March 08, 2021, 01:21:08 AM
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Topic: Question on How to calculate ln k using the Arrhenius Equation from a graph  (Read 34507 times)

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Offline slu1986

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I am trying to figure out how to calculate ln k using the arrhenius equation b/c I have to graph  ln k vs. 1/T

The temperature is 24.7 C  and in Kelvin = 298 K

These are the values that that my graph gave for a first order reaction

m(slope) = -0.0008313
b (Y intercept) = -0.8794


Arrhenius Equation:  ln k = -Ea/R (1/T) + ln (A)<----- this is the y = mx + b  form of the equation, however I am having trouble understanding how to solve it.

 ln k = - 0.0008313/8.314 J/mol K (1/298 K) + ln (-0.8794) <----this is how I set up the numbers but I dont think its right...Could someone please help me understand how to do this so I can solve for my three rate constants and make my graph of  ln k vs. 1/T

Offline Astrokel

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hi.
Quote
These are the values that that my graph gave for a first order reaction

m(slope) = -0.0008313
b (Y intercept) = -0.8794

Graph of ln K against 1/T? if it is, then you subsitute your value in wrongly. ln A should be  = -0.8794. same for the gradient.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline slu1986

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that's what i put for ln A. Can someone help me understand how to get the ln k from my data using the arrhenius equation?

Offline Astrokel

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that's what i put for ln A.

But in your first post, you put c = ln (-0.8794), where it should be c= ln A = -0.8794, no?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline slu1986

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I am so confused about what you are saying. 

Offline Astrokel

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hi,
Quote
m(slope) = -0.0008313
b (Y intercept) = -0.8794


Arrhenius Equation:  ln k = -Ea/R (1/T) + ln (A)<----- this is the y = mx + b  form of the equation, however I am having trouble understanding how to solve it.

 ln k = - 0.0008313/8.314 J/mol K (1/298 K) + ln (-0.8794) <----this is how I set up the numbers but I dont think its right...


What is your y-intecept? It's defined by you as b right, and in actual b = ln A, where A is the pre-exponential factor.
So when you obtain the y-intecept from the graph is -0.8794, right? and should be equivalent to b = ln A = -0.8794
But look at the math which i bolded it, instead of substiuting the whole y-intecept as -0.8794, you actual ln (y-intecept) which is wrong ye?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline jotakabe

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In an Arrhenius plot, the slope is -Ea/R and the intercept is lnA.  If you have a slope and y intercept, you are done!  Just do these calculations.  However, your intercept looks a bit screwy.  The frequency factor is usually a pretty large number, so your y intercept looks to be in error.

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