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Delta H for the reaction. HCN(g) + 2H...

`Delta H` for the reaction.
`HCN(g) + 2H_(2)(g) to CH_(3) NH_(2)(g)` is -150 kJ.
Calculate the bond energy of -C = N bond.
[Given bond energies of `C-H = 414 kJ mol^(-1)` H-H = `435 kJ mol^(-1)` , C-N = `293 kJ mol^(-1)` N-H = `396 kJ mol^(-1)`

Text Solution

Verified by Experts

`H - C=N + 2H - H to H -overset(H)overset(|)underset(H)underset(|)(C)-N- H`
`DeltaH_("reaction") = sumBE_(R) - sumBE_(R)`
`- 150 = [BC (C-H) + BE (C=N) + 2 xx BE (H-H)] - [ 3 xx BE (C-H) + BE (C-N) + 2 xx`
`BE (N-H) - 150 = [414 + xx + 2435] - [3 xx 414 + 293+2 xx 396]`
`therefore x = 893 kJ mol^(1)` or `BR(C=N) = 893 kJ mol^(-1)`
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