Given the bond energies N ≡ N, H − H and N

Question

Given the bond energies

N \equiv N

H - H

and

N - H

bonds are

945, 436

and

391 k J m o l e^{- 1}

respectively, the enthalpy of the following reaction

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Options:

A . -93 kJ

B . 102 kJ

C . 90 kJ

D . 105 kJ

Answer: Option A
:
A

N \equiv N + 3 H - H \to 2 N | H | H - H

945 + 3 x 436

2 \times (3 \times 391) = 2346

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E n e r g y r e l e a s e d     

Net. Energy released =

2346 - 2253 = 93 k J

i.e.,

Δ H = - 93 k J

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