Enthalpy vs Internal Energy question inside

[JFK90787]

Explanation 1:

OK here's my question. I think 'energy of reaction' in the question above just means ΔInternal Energy, for which the equation is

ΔInternal Energy=q+work (where work = -PΔV)

Yes? And enthalpy is just ΔH=q at constant pressure conditions, yes?


Explanation 2:

OK so maybe you're seeing where I'm going with this. Is there a contradiction in the example above where one explanation says enthalpy is a measurement with no work done during a rxn, and that other explanation saying enthalpy is measured at a constant pressure? More moles of gas are made in the product than the reactants in answer A for this question, so don't the products necessarily need to do work on the system in order to keep the pressure constant according to PV=nRT (violating explanation 1)? And vice versa, if no work is able to be done, pressure would need to increase (violating explanation 2)?

tl;dr Basically for the past 6 months I've thought enthalpy was just a measurement of internal energy under constant pressure but now I realize that would just give it the same value as the internal energy equation above and now I'm severely confused and worried I'm only figuring this out 2 weeks before the test

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[JFK90787]

OK upon further investigation I think I might have my definitions of things screwed up

Enthalpy is just H, with H=Internal Energy + PV where pressure and volume are constant.

Change in enthalpy / heat of reaction / energy of reaction is ΔH, with ΔH=ΔInternal Energy +ΔPV, which becomes ΔH=q - work done on the gas + work done on the gas. The works cancel out, and ΔH=q, so ΔH just means the change of heat in a reaction thta took place as if it was in a container where no work was possible. Am I right?

 

[JFK90787]

OK upon further investigation I think I might have my definitions of things screwed up

Enthalpy is just H, with H=Internal Energy + PV where pressure and volume are constant.

Change in enthalpy / heat of reaction / energy of reaction is ΔH, with ΔH=ΔInternal Energy +ΔPV, which becomes ΔH=q - work done on the gas + work done on the gas. The works cancel out, and ΔH=q, so ΔH just means the change of heat in a reaction thta took place as if it was in a container where no work was possible. Am I right?

 

[BennieBlanco]

OK upon further investigation I think I might have my definitions of things screwed up

Enthalpy is just H, with H=Internal Energy + PV where pressure and volume are constant.

Change in enthalpy / heat of reaction / energy of reaction is ΔH, with ΔH=ΔInternal Energy +ΔPV, which becomes ΔH=q - work done on the gas + work done on the gas. The works cancel out, and ΔH=q, so ΔH just means the change of heat in a reaction thta took place as if it was in a container where no work was possible. Am I right?

Ok.

You are right in saying that the Enthalpy is equal to the q(p) - (heat at constant pressure). That is why we define ΔH=ΔInternal Energy +ΔPV at constant pressure.

The easiest way of looking at it is this:

ΔInternal Energy = q + w
ΔPV = -w

Therefore: ΔH=ΔInternal Energy +ΔPV IS EQUIVALENT TO: ΔH = q(p) + w - w = q(p)

The works cancel out, and ΔH=q, so ΔH just means the change of heat in a reaction thta took place as if it was in a container where no work was possible. Am I right?

Yes, it is the change in heat that took place during the reaction. It is the heat flow. Enthalpy comes from a greek word meaning "to warm".

The only difference is the -PΔV work. Therefore when the volume change is negligible (as in water), the ΔV nears zero and therefore ΔH=ΔInternal Energy. In the question above the volume change is NOT negligible, the volume increases and therefore the ΔInternal Energy is actually LARGER than the heat flow we would measure in the reaction... because some of that energy went to changing the volume.

Enthalpy is just H, with H=Internal Energy + PV where pressure and volume are constant.

The volume is NOT constant, but typically is negligibly small. Pressure is constant. Again, this is why we state that internal energy is typically equal to the enthalpy but not always.

Look at it like this...

The question is asking:

when are ΔH and ΔInternal Energy different?

The only solution in which work can be performed by the change in volume is A which adds to the internal energy.

 

[JFK90787]

Thank you, that is exactly what I wanted to read.

I was seriously sitting on the train yesterday for an hour just going over those formulas in my head and wondering to myself what the hell the point was in adding negative work in the ΔH one.

 

[JFK90787]

OK one last question

In Examkrackers there was a question: "Heat transfer in a coffee cup calorimeter (I.E. a constant pressure calorimeter) corresponds to ____, and heat trasnfer in a bomb calorimeter (I.E. constant volume calorimeter) corresponds to _____."

The first blank is enthalpy change, the second blank was energy change. I understand why the first blank had to be enthalpy, since heat is lost as work at constant pressure. But for the second blank, the change in heat should be the same whether using enthalpy or internal energy calculations since the change in volume is 0, yes? Is it just a formality or something to say constant volume calorimeters measure internal energy as opposed to enthalpy?

 

[Eye Cue]

delta E = q - (P)(delta V)
At constant volume, no PV work done, so PV = 0, so the change in internal energy must equal q.

Someone else could probably answer your question better than I could, but this is how I reason it out.

 

[BennieBlanco]

Thank you, that is exactly what I wanted to read.

I was seriously sitting on the train yesterday for an hour just going over those formulas in my head and wondering to myself what the hell the point was in adding negative work in the ΔH one.

I'm glad you asked because I hadn't thought deeply about it until I answered the Q.

 

[BennieBlanco]

OK one last question

In Examkrackers there was a question: "Heat transfer in a coffee cup calorimeter (I.E. a constant pressure calorimeter) corresponds to ____, and heat trasnfer in a bomb calorimeter (I.E. constant volume calorimeter) corresponds to _____."

The first blank is enthalpy change, the second blank was energy change. I understand why the first blank had to be enthalpy, since heat is lost as work at constant pressure. But for the second blank, the change in heat should be the same whether using enthalpy or internal energy calculations since the change in volume is 0, yes? Is it just a formality or something to say constant volume calorimeters measure internal energy as opposed to enthalpy?

So a bomb calorimeter is by definition one held at constant volume.

Remember what we just defined internal energy and enthalpy as? We said enthalpy's equation, while complex, simplifies down to heat flow at constant pressure. We defined that the internal energy isn't the same, but instead it canperform work AND exhibit heat flow.

ΔE = q + w (see both work/heat)
ΔH = q (only heat BUT at constant pressure)

So why the heck do we use a bomb calorimeter? Why do an experiment to keep volume constant? Well, it allows us to easy isolate q! You see, if I keep my volume constant then the work will go to zero. (w = pΔV = 0)

Now I can relate 2 quantities (ΔH and ΔE) by the same measurement, which is q.

So now: ΔE = q (constant V) ΔH = q (constant P)

If both of these q's are the same, then the ΔE = ΔH.

Finishing up, why does the Bomb C. correspond to ONLY ΔE and not ΔH? Because by using the bomb C. I prevent any PV work from being done and ALL the energy goes to HEAT (q)... If I didn't use a bomb C. and there was a volume change, then the internal energy would be difficult to measure because we can measure heat very easily BUT that small amount of work done by the change in volume is hard to measure.

But for the second blank, the change in heat should be the same whether using enthalpy or internal energy calculations since the change in volume is 0, yes? [/B]Is it just a formality or something to say constant volume calorimeters measure internal energy as opposed to enthalpy?

It is not just a formality, it is essential to understanding the concept.

Simple example:

1. A bomb C has a rxn A produces heat of 100. This is all internal energy, it is NOT the same as enthalpy.

2. A rxn A in a coffe C produces heat of 99. This is solely the heat flow.

The situation #1 produced more heat, because at constant volume there was no ability to perform work so that energy converted to heat.

Situation #2 produced less heat because enthalpy only accounts for the heat flow.

 

[BennieBlanco]

delta E = q - (P)(delta V)
At constant volume, no PV work done, so PV = 0, so the change in internal energy must equal q.

Someone else could probably answer your question better than I could, but this is how I reason it out.

👍

 

[Kalyx]

So a bomb calorimeter is by definition one held at constant volume.

Remember what we just defined internal energy and enthalpy as? We said enthalpy's equation, while complex, simplifies down to heat flow at constant pressure. We defined that the internal energy isn't the same, but instead it canperform work AND exhibit heat flow.

ΔE = q + w (see both work/heat)
ΔH = q (only heat BUT at constant pressure)

So why the heck do we use a bomb calorimeter? Why do an experiment to keep volume constant? Well, it allows us to easy isolate q! You see, if I keep my volume constant then the work will go to zero. (w = pΔV = 0)

Now I can relate 2 quantities (ΔH and ΔE) by the same measurement, which is q.

So now: ΔE = q (constant V) ΔH = q (constant P)

If both of these q's are the same, then the ΔE = ΔH.

Finishing up, why does the Bomb C. correspond to ONLY ΔE and not ΔH? Because by using the bomb C. I prevent any PV work from being done and ALL the energy goes to HEAT (q)... If I didn't use a bomb C. and there was a volume change, then the internal energy would be difficult to measure because we can measure heat very easily BUT that small amount of work done by the change in volume is hard to measure.

It is not just a formality, it is essential to understanding the concept.

Simple example:

1. A bomb C has a rxn A produces heat of 100. This is all internal energy, it is NOT the same as enthalpy.

2. A rxn A in a coffe C produces heat of 99. This is solely the heat flow.

The situation #1 produced more heat, because at constant volume there was no ability to perform work so that energy converted to heat.

Situation #2 produced less heat because enthalpy only accounts for the heat flow.

Thank you, that was a fantastic explanation! 🙂

~Kalyx