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Heat of formation of `2 mol` of `NH_(3)(g)` is `= -90 kJ`, bond energies of `H-H` and `N-H` bonds are `435kJ` and `390 kJ mol^(-1)`, respectively. The value of the bond enegry of `N-=N` will beA. `-472.5kJ`B. `-945kJ`C. `472.5kJ`D. `945kJ" "mol^(-1)` |
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Answer» Correct Answer - D `N-=+3(H-H)to2underset(H)underset(|)overset(H)overset(|)(N)-H," "DeltaH=-90kJ` `DeltaH_("reaction")=sum(BE)_("reactants")-sum(BE)_("products")` `-90=[(BE)_(N-=N)+3(BE)_(H-H)]-[6(BE)_(N-H)]` `-90=x+3xx435-6xx390` `x=945" kJ "mol^(-1)`. |
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