relation between Q and deltaH

[mcgill2012]

Advertisement - Members don't see this ad

can someone clarify what the relation is between Q and enthalpy? ie: in calorimetry, are they measuring enthalpy? and what is the difference between what is measured in constant pressure vs constant volume calorimetry?

 

[Rabolisk]

In your typical thermochemistry problems and applications, enthalpy change is same thing as heat. In more technical terms...

H (enthalpy) = U (internal energy) + pV
dH = dU + d(pV)
dH = dq - pdV + pdV at constant pressure
dH = dq

Enthalpy equals heat at constant pressure.

In constant pressure calorimetry, enthalpy is directly measured. In constant volume calorimetry, heat is measured, but the difference between heat and enthalpy are usually small. Both measure these values indirectly by measuring the temperature change of a substance with some known heat capacity.

 

[mcgill2012]

thank you!
im still confused about constant volume calorimetry. am i allowed to consider heat measured as enthalpy? or is it better to consider it the internal energy?

 

[Rabolisk]

Technically it is internal energy in that case, but don't sweat it if the question asks you about enthalpy of combustion rather than heat of combustion.

 

[sangria1986]

thank you!
im still confused about constant volume calorimetry. am i allowed to consider heat measured as enthalpy? or is it better to consider it the internal energy?

In constant volume calorimetry...enthalpy cannot be measured. However, as you mentioned, q = energy. The fact is...the values are about the same, its just different ways of coming to the same answer because of differences in what is kept constant : pressure or volume.

Normally, constant volume calorimetry is also known as bomb calorimetry where the container is a rigid solid. this is useful for reactions that are gaseous but to ensure that all the energy transformed is in the form of heat and not work(pV where volume has to increase) the volume is kept constant by placing the reaction in a rigid container. Thus, whatever energy is measured (heat capacity *temp) = the heat of the reaction.

sometimes it maybe useful to use constant pressure calorimetry. this is when the pressure is kept constant and is useful for reactions that are solids or in solution (ie no gas evolves so no change in volume or pressure). In these cases, the heat evolved equals enthalpy which equals the energy. E=q=H. These reactions are solved by moles (or grams depending on units) * the specific or molar heat capacity * temperature. again both constant volume and constant pressure calorimetry give you the heat evolved...the goal is to determine the amount of heat evolved in a reaction.

There are instances where in a constant pressure calorimetry both energy and heat values are different. in these cases, gasses are probably reacting and the overall reaction changes the pressure of the reaction. This means that E does not equal q but E=q+w. I doubt such questions will arise in the MCAT. but for these equations, E= q+w =H+ -nRT where n is the change in moles in the reaction.

 

[mcgill2012]

wow, thank you!

 

[vsl5]

so if delta H = delta U + pdeltaV

then at constant volume (eg. bomb calorimeter) which delta U = q
doesnt delta H = q as well since delta V = 0 and delta U= q?

 

[Rabolisk]

No, not exactly. dH = dU + d(pV). Only at constant pressure can we say that dH = dU + pdV. Note the subtle differences. At constant volume, dU = q, but dH = q + Vdp.

 

[sangria1986]

No, not exactly. dH = dU + d(pV). Only at constant pressure can we say that dH = dU + pdV. Note the subtle differences. At constant volume, dU = q, but dH = q + Vdp.

huh? i thought its U that equals h + pv.... that during constant pressure, dU= dH+ pdV and constant volume, dU= q only.... also i dont think we need to know dH during constant volume...its probably not tested on MCAT right?

 

[Rabolisk]

Most of the things stated in this thread are not tested on the MCAT. However, U does not equal h + pv. dU = q + w (sign convention for work varies among textbooks). H refers to enthalpy, not heat.

 

[Chocolatebear89]

sometimes it maybe useful to use constant pressure calorimetry. this is when the pressure is kept constant and is useful for reactions that are solids or in solution (ie no gas evolves so no change in volume or pressure). In these cases, the heat evolved equals enthalpy which equals the energy. E=q=H. These reactions are solved by moles (or grams depending on units) * the specific or molar heat capacity * temperature. again both constant volume and constant pressure calorimetry give you the heat evolved...the goal is to determine the amount of heat evolved in a reaction.

If E=Q=H for constant pressure, when do we use U=q+PdV? Since pressure isn't changing, why can't we just use it?

 

[Rabolisk]

If E=Q=H for constant pressure, when do we use U=q+PdV? Since pressure isn't changing, why can't we just use it?

We can use it, but it wouldn't be useful. What we can measure is q or dH through dT. Also E does not equal merely heat at constant pressure, but when it is also constant volume. All this is rather academic; the important thing is to understand the purpose of calorimetry.

 

[Chocolatebear89]

So finding dH is more useful, because it is independent of pathway?

 

[Rabolisk]

You find heat, q, through calorimetry. Enthalpy corresponds to heat at constant pressure. You typically find change in internal energy when dealing with ideal gases expanding or compressing.

 

[fizzgig]

ok so lemme get this straight, maybe.

dH = dU + d(pV) by definition

dH = dU + PdV at constant pressure with volume changing (this is the case in neither calorimeter)

dH = dU when you have constant pressure and the volume also does not change (if you are working with a coffee cup calorimeter with a liquid solution that doesn't evolve gas, this is representative, and my understanding is that's what the ccupcal is used for). here E=q+w, but in a closed system U=q+w and no mass is leaving here so that is valid. we've already said there's essentially no volume change here, so U=q since no pv work is done. so dH=dU and dU = q so dH=q. we directly measure heat by q=mcdT and that's how we get an enthalpy of reaction.

dH = dU + VdP when you have constant volume only (this is the bomb calorimeter). still no mass escaping, so E=U=q+w, and again no volume change means U=q. now we have dH= q + VdP. all of this looks measurable, so i take it that from here we find q via mcdT, measure V, Pinit, Pfinal, and directly calculate dH?

does all that jazz sound about right? thanks, my thermo people!

 

[sangria1986]

ok so lemme get this straight, maybe.

dH = dU + d(pV) by definition

dH = dU + PdV at constant pressure with volume changing (this is the case in neither calorimeter)

dH = dU when you have constant pressure and the volume also does not change (if you are working with a coffee cup calorimeter with a liquid solution that doesn't evolve gas, this is representative, and my understanding is that's what the ccupcal is used for). here E=q+w, but in a closed system U=q+w and no mass is leaving here so that is valid. we've already said there's essentially no volume change here, so U=q since no pv work is done. so dH=dU and dU = q so dH=q. we directly measure heat by q=mcdT and that's how we get an enthalpy of reaction.

dH = dU + VdP when you have constant volume only (this is the bomb calorimeter). still no mass escaping, so E=U=q+w, and again no volume change means U=q. now we have dH= q + VdP. all of this looks measurable, so i take it that from here we find q via mcdT, measure V, Pinit, Pfinal, and directly calculate dH?

does all that jazz sound about right? thanks, my thermo people!

i dont know about the last part.... im not sure that the pressure values need be calculated for constant volume calorimetry. its really done the same as above...just q= mcdT...honestly the only difference i notice with constant pressure and constant volume calorimetry is that the constant used for heat capacity doesnt include grams or moles for constant volume calorimetry...its just Joules/Kevin or Joules per celsius with no use of mass or moles...im not sure why this is the case.

 

[Rabolisk]

ok so lemme get this straight, maybe.

dH = dU + d(pV) by definition

dH = dU + PdV at constant pressure with volume changing (this is the case in neither calorimeter)

dH = dU when you have constant pressure and the volume also does not change (if you are working with a coffee cup calorimeter with a liquid solution that doesn't evolve gas, this is representative, and my understanding is that's what the ccupcal is used for). here E=q+w, but in a closed system U=q+w and no mass is leaving here so that is valid. we've already said there's essentially no volume change here, so U=q since no pv work is done. so dH=dU and dU = q so dH=q. we directly measure heat by q=mcdT and that's how we get an enthalpy of reaction.

dH = dU + VdP when you have constant volume only (this is the bomb calorimeter). still no mass escaping, so E=U=q+w, and again no volume change means U=q. now we have dH= q + VdP. all of this looks measurable, so i take it that from here we find q via mcdT, measure V, Pinit, Pfinal, and directly calculate dH?

does all that jazz sound about right? thanks, my thermo people!

Seems right to me. Standard heat of reaction/formation/combustion is the same as standard enthalpy of reaction/formation/combustion because standard state is considered to be constant 1 atm. Calorimetry is generally not standard state.

 

[fizzgig]

ok so with regard to the last two answers...

for the bomb calorimeter you do have dH=q+VdP right? volume is constant, but pressure IS changing so i can't get rid of that term can I? i have to keep VdP in the equation, measure them, and use them in calc?

i don't see mathematically how i get to drop out VdP for that one.

 

[Rabolisk]

Yep. With any noticeable change in pressure, enthalpy will be different from heat. Generally though, I don't think we care too much.

 

[fizzgig]

ok, i think i'm on the train then now.

thanks a lot!

 

[Don Draper]

I would guess this thread is done 12-18 times per year.

And the MCAT basically never tests it. 15 FLs and hundreds of chemistry practice passages (~700 questions) and I never came across anything with this type of depth.

This is where prep companies (or individuals) forget what the MCAT is about and go on long unhelpful excursions.

 

[Don Draper]

change of volume is insignificant.

reactions are done in water typically so the change is negligible.

This reminds me of the time when Berkeley Review spent 7 pages to go over heat engines (not an MCAT topic), or gave a list of ~30 weak acids to know.

This is why some people prep in 1/2 the time of others.

 

[fizzgig]

maybe, but it took me 2 posts and 10-20 minutes grabbing notes and checking a few sites, and getting the basics makes me more confident if i see anything even remotely relating to this, so i'm fine with it.

some people have a disease where they need to understand things kinda fully or they have trouble using them at all 🙂

 

[Don Draper]

some people have a disease where they need to understand things kinda fully or they have trouble using them at all 🙂

I understand. I am the same way. You will see me writing in posts like these sometimes.

I'm trying to help you.

If you do that for the MCAT, you will waste lots of time. The goal is to get the correct answers and master the fundamentals. This topic is not tested (watch as you do practice problems). How would you make a multiple choice question testing this?

Good for science, bad for MCAT.👍

Simplify.

 

[fizzgig]

totally understood, don, and i'm not gettin on your back either.

sometimes i push because i just wanna get it, and i understand that it's not a do or die topic/depth of topic for the mcat. and that's fine if that's how i wanna spend that time i guess.

other times i may well be panicking with the herd and think i neeeeeeed this for 10 points on the mcat oh nooooes!! hahaha... and having people let me know what's key and what's not is a GOOD thing.

party on, wayne.