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Given that Bond energy of `H_(2)` and `N_(2)` are `400KJmol^(-1)` and `240KJmol^(-1)` respectively and `DeltaH_(f)` of `NH_(3)` is `-120KJmol^(-1)` calculate the bond energy of `N-H` bond :-A. `300KJmol^(-1)`B. `250KJmol^(-1)`C. `410KJmol^(-1)`D. `280KJmol^(-1)` |
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Answer» Correct Answer - D `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)toNH_(3)(g)` `DeltaH_rxn=Delta_(f)(NH_(3))=(1)/(2)B.E(N-=N)+(3)/(2)B.E.(H-H)-3B.E.(N-H)-120=(1)/(2)xx240+400-3B.E(N-H)` `B.E.(N-H)=280KJ//mol.` |
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