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

Question

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

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 be
A. `-472.5kJ`
B. `-945kJ`
C. `472.5kJ`
D. `945kJ" "mol^(-1)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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)`.

5 views Answered 2023-02-05 15:29