Bond dissociation enthalpies of H2(g)andN2(g) are 436.0kJmol−1 and 941.3kJmol−1 respectively. Enthalpy of formation of NH3(g) is −46.0kJmol−1. What is the enthalpy of atomization ofNH3.
A
490.2kJ/mol
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B
−490.2kJ/mol
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C
390.2kJ/mol
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D
−390.2kJ/mol
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Solution
The correct option is C390.2kJ/mol We are given, B.E. of H2=436.0kJ/mol B.E. of N2=941.3kJ/mol
Reaction taking place is: 12N2+32H2→NH3
ΔHo(NH3)=B.E.ofreactants−B.E.ofproducts =12(B.E.ofN2bond)+32(B.E.ofH2bond)−3(B.E.ofN−Hbond putting all the values, −46=12(941.3)+32(436.0)−3(B.E.ofN−Hbond
solving this, we get; ⇒B.E.ofN−Hbond=390.2kJ/mol