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Bond Order and Enthalpy

The heat of f...

Question

The heat of formation of NO from its elements is $+ 90 k j {m o l}^{- 1}$ . What is the approximate bond dissociation energy of the bond in NO?
Bond Bond energy Bond Bond energy
N=N 941kj ${m o l}^{- 1}$ O=O 499kJ ${m o l}^{- 1}$

630 kJ.

{m o l}^{1}

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720 k.J.

{m o l}^{1}

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765 k.J.

{m o l}^{1}

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810 k.J.

{m o l}^{1}

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Solution

The correct option is A 630 kJ. ${m o l}^{1}$
$\frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⟶ N O$
Heat of formation = Bond dissociation energy of reactants-Bond dissociation energy of products
$90 = \frac{941}{2} + \frac{499}{2} - (N O)$ bond dissociation energy
(NO) bond dissociation energy
= $\frac{1400}{2} - 90 = 720 - 90 = 630 k J$

Suggest Corrections

Similar questions

N O

+ 90 k J m o l^{- 1}

N O

B E_{N == N} = 941 k J m o l^{- 1}

B E_{O == O} = 499 k J m o l^{- 1}

Δ H (r e a c t i o n) = \sum

- \sum

N O

+ 90 k J m o l^{- 1}

N O

B E_{N \equiv N} = 941 k J m o l^{- 1}

B E_{O = O} = 499 k J m o l^{- 1}

C - H = 413.4 k J {m o l}^{- 1}

C - C = 347.0 k J {m o l}^{- 1}

C = O = 728.0 k J {m o l}^{- 1}

O = O = 495.0 k J {m o l}^{- 1}

H - H = 435.8 k J {m o l}^{- 1}

Δ H_{s u b} C (s) = 718.4 k J {m o l}^{- 1}

N_{2} O

△ H^{\circ} (N_{2} O) = 82 k J m o l^{- 1}

N^{Θ} = N^{⨁} = O

(N \equiv N) b o n d = 950 k J m o l^{- 1}

(N = N) b o n d = 420 k J

^{- 1}

O = O) b o n d = 500 k J m o l^{- 1}

(O = N) b o n d = 610 k J m o l^{- 1}

C H_{4} is 360 k J m o l^{- 1}

C_{2} H_{6} is 620 k J m o l^{- 1}

C - C

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