Calculate resonance energy of N2O from the following data ΔfH∘N2O=82KJmol−1 Assume structure NΘ=N⊕=O BE of (N ≡ N) bond = 950 kJ mol-1(N=N) bond =420kJmol−1(O=O) bond =500kJmol−1(O=N) bond =610kJmol−1

We calculate theoretical value of ΔfH∘ of N2O. Difference of experimental and theoretical value gives resonance energyN2(g)+12O2(g)⟶NΘ=N=O(g)ΔfH∘N2O=(BE)N≡N+12(BE)O=0−(BE)N=N+(BE)N=0=950+5002−[420+610]=170kJΔfH∘experimental =82 KJ mol−1 Thus, resonance energy of N2O=82−170=−88kJmol−1

Calculate resonance energy of N2O from the following data ΔfH∘N2O=82KJmol−1 Assume structure NΘ=N⊕=O BE of (N ≡ N) bond = 950 kJ mol-1(N=N) bond =420kJmol−1(O=O) bond =500kJmol−1(O=N) bond =610kJmol−1

We calculate theoretical value of ΔfH∘ of N2O. Difference of experimental and theoretical value gives resonance energyN2(g)+12O2(g)⟶NΘ=N=O(g)ΔfH∘N2O=(BE)N≡N+12(BE)O=0−(BE)N=N+(BE)N=0=950+5002−[420+610]=170kJΔfH∘experimental =82 KJ mol−1 Thus, resonance energy of N2O=82−170=−88kJmol−1

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Calculate resonance energy of N₂O from the following data $Δ_{f} H^{\circ} (N_{2} O) = 82 KJ {mol}^{- 1}$ Assume structure $\overset{Θ}{N} = \overset{\oplus}{N} = O$ BE of (N $\equiv$ N) bond = 950 kJ mol^-1
$\begin{aligned} (N = N) bond = 420 kJ {mol}^{- 1} \\ (O = O) bond = 500 kJ {mol}^{- 1} \\ (O = N) bond = 610 kJ {mol}^{- 1} \end{aligned}$

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- 170 kJ mol^-1

-356 kJ mol^-1

-88 kJ mol^-1

-190 kJ mol^-1

answer is C.

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Detailed Solution

We calculate theoretical value of $Δ_{f} H^{\circ} of N_{2} O$ . Difference of experimental and theoretical value gives resonance energy

$\begin{array}{l} N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ \overset{Θ}{N} = N = O (g) \\ Δ_{f} H^{\circ} (N_{2} O) = [(B E)_{N \equiv N} + \frac{1}{2} (B E)_{O = 0}] \\ - [(B E)_{N = N} + (B E)_{N = 0}] \\ = [950 + \frac{500}{2}] - [420 + 610] = 170 kJ \end{array}$