Equilibrium Constant and Temperature

[imn0tokei]

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Can someone please explain to me what happens to the equilibrium constant when the temperature changed.
This was the question in one of Chad's quizzes:

Which of the following will increase the value of the equilibrium constant for the following reaction?
CaCO3(s) ↔ CaO(s) +CO2(g) ΔH = 178kJ

The answer was increasing the temperature. But I don't quiet understand why.

 

[qwerty934]

constants like Ka Kb Keq Ksp will not be affected unless you change the temperature.

 

[whitepapertree]

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[DentalGirl5]

Okay, so first write out the equilibrium constant expression. We know that solids and liquids are not included in the expression so it only incorporates the CO2 gas.
Ksp = [CO2]. The only thing that can affect the value of an equilibrium constant is a change in temperature. And we know that an increase in temperature ALWAYS favors the endothermic reaction. So given this knowledge, we can look at the given value of deltaH and see that it's positive, which means that the forward reaction is indeed endothermic. Since an increase in temperature favors the endothermic reaction, the reaction will be shifted to the RIGHT with an increase in temperature. Shifting it to the right means a higher CO2 concentration which, given our Ksp expression, means a higher equilibrium constant. Hope that helps!!

 

[tommyngu]

Or to put it in simple terms, the reaction is endorthermic according to the information provided (enthalphy change is positive). (heat would be written on the left) Since the equilibrium constant would be the product of valid terms from the products divided by the product of valid terms from the reactants, you would have to increase the numerator (products) or decrease the denominator (reactants) in order to increase the equilibrium constant. Increasing heat (which is like a reactant) would push the equilibrium to the right thereby increasing the products and decreasing the reactants like we wanted.