Can't Understand this logic from NextStep

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The8

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Just want to get people's opinions on this statement from NextStep. I understand what they are trying to say and I even got the question correct regarding this, but isnt this plain wrong?

"The RATE of cation formation through deprotonation is directly related to the ENERGY required to deprotonate the compound."

I thought rate and energy were not related....
 
When they said "energy" they probably meant "raising temperature" because that is the only practical way to supply "energy" to the reaction.

But my concern is that deprotonation will not result in cations....

Their "lel-gic" must be along the line of "protonation results in cations. Deprotonation opposes that. The more energy required for deprotonation the easier for the reverse (protonation) to happen." and thus, that statement. Seriously...?
 
"The RATE of cation formation through deprotonation is directly related to the ENERGY required to deprotonate the compound."

That's because it's NS logic. Their statement is plain wrong on multiple fronts. First, the rate of X formation is in fact related to the energy required to deprotonate the compound because what they're trying to say with the second half of the sentence is activation energy. If they were referring to the thermodynamic start and end states of the system, then they're just plain wrong. Second, although the rate of reaction is related to the kinetic barrier to reaction, it's not a direct relationship - it's inverse. The higher the kinetic barrier, the lower the rate of reaction. Third, the reason I inserted X above is because deprotonation will never leave you with a cation unless there were already two positive charges on the compound, i.e. it was already a cation. Otherwise, it will leave you with a neutral or anionic compound due to conservation of charge (charge is energy and energy must be conserved in all cases).
 
Thanks guys, very good responses all around. Thats what I thought too.... was questioning my own sanity haha. Ok, so regarding calorimeters, how do these devices give an accurate reading of enthalpy/heat of combustion if activation energy is included. Heats of formations and such should not have activation energy included as they are state functions but when measurements of heat are being made it would be hard if not impossible to separate the two.
 
It's very possible to separate the two. Say your reaction is exothermic. If you put your reaction mixture in a calorimeter and heat it up so that the reaction goes (i.e. to the activation energy), then on the downward side of the reaction coordinate, you release all the activation energy that you put in thereby getting it back and also release the delta H of the reaction.
 
Another nextStep answer.... "A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm.  This results in a section in the middle of each chromosome being exchanged."

The middle section is not being exchanged.... it is simply having two parts switched around it, only two alleles specifically. The rest of the chromosomes are intact. Am i right?
 
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