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According to Le Chatelier's principle, chemical reactions adjust so as to minimize the effect of changes. Therefore, I believe decreasing the temperature will increase the K value (more products) for an exothermic reaction and decrease the K value for an endothermic reaction. Please correct me if I am wrong.
Victor Li 2A wrote:According to Le Chatelier's principle, chemical reactions adjust so as to minimize the effect of changes. Therefore, I believe decreasing the temperature will increase the K value (more products) for an exothermic reaction and decrease the K value for an endothermic reaction. Please correct me if I am wrong.
This is correct. In an exothermic reaction, heat is being released. When temperature decreases, you're decreasing heat, and so the reaction will move in the forward direction, thereby increasing the value of K. Similarly, if the reaction is endothermic, heat is needed for the reaction to move forward, and if you decrease temperature, the reaction will move in the reverse direction, thereby decreasing the value of K.
Hope this helps!
On the SAPLING with the questions about temp changes and k changes, I found it helpful to think about which direction the reaction will shift based on whether the forward or reverse was endothermic, then whether K would have to increase or decrease to cause a shift in this direction when compared to Q... so by that logic a decrease in temperature will increase the K value of an exothermic reaction because the reaction shifts to products, so Q<K, meaning K is larger than it was before the temp decrease.
This difference is due to the place where heat is being added in the reaction. Because heat is in the reactants side for endo, when decreased reactants are favored. On the other hand, since heat is on the product's side for exo, when decreased products are favored.
An exothermic reaction means heat is being released. During a step - up session the UA described heat as one of the products of the reaction which I found helpful. So, as more heat is released and the temperature decreases, there are essentially more "products", so the K value will increase.
I'm fairly certain you are correct. Endothermic reactions require heat to react, so if the temperature decreases, less products would be made, and therefore K would decrease. I'm not entirely sure about the logic of exothermic reactions, but the decrease in temperature probably increases products being made, so K would increase.
It is important to remember that in equilibrium reactions, if the forward reaction is endothermic, the reverse is exothermic, and vice versa. For this reason, if temperature decreases: an endothermic reaction will shift left and an exothermic reaction will shift right.
I was also a little confused about this on Sapling. From what I understand, if temperature is decreased in an exothermic reaction, more products would be formed since heat is a product. This will increase the value of K, resulting in Q < K. Similarly, if temperature is decreased in an endothermic reaction, more reactants would have to be formed since heat is a reactant. This will decrease the K value, causing Q to be greater than K.
I find it helpful to think about heat as a reactant or product. In an endothermic reaction, heat is being absorbed and is a reactant. Therefore, adding heat will cause the equilibrium to shift towards the products. In an exothermic reaction, heat is being released and is a product. Therefore, adding heat will cause the equilibrium to shift towards the reactants.
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