Chapter 21, Problem 21.69QP

Interpretation Introduction

Interpretation:

The longest wavelength required for the dissociation of ${\text{NO}}_{\text{2}}$  in the given reaction should be calculated.

Concept Introduction:

• Standard enthalpy of reaction: ( ${\text{ΔHº}}_{\text{reaction}}$) is equal to its enthalpy change $\text{ΔH}$ when one mole of compound is formed at 25°C and 1 atm from elements in their stable form. All the reactant and product in the reaction are in their standard state also.

${\text{ΔHº}}_{\text{reaction}}\text{=}\text{Σ}{\text{nΔH}}_{\text{f}}{}^{\text{o}}\text{(products)}\text{-}\text{Σ}{\text{mΔH}}_{\text{f}}{}^{\text{o}}\text{(reactants)}$

Where ${\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(products)}$ and ${\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(reactants)}$ are the standard enthalpy of formation of product and reactant respectively. n and m are the co-efficient.

• Internal energy is one of the thermodynamic properties of a system; which refers the sum of kinetic and potential energies of the particles that form the system. At standard conditions the internal energy ${\text{ΔE}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{PΔV}$

  Where ${\text{ΔH}}^{\text{o}}$ is the standard enthalpy change, P and $\text{ΔV}$ are the pressure and change in volume respectively.

• Ideal gas equation, $\text{PV}\text{=}\text{nRT}$ Where P is the pressure, V is the volume, n, R and T are the amount of substance, universal gas constant and temperature respectively.
• Energy of photon E $\text{=}\text{hν}\text{=}\frac{\text{hc}}{\text{λ}}$

h is the Plank’s constant its value is $\text{6}\text{.63 ×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$

$\text{ν}$ is the frequency of radiation.

C is the velocity of light it equal to $\text{3×}{\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}$

$\text{λ}$ is wavelength of radiation.

• Bond energy: It is the amount of energy required to pull two bonded atoms apart which is the same as the amount of energy released when two atoms are brought together into a bond.

To determine: standard enthalpy of   ${\text{NO}}_{\text{2}}$ dissociation reaction.

Answer to Problem 21.69QP

The longest wavelength that can dissociate ${\text{NO}}_{\text{2}}$ is $\text{3}{\text{.94×10}}^{\text{-7}}\text{m}=\text{394}\text{nm}$ .

Explanation of Solution

The given dissociation reaction of ${\text{NO}}_{\text{2}}$ is,

${\text{NO}}_{\text{2}}\to \text{NO}\text{+}\text{O}$

Here the nitrogen dioxide is dissociated into nitrogen monoxide and oxygen.

Standard enthalpy of reaction: ( ${\text{ΔHº}}_{\text{reaction}}$) is equal to its enthalpy change ( $\text{ΔH}$) when one mole of compound is formed at 25°C and 1 atm from elements in their stable form. All the reactant and product are in their standard state.

${\text{ΔHº}}_{\text{reaction}}\text{=}\text{Σ}{\text{nΔH}}_{\text{f}}{}^{\text{o}}\text{(products)}\text{-}\text{Σ}{\text{mΔH}}_{\text{f}}{}^{\text{o}}\text{(reactants)}$

Where ${\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(products)}$ and ${\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(reactants)}$ are the standard enthalpy of formation of product and reactant respectively. n and m are the co-efficient.

According with the above mentioned equation, standard enthalpy of ${\text{NO}}_{\text{2}}$ dissociation reaction can be calculated as follows,

Dissociation reaction of ${\text{NO}}_{\text{2}}$ is,

${\text{NO}}_{\text{2}}\to \text{NO}\text{+}\text{O}$

${\text{ΔHº}}_{\text{reaction}}\text{=}\text{Σ}{\text{nΔH}}_{\text{f}}{}^{\text{o}}\text{(products)}\text{-}\text{Σ}{\text{mΔH}}_{\text{f}}{}^{\text{o}}\text{(reactants)}$

${\text{ΔHº}}_{\text{reaction}}\text{=}{\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(NO)}+{\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(O)}-{\text{ΔH}}_{\text{f}}{}^{\text{o}}{\text{(NO}}_2\text{)}$

$\begin{array}{l}{\text{ΔHº}}_{\text{reaction}}\text{=}\text{(1)(90}\text{.4kJ/mol)+(1)(249}\text{.4kJ/mol)-(1)(33}\text{.85kJ/mol)}\\ \text{=}\text{306}\text{.0}\text{kJ/mol}\end{array}$

Therefore ${\text{ΔHº}}_{\text{reaction}}$ of the given reaction is $\text{306}\text{.0}\text{kJ/mol}$.

To determine: internal energy of the given dissociation reaction of ${\text{NO}}_{\text{2}}$.

Internal energy is one of the thermodynamic properties of a system; which refers the sum of kinetic and potential energies of the particles that form the system. At standard conditions the internal energy ${\text{ΔE}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{PΔV}$

Where ${\text{ΔH}}^{\text{o}}$ is the standard enthalpy change, P and $\text{ΔV}$ are the pressure and change in volume respectively.

${\text{ΔE}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{PΔV}$

According with the ideal gas equation, $\text{PV}\text{=}\text{nRT}$, $\text{RTΔn}$ can be used instead of $\text{PΔV}$ in the above equation. Then,

${\text{ΔE}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}\text{RTΔn}$

Substituting the values of ${\text{ΔH}}^{\text{o}}{}_{\text{reaction}}\text{,}\text{R,T}\text{and}\text{n}$ to calculate ${\text{ΔE}}^{\text{o}}$. Standard enthalpy change value is converted from kilo joule to joule by multiplying it with ${10}^3$.

Therefore,

${\text{ΔE}}^{\text{o}}\text{=}\text{(306}{\text{.0×10}}^{\text{3}}\text{J/mol)}\text{-}\text{(8}\text{.314}\text{J/mol}\text{. K)(298}\text{K)}\text{(1)}$

${\text{ΔE}}^{\text{o}}\text{=}{\text{304×10}}^{\text{3}}\text{J/mol}$

This value of internal energy is equal to the energy needed to dissociate 1 mole of ${\text{NO}}_{\text{2}}$ also.

To determine: the energy required to dissociate one molecule of ${\text{NO}}_{\text{2}}$

The amount of energy required to pull two bonded atoms apart which is the same as the amount of energy released when two atoms are brought together into a bond is known as bond energy and which is considered as the bond dissociation energy.

The value of internal energy is equal to the energy needed to dissociate 1 mole of ${\text{NO}}_{\text{2}}$ . The energy needed to dissociate one molecule of ${\text{NO}}_{\text{2}}$ can be calculated as follows,

$\frac{{\text{304×10}}^{\text{3}}\text{J}}{\text{1}\text{mol}{\text{NO}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}{\text{NO}}_{\text{2}}}{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}{\text{NO}}_{\text{2}}}\text{=}\text{5}\text{.05}\text{×}{\text{10}}^{\text{-19}}\text{J/molecule}$

Therefore the energy needed to dissociate one molecule of ${\text{NO}}_{\text{2}}$ is $\text{5}{\text{.5×10}}^{\text{-19}}\text{J/molecule}$

To determine: the longest wavelength that can dissociate ${\text{NO}}_{\text{2}}$

From the equation to find the energy of photon that is

Energy of photon E $\text{=}\text{hν}\text{=}\frac{\text{hc}}{\text{λ}}$

Wavelength can easily calculated by using equation  $\text{λ}\text{=}\frac{\text{hc}}{\text{E}}$ .

Therefore the longest wavelength that can dissociate ${\text{NO}}_{\text{2}}$ is,

$\text{λ}=\frac{\text{hc}}{\text{E}}=\frac{\left(\text{6}\text{.63}{\text{×10}}^{\text{-34}}\text{J}\text{.s}\right)\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\text{5}{\text{.05×10}}^{\text{-19}}\text{J}}$

$\begin{array}{l}\text{=}\text{3}\text{.94}{\text{×10}}^{\text{-7}}\text{m}\\ \text{=}\text{394}\text{nm}\end{array}$

Conclusion

The longest wavelength required for the dissociation of ${\text{NO}}_{\text{2}}$  in the given reaction is calculated.