Thermodynamics: internal energy change

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SUMMARY

The discussion focuses on calculating the internal energy change (ΔU) during the dissociation of molecular hydrogen (H2) into atomic hydrogen (H). The enthalpy of formation for atomic hydrogen is given as 218 kJ. The calculation involves the equation ΔU = ΔH - Δ(PV), where Δ(PV) is derived from the ideal gas law. The final calculated ΔU is 216 kJ, assuming the process occurs at constant pressure and temperature (25°C). The participants clarify that the system's potential energy increases during dissociation, which aligns with the positive enthalpy of formation.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first law of thermodynamics.
  • Familiarity with the concepts of enthalpy (ΔH) and internal energy (ΔU).
  • Knowledge of the ideal gas law (PV = nRT).
  • Ability to perform logarithmic calculations related to gas volume changes.
NEXT STEPS
  • Study the derivation and application of the first law of thermodynamics in chemical reactions.
  • Learn about the relationship between enthalpy and internal energy in various thermodynamic processes.
  • Explore the implications of constant pressure vs. constant volume conditions in thermodynamic calculations.
  • Investigate the concept of potential energy in chemical systems and its relation to reaction spontaneity.
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Chemistry students, thermodynamics enthusiasts, and professionals involved in chemical engineering or energy systems will benefit from this discussion.

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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guys...
 
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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