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So I’m starting to understand a little bit better about endothermic and exothermic reactions, but I wanted a little clarification here? As temperature increases, so does the equilibrium constant? Sorry, just wanted to have a clear understanding if anything.
The equilibrium constant will only rise if your reaction is endothermic and you add heat. If you add heat and it's exothermic the equilibrium constant will decrease (this is because your reverse reaction is endothermic and so adding heat will cause a reaction to proceed in reverse direction.
Temperature is the only condition that will alter the K value. In an exothermic reaction, where heat is released, your constant will decrease. If the reaction is endothermic, where heat is absorbed, your constant will increase.
Michelle N - 2C wrote:So I’m starting to understand a little bit better about endothermic and exothermic reactions, but I wanted a little clarification here? As temperature increases, so does the equilibrium constant? Sorry, just wanted to have a clear understanding if anything.
Yeah, as temperature increases, so will the equilibrium constant.
To figure out the reasoning behind the change in equilibrium constant, figure out if heat is a product (exothermic reaction) or a reactant (endothermic reaction). When heat is added, the reaction will shift towards the opposite side. Then, based on whether the reactant or product side is favored, you can determine whether the equilibrium constant increases or decreases.
You can think of this in another way: imagine putting heat as a reactant or product based on if the reaction is endothermic or exothermic. Then you can use le chatelier's principle to determine how K will change. For an exothermic reaction, the heat is on the products side because it is released. If you add heat you will then in turn favor the reverse reaction, lowering K and on the flip side, if you decrease heat the reaction will favor the products side and K will increase. If the reaction is endothermic then the heat sits on the reactant side. If you increase heat you will favor the forward reaction, increasing K, and if you lower heat then you favor the reverse reaction, lowering K.
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