2008 Destroyer GCHEM #57

[r8 dds]

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Consider the following:

H20(l) + H2O(l) ---> OH-(aq) + H30+(aq)

When the temperature is decreased from 25C, it is found that the Kw has decreased. Which of the following is true?

a) The reaction must be exothermic
b) The reaction must be endothermic
c) The reaction is thermoneutral
d) Not possible. Kw = 1x10^-14 at all temperatures
e) None of these.

Answer: B - the reaction must be endothermic

I don't understand why it's endothermic? When the temperature is decreasing, doesn't that mean you're releasing heat, and thus be exothermic? And when exothermic, the solubility of the water should decrease (ex. making bonds).

 

[nze82]

Consider the following:

H20(l) + H2O(l) ---> OH-(aq) + H30+(aq)

When the temperature is decreased from 25C, it is found that the Kw has decreased. Which of the following is true?

a) The reaction must be exothermic
b) The reaction must be endothermic
c) The reaction is thermoneutral
d) Not possible. Kw = 1x10^-14 at all temperatures
e) None of these.

Answer: B - the reaction must be endothermic

I don't understand why it's endothermic? When the temperature is decreasing, doesn't that mean you're releasing heat, and thus be exothermic? And when exothermic, the solubility of the water should decrease (ex. making bonds).

Kw is the dissociation constant for the forward reaction (dissociation of water). Kw decreases, when temperature decreased. Why does the decrease in Kw represent?
It shows that the forward reaction is slowing down (i.e. the reverse reaction is favored over the forward reaction).
How do we get from here to concluding that the reaction is endothermic?
The question is essentially saying that if you lower the temperature, the reaction moves to the left. Subsequently, heat my be on the product side of the reaction (Endothermic). Why?
When you lower the temperature, while you have heat on the reactant side, it's essentially like removing one of the reactants. Under such circumstances the reaction must shift toward the reactant side to re-establish the equilibrium.