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If the forward reaction is exothermic, and the reverse is obviously endothermic. I getting rather confused. If the temperature is increased the equilibrium will want to decrease the temperature. However, (this is the bit im struggling with) if I want to decrease the temperature wouldn't you shift it to the left? As the reverse reaction is endothermic, and an endothermic reaction causes a decrease in the temperature or is this incorrect?

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    $\begingroup$ True, the reverse reaction would be favored, hence the equilibrium would shift somewhat to the left. Also, welcome to Chem.SE. $\endgroup$ – Ivan Neretin Apr 12 '17 at 15:53
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In order to understand an answer to this you need to be able to differentiate between system and surroundings The system is the reactant mixture , now for an endothermic reaction the temperature of the reactant mixture will increase , however the temperature of the surroundings will decrease since heat is lost by the surroundings to the system , let's assume that an endothermic reaction took place in a test tube , if we put our hand around the test tube our palm will feel cooler since heat was transferred from our palm , however that heat was transferred to the reaction mixture hence the temperature of the mixture was raised . Hence this increase in temperature means we have altered the equilibrium conditions and thus excess heat needs to be lost therefore the eq will shift towards the exothermic reaction . However a decrease in temperature means the eq needs to gain heat which is in the form of an endothermic reaction

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