Hi.

As a general rule, increasing the temperature will increase the reaction rate (for exothermic and endothermic) reactions simply because it means more energy available in the system. However, if you have a reversible exothermic reaction, there's a range of temperatures where this might not be true.

Picture the reaction A+B⇌C+D (- ΔH)

The reaction rate (relativelly to A, for example) can be defined as: -rA= K1[A]^a[B]^b - K2[C]^c[D]^D

Both K1 and K2 values depend on the temperature (and the induvidual activation energies for the reactions). So, if for some temperature and concentration ranges, the term K2[C]^c[D]^D is bigger than K1[A]^a[B]^b, the reaction can even occur from the products to the reactants. The difference between the terms is an indication of the direction and rate.

A well known example is the reaction between hydrogen and nitrogen to make ammonia. Very slow at room temperature, increasing rate with rising temperature up to the point where the balance at equilibrium shifts back towards reactants.

So, study well the kinetics of your reaction and you be the judge of the possibility.

Hope this helps!

Hi.

As a general rule, increasing the temperature will increase the reaction rate (for exothermic and endothermic) reactions simply because it means more energy available in the system. However, if you have a reversible exothermic reaction, there's a range of temperatures where this might not be true.

Picture the reaction A+B⇌C+D (- ΔH)

The reaction rate (relativelly to A, for example) can be defined as: -rA= K1[A]^a[B]^b - K2[C]^c[D]^D

Both K1 and K2 values depend on the temperature (and the induvidual activation energies for the reactions). So, if for some temperature and concentration ranges, the term K2[C]^c[D]^D is bigger than K1[A]^a[B]^b, the reaction can even occur from the products to the reactants. The difference between the terms is an indication of the direction and rate.

A well known example is the reaction between hydrogen and nitrogen to make ammonia. Very slow at room temperature, increasing rate with rising temperature up to the point where the balance at equilibrium shifts back towards reactants.

So, study well the kinetics of your reaction and you be the judge of the possibility.

Hope this helps!

I've no ideas about metal hydride reactions. Just general comments. If you define the reaction rate as the rate of consumption (accumulation) of a reagent (product), then the reaction may proceed in different directions with an increase of temperature for a reversible process (as mentioned by Mário Silva ). Keep in mind that theoretically all reactions are reversible.

yes all the exothermic reactions are influenced by the increase of the temperature, adsorption and refoiredissement to maintain the temperature of constant reaction medium avoids the parasitic reactions and aumgente the yield.

Generally, increase in temperature lead to increase in rate of reaction

I agree with Mohammed Jaafar Ali Alatabe

The question is interesting and well formulated. Rahmani Youcef Mohammed Jaafar Ali Alatabe Ahmed Alaa Hussein Have you read it? Your answers are evident text book citations

Zahraa S. Mahdi Can you rationalize your recommendation of Ahmed's post.

I have just opened a discussion about the post such as "following," and "I agree with"

Mário Silva Yurii V Geletii thanks for your valuable comments.

I agree with Mário Silva's comments. Yes, it is possible. When you have one reversible exothermic reaction; as the conversion goes on; first, you have rate increasing by temperature just near thermodynamics equilibrium; then, the reversal products into reactants takes advantage.

In my opinion, the question was incorrectly formulated. The rate of chemical reaction changes during its course. Begin by specifying the reaction system and give how you define the chemical reaction rate. Regards,

If you really need help, share your kinetic data. Otherwise, it's just guessing. Regards

Arun,

A very good question and the answer is yes! Formation of metal hydrides is exothermic so when you are forming a metal hydride, by exposing the metal powder to high pressure hydrogen and some heat (i.e., absorption stage) there is a lot of heat that is given off (depending on the enthalpy of the reaction), therefore the temperature will rise. This will in turn slow down the reaction - it is a thermodynamic effect. At a higher temperature (T2 > T1), you require higher pressure (P2 > P1) for the hydrogen absorption to proceed, otherwise reaction will slow down and basically stop. This is explained by the Van Hoff principle and can be illustrated by the respective isotherms.

In metal hydride beds, heat dissipation during the hydriding phase is quite tricky and higher enthalpy materials are more difficult to deal with. Not to mention that for the desorption stage of higher enthalpy materials (i.e., the endothermic release of hydrogen) you would need to heat them to very high temperatures as well.

Best,

Arman

In autocatalytic reaction, the rate increases and then decreases. Additional temperature increases may or may not result in a temporary increase in its rate.

Hi Miroslaw,

Arun's question is specific to metal hydrides. I am not aware of any metal hydride which shows autocatalytic behavior. In fact, most metal hydrides (simple or complex) suffer from poor kinetics and require some sort of a catalyst, which usually consist of transition metals. When small quantities of metal hydrides are measured (e.g., in a Sievert apparatus), heat dissipation may be sufficient enough as to not impact the absorption rate. In metal hydride beds, designed to store kilograms of hydrogen, however, heat removal is not trivial and active cooling may most likely be required - particularly if fast refilling is required. The literature on this topic is abundant.

Best,

Arman