General Chem Question

[farmin]

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On EK General Chem 30min exam lecture 6 question 126 it says:
Why can the relative strength of HCl and HClO4 be determined in acetic acid but not water?
A. acetic acid's a weaker acid than H3O+
B. acetic acid's a stronger acid than H3O+
C. acetic acid's weaker bronsted lowry base than H2O
D. acetic acid's a stronger bronsted lowry base than H2O

Answer was C. I read the explanation but it was not useful.

What does strong/weak acid have to do with anything? Honestly, I had no idea what the question was even talking about, nevermind how to answer it.

Another question. What is the difference between ΔU and ΔE when we're talking about calorimeters? Like ΔH=ΔU+PΔV and ΔE=q+w. But ΔH=q when P is constant. How the heck does that work? I understand why ΔE=q when ΔV is constant b/c PΔV would equal 0.

Thanks!!

 

[Like]

Both of these acids are strong acids meaning that they would both dissociate completely in water. Using acetic acid as the solvent instead, would prevent these acids from completely dissociating because there aren't as many nucleophilic molecules to take up all the protons as there are in water.


Delta U and delta E refer to the same thing lol.🙄
I think you can figure it out from there.

 

[farmin]

So then it is because H2O can accept H+ to become H3O+ easier than CH3COOH can accept H+ to become CH3COOH2?

And I kinda figured out the ΔH=ΔU+PΔV and ΔE=q+w now. So ΔH=ΔU+PΔV is just ΔH=ΔE+w which is just ΔE=ΔH-w which is just ΔE=q-w. So ΔE=q+w is about doing work on the system but ΔE=q-w is about the system doing work. OK but why would the EK book say ΔH=q only when pressure is constant? Isn't ΔH=q by definition? Is there a situation where ΔH≠q?

 

[mehc012]

So then it is because H2O can accept H+ to become H3O+ easier than CH3COOH can accept H+ to become CH3COOH2?

And I kinda figured out the ΔH=ΔU+PΔV and ΔE=q+w now. So ΔH=ΔU+PΔV is just ΔH=ΔE+w which is just ΔE=ΔH-w which is just ΔE=q-w. So ΔE=q+w is about doing work on the system but ΔE=q-w is about the system doing work. OK but why would the EK book say ΔH=q only when pressure is constant? Isn't ΔH=q by definition? Is there a situation where ΔH≠q?

ΔH=q only when pressure is constant.

So, the confusing part here is that the first equation is really ΔH = ΔU + Δ(PV)
While this may look the same, it's actually different in that changes in pressure also affect enthalpy. Under constant pressure, it turns out to be the equation you're used to (the one chemists use most often because reactions done on the bench are isobaric):

ΔH = ΔU + Δ(PV)_____Δ(PV) = PΔV if P is constant
ΔH = ΔU + PΔV______ PΔV = W in isobaric conditions
ΔH = ΔU + W
ΔH = Q

However, starting from the full equation yields different results under, say, isochoric conditions:

ΔH = ΔU + Δ(PV)_____Δ(PV) = VΔP if V is constant
ΔH = ΔU + VΔP______ If volume is constant, no work is done, W = 0
ΔH = Q + Δ(PV)_____)) Remember that if Q = ΔU + W and W=0, Q = ΔU

Does that make at least a little bit more sense now?

 

[rfvbnmju]

So then it is because H2O can accept H+ to become H3O+ easier than CH3COOH can accept H+ to become CH3COOH2?

Yep that is correct.

Because acetic acid doesn't accept too H well. Only the stronger acid (HClO4 if I recall correctly due to electron withdrawing groups that make the H on there even more polarizable/acidic) would dissociate more.

 

[farmin]

Thanks everyone (especially to mehc for the detailed explanation) it all makes sense now!