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Question: Use the given data to calculate resonance energy of \({{\rm{N}}_{\rm{2}}}{\rm{O}}\). \(\left( {\Delt...

Use the given data to calculate resonance energy of N2O{{\rm{N}}_{\rm{2}}}{\rm{O}}. (ΔHf0 of N2O=82 kJ  mol−1)\left( {\Delta {H_f}^0\,{\rm{of}}\,{{\rm{N}}_{\rm{2}}}{\rm{O}} = {\rm{82}}\,{\rm{kJ}}\,\,{\rm{mo}}{{\rm{l}}^{ - 1}}} \right), Bond energies of N≡N{\rm{N}} \equiv {\rm{N}}, N=N{\rm{N}} = {\rm{N}}, O=O{\rm{O = O}} and N=O{\rm{N}} = {\rm{O}} are 946, 418, 498 and 607 kJ mol−1{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}.
A. 88  kJ mol−188\,\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
B. 77  kJ mol−177\,\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
C. −88 kJ  mol−1 - 88\,{\rm{kJ}}\,\,{\rm{mo}}{{\rm{l}}^{ - 1}}
D. −77  kJ  mol−1 - 77\,\,{\rm{kJ}}\,\,{\rm{mo}}{{\rm{l}}^{ - 1}}

Explanation

Solution

We know that the resonance energy is the difference in the energy of the most stable contributing structure and the resonance hybrid. Here, first we have to calculate the delta heat of formation of N2O{{\rm{N}}_{\rm{2}}}{\rm{O}}. Then we have to subtract the calculated value from the given value of Heat of formation of ΔHf0\Delta {H_f}^0.

Complete step by step answer:
N≡N(g)+12O=O(g)→N=N+N=O{\rm{N}} \equiv {\rm{N}}\left( g \right) + \dfrac{1}{2}{\rm{O = O}}\left( g \right) \to {\rm{N}} = {\rm{N}} + {\rm{N}} = {\rm{O}}.
Now, the expression of heat of formation of N2O{{\rm{N}}_{\rm{2}}}{\rm{O}} is,
ΔfH0{\Delta _f}{H^0} = Bond energy N≡N{\rm{N}} \equiv {\rm{N}}+12\dfrac{1}{2} Bond energy (O=O)\left( {{\rm{O}} = {\rm{O}}} \right) −-Bond Energy(N=N)\left( {{\rm{N}} = {\rm{N}}} \right)+ Bond energy (N=O)\left( {{\rm{N}} = {\rm{O}}} \right)
Given values are as follows:
N≡N{\rm{N}} \equiv {\rm{N}} = 946 kJ mol−1{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
N=N{\rm{N}} = {\rm{N}} = 418 kJ mol−1{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
O=O{\rm{O = O}} = 498 kJ mol−1{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
N=O{\rm{N}} = {\rm{O}} = 607 kJ mol−1{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
ΔfH0{\Delta _f}{H^0} = (946 +12×498 )−(418+607)\left( {{\rm{946}}\, + \dfrac{1}{2} \times 498\,} \right) - \left( {418 + 607} \right)
⇒ΔfH0=(946+249)−1025=170 kJ mol−1\Rightarrow {\Delta _f}{H^0} = \left( {946 + 249} \right) - 1025 = 170\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
We know that the resonance energy is the difference of observed heat of formation and calculated heat of formation.
The calculated heat of formation is 170 kJ mol−1170\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}
and the observed heat of formation is 82 kJ mol−182\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}.
So, the resonance energy will be:
Resonance energy = Observed ΔfH0{\Delta _f}{H^0}-Calculated ΔfH0{\Delta _f}{H^0}
⇒\RightarrowResonance energy = 82−170=−88  kJ mol−182 - 170 = - 88\,\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}

So, the correct answer is Option C.

Additional Information:
Resonance is the phenomenon due to which a molecule can be expressed in different forms and none of the forms can explain all the properties of the molecule. Resonance hybrid is the real structure of the molecule. Resonance structures are only hypothetical structures in the sense that none of them can be prepared in the laboratory. The form in which the molecule exists is the hybrid form.

Note: It is to be noted that greater the value of resonance energy more will be the stability of the hybrid. Relative stabilities of different hybrids can be compared in terms of the resonance energy.