Chapter 20, Problem 20.74QP

Ozone in the troposphere is formed by the following steps:

${\text{NO}}_2\to \text{NO}+\text{O}$ (1)

$\text{O}+{\text{O}}_2\to {\text{O}}_3$     (2)

The first step is initiated by the absorption of visible light (NO₂ is a brown gas). Calculate the longest wavelength required for step (1) at 25°C. [Hint: You need to first calculate ΔH and hence ΔU for (1). Next, determine the wavelength for decomposing NO₂ from ΔU.]

Interpretation Introduction

Interpretation:

The maximum wavelength required to decompose ${\text{NO}}_{\text{2}}$ has to be calculated.

Concept introduction:

Standard enthalpy of reaction (${\text{ΔHº}}_{\text{reaction}}$):

It is the change in enthalpy occurs when materials involved in reaction react under standard state condition.

$\Delta H{º}_{reaction}={\displaystyle \sum n}\Delta H_f{}^o(products)-{\displaystyle \sum m}\Delta H_f{}^o(reac\tan ts)$

$\Delta H_f{}^o(products)$ and $\Delta H_f{}^o(reac\tan ts)$ are the standard enthalpy of formation of product and reactant respectively. n and m are co-efficients.

Internal energy:

It is the sum of kinetic and potential energies of the particles that form the system. At standard conditions the internal energy $\Delta E^o=\Delta H^o-P\Delta V$

The mathematical equation of various characteristics of light wave is

$\text{c = νλ}$

c is the speed of light ($\text{3}{\text{.00 × 10}}^{\text{8}}\text{ m/s}$).

$\text{ν}$ is the frequency

$\text{λ}$ is wavelength.

$\text{E}\text{=}\text{hν}$

h is Planck’s constant ($\text{6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}$ ) which relates energy and frequency.

$\text{ν}$ is the frequency.

E is the energy of light particle

Answer to Problem 20.74QP

The maximum wavelength required to decompose ${\text{NO}}_{\text{2}}$ is $\text{394}\text{nm}$

Explanation of Solution

Given,

${\text{ΔH}}_{\text{f}}{}^{\text{o}}$ values in appendix three are given as,

$\begin{array}{l}{\text{ΔH}}_{\text{f}}{{}^{\text{o}}}_{\text{(NO)}}\text{=}\text{90}\text{.4}\text{KJ/mol}\\ \\ {\text{ΔH}}_{\text{f}}{{}^{\text{o}}}_{\text{(O)}}\text{=}\text{249}\text{.4}\text{KJ/mol}\\ \\ {\text{ΔH}}_{\text{f}}{{}^{\text{o}}}_{{\text{(NO}}_{\text{2}}\text{)}}\text{=}\text{33}\text{.85}\text{KJ/mol}\end{array}$

The standard enthalpy of decomposition of ${\text{NO}}_{\text{2}}$ is calculated as

$\begin{array}{l}{\text{NO}}_{\text{2}}\underset{}{\overset{}{\to }}\text{NO}\text{+}\text{O}\\ \\ {\text{ΔHº}}_{\text{reaction}}\text{=}{\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(NO)}+{\text{ΔH}}_{\text{f}}{}^{\text{o}}\text{(O)}-{\text{ΔH}}_{\text{f}}{}^{\text{o}}{\text{(NO}}_2\text{)}\\ \\ {\text{ΔHº}}_{\text{reaction}}\text{=}\text{(1)(90}\text{.4kJ/mol)+(1)(249}\text{.4kJ/mol)-(1)(33}\text{.85kJ/mol)}\\ \\ \text{=}\text{306}\text{.0}\text{kJ/mol}\end{array}$

The energy involved in decomposition of ${\text{NO}}_{\text{2}}$ is calculated as,

$\begin{array}{l}\Delta E^o=\Delta H^o-P\Delta V\\ \\ \mathrm{Ideal}gasequationPV=nRT\\ \\ \Delta E^o=\Delta H^o-RT\Delta n\\ \\ \Delta E^o=(306.0\times 10^{3}J/mol)-(8.314J/mol.K)(298K)(1)\\ \\ \Delta E^o=304\times 10^{3}J/mol\end{array}$

The energy needed to decompose one molecule of ${\text{NO}}_{\text{2}}$ is calculated using Avogadro number as,

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

The maximum wavelength that can decompose ${\text{NO}}_{\text{2}}$ is determined as,

$\begin{array}{l}\text{λ}\text{=}\frac{\text{hc}}{\text{E}}\text{=}\frac{\left(\text{6}\text{.63}{\text{×10}}^{\text{-34}}\text{J}\text{.s}\right)\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\text{5}{\text{.05×10}}^{\text{-19}}\text{J}}\\ \\ \text{=}\text{3}\text{.94}{\text{×10}}^{\text{-7}}\text{m}\text{1m}\text{=}{\text{10}}^{\text{9}}\text{nm}\\ \\ \text{=}\text{394}\text{nm}\end{array}$