Chapter 21, Problem 69AP

Interpretation Introduction

Interpretation:

The longest wavelength required to dissociate the ${\text{NO}}_2$ molecule is to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the difference between the sum of standard enthalpy of products and the sum of the standardenthalpy of reactants.

$\Delta H_{rxn}^{\circ }\text{=}{\displaystyle \sum \Delta H_{product}^{\circ }}-{\displaystyle \sum \Delta H_{reac\tan t}^{\circ }}$.

The wavelength is represented by the following expression:

$\lambda =\frac{\text{hc}}{E}$

Here, h is the Planck’s constant, $E$ is the energy of molecules, and $\text{c}$ is the speed of light.

Answer to Problem 69AP

Solution: $\text{394 nm}$

Explanation of Solution

Given information: The given reaction is as follows:

$\begin{array}{l}{\text{NO}}_2\to \text{ NO }+\text{ O}\\ \text{O }+{\text{ O}}_2\to {\text{ O}}_3\end{array}$

The reaction is as follows:

${\text{NO}}_2\stackrel{}{\to }\text{ NO + O}$

The standard enthalpy of a reaction is calculated as follows:

$\Delta H_{rxn}^{\circ }\text{=}{\displaystyle \sum n\Delta H_{\text{f}}^{\circ }}\left(products\right)-{\displaystyle \sum n\Delta H_{\text{f}}^{\circ }\left(reac\tan ts\right)}$

For the given reaction, the enthalpy change is given by the following expression:

$\Delta H^{\text{o}}=\Delta H_{\text{f}}^{\text{o}}\left(\text{NO}\right)+\Delta H_{\text{f}}^{\text{o}}\left(\text{O}\right)-\Delta H_{\text{f}}^{\text{o}}\left({\text{NO}}_2\right)$

Substitute the values of $\Delta H_{\text{f}}^{\text{o}}$, using appendix $2$, in the above equation:

$\begin{array}{l}\Delta H^{\text{o}}=\left(1\right)\left(90.4\text{ kJ/mol}\right)+\left(1\right)\left(249.4\text{ kJ/mol}\right)-\left(1\right)\left(33.85\text{ kJ/mol}\right)\\ \Delta H^{\text{o}}=306.0\text{ kJ/mol}\end{array}$

Thus, the standard enthalpy of reaction is $306.0\text{ kJ/mol}$.

The energy change is calculated as follows:

$\Delta E^{\text{o}}=\Delta H^{\text{o}}-P\Delta V$ …… (1)

The ideal gas equation is as follows:

$P\Delta V=\Delta n\text{R}T$

Here, $P$ is the pressure, $n$ is the number of moles, $R$ is the universal gas constant, $V$ is the volume, and $T$ is the absolute temperature.

Substitute the value of $P\Delta V$ in equation (1),

$\Delta E^{\text{o}}=\Delta H^{\text{o}}-RT\Delta n$

Substitute the values in the above equation:

$\begin{array}{l}\Delta E^{\text{o}}=\left(306.0\times {10}^3\text{ J/mol}\right)-\left(8.314\text{ J/mol}\text{.}\cancel{\text{K}}\right)\left(298\cancel{\text{K}}\right)\left(1\right)\\ \Delta E^{\text{o}}=304\times {10}^3\text{ J/mol}\\ \end{array}$

Therefore, the amount of energy required to dissociate $1$ mole of ${\text{NO}}_2$ is $304\times {10}^3\text{ J/mol}$

The amount of energy required to dissociate $1$ molecule of ${\text{NO}}_2$ is calculated as follows:

$\begin{array}{l}{M}_{{\text{NO}}_2}=\left(\frac{304\times {10}^3\text{ J}}{\cancel{1{\text{ mol NO}}_2}}\times \frac{\cancel{1{\text{ mol NO}}_2}}{6.022\times {10}^{23}{\text{ molecule NO}}_2}\right)\\ M_{{\text{NO}}_2}=5.05\times {10}^{-19}\text{ J/molecule}\end{array}$

Thus, amount of energy required to dissociate $1$ molecule of ${\text{NO}}_2$ is $5.05\times {10}^{-19}\text{ J/molecule}$.

The longest wavelength is calculated as follows:

$\lambda =\frac{\text{hc}}{E}$

Here, h is the Planck’s constant, $\text{c}$ is the speed of light, and $E$ is the energy of molecules.

Substitute the values of speed of light, energy, and Planck’s constant in the above equation:

$\begin{array}{l}\lambda =\frac{\left(6.63\times {10}^{-34}\cancel{\text{J}}\text{.}\cancel{\text{s}}\right)\left(3.00\times {10}^8\text{ m/}\cancel{\text{s}}\right)}{5.05\times {10}^{-19}\cancel{\text{J}}}\\ \lambda =3.94\times {10}^{-7}\text{ m}\\ \lambda =\text{394 nm}\end{array}$

Conclusion

The longest wavelength required to break ${\text{NO}}_2$ is $\text{394 nm}$.