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We know that activation energy is independent of temperature. But during the dehydration of alcohol or any reaction, activation reaction can be increased by following the thermodynamically controlled product, which is favourable at high temperature. So how is it possible?

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    $\begingroup$ The thermodynamically controlled product is irrelevant to activation energy. $\endgroup$ Feb 15 '20 at 7:31
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There are two misconceptions in the OP's question

We know that activation energy is independent of temperature.

This is not true in general. Granted, over fairly small temperature intervals (for example, going from zero to hundred degrees Celsius is only about 25% change in absolute temperature terms) Arrhenius plots mostly show a fairly straight line (i.e. constant activation energy). However, there are examples of significant non-Arrhenius behavior, e.g. https://pubs.acs.org/doi/pdf/10.1021/j100310a026.

See also: Is activation energy temperature-independent?

[...] activation reaction can be increased by following the thermodynamically controlled product, which is favourable at high temperature

As pointed out by Ivan in the comments, once you have a thermodnamically controlled product, the kinetics no longer determine the outcome. So the explanation for the different product ratios is not that the activation energies of the two competing reactions have different temperature dependence; instead, at the high temperatures the reverse reactions also take place, and the product of lower Gibbs energy forms (i.e. the product ratio is determined by the equilibrium constants of the two processes).

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Activation energy is an experimental result, obtained from the slope of the log of the rate constant vs $1/T$. It is usually determined over a rather small domain of temperature, much less than $100\ \mathrm{^\circ C}$. Who knows whether this slope is dependent on $T$, if it could be measured between, say, $10\ \mathrm K$ and $1000\ \mathrm K$?

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