Heat enthelpy of molecular hydrogen

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SUMMARY

The discussion focuses on calculating the enthalpy of formation of atomic hydrogen from molecular hydrogen (H2). The enthalpy change for the dissociation of H2 is determined to be -217.97 kJ, with the equation delta(H) = delta(U) + P*delta(V) applied. The pressure-volume work is calculated using PV = RT, yielding a value of 2.5 kJ. The final result for delta(U) is found to be -220.5 kJ, which is then converted to electron-volts for the final answer.

PREREQUISITES
  • Understanding of thermodynamics, specifically enthalpy and internal energy
  • Familiarity with the ideal gas law and its applications
  • Knowledge of unit conversions, particularly between kJ and eV
  • Proficiency in using standard thermodynamic tables
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  • Research the enthalpy of formation for other diatomic molecules
  • Learn about the relationship between enthalpy and Gibbs free energy
  • Explore the concept of bond dissociation energy and its calculations
  • Study the implications of temperature and pressure on enthalpy changes
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Homework Statement


Look up the enthalpy of formation of atomic hydrogen in the back of the book. This is the enthalpy change when a mole of atomic hydrogen is formed by dissociating 1/2 mole of molecular hydrogen (the more stable state of the element). From this number, determinethe energy needed to dissociate a single H2 molecule , in electron-volts.


Homework Equations


PV=3100 J
delta(H)=delta(U)+P*delta(V)
H=U+PV

The Attempt at a Solution


find delta(U)
delta(U)=delta(H)-P*delta(V)

found delta(H) to be delta(H)=-217.97 kJ

According to the standard table T=298 K and P=1 bar=1atm?
R=8.31 J/K

PV=RT=(8.31 J/K)(298 K)=2.5 kJ

H-PV=(-217.97 kJ)-(2.5 kJ)=-220.5 kJ

Did I find delta(U) correctly?
 
Physics news on Phys.org
This is the enthalpy of formation of atomic hydrogen, so the answer needs to be doubled (then the units changed to eV).
 

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