Question

NO has equilibrium bond length of 1.154 Å and bond dissociation energy of 629 kJ/mol, Compare the effective potential curve for NO with those for ${\mathbf{N}}_{\mathbf{2}} \mathbf{and} {\mathbf{O}}_{\mathbf{2}}$ in problem 18.

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${\text{8.2×10}}^{\text{- 19}}\text{J}$

Energy of a single molecule of NO is ${\text{1.04×10}}^{\text{- 18}}\text{J}$

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: 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,

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

Similarly, Energy of a single molecule of oxygen,

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

role="math" localid="1663670248874" $$\begin{gathered} \text{Energy} \text{of} \text{single} \text{molecule} \text{of} \text{HF =}\frac{\text{Energy} \text{in} \text{kJ/mol}}{\text{Avogadro's} \text{number}} \\ \text{Energy} \text{of} \text{single} \text{molecule} \text{of} \text{HF =}\frac{629}{\text{6.023×} {\text{10}}^{\text{23}}}\text{=} 104.43 \times {\text{10}}^{\text{- 23}}{\text{kJ = 1.04×10}}^{\text{- 18}}\text{J} \end{gathered}$$

Step 3: Effective Potential Energy Curve

${\text{N}}_{\text{2}}$ 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 and NO has equilibrium bond length of 1.154 Å and bond dissociation energy of 629 kJ/mol To compare the effective potential energy curve ${\text{V}}_{\text{eff}}$ can be given as below-