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Question: Calculate resonance energy of \(N_{2}O\) from the following data. observed\(\Delta H_{f}^{o}(N_{2}O)...

Calculate resonance energy of N2ON_{2}O from the following data. observedΔHfo(N2O)=82kJmol−1\Delta H_{f}^{o}(N_{2}O) = 82kJmol^{- 1}

B.E. of N≡N⇒946kJmol−1;B.E.N \equiv N \Rightarrow 946kJmol^{- 1};B.E. of

N=N⇒418kJmol−1N = N \Rightarrow 418kJmol^{- 1}

B.E.B.E. ofO=O⇒498kJmol−1;B.E.O = O \Rightarrow 498kJmol^{- 1};B.E. of

N=0⇒607kJmol−1N = 0 \Rightarrow 607kJmol^{- 1}

A

– 88 kJ mol–1

B

– 44 kJ mol–1

C

– 22 kJ mol–1

D

None of these

Answer

– 88 kJ mol–1

Explanation

Solution

N2(g)+12O2(g)→N2ON_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow N_{2}O

N≡N+12O=O→N=N=O\mathbf{N}\mathbf{\equiv}\mathbf{N +}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{O = O}\mathbf{\rightarrow}\mathbf{N = N = O}

Calculated ΔHfo(N2O)=[B.E.(N≡N)+12B.E.(O=O)]−\mathbf{\Delta}\mathbf{H}_{\mathbf{f}}^{\mathbf{o}}\mathbf{(}\mathbf{N}_{\mathbf{2}}\mathbf{O) = \lbrack B.E}\mathbf{.}_{\mathbf{(N}\mathbf{\equiv}\mathbf{N)}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{B.E.(O = O)\rbrack}\mathbf{-}\mathbf{ } $$\mathbf{\lbrack B.}\text{E}\text{.}{\mathbf{(}\mathbf{N}\mathbf{=}\mathbf{N}\mathbf{)}}\mathbf{+ B.E}\mathbf{.}{\mathbf{N = O}}\mathbf{\rbrack}

\mathbf{=}\left\lbrack \mathbf{946 +}\frac{\mathbf{498}}{\mathbf{2}} \right\rbrack\mathbf{-}\left\lbrack \mathbf{418 + 607} \right\rbrack\mathbf{= + 170kJ/mole}

Resonance energy = observed ΔHf0−\Delta H_{f}^{0} - calculated ΔHf0\Delta H_{f}^{0}

=82−170=−88kJmol−1= 82 - 170 = - 88kJmol^{- 1}