Thermodynamics: internal energy change

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Homework Help Overview

The discussion revolves around calculating the change in internal energy (ΔU) during the dissociation of molecular hydrogen into atomic hydrogen, using the enthalpy of formation and relevant thermodynamic equations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between enthalpy change and internal energy change, referencing the equation ΔH = ΔU + Δ(PV). There are attempts to calculate ΔU using given values and equations, with some participants questioning the correctness of their results.

Discussion Status

There is ongoing exploration of the assumptions regarding temperature and pressure conditions during the dissociation process. Some participants suggest that the process occurs at constant pressure, while others raise concerns about the sign of the calculated ΔU and the implications of potential energy changes in the system.

Contextual Notes

Participants note the importance of knowing the temperature at which the dissociation occurs, with references to standard conditions. There is also mention of the need to consider the effects of pressure on volume changes during the process.

Dr. Science
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Homework Statement



Enthalpy of formation of a mole of atomic hydrogen = 218kJ. Enthalpy changes when a mole of atomic hydrogen is formed by dissociating half a mole of molecular hydrogen. Calculate ΔU of the process of molecular hydrogen dissociation.

Homework Equations



ΔH = ΔU + Δ(PV)
PV=nRT

The Attempt at a Solution



ΔU = ΔH - Δ(PV)
ΔU = ΔH - Δ(nRT)
ΔU = 218kJ - (1/2 mol)(8.31 J/Kmol)(298K)
ΔU = 216kJ

I really don't think my answer is correct.
 
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Dr. Science said:

Homework Statement



Enthalpy of formation of a mole of atomic hydrogen = 218kJ. Enthalpy changes when a mole of atomic hydrogen is formed by dissociating half a mole of molecular hydrogen. Calculate ΔU of the process of molecular hydrogen dissociation.

Homework Equations



ΔH = ΔU + Δ(PV)
PV=nRT

The Attempt at a Solution



ΔU = ΔH - Δ(PV)
ΔU = ΔH - Δ(nRT)
ΔU = 218kJ - (1/2 mol)(8.31 J/Kmol)(298K)
ΔU = 216kJ

I really don't think my answer is correct.
You will need to know the temperature at which the disassociation occurs. So I assume you are referring to the standard enthalpy of formation, which occurs T = 25C.

dU = dH - d(PV) = dH - PdV -VdP

If the process occurs at constant pressure (eg. at atmospheric pressure) the VdP term is zero. Alternatively, if it occurs at constant volume, the PdV term is zero. Let's assume that it occurs at constant pressure of one atmosphere (and temperature of 25C = 298K).

So:

dU = dH - PdV

Since there are double the number of molecules after disassociation, the volume must double, as you note. I think you should be able to take it from there.

AM
 
so then

ΔU = ΔH -P(dv)
ΔU = ΔH - ∫ nRT/V dv
ΔU = ΔH - nRT ln(vf/vi)
ΔU = 218kJ - 1 mol * 8.21 J/k Mol *298 K * ln2
ΔU = 216kJ ?
 
Last edited:
right?
 
ΔH formation is +'ve

shouldn't my answer be negative since the system has less potential energy.
1/2 H2 -> H
 
Dr. Science said:
so then

ΔU = ΔH -P(dv)
ΔU = ΔH - ∫ nRT/V dv
ΔU = ΔH - nRT ln(vf/vi)
ΔU = 218kJ - 1 mol * 8.21 J/k Mol *298 K * ln2
ΔU = 216kJ ?
Since pressure is constant, ∫PdV = PΔV

AM
 
Dr. Science said:
ΔH formation is +'ve

shouldn't my answer be negative since the system has less potential energy.
1/2 H2 -> H
The system has greater potential energy since it takes energy to separate the hydrogen atoms. (In forming H2 from H atoms, heat would be given off).

AM
 

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