Question

${\mathbf{N}}_{\mathbf{2}}$has equilibrium bond length of 1.100 Å and bond dissociation energy of 942 kJ/mol, whereas ${\mathit{O}}_{\mathbf{2}}$has equilibrium bond length of 1.211 Å and bond dissociation energy of 495 kJ/mol. On the same graph show qualitative sketches of the effective potential energy curve ${\mathbf{V}}_{\mathbf{eff}}$ for an ${\mathbf{N}}_{\mathbf{2}}$and an ${\mathit{O}}_{\mathbf{2}}$molecule. In your solution, show the conversion from kJ/mol to the energy of a single molecule.

Short answer

Energy of a single molecule of Nitrogen is${\text{1.56×10}}^{\text{- 18}}\text{J}$

Energy of a single molecule of oxygen is ${\text{8.2×10}}^{\text{- 19}}\text{J}$

Step-by-step solution

Introduction

The distance between the nuclei of two atoms which are chemically bonded to each other, is called bond length and the energy required to break these bonds is called as bond energy. If the bond length is less, atoms are bonded strongly and more energy is required to break them so bond energy is high. On the other hand, if bond length has a higher magnitude, energy required to break the bond is less so molecules having high bond length have lower bond energy.

Step 2: Effective Potential Energy Curve

${\text{N}}_{\text{2}}$has equilibrium bond length of 1.100 Å and bond dissociation energy of 942 kJ/mol, whereas $O_{\text{2}}$has equilibrium bond length of 1.211 Å and bond dissociation energy of 495 kJ/mol for which effective potential energy curve${\text{V}}_{\text{eff}}$ can be given as below-

Step 3: Conversion from kJ/mol to the energy of a single molecule

To convert from kJ/mol into the energy of a single molecule, divide the energy in kJ/mol by Avogadro’s number.

Energy of a single molecule of nitrogen,

$$\begin{gathered} \text{Energy} \text{of} \text{single} \text{molecule} \text{of} Nitro\text{gen =}\frac{\text{Energy} \text{in} \text{kJ/mol}}{\text{Avogadro's} \text{number}} \\ \text{Energy} \text{of} \text{single} \text{molecule} \text{of} Nitro\text{gen =}\frac{\text{942}}{{\text{6.023×10}}^{\text{23}}}\text{=} 156.4\times {\text{10}}^{\text{- 23}}{\text{kJ = 1.56×10}}^{\text{- 18}}\text{J} \end{gathered}$$

Similarly, Energy of a single molecule of oxygen,

$$\begin{gathered} \text{Energy} \text{of} \text{single} \text{molecule} \text{of} \text{Oxygen =}\frac{\text{Energy} \text{in} \text{kJ/mol}}{\text{Avogadro's} \text{number}} \\ \text{Energy} \text{of} \text{single} \text{molecule} \text{of} \text{Oxygen =}\frac{\text{495}}{\text{6.023×} {\text{10}}^{\text{23}}}\text{=} 82.18\times {\text{10}}^{\text{- 23}}{\text{kJ = 8.2×10}}^{\text{- 19}}\text{J} \end{gathered}$$