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Question

Use the following data to calculate second electron affinity of oxygen for the process
O(g)+e(g)O2(g).
Heat of sublimation of Mg(s)=147.7kJ/mol
Ionisation energy of Mg(g) to form
Mg2+(g)=2189 kJ/mol
Bond dissociation energy for O2 =498.4 kJ/mol
First electron affinity of O(g)=141 kJ/mol
Heat formation of MgO(s)=601.7 kJ/mol
lattice energy of MgO=3791 kJ/mol

A
601.7 kJ/mol
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B
744.4 kJ/mol
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C
1346.1 kJ/mol
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D
147.7 kJ/mol
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Solution

The correct option is B 744.4 kJ/mol
Given :
subHMg=147.7 kJmol1
dissHO2=498.4 kJmol1
but we require bond dissociation energy for
12O2(g)O(g)
so =498.4/2=249.2 kJ/mol

IEHMg=2189 kJmol1

First electron affinity EAHO=141 kJmol1

fHMgO=601.7 kJmol1
lHMgO=3791 kJmol

to find second electron gain enthalpy of O i.e:
O(g)+e(g)O2(g) ΔH=x

By Born Haber Cycle (Based on Hess's Law) :
601.7=3791+147.7+249.2+2189141+x
x=744.4 kJ/mol

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