Help Determining the Activation Energy

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    Activation Energy
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The discussion centers on determining the activation energy (Ea) for the reaction X + Y --> Z, which is first order in X, using temperature and rate constant data. The provided data shows the rate constants (k) at various temperatures. To find Ea, the Arrhenius equation, k = Ae^(-Ea/RT), is utilized. By taking the natural logarithm of both sides, the equation transforms into ln(k) = -(Ea/R)(1/T) + ln(A), resembling the linear equation y = mx + b. This indicates that plotting ln(k) against 1/T will yield a straight line, allowing for the calculation of Ea from the slope. Emphasis is placed on using SI units, specifically Kelvin and Joules, to ensure accuracy in calculations. The activation energy was noted to be 30.3 Joules, which the original poster struggled to confirm until the linear relationship was clarified.
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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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