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Question

Consider the following data: ΔfH(N2H4, l)=50kJ/mol, ΔfH(NH3, g)=46kJ/mol, B.E(NH)=393 kJ/mol and B.E(HH)=436 kJ/mol, ΔvapH(N2H4 l)=18 kJ/mol. Calculate the N-N bond energy in kJ/mol for N2H4.

A
190 kJ/mol
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B
190 kJ/mol
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C
95 kJ/mol
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D
95 kJ/mol
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Solution

The correct option is A 190 kJ/mol

Enthalpy of formation:
For one mole of product
ΔH= (
Enthalpy required to break reactants into gaseous atoms) (Enthalpy released to form products from the gaseous atoms)

ΔfHN2H4(l)=50

N2(g)+2H2N2H4(l) (1)

N2H4(l)N2H4(g)= (2)
On addition:
N2(g)+2H2N2H4(g) ΔH=68kJ (3)
For above expression enthalpy in term of bond energy is:
68=(ENN+2EHH)(ENN+4ENH)
68=ENN+2×436(ENN+4×393) (4)

And ΔfHNH3(g)=46kJ
12N2+32H2NH3
Enthalpy in terms of bond energy is:
46=12ENN+(32EHH)(3×ENH)
46=12ENN525
ENN=958kJ
Substituting in equation 4
We get ENN=190 kJ/mol


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