Calculating Activation Energy Using Slope and the Arhennius Equation

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The discussion revolves around the application of the Arrhenius equation, specifically how to plot temperature against the rate constant (k) to derive the activation energy (Ea). Participants clarify that by taking the natural logarithm of both sides of the Arrhenius equation, they can linearize the relationship into the form of y = mx + b, where ln(k) is plotted against 1/T. It is emphasized that the slope of this linear plot is negative, and the absolute value of the slope corresponds to Ea multiplied by the gas constant R. The conversation highlights the importance of correctly interpreting the slope and ensuring clarity in communication regarding the constants involved in the equation.
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For some reason, this isn't working out.

The Arhennius equation, k = A*e^(-Ea/RT)

I have tables of temperature and k. The book says just calculate it by having the x-axis be 1/T and y-axis be ln(k)

Ok, take Ln of both sides
ln(k) = ln(A*e^(-Ea/RT))
ln(k) = -Ea/RT + ln(A)

Oh, that looks like y = mx + b
 
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So there's actually no problem.It's a standard way of linearizing exponential laws...

Daniel.
 
Ya, I was just being stupid.
 
Note that Ea will be the absolute value of the slope. The slope will be negative, Ea must be positive.
 
Gokul,if you wanted to be strict,u should have mentioned that Ea is the absolute value of the slope TIMES THE GAS CONSTANT "R"...

Daniel.
 
dextercioby said:
Gokul,if you wanted to be strict,u should have mentioned that Ea is the absolute value of the slope TIMES THE GAS CONSTANT "R"...

Daniel.

I should have.

It's common to make the x-axis 1/RT or \beta (when dealing with numbers instead of moles).
 
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