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Fluorides of Xenon

Calculate the...

Question

Calculate the enthalpy of formation of ammonia from the following bond energy data:
$(N - N)$ bond $= 398 k J {m o l}^{- 1}$ ; $(H - H)$ bond $= 435 k J {m o l}^{- 1}$ and $(N \equiv N)$ bond $= 945.36 k J {m o l}^{- 1}$ .

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Solution

$Δ H = [Δ H_{(N \equiv N)} + 3 \times Δ H_{(H - H)}] - [6 Δ H_{(N - H)}]$
$945 + 3 \times 435.0 - 6 \times 389.0 = - 83.64 k J$
Heat formation of ${N H}_{3} = \frac{Δ H}{2} = - \frac{83.64}{2} = - 41.82 k J {m o l}^{- 1}$

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