Destroyer GC#41, & 46

[Mrbubbles]

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41. H20 (l) + H20 (l)--> OH-aq + H30+ (aq)

When the temp is decreased from 25C, it is found that Kw has decreased, WHich is true?

a. rxn must be exo
b. rxn must be endo
c. rxn is thermonetutral
d. not possible . kw is always 1x10^-14
e. none of these.

the answer is B. and it talks some nonsense about increasing temp favors endothermic rxn and opposite favors exo. Since rxn temp is decreasing, but Kw is smaller, means that rxn doens't favor exo rxn.

WTH??! WHERE did kaplan talk about this??

for question $46,
c) Hydrogen bonding will cause the solid-liquid line to be skewed to the left in a phase diagram P vs. T.

ok, to me, H-bonding increases the boiling point, therefore decreases vapor pressure...so looking at a linear graph of P vs. T (cuz as P increases, T increase..right?), shouldn't the line be more flat, cuz as temp increases, the h-bonding decreases vapor pressure. so it's not a 1:1 P:T anymore, but a 1:3 ratio..sorry if i confused u, maybe i'm wrong, but please explain how YOU understand it..thanx SDN!!


DAT next friday!!!!!!! ahhh

 

[winkipink]

41. H20 (l) + H20 (l)--> OH-aq + H30+ (aq)

When the temp is decreased from 25C, it is found that Kw has decreased, WHich is true?

a. rxn must be exo
b. rxn must be endo
c. rxn is thermonetutral
d. not possible . kw is always 1x10^-14
e. none of these.

the answer is B. and it talks some nonsense about increasing temp favors endothermic rxn and opposite favors exo. Since rxn temp is decreasing, but Kw is smaller, means that rxn doens't favor exo rxn.

WTH??! WHERE did kaplan talk about this??

destroyer is definitely more in depth than kaplan so don't be surprised. yes increasing temp = endo bc it takes energy to, decreasing temp is exo.
remember K = products/ reactants? small K means more reactants, which means the reverse reaction is favored. if lowering the temperature will shift the equilibrium towards the left by le chatelier, it can be concluded the reverse reaction is endothermic (solution will replace the heat lost by shifting to the left). put heat on the left side of the reaction and you will see. i hope this helps.

 

[lemoncurry]

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