Question

# Calculate the resonance energy of $$N_2O$$ from the following data:                   $$\Delta H ^{\circ}_f$$ of $$N_2O=82 kJ mole^{-1}$$Bond energies of $$N\equiv N$$, $$N=N$$, $$O=O$$ and $$N=O$$ bonds are 946, 418, 498  and 607 kJ $$mole^{-1}$$ respectively.

Solution

## $$N_2+\dfrac{1}{2}O_2 \rightarrow N_2O\ (N=N=O)$$$$\Delta H_{W.O.R}=(B.E_{N \equiv N}+\frac{1}{2} B.E_{O=O})-(B.E_{N=N}+B.E_{N=O})$$$$\Delta H_{W.O.R}=(946+\dfrac{1}{2} \times 498)-(418+607)=1195-1025=170$$ kJ/molealso$$\Delta H_{W.R}=\Delta H_{W.O.R} - R.E _{Reactant}+ R.E_{product}$$$$82=170-0+R.E _{product}$$$$R.E _{product}=82-170=-88kJ$$$$R.E _{N_2O}=82-170=-88kJ$$Chemistry

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