before is

k=Ae^-(Ea/RT).

What happens when you take the natural log of both sides?
you get from simple algebra-

lnK=-(Ea/RT) + lnA

Doesn't this look exactly almost like y=mx+b? (here y is the lnk, m is -Ea/R, and x is 1/T) which means this equation is a straight line.

The variables are lnk, and 1/T so if you plot lnk vs. 1/T you will get a straight line. It should be clear how to find Ea now.

MAKE SURE YOU ALWAYS ALWAYS WORK IN SI UNITS! Make sure to use kelvins,joules, etc.
 
Help Determining the Activation Energy Reply to Thread

Ahh, yes that's what I was missing. Thank you very much for helping me out!
 

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