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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

Question ID
533398

Created
April 19, 2011 2:11pm UTC

Rating
3

URL
https://questions.llc/questions/533398

Answers
2

Views
12948

2 answers

  1. 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

    Answer ID
    1280004

    Created
    October 25, 2015 10:01pm UTC

    Rating
    6

    URL

  2. 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

    Answer ID
    1280013

    Created
    October 25, 2015 10:06pm UTC

    Rating
    2

    URL

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