How does a change in temperature affect the Kc value when the forward reaction is exothermic?

To explain this I will use a made up example question: A_2(g) + B_2(g) <-> 2AB Change in enthalpy= Negative remember: according to Le Chatlier's principle, if a system in equilibrium is subjected to change, process occur to counteract the change imposed.The forward reaction here is the one that the reactants (A +B) form the product (AB). It is an exothermic reaction, which gives out heat to its surroundings.The system will then increase in temperature. Does Le Chatlier like that? No, it wants to decrease that temperature again back to normal.If the exothermic reaction is the one going forward, the reverse is the endothermic reaction (endothermic takes in heat) The endothermic reaction will absorb the added heat.The position of the equilibrium will move to the left to absorb the added heat and increase the concentration of the reactants.Remember: The equilibrium constant (Kc) is calculated from the concentrations of products over the concentrations over the reactants. (mol dm^-3 ) Kc= Products divided by the reactants.In this situation: Kc= [AB] divided by [A₂][B₂]. The equilibrium will have moved to the left to increase the concentrations of the reactants and decrease the concentrations of the products. The value of the equilibrium constant will then be less.