Confusion about the graph plotted from the Arrhenius equation

According to Arrhenius equation $k = A\mathrm e^{(E/RT)}$, where $k$ is the rate constant, $E$ is the activation energy, $R$ is the universal gas constant and $T$ is the temperature.


Taking log both sides,  

$$\log k = \log A - \frac{E}{2.303RT}$$
Why is the graph plotted between $\log k$ and $1/T$ linear? We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.




[Arrhenius equation](https://en.wikipedia.org/wiki/Arrhenius_equation) 


$k = A \exp{\dfrac{-E}{RT}}$


where:  




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> > k is the rate constant  
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> >  E is the activation energy  
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> >  R is the universal gas constant  
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> >  T is the temperature
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**The line**



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> > *Why is the graph plotted between $\log k$ and $1/T$ linear?*
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Taking log both sides,


$\text{log}(k) = \text{log}(A)-\dfrac{E}{2.303RT}$


A straight line for $y$ in terms of $x$ is:


$y = a + bx$


so:



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> $y$ is $\text{log}(k)$  
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>  $a$ is $\text{log}(A)$  
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>  $b$ is $-\dfrac{E}{2.303R}$  
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>  $x$ is $\dfrac{1}{T}$
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so plotting $\text{log}(k)$ versus $\dfrac{1}{T}$ gives a straight line. 


**E "confusion"**



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> > *We know $E =$ threshold energy—average energy of the molecules. Now average energy is given by $\sqrt{ (8RT/\pi M)}$, it is itself a function of temperature. So the slope is variable and hence should have been some other curve.*
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As you note there is some temperature dependency. But from the Wikipedia article:



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> > "Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature." 
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It is sort of circular reasoning but look at the Arrhenius equation in log form. 



> 
> $\text{log}(k) = \text{log}(A)-(\dfrac{E}{2.303R})\dfrac{1}{T}$
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then



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> $\dfrac{d(\text{log}(k))}{d(1/T)} = -\dfrac{E}{2.303R} $= constant
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if $\text{log}(k)$ and $\dfrac{1}{T}$ are not linear over the temperature range of interest because of a temperature dependency, then the reaction isn't of the "Arrhenius equation" type. Such plots are in fact used to verify confirmatory to the Arrhenius equation over the temperature range of interest.


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The equation you gave for E is the average energy of the molecules, not the **activation energy of the reaction**.


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