The isolated O 2− ion is unstable so it is not possible to measure the electron affinity of the O − ion directly. Show how you can calculate its value by using the lattice energy of MgO and the Born-Haber cycle. [Useful information: Mg( s ) → Mg( g ) Δ H ° = 148 kJ/mol.]
The isolated O 2− ion is unstable so it is not possible to measure the electron affinity of the O − ion directly. Show how you can calculate its value by using the lattice energy of MgO and the Born-Haber cycle. [Useful information: Mg( s ) → Mg( g ) Δ H ° = 148 kJ/mol.]
The isolated O2− ion is unstable so it is not possible to measure the electron affinity of the O− ion directly. Show how you can calculate its value by using the lattice energy of MgO and the Born-Haber cycle. [Useful information: Mg(s) → Mg(g) ΔH° = 148 kJ/mol.]
Definition Definition Change in energy of a neutral gaseous atom when an electron is added to the atom to form a negative ion.
Expert Solution & Answer
Interpretation Introduction
Interpretation:
Using the Born-Haber cycle for MgO lattice energy of O− has to be calculated.
Concept Introduction:
Born-Haber cycle is based on Hess’s law to calculate the lattice enthalpy of ionic compounds and deals with energy changes in formation of ionic compounds.
The energy released when gaseous state ions of unlike charges that are infinitely farther apart combine to form a stable ionic solid is called Lattice energy. Conversely, the energy required to break the electrostatic force of attraction between the ions of unlike charges in the ionic solid and revert them to gaseous state is also termed as Lattice energy of an ionic solid.
Electron affinity of an atom refers to the energy released when one electron is added to neutral atom in gaseous state.
Hess’s law is applied to calculate the enthalpy changes in a reaction. According to Hess’s law – “The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes involving in each and every individual steps in the reaction.” Thus if a reaction involves ‘n’ steps then enthalpy change ΔH° of the reaction is,
ΔH°=ΔH1°+ΔH2°+ΔH3°....+ΔHn°
Answer to Problem 9.141QP
Electron affinity of O− is calculated as −844kJ/mol.
Explanation of Solution
Given data:
heat of sublimation of Mg= 148 kJ/mol
The first step of Born-Haber cycle involves sublimation of solid Mg into gaseous Mg
Mg(s)→Mg(g)ΔH1°=148kJ/mol
The second step of Born-Haber cycle involves dissociation of gaseous O2 into gaseous O atoms.
12O2(g)→O(g)ΔH2°=12(498.7)kJ/mol
The third step of Born-Haber cycle is ionization of gaseous Mg into gaseous Mg2+ ions.
The fifth and final step of Born-Haber cycle is formation of solid NaCl as a result of binding gaseous Na+ and Cl− ions together by electrostatic force of attraction.
Mg2+(g)+O(g)2−→ MgO(s)ΔH5°=−3890kJ/mol
ΔH4°'' corresponding to electron affinity of O− ion is calculated by Hess’s law as follows,
Use the vapor-liquid equilibrium data at 1.0 atm. for methanol-water (Table 2-8 ) for the following:
If the methanol vapor mole fraction is 0.600, what is the methanol liquid mole fraction?
Is there an azeotrope in the methanol-water system at a pressure of 1.0 atmospheres?
If water liquid mole fraction is 0.350, what is the water vapor mole fraction?
What are the K values of methanol and of water at a methanol mole fraction in the liquid of 0.200?
What is the relative volatility αM-W at a methanol mole fraction in the liquid of 0.200?
Check the box under each structure in the table that is an enantiomer of the molecule shown below. If none of them are, check the none of the above box under
the table.
||
|II*****
Molecule 1
|
Molecule 4
none of the above
Molecule 2
Molecule 3
Х
mm...
C
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***
Molecule 5
Molecule 6
is SiBr4 Silicon (IV) tetra Bromine?
is KClO2 potassium dihypochlorite ?
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell