Chapter 8, Problem 142AP

The hydroxyl radical $\left(\text{OH}\right)$ plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus $\text{OH}$ is sometimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction:

$\text{OH}\left(g\right){\text{ + CH}}_{\text{4}}\left(g\right)\underset{}{\to }{\text{CH}}_{\text{3}}\left(g\right){\text{ + H}}_{\text{2}}\text{O}\left(g\right)$

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an $\text{O—H}$ bond in ${\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

Interpretation:

The Lewis structure of hydroxyl radical is to be drawn; the reason for radical’s high affinity for H atoms, enthalpy change for the given reaction, and the maximum wavelength required to break given bond in given compound are to be calculated.

Concept introduction:

The standard enthalpy of a reaction is the enthalpy change that occurs under standard conditions.

The standard enthalpy of a reaction is determined using the equation as given below:

$\Delta H{°}_{rxn}={\displaystyle \sum n}\Delta H_f^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_f^{°}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for reactants and n for products, and enthalpy of formation at standard conditions is represented by $\Delta H_f^{°}$.

The standard enthalpy of formation is the amount of heat change that arises as a result of one mole of compound formation at the standard states from its integral elements.

The value of enthalpy of formation of an element is zero at its most stable state.

Bond Enthalpy is the energy required to break a bond of 1 mole of a substance. It is given in $\text{KJ}/\text{mol}$.

The Lewis dot symbols contain dots, which give information about valence electrons.

According to Planck’s energy–frequency law, energy is given by:

$E=h\frac{c}{\lambda }$

Here, h is Planck’s constant, $c$ is speed of light, and $\lambda$ is wavelength of radiation.

$\begin{array}{l}\text{h = }6.626\times {10}^{-34}\text{Js}\\ \text{c = 3}.0\times {10}^8\text{ m/s}\end{array}$

Answer to Problem 142AP

Solution:

(a)

(b) The reason for radical $\text{OH}$ to have a strong affinity toward hydrogen is the bond enthalpy value of oxygen–hydrogen bond, which is very high.

(c) $-46\text{ KJ}/\text{mol}$

(d) $260\text{ nm}$

Explanation of Solution

a) Lewis structure for radical (OH)

The skeletal structure for $\text{OH}$ radical is as follows:

$\text{O}-\text{H}$

The number of valence electrons is as follows:

$\begin{array}{c}n=\left(\text{O atom}\right)+\left(\text{H atoms}\right)\\ =6+1\\ =7\end{array}$

Now, only two electrons are used in bond formation. So, the number of remaining electrons is 5.

The oxygen atom has only one electron used in bond formation, so it will have 5 remaining electrons.

So, the Lewis structure for $\text{OH}$ radical is as follows:

b) The radical has high affinity for H atom, refer table 8.6.

From table 8.6, the bond enthalpy of oxygen–hydrogen bond: $\text{BE}\left(\text{O}-\text{H}\right)=460\text{ kJ}/\text{mol}$

Since the value of oxygen–hydrogen bond is very high, the radical $\text{OH}$ has very strong affinity toward hydrogen to complete its octet.

Hence, the reason for radical $\text{OH}$ to have a strong affinity toward hydrogen is the high bond enthalpy value of oxygen–hydrogen bond.

c) The enthalpy change for the following given reaction:

$\text{OH}\left(g\right)+{\text{CH}}_4\left(g\right)\to {\text{CH}}_3\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

The enthalpy of reaction can be calculated as follows:

$\Delta H{°}_{rxn}={\displaystyle \sum \text{BE}}\left(\text{reactants}\right)-{\displaystyle \sum \text{BE}}\left(\text{products}\right)$

From table 8.6, the enthalpy of formation values are as follows:

$\begin{array}{l}\text{BE}\left[\text{C}-\text{H}\right]=414\text{ KJ}/\text{mol}\\ \text{BE}\left[\text{O}-\text{H}\right]=460\text{ KJ}/\text{mol}\end{array}$

Now, the standard enthalpy of the given reaction is as follows:

$\begin{array}{l}\Delta H{°}_{rxn}=\left\{\text{4BE}\left[\text{C}-\text{H}\right]\text{ + BE}\left[\text{O}-\text{H}\right]\right\}-\left\{\text{3BE}\left[\text{C}-\text{H}\right]\text{+ 2BE}\left[\text{O}-\text{H}\right]\right\}\\ \text{ = }\left\{\text{BE}\left[\text{C}-\text{H}\right]\right\}-\left\{\text{BE}\left[\text{O}-\text{H}\right]\right\}\end{array}$

Substitute $414\text{ KJ}/\text{mol}$ for $\text{BE}\left[\text{C}-\text{H}\right]$ and $460\text{ KJ}/\text{mol}$ for $\text{BE}\left[\text{O}-\text{H}\right]$ in the above equation

$\begin{array}{c}\Delta H{°}_{rxn}=\left[414\text{ KJ}/\text{mol}\right]-\left[460\text{ KJ}/\text{mol}\right]\\ =-46\text{ KJ}/\text{mol}\end{array}$

Hence, the enthalpy of the given reaction from the bond enthalpy values is $-46\text{ KJ}/\text{mol}$.

d) The maximum wavelength (in nm) required to break an $\text{O}-\text{H}$ bond in water.

From table 8.6, the bond enthalpy for oxygen–hydrogen bond is as follows:

$\text{BE}\left(\text{O}-\text{H}\right)=460\text{ kJ}/\text{mol}$

This can be written as the energy of oxygen–hydrogen bond as

$\begin{array}{c}\text{E}\left(\text{O}-\text{H}\right)=460\times {10}^3\text{ J}/\text{mol}\times \frac{1}{6.022\times {10}^{23}\text{ bonds}/\text{mol}}\\ =\frac{460\times {10}^3}{6.022\times {10}^{23}}\text{ J}/\text{bonds}\\ =76.39\times {10}^{-20}\text{ J}/\text{bonds}\\ =7.64\times {10}^{-19}\text{ J}/\text{bonds}\end{array}$

According to Planck’s energy–frequency law, energy is given as follows:

$E=h\frac{c}{\lambda }$

This can be written as follows:

$\lambda =h\frac{c}{E}$

Substitute $c$ for $3\times {10}^8\text{m}/\text{s}$, $6.626\times {10}^{-34}\text{Js}$ for $h$, and $7.64\times {10}^{-19}\text{ J}$ for $E$ in the above equation

$\begin{array}{c}\lambda =\left(6.626\times {10}^{-34}\text{Js}\right)\times \frac{3\times {10}^8\text{ m}/\text{s}}{\left(7.64\times {10}^{-19}\text{ J}\right)}\\ =\frac{19.9\times {10}^{-26}}{7.64\times {10}^{-19}}\text{ m}\\ =2.6\times {10}^{-7}\text{ m}\\ =260\text{ nm}\end{array}$

Hence, the maximum wavelength required to break the oxygen–hydrogen bond is $2.6\times {10}^{-7}\text{ nm}$.