Thermodynamics - Heat of Reaction

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Discussion Overview

The discussion revolves around the heat of reaction in thermodynamics, specifically comparing the heat of reaction at constant pressure and constant volume for a given chemical reaction at 298 K. Participants explore the definitions and implications of state functions, internal energy, and enthalpy in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to calculate the difference between heat of reaction at constant pressure (ΔHp) and at constant volume (ΔHv) using the relationship ΔHp = ΔUp + PΔV and questions the equality of ΔUp and ΔUv.
  • Another participant asserts that ΔU is a state function and is zero for a process between two states, suggesting that ΔUp = ΔUv.
  • Some participants argue that since ΔH is a state function, ΔHp should equal ΔHv, questioning why this is not the case.
  • There is a discussion about the thermodynamic state variables that determine internal energy for an ideal gas and how they relate to constant pressure and constant volume processes.
  • Several participants clarify that ΔH is defined as the enthalpy change at constant pressure and that the terms ΔHp and ΔHv are not standard, suggesting that Qp and Qv might be more appropriate terms.
  • One participant notes that while ΔH can be calculated for processes not at constant pressure, it is conventionally defined at constant pressure.
  • There is a mention that performing a reaction at constant temperature does not yield the same final state as at constant volume, affecting the final pressure and volume.
  • A participant questions whether Δ(PV) equals zero for the reaction at constant volume.
  • The discussion touches on the procedure for measuring the heat of reaction, emphasizing the importance of maintaining constant temperature and pressure during the measurement process.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between ΔHp and ΔHv, the definitions of ΔH, and the implications of state functions. There is no consensus on whether ΔHp equals ΔHv, and the discussion remains unresolved regarding the definitions and calculations involved.

Contextual Notes

Participants highlight potential confusion regarding the definitions of ΔHp and ΔHv, as well as the conditions under which ΔH is calculated. There are also unresolved questions about the behavior of state functions in different thermodynamic processes.

Prathamesh
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Moved from a technical forum, so homework template missing
Calculate the difference between heat of reaction at constant pressure and at constant volume for following reaction at 298 K.
N2(g)+3H2(g)→2NH3(g)

My attempt ~
At constant pressure
ΔHp=ΔUp+PΔV

At constant volume
ΔHv=ΔUv

∴ΔHp-ΔHv
=ΔUp-ΔUv+ PΔV
And PΔV = ΔnRT
But what to do of ΔUp-ΔUv?
In solution,
For both , at constant pressure and volume
ΔU is mentioned and
ΔU-ΔU =0 is given..
But how is it possible that
ΔUp=ΔUv ?
 
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Well, if you move from one state to another, it doesen't matter how you move. ΔU is zero anyway. So yes, ΔUp = ΔUv.
 
CrazyNinja said:
Well, if you move from one state to another, it doesen't matter how you move. ΔU is zero anyway. So yes, ΔUp = ΔUv.

so, ΔHp should also be equal to ΔHv since it is a state function like internal energy U , why it is not so?
 
What thermodynamic state variables determine the internal energy of an ideal gas?

If you know this, then you should be able to compare the final internal energy for the constant pressure process with the final internal energy for the constant volume process.
 
Prathamesh said:
so, ΔHp should also be equal to ΔHv since it is a state function like internal energy U , why it is not so?
ΔH is defined as the enthalpy change of a reaction at constant pressure and is a state function.. There is no such thing as ΔHp and ΔHv. What you must mean is Qp and Qv.
 
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CrazyNinja said:
ΔH is defined as the enthalpy change of a reaction at constant pressure and is a state function.. There is no such thing as ΔHp and ΔHv. What you must mean is Qp and Qv.
ΔH is defined for any process between any two equilibrium states. As you say, H is a state variable. The process need not be at constant pressure in order to calculate ΔH.
 
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TSny said:
ΔH is defined for any process between any two equilibrium states. As you say, H is a state variable. The process need not be at constant pressure in order to calculate ΔH.

To directly calculate ΔH, I agree the process need not be done at constant pressure. But ΔH is defined as such.
 
CrazyNinja said:
To directly calculate ΔH, I agree the process need not be done at constant pressure. But ΔH is defined as such.

ΔH is not defined as the heat produced at constant pressure and temperature, though it is true that ΔH = q for such a process. H is most rigorously defined as H = U + PV.

Also a helpful note. Performing the reaction at constant temperature does not lead to the same final state as performing the reaction at constant volume. In one case, the final volume will be different and in the other case the final pressure will be different.
 
Prathamesh said:
At constant volume
ΔHv=ΔUv
Is this true?

ΔH = Δ(U + PV) = ΔU + Δ(PV).

Does Δ(PV) equal 0 for the reaction at constant volume?
 
  • #10
Ygggdrasil said:
ΔH is not defined as the heat produced at constant pressure and temperature

Constant pressure and temprature @Ygggdrasil ?
 
  • #11
CrazyNinja said:
Constant pressure and temprature @Ygggdrasil ?
Yes. For tabulated heats of reaction, both the temperature and the pressure are specified to not change between the initial state of pure reactants and for the final state of pure products.
 
  • #12
Chestermiller said:
Yes. For tabulated heats of reaction, both the temperature and the pressure are specified to not change between the initial state of pure reactants and for the final state of pure products.
So basically we start at some T,P... do the reaction with reactant A... get product B... bring product B to the same state T,P... measure the heat exchanged... and label this as the heat of the reaction? OK, that actually makes sense.
 
  • #13
CrazyNinja said:
So basically we start at some T,P... do the reaction with reactant A... get product B... bring product B to the same state T,P... measure the heat exchanged... and label this as the heat of the reaction? OK, that actually makes sense.
Yes. That's basically right.
 

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