akimhaneul

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does increasing temperature allow the reaction to increase its rate?
 

theonlytycrane

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yeah :) heats those molecules up to move around faster and bump into each other and react
 
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akimhaneul

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yeah :) heats those molecules up to move around faster and bump into each other and react

For this TBR question, why is the answer D? I thought activation energy of a reaction can be determined only from a graph for free energy?

Also, isn't Y releasing more heat as an exothermic reaction? If the temperature increases, then wouldn't that push reaction Y backward more than reaction X?

This is the question. The answer is D.

At higher temperature what is true about pathway X and pathway Y of reaction 1?
A) Pathway X is the kinetic pathway and is more probable than Pathway Y.
B) X is the thermodynamic pathway and is more probable than Y
C) Y is the kinetic pathway and is more probable than X
D) Y is the thermodynamic pathway and is more probable than X.



The attached graph is kinda bad quality...Y axis is enthalpy(H) and X axis is Reaction coordinate. Y is the dotted line and X is the solid line. Y ends at a lower enthalpy than X.

Can anyone explain the thermodynamic vs kinetic pathway of a reaction? Thanks!
 

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aldol16

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Can anyone explain the thermodynamic vs kinetic pathway of a reaction? Thanks!
Also, isn't Y releasing more heat as an exothermic reaction? If the temperature increases, then wouldn't that push reaction Y backward more than reaction X?
You're confusing thermodynamics with kinetics. Exo- vs. endothermic is thermodynamics. Increasing the rate of a reaction by a temperature increase doesn't affect that. It increases the speed and frequency of collisions and therefore increases the probability that a given collision will have enough energy to reach an activated complex and go over the kinetic barrier.

The thermodynamic pathway is the one that leads to the lowest-energy products. The kinetic pathway, if present, is the pathway that has the lowest activation barrier. So in some instances, you might have some molecule on a reaction coordinate that can go left, with a low kinetic barrier but to products with high energy, or right, facing a higher kinetic barrier but to products with comparatively lower energy. The left side would be the kinetic products and the right side would be the thermodynamic products. Tuning reaction temperature will allow you to choose which one you get - if you increase the temperature, the system will settle to the thermodynamic products.

A good example of this is kinetic vs. thermodynamic enolates. If the option is given, the more-substituted enolate will form if the system has enough energy. Therefore, if you want to form the less-substituted enolate, you do the reaction at -70 to prevent the system from reaching the activation barrier for the thermodynamic pathway.
 
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akimhaneul

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You're confusing thermodynamics with kinetics. Exo- vs. endothermic is thermodynamics. Increasing the rate of a reaction by a temperature increase doesn't affect that. It increases the speed and frequency of collisions and therefore increases the probability that a given collision will have enough energy to reach an activated complex and go over the kinetic barrier.
.

Thanks! so the reaction being pushed back or forward when temp changes because of le chatelier when heat is absorbed or released in a reaction is unrelated to kinetic vs thermodynamic concept and I shouldn't really think about those two concepts together?


What if the reaction was like this?
 
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aldol16

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Thanks! so the reaction being pushed back or forward when temp changes because of le chatelier when heat is absorbed or released in a reaction is unrelated to kinetic vs thermodynamic concept and I shouldn't really think about those two concepts together?
What if the reaction was like this?
The reaction being pushed back or forward based on Le Chatelier refers to a thermodynamic effect. If the reaction is reversible and exothermic, then adding heat will shift equilibrium to the left. But it is not a kinetic effect. Kinetically, increasing temperature will always increase the rate of reaction - and for a reversible reaction, it will increase the rate at which equilibrium is achieved - because it increases molecular speed, thereby increasing the chance that a given collision will have enough energy to go over the energetic barrier.

If the reaction is as drawn, then this has nothing to do with kinetic vs. thermodynamic effect because both products are reached by overcoming the same kinetic barrier. Thus, the product distribution will likely be determined by thermodynamics alone. There's no reason for the system to go to B because A has a lower energy with the same activation barrier.
 
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