Question

Calculate the resonance energy of ${{N}_{2}}O$ from the following data:
$\Delta {{H}_{f}}^{\circ }\,$of ${{N}_{2}}O$ = $134KJ\,mo{{l}^{-1}}$
Bond energy of $N-O$, $N=O$, $N\equiv N$ and $O=O$ bonds are 222, 607, 946 and
498 $kJ\,mo{{l}^{-1}}$ respectively.

Answer 1

Hint : Many compounds exhibit resonance. Due to resonance they exist in a structure which is different from the expected one and more stable. Resonance energy can be calculated by the formula:
Resonance energy = $\Delta {{H}_{f}}^{\circ }(actual)-\Delta {{H}_{f}}^{\circ }(calculated)$

Complete step by step solution :
Before we start solving let us define some basic terms:
The standard enthalpy change of formation ($\Delta {{H}_{f}}^{\circ }\,$) of a compound is defined as the heat change which takes place when one mole of the compound is formed from its elements under standard conditions, usually at 25℃ and 1 atm.
For example:
For 1 mole of water
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O\,\,\,\,\,\,\,\,\,\,\Delta H_{f}^{\circ }=-286\,kJ\,mo{{l}^{-1}}$
The standard enthalpy change of a reaction ($\Delta H_{r}^{\circ }$) is defined as the heat change which takes place when equation quantities of materials react under standard conditions, usually at 25℃ and 1 atm, and with everything in its standard state.
For example:
$2{{H}_{2}}+{{O}_{2}}\to 2{{H}_{2}}O\,\,\,\,\,\,\,\,\,\,\Delta H_{r}^{\circ }=-572\,kJ\,mo{{l}^{-1}}$
It is mathematically given as:
$\Delta H_{r}^{\circ }\text{=}\sum{\text{Bond enthalpy of}\,\text{product-}\sum{\text{Bond enthalpy}\,\text{of}\,\text{reactants}}}$

We have been provided in the question that the:
$\Delta {{H}_{f}}^{\circ }\,$of ${{N}_{2}}O$ = $134KJ\,mo{{l}^{-1}}$
Bond energy of $N-O$= 222 $kJ\,mo{{l}^{-1}}$
Bond energy of $N=O$= 607 $kJ\,mo{{l}^{-1}}$
Bond energy of $N\equiv N$= 946 $kJ\,mo{{l}^{-1}}$
Bond energy of $O=O$= 498 $kJ\,mo{{l}^{-1}}$
The chemical reaction for the formation of one mole of ${{N}_{2}}O$ will be,
$N\equiv N+\dfrac{1}{2}(O=O)\to N=N=O$
We know that in a reaction,
$\Delta H_{r}^{\circ }\text{=}\sum{\text{Bond enthalpy of}\,\text{product-}\sum{\text{Bond enthalpy}\,\text{of}\,\text{reactants}}}$
$\Delta H_{f({{N}_{2}}O)}^{\circ }=\left[ \Delta {{H}_{(N\equiv N)}}+\dfrac{1}{2}\Delta {{H}_{(O=O)}} \right]-\left[ \Delta {{H}_{(N=N)}}+\Delta {{H}_{(N=O)}} \right]$
Putting the values in the above equation,

& \Delta H_{f({{N}_{2}}O)}^{\circ }=\left[ 946+\dfrac{1}{2}\times (498) \right]-\left[ 607+418 \right] \\\ & \,\,\,\,\,=\,1195-1025 \\\ & \,\,\,\,\,=\,170\,kJ\,mo{{l}^{-1}} \\\ \end{aligned}$$ So, the calculated heat of formation is $$170\,kJ\,mo{{l}^{-1}}$$. Using formula provided in the hint, Resonance energy = $$\Delta {{H}_{f}}^{\circ }(actual)-\Delta {{H}_{f}}^{\circ }(calculated)$$ = 82-170 = -88 $$kJ\,mo{{l}^{-1}}$$ Therefore, the resonance energy is calculated as -88 $$kJ\,mo{{l}^{-1}}$$. **Note** : In the case of Nitrous oxide, the electronegativity of oxygen is higher than nitrogen, meaning a negative charge on oxygen atom will be more stable than a negative charge on nitrogen atom. Stability order: II>I>III