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Chemistry 18 Online
OpenStudy (anonymous):

For an endothermic reaction, how will the value for Keq change when the temperature is increased?

OpenStudy (anonymous):

a) Keq is not effected by temperature. b) Keq will decrease. c) Keq will increase. d) It will be reduced by half. I know it's either B or C .

OpenStudy (jfraser):

since the reaction is endothermic, "heat" is treated just like a \(reactant\), so it will follow the same rules of shifting equilibrium

OpenStudy (anonymous):

I understand that because the temp is increased it will shift towards the product. I just don't know if that means Keq will increase or decrease?

OpenStudy (anonymous):

I think the answer is C ?

OpenStudy (anonymous):

@aaronq @amistre64 do you guys know?

OpenStudy (aaronq):

Here's a hint: \(K_{eq}\) is the equilibrium constant, the ratio of the products to reactants. \(\sf K_{eq}=\dfrac{[products]}{[reactants]}\)

OpenStudy (anonymous):

I understand all that. I know that when the temp increases Keq is shifted towards the products. I just don't know weather or not that means Keq will increase or decrease?

OpenStudy (aaronq):

I'll make it more obvious. if initially at given temperature, [products]=1 and [reactants]=1, then \(\sf K_{eq}=\dfrac{[products]}{[reactants]}=\dfrac{1}{1}=1\) at a higher temp [products]=1.5 and [reactants]=0.5, then \(\sf K_{eq}=\dfrac{[products]}{[reactants]}=\dfrac{1.5}{0.5}=3\)

OpenStudy (anonymous):

thank you! It is C, correct? hahah, I thought it was C.

OpenStudy (aaronq):

yeah. next time try to use numbers, it makes things clearer

OpenStudy (aaronq):

no prob by the way

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