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#### slu1986

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##### Question on How to calculate ln k using the Arrhenius Equation from a graph
« on: October 09, 2008, 11:01:39 AM »

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
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#### Astrokel

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #1 on: October 09, 2008, 04:42:57 PM »

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.
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#### slu1986

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #2 on: October 10, 2008, 07:59:06 AM »

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?
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#### Astrokel

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #3 on: October 10, 2008, 08:24:55 AM »

Quote
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?
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#### slu1986

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #4 on: October 10, 2008, 10:33:50 AM »

I am so confused about what you are saying.
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#### Astrokel

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #5 on: October 10, 2008, 01:32:29 PM »

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?
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#### jotakabe

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##### Re: Question on How to calculate ln k using the Arrhenius Equation from a graph
« Reply #6 on: March 11, 2009, 02:11:52 PM »

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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