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Q.

Given the bond energies of N º N, H - Hand N - H bonds are 945, 436 and 391 KJ/mole respectively. The enthalpy of the reaction; N₂ (g) + 3H₂ (g) ® 2NH₃ (g) is

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a

-93KJ

b

-102 KJ

c

93 KJ

d

102 KJ

answer is A.

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Detailed Solution

Enthalpy of reaction is DH^°=Bond dissociation energy – Bond formation energy

$∆ H_{r x n}^{°} = [945 + (3 x 436)] - 6 x 391$

=2253-2346

$∴ H_{rxn}^{°} = - 93 kJ$

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