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Arrhenius equation 
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#1
Apr1514, 08:50 AM

P: 343

I wanted to see how the rate of change of rate constant with temperature (dk/dT) changes with activation energy. I tried to do this with differentials: k=A*e^{EA/RT} so
[tex]\frac{dk}{dT} = \frac{A E_A \cdot e^{\frac{E_A}{RT}} } {RT^2}[/tex] and then [tex]\frac{d(\frac{dk}{dT})}{dE_A} = A \cdot e^{\frac{E_A}{RT}} \cdot (\frac{1}{RT^2}  \frac{1}{R^2T^3})[/tex] So long as E_{A}>RT, it appears that rate of change of rate constant with temperature would decrease for increasing activation energy (i.e. if you increase activation energy, then the rate constant is less susceptible to increasing when you change the temperature). E_{A}>RT would be almost universally the case I imagine  at 298 K that's less than 2.5 kJ/mol. But this doesn't tally up with what we know, which is that for a given temperature increase, a greater activation energy leads to a greater increase in rate constant. e.g. if we write wellknown k_{2}/k_{1} = exp( E_{A}/R * (1/T_{2}^{2}  1/T_{1}^{2}) ) and try some values we get this conclusion. So what went wrong or what is wrong with my idea to do it by differentiation? 


#2
Apr1814, 07:17 AM

P: 343

k_{2}/k_{1} = exp( E_{A}/R * (1/T_{1}  1/T_{2}) ) No squared terms. 


#3
Apr1814, 07:57 PM

Mentor
P: 5,402




#4
Apr2014, 03:53 PM

P: 343

Arrhenius equation
[tex]\frac{d(\frac{dk}{dT})}{d(E_A)} = \frac{A}{RT^2} \cdot e^{\frac{E_A}{RT}} \cdot (1  \frac{E_A}{RT})[/tex] But it seems that my original conclusion still holds true  that E_{A} < RT (incredibly small activation energy) is necessary for rate of change of rate constant with temperature to increase at a higher activation energy, which however is the trend which the expression for k_{2}/k_{1} indicates is true for all E_{A} (it seems to me)? 


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