[AdiDex]

From High School , I've been reading that Elimination reaction are favorable at high temperature because Their Entropy change is positive , so at higher temperature the Gibbs free energy will be more negative so the equilibrium will be more towards product side .

but I've encountered with a statement in Clayden Organic Chemistry .

"This explanation is simplified because what matters is the rate of the reaction, not the stability of the products. A detailed discussion is beyond the scope of the book, but the general argument still holds. "

So what the reason behind this ? Why Substitution reaction are not that much faster at High Temperature than that of Elimination ?
I have no idea , any Suggestions please . 

[Babcock_Hall]

I am not sure what the answer is, and I am not certain that the reactions are under kinetic control.  However, the entropies of activation of the two processes are likely to be very different.  Therefore, the two processes might have different responses to temperature.

[Irlanur]

Well it says that the explanation is "simplified", which means that it is not the whole truth, and it isn't. Not all reactions necessarily follow thermodynamic control, and if they don't one can not only look at the starting and end products.

[orgopete]

"This explanation is simplified because what matters is the rate of the reaction, not the stability of the products. A detailed discussion is beyond the scope of the book, but the general argument still holds. "

So what the reason behind this ? Why Substitution reaction are not that much faster at High Temperature than that of Elimination ?
I have no idea , any Suggestions please .

In my opinion, this is a little complicated.

Re: kinetic v thermodynamic
This requires a reaction to be reversible. If there is a step that is essentially irreversible, then the reaction will only give the kinetic product. A reaction of butyl chloride with NaI in acetone will give butyl iodide and NaCl. Butyl iodide is the weaker and more easily broken bond. The reaction is done in acetone because NaCl precipitates making the reaction one way. The reverse reaction is slowed due to a low concentration of chloride. If the reaction were done in a solvent in which NaCl and NaI are soluble, then a thermodynamic mixture results. The thermodynamic mixture is a function of bond strength and the concentrations of the halides. In order to get a thermodynamic product, the forward and reverse reaction rates cannot be greatly different. (I'm guessing one could calculate limits for K_eq for a reaction to give a thermodynamic v a kinetic product.)

Re: substitution v elimination
I have a different opinion here as well. It has to do with concertedness. Reaction kinetics tell us what steps are part of the rate determining step, but I argue there are micro-steps involved also. For example in a substitution reaction, if the collision were totally synchronous, then bond making and breaking would be entirely coordinated. If this were billiards, then the object ball would begin to move before the collision ball hits the intermediate ball. If a reaction occurred like this, then a collision would have to precede bond cleavage. The analogy for an SN1/E1 reaction would be the bond would have to break entirely before any interaction with a nucleophile/base occurs. Real reactions are usually somewhere in between. I believe this concept in embodied in the Hammond postulate in which reaction can be reactant like or product like, meaning there is an intermediate timing to reactions.

With the above proviso, then substitution/elimination reactions can have different rates of collision/breakage of bonds. If a reaction conditions increased bond breaking, then we may see the impact of breakage. For this, I like to use a conjugative/hyperconjugative model. If you increase bond breakage, then you will increase the degree to which neighboring electrons can interact in a back-side attack manner. If the electrons are attached to a carbon, then rearrangement is a result. If the electrons are attached to a hydrogen, then a rearrangement or elimination can result. I also reason that breakage and back-side attack will impede nucleophilic attack.

I reason that increasing the temperature will create a greater level of bond breakage to compete with a substitution reaction. That is, even though the rate of a substitution reaction should also increase, the bond breaking and its effect will alter the substitution reaction rate and elimination can become faster as the temperature increases.

[AdiDex]

Someone Gave me This Answer, It's acceptable to me  -


With temperature, the rates of substitution reactions and elimination reactions both increase.

For the purpose of answering this question, let us consider S_N2 and E2 type reactions.

For SN2 type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.

For E2 type reactions, the base attacks the α-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the σ* orbital of the C−H bond, and then the σ orbital interacts with the σ* orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be anti-periplanar with respect to the proton.

With an increase in temperature, the "rate of bond rotation" increases and hence, the probability that a collision with a proper orientation will occur in a given time frame increases. This increase is more prominent in the case of the elimination reaction because it is quite dependent on the orientation of bonds (2 constraints vs. 1 constraint) compared to the substitution reaction.

So, a rise in the temperature will not only affect the thermodynamics of the equation, but it will also directly affect the rate of reaction.

P.S.
I have not spend that much time with chemistry so I am not able to figure out if something is wrong . If there is then please mention it .