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Enthalpy

Heat of forma...

Question

Heat of formation of $2$ moles of ${N H}_{3} (g)$ is $- 90 k J$ ; bond energies of $H - H$ and $N - N$ bonds are $435 k J$ and $390 k J$ ${m o l}^{- 1}$ respectively. The value of the bond energy of $N \equiv N$ will be:

- 472.5 k J

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- 945 k J

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472.5 k J

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945 k J

{m o l}^{- 1}

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Solution

The correct option is D $945 k J$ ${m o l}^{- 1}$
$Δ H_{r e a c t i o n} = \sum {(B E)}_{r e a c t a n t s} - \sum {(B E)}_{p r o d u c t s}$
$- 90 = x + 3 \times 435 - 6 \times 390$
$x = 945 k J$ ${m o l}^{- 1}$

Suggest Corrections

Similar questions

(N - N)

= 398 k J {m o l}^{- 1}

(H - H)

= 435 k J {m o l}^{- 1}

(N \equiv N)

= 945.36 k J {m o l}^{- 1}

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