Metabolics

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So the equilibrium constant is only affected by temperature.

I understand that if I have an exothermic reaction, and I remove heat, the reaction goes forward. This is because the removal of thermal energy from the system INCREASES the kf/kr ratio to give a greater K.

However, what if I had an exothermic reaction and I increased the temperature? I know that the reaction would proceed in reverse, but is this due to a decrease in the equilibrium constant, or due to the system compensating for the heat in some other way?

How can temperature increase favor the reverse reaction by increasing one rate over the other? Isn't it according to Arrhenius equation that temperature will always increase rates of reaction, so the ratio should be unchanged?
 

MedPR

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So the equilibrium constant is only affected by temperature.

I understand that if I have an exothermic reaction, and I remove heat, the reaction goes forward. This is because the removal of thermal energy from the system INCREASES the kf/kr ratio to give a greater K.

However, what if I had an exothermic reaction and I increased the temperature? I know that the reaction would proceed in reverse, but is this due to a decrease in the equilibrium constant, or due to the system compensating for the heat in some other way?

How can temperature increase favor the reverse reaction by increasing one rate over the other? Isn't it according to Arrhenius equation that temperature will always increase rates of reaction, so the ratio should be unchanged?
When doing Le Chatelier problems it is best to think of heat the same way you think of any reactant or any product (depending on if it is exo or endo). Even if you change the rate constant, the reactants/products remain the same. Heat is a reactant in endothermic and a product in exothermic. The reaction may function at a different rate or different energy level, but the directionality will not change.
 

Morsetlis

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Cri du chat syndrome is a deletion of a sizable portion of the short arm of chromosome 5.

;p