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Topic: Calculations of ΔH and ΔU  (Read 9429 times)

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Offline Big-Daddy

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Calculations of ΔH and ΔU
« on: January 07, 2013, 11:55:13 AM »
I know that ΔU=Q+W so for any chemical system ΔU=Q-pΔV (ΔV is V2-V1, so an increase in volume will cause a decrease in internal energy which is what you'd expect from an expansion).

I've also heard that the definition of ΔH is that ΔH=ΔU+pΔV. So this would mean that ΔH=Q?

And in conceptual terms, what is meant by ΔH? If a question says "calculate change in energy" or "calculate amount of energy released" or "calculate amount of heat released", do these each refer to ΔH or ΔU (or Q, if that's somehow different)?

Offline curiouscat

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Re: Calculations of ΔH and ΔU
« Reply #1 on: January 07, 2013, 12:32:21 PM »
And in conceptual terms, what is meant by ΔH?

Enthalpy change.

Quote
If a question says "calculate change in energy"

U Internal energy. In chemistry. Otherwise could be a lot of different things.

Quote
or "calculate amount of energy released"

I'd say ambiguous.

Quote
or "calculate amount of heat released",

Q

Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #2 on: January 07, 2013, 03:17:14 PM »
And in conceptual terms, what is meant by ΔH?

Enthalpy change.

Quote
If a question says "calculate change in energy"

U Internal energy. In chemistry. Otherwise could be a lot of different things.

Quote
or "calculate amount of energy released"

I'd say ambiguous.

Quote
or "calculate amount of heat released",

Q

And, according to what I showed above, ΔH=Q, so ΔH refers to "amount of heat released"? (I assume we're talking about Q being heat released per mole of reaction, since total Q would be to multiply this by the number of moles of the reaction occurring.)

Offline curiouscat

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Re: Calculations of ΔH and ΔU
« Reply #3 on: January 07, 2013, 03:19:48 PM »
And, according to what I showed above, ΔH=Q, so ΔH refers to "amount of heat released"?

Not always. You are talking of a particular special case.

Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #4 on: January 09, 2013, 10:21:37 AM »
And, according to what I showed above, ΔH=Q, so ΔH refers to "amount of heat released"?

Not always. You are talking of a particular special case.

OK, so I have a new understanding now. Let me give it a try:

ΔU=ΔH-pΔV
ΔU=ΔH-ΔngasRT

(Always)

The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

Correct?

If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

What happens if both volume and pressure change? What equations hold the system together then?

Offline curiouscat

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Re: Calculations of ΔH and ΔU
« Reply #5 on: January 09, 2013, 11:31:40 AM »
OK, so I have a new understanding now. Let me give it a try:

ΔU=ΔH-pΔV
ΔU=ΔH-ΔngasRT

(Always)

The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

Correct?


Yes, I think that's right now.

Not (always) for your 2nd Eq. Only Ideal Gases.

Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #6 on: January 27, 2013, 05:07:44 PM »
OK, so I have a new understanding now. Let me give it a try:

ΔU=ΔH-pΔV
ΔU=ΔH-ΔngasRT

(Always)

The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

Correct?


Yes, I think that's right now.

Not (always) for your 2nd Eq. Only Ideal Gases.

Ok ... but what happens if both volume and pressure change? Let us say we can determine Q experimentally (from the amount of heat evolved: Q=-mcΔT/n where n is the number of moles of the reaction that occurred and m is the mass of the reactants), or ΔH experimentally (from a Hess' cycle). How then do we find ΔU and W?

Offline curiouscat

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Re: Calculations of ΔH and ΔU
« Reply #7 on: January 27, 2013, 10:49:16 PM »
Here's another mistake in your analysis:

Quote
If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

If neither P nor V change question of Δ does not arise.

Yes, ΔU=ΔH=Q but only trivially because they will all be zero.  (for a closed, constant n system)


Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #8 on: January 28, 2013, 03:30:55 PM »
Here's another mistake in your analysis:

Quote
If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

If neither P nor V change question of Δ does not arise.

Yes, ΔU=ΔH=Q but only trivially because they will all be zero.  (for a closed, constant n system)

Surely in a reaction that occurs purely in aqueous solution, for instance, you can safely neglect P and V but not Q, ΔU or ΔH? This is the approach my textbook takes.

Offline curiouscat

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Re: Calculations of ΔH and ΔU
« Reply #9 on: January 28, 2013, 03:37:50 PM »
Surely in a reaction that occurs purely in aqueous solution, for instance, you can safely neglect P and V but not Q, ΔU or ΔH? This is the approach my textbook takes.

My bad. I was assuming non reactive gaseous systems.

Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #10 on: January 29, 2013, 12:54:14 PM »
Surely in a reaction that occurs purely in aqueous solution, for instance, you can safely neglect P and V but not Q, ΔU or ΔH? This is the approach my textbook takes.

My bad. I was assuming non reactive gaseous systems.

So the equations I wrote above all still work, for any system? I'll recap:

ΔU=ΔH-pΔV (always)

The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

Of course ΔU=ΔH-ΔngasRT is still true in all cases (as the volume of liquid or aqueous solution in your system cannot change if it all remains liquid or in solution, right?).

Offline Yggdrasil

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Re: Calculations of ΔH and ΔU
« Reply #11 on: January 29, 2013, 02:17:27 PM »
Surely in a reaction that occurs purely in aqueous solution, for instance, you can safely neglect P and V but not Q, ΔU or ΔH? This is the approach my textbook takes.

My bad. I was assuming non reactive gaseous systems.

So the equations I wrote above all still work, for any system? I'll recap:

ΔU=ΔH-pΔV (always)

No, ΔU=ΔH-Δ(pV).  The case you give is valid only for cases of constant pressure.

Quote
The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

Of course ΔU=ΔH-ΔngasRT is still true in all cases (as the volume of liquid or aqueous solution in your system cannot change if it all remains liquid or in solution, right?).

The rest looks good.

Offline Big-Daddy

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Re: Calculations of ΔH and ΔU
« Reply #12 on: January 29, 2013, 07:48:19 PM »
Surely in a reaction that occurs purely in aqueous solution, for instance, you can safely neglect P and V but not Q, ΔU or ΔH? This is the approach my textbook takes.

My bad. I was assuming non reactive gaseous systems.

So the equations I wrote above all still work, for any system? I'll recap:

ΔU=ΔH-pΔV (always)

No, ΔU=ΔH-Δ(pV).  The case you give is valid only for cases of constant pressure.

Quote
The identity of Q (Q=amount of heat released) changes depending on the conditions. If the pressure is constant, Q per mole = ΔH. If the volume is constant, Q per mole = ΔU = change in internal energy as well.

If neither pressure nor volume change, then from the definition of ΔH, ΔU=ΔH=Q (amount of heat released = change in internal energy = enthalpy change).

Of course ΔU=ΔH-ΔngasRT is still true in all cases (as the volume of liquid or aqueous solution in your system cannot change if it all remains liquid or in solution, right?).

The rest looks good.

Thank you, you have clarified a lot!

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