Question

# Consider the following data: ΔfH∘(N2H4, l)=50kJ/mol, ΔfH∘(NH3, g)=−46kJ/mol, B.E(N−H)=393 kJ/mol and B.E(H−H)=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/molEnthalpy 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)+2H2→N2H4(l) (1) N2H4(l)→N2H4(g)= (2) On addition: N2(g)+2H2→N2H4(g) ΔH=68kJ (3) For above expression enthalpy in term of bond energy is: 68=(EN≡N+2EH−H)−(EN−N+4EN−H) 68=EN≡N+2×436–(EN−N+4×393) (4) And ΔfHNH3(g)=−46kJ 12N2+32H2→NH3 Enthalpy in terms of bond energy is: −46=12EN≡N+(32EH−H)−(3×EN−H) −46=12EN≡N−525 EN≡N=958kJ Substituting in equation 4 We get EN−N=190 kJ/mol

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