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Given the bond energirs N=N,H-H and N-H ...

Given the bond energirs `N=N,H-H` and `N-H` bonds are `945,436` and `391KJmol^(-1)` respectively, the enthalpy of the follwing reactlon `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

A

`-93kJ`

B

102 kJ

C

90 kJ

D

105kJ

Text Solution

Verified by Experts

The correct Answer is:
A
Enthalpy of reaction = Bond energy of reactants- Bond energy of products `=945 + 3(436) -2 xx 3 xx 391= -93kJ`
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