Calculate resonance energy of N2O from the following data
Hence, △H∘(N2O)=82kJmol−1
Assume structure NΘ=N⨁=O
BE of (N≡N)bond=950kJmol−1 (N=N)bond=420kJ mol−1 O=O)bond=500kJmol−1 (O=N)bond=610kJmol−1
A
−170kJmol−1
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B
−356kJmol−1
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C
−88kJmol−1
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D
−190kJmol−1
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Solution
The correct option is C−88kJmol−1 We calculate theoretical value of △fH∘ of N2O. Difference of experimental and theoretical value gives resonance energy N2(g)+12O2(g)→NΘ=N⨁=O(g)△fH∘(N2O)=[(BE)N≡N+12(BE)O=O]−[(BE)N=N+(BE)N=O]=[950+5002]−[420+610]=170kJ △fH∘ (experimental) = 82kJ
Thus resosance energy of N2O=82−170=−88kJmol−1