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deleted987519
View attachment 289770
Can someone explain what I'm missing here?
The only thing I can think of is that by "the reaction" they are strictly speaking about the chemical reaction and NOT about the temperature it was performed at, even though they wouldn't be able to provide the DeltaG without it...
Thanks
Actually, temperature IS the deal-breaker here for whether the reaction will be spontaneous or not.
dG = dH - TdS
Under standard conditions, the reaction is evidently spontaneous (dG is negative). The "naught" symbol denotes standard conditions.
However, the entropy of the reaction should be negative (two molecules combining into one reduces dispersal of energy) - - which is unfavorable. (You also know this since choice B is not correct)
So, since the entropy is unfavorable for the reaction, but we know under standard conditions the reaction is spontaneous, the enthalphy - dH - must be favorable (-) (If the enthalphy were also unfavorable - an endothermic (dH = +) reaction requiring an input of heat), this reaction would never be spontaneous.
Since entropy if unfavorable, but the enthalphy is favorable, the reaction could be spontaneous or nonspontaneous depending on the temperature. You can play around with the expression dG = dH - TdS and plug in numbers and switch signs and whatnot to prove this to yourself too.