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Given the bond energies N-N , N-H and H-...

Given the bond energies `N-N` , `N-H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

A

`-93 kJ`

B

`+93 kJ`

C

`+102 kJ`

D

`-102 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
` N equiv N+3H-Hrarroverset(H)overset(|)underset(H)underset(|)2N-H`

Net energy released `=2346-2253 kJ=93 kJ`
`therefore Delta H=-93 kJ`
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