CHEM please!!! help

posted by .

Calculate the lattice energy of potassium oxide from the following data:

Enthalpy of sublimation of potassium: +89.24 kJ/mol
Bond energy of oxygen: +498 kJ/mol
First ionization energy of potassium: +419 kJ/mol
1st electron affinity of oxygen: -141 kJ/mol
2nd electron affinity of oxygen: +744 kJ/mol
deltaHf potassium oxide: −363.17 kJ/mol


the answer is - 2232 kJ/mol. i do not know how to go about getting this answer. i know youre supposed to add everything and have it equal -363.17 but i don't know if there is something that i should exclude or something that i'm supposed to multiply by two. i just really need help please

  • CHEM please!!! help -

    2K(s) + 1/2O2(g)---->K2O deltaH=−363.17 kJ or better still,
    K2O --->2K(s) + 1/2O2(g)deltaH=363.17
    take note that the equation is reversed above.
    2K(s)---->2K(g) deltaH=2(+89.24 kJ/mol )=178.48kJ
    2K(g)---->2K+(g) deltaH=2(+419 kJ/mol )=838kJ
    1/2O2(g)---->O(g) deltaH=1/2(498 kJ/mol )=249kJ
    O(g)---->O-(g) deltaH=-141kJ
    O-(g)---->O2-(g) deltaH=744
    as such, from the energy cycle (Born Haber cycle for K2O, the LE=-(363.17+178.48+838+249-141+744)=2231.65~2232kJ/mol

  • CHEM please!!! help -

    2K(s) + 1/2O2(g)---->K2O deltaH=−363.17 kJ or better still,
    K2O --->2K(s) + 1/2O2(g)deltaH=363.17
    take note that the equation is reversed above.
    2K(s)---->2K(g) deltaH=2(+89.24 kJ/mol )=178.48kJ
    2K(g)---->2K+(g) deltaH=2(+419 kJ/mol )=838kJ
    1/2O2(g)---->O(g) deltaH=1/2(498 kJ/mol )=249kJ
    O(g)---->O-(g) deltaH=-141kJ
    O-(g)---->O2-(g) deltaH=744
    as such, from the energy cycle (Born Haber cycle for K2O, the LE=-(363.17+178.48+838+249-141+744)=-2231.65~-2232kJ/mol

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    What is the lattice energy of NaH? Use the given information below: Heat of formation for NaH = -56.442 kJ/mol Head of sublimation for Na = 107.3 kJ/mol Ionization energy for Na = 496.00 kJ/mol Bond dissociation energy for H2 = 435.9kJ/mol
  2. Chemistry

    What is the lattice energy of NaH? Use the given information below: Heat of formation for NaH = -56.442 kJ/mol Head of sublimation for Na = 107.3 kJ/mol Ionization energy for Na = 496.00 kJ/mol Bond dissociation energy for H2 = 435.9kJ/mol
  3. ap chem

    calculate the lattice enthalpy of potassium fluoride from the following data: enthalpy of formation of K(g): +89 kJ · mol−1 first ionization energy of K(g): +418 kJ · mol−1 enthalpy of formation of F(g): +79 kJ · mol−1 …
  4. chemistry

    What is the lattice energy of NaBr? heat of formation of NaBr = -362 kJ/mol heat of sublimation for Na = 107.30 kJ/mol ionization energy for Na = 496 kJ/mol bond dissociation energy for Br2 = 190 kJ/mol electron affinity of Br = -325
  5. chemistry please help

    Calculate the lattice energy of potassium oxide from the following data: Enthalpy of sublimation of potassium: +89.24 kJ/mol Bond energy of oxygen: +498 kJ/mol First ionization energy of potassium: +419 kJ/mol 1st electron affinity …
  6. chemistry

    calculate the lattice energy of sodium oxide (Na2O) from the following data: Ionization energy of Na(g): 495 kJ/mol Electron affinity of O2 for 2e: 603 kJ/mol Energy to vaporize Na(s): 109 kJ/mol O2(g) bond energy: 499 kJ/mol Energy …
  7. Chemistry Problem

    Consider an ionic compound, MX, composed of generic metal M and generic halogen X. The enthalpy of formation of MX is ÄHf° = –457 kJ/mol. The enthalpy of sublimation of M is ÄHsub = 121 kJ/mol. The ionization energy of M is IE …
  8. Chemistry

    What is the lattice energy of NaI? Use the given information below. Heat of formation for NaI = -288.0 kJ/mol Heat of sublimation for Na = 107.3 kJ/mol Ionization energy for Na = 496.0 kJ/mol Bond dissociation energy for I2 = 149 kJ/mol
  9. chemistry

    The enthalpy of formation of MX is ΔHf° = –493 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 143 kJ/mol. The ionization energy of M is IE = 437 kJ/mol. The electron affinity of X is ΔHEA = –317 kJ/mol. (Refer …
  10. Chemistry

    Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX2 is ΔHf° = –799 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 157 kJ/mol. The first and …

More Similar Questions