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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.

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  • $\begingroup$ Activation energy can vary but it isn't average energy, but but certain value depending on mechanism of reaction. BTW A varies with temp. but this effect is minor for your calculation $\endgroup$ – Mithoron Nov 16 '15 at 0:08
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Arrhenius equation

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

where:

k is the rate constant
E is the activation energy
R is the universal gas constant
T is the temperature

The line

Why is the graph plotted between $\log k$ and $1/T$ linear?

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:

$y$ is $\text{log}(k)$
$a$ is $\text{log}(A)$
$b$ is $-\dfrac{E}{2.303R}$
$x$ is $\dfrac{1}{T}$

so plotting $\text{log}(k)$ versus $\dfrac{1}{T}$ gives a straight line.

E "confusion"

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.

As you note there is some temperature dependency. But from the Wikipedia article:

"Given the small temperature range kinetic studies occur in, it is reasonable to approximate the activation energy as being independent of the temperature."

It is sort of circular reasoning but look at the Arrhenius equation in log form.

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

then

$\dfrac{d(\text{log}(k))}{d(1/T)} = -\dfrac{E}{2.303R} $= constant

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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    $\begingroup$ I think, OP is asking ‘isn’t $E = f(T)$?’, and ‘Shouldn’t that change the graph?’ $\endgroup$ – Jan Nov 15 '15 at 23:41
  • $\begingroup$ This doesn't provide answer, you probably didn't understood what he's asking. Please be more careful with answering. $\endgroup$ – Mithoron Nov 16 '15 at 0:02
  • $\begingroup$ Jan & @Mithoron - Ok I think I fixed the problem with my answer. Thanks for pointing out the problem. Based on the wording of the question I should have known that the OP was smart enough to know what a "line" was. ;-) $\endgroup$ – MaxW Nov 16 '15 at 0:56
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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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