Question

Using data from Appendix $\mathbf{4}$, calculate ${\mathbf{\text{ΔH}}}^{\mathbf{\text{o}}}{\mathbf{\text{,ΔG}}}^{\mathbf{\text{o}}}$, and${\mathbf{\text{K}}}_{\mathbf{\text{p}}}$ (at $\mathbf{\text{298K}}$) for the production of ozone from oxygen:

role="math" localid="1663835974541" ${\mathbf{\text{3O}}}_{\mathbf{\text{2}}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{2O}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

At$\mathbf{\text{30km}}$ above the surface of the earth, the temperature is about$\mathbf{\text{230K}}$ and the partial pressure of oxygen is about $\mathbf{\text{1.0}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{-}\mathbf{3}}\mathbf{\text{atm}}$. Estimate the partial pressure of ozone in equilibrium with oxygen at$\mathbf{\text{30km}}$ above the earth’s surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

Short answer

Using the given condition, the equilibrium is probably not maintained. This is because when only two ozone molecules are in a volume of ${\text{9.5×10}}^{\text{17}}\text{L}$, the reaction is not at equilibrium. Under these conditions, $\text{Q>K}$, hence, the reaction shifts to the left. But, with this huge volume, it is implausible that the$2$ ozone molecules will collide. Therefore, in these conditions, the concentration of ozone is not large enough to maintain equilibrium.

Step-by-step solution

Concept Introduction:

Chemical equilibrium is a state in which there is no net change in the amounts of reactants and products during a reversible chemical process. A reversible chemical reaction occurs when the products react with the original reactants as they are formed.

Information Provided:

The reaction is,

${\text{3O}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{2O}}_{\text{3}}\left(g\right)$

The location of the reaction is$\text{30km}$ above the earth's surface.

The temperature is $\text{230K}$.

The partial pressure of oxygen is $1.0\times {10}^{-3}\text{atm}$.

Appendix $4$is:

Calculate the values for ΔHo, ΔGo, and Kp:

Solve for the values of ${\text{ΔH}}^{\text{o}}{\text{,ΔG}}^{\text{o}}$,${\text{K}}_{\text{p}}$and (at $\text{298K}$).

The calculation for$\text{ΔH}°$is as follows –

$\begin{array}{rcl}\text{ΔH}°& =& {\text{Σn}}_{\text{p}}{\text{ΔH}}_{\text{fproducts}}^{\text{°}}{\text{-Σn}}_{\text{p}}{\text{ΔH}}_{\text{freactants}}^{\text{°}}\\ & =& \text{2mol}\left(\text{143}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)\text{-3mol}\left(\text{0}\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\right)\\ & =& \text{286kJ}\end{array}$

The calculation for ${\text{ΔG}}^{\text{o}}$is as follows.

$\begin{array}{rcl}\text{ΔG}°& =& {\text{Σn}}_{\text{p}}{\text{ΔG}}_{\text{fproducts}}^{\text{°}}{\text{-Σn}}_{\text{p}}{\text{ΔG}}_{\text{freactants}}^{\text{°}}\\ & =& \text{2mol}\left(\text{163}\raisebox{1ex}{${\displaystyle \text{kJ}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{mol}}$}\right.\right)\text{-3mol}\left(\text{0}\raisebox{1ex}{${\displaystyle \text{kJ}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{mol}}$}\right.\right)\\ & =& \text{326kJ}\end{array}$

The calculation for ${\text{K}}_{\text{p}}$is as follows.

role="math" localid="1663837612736" $\begin{array}{rcl}\text{lnK}& =& \frac{-\text{ΔG}°}{\text{RT}}\\ & =& \frac{-{\text{326×10}}^{\text{3}}\text{J}}{\left(\text{8.314}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.\right)\left(\text{298K}\right)}\end{array}$

$\text{K}=7.22\times {10}^{-58}$

Calculate the values for K1 and K2:

Calculate for the ${\text{K}}_{\text{p}}$at $\text{230K}$, then use the formula.

$\text{lnK}=\frac{-\Delta \text{H}°}{\text{RT}}+\frac{\Delta \text{S}°}{\text{R}}$

For two sets of $\text{K}$and $\text{T}$, it can be written as,

$$\begin{gathered} {\text{lnK}}_{\text{1}}=\frac{-\text{ΔH°}}{{\text{RT}}_{\text{1}}}+\frac{\text{ΔS°}}{\text{R}} \\ {\text{lnK}}_{\text{2}}=\frac{-\text{ΔH°}}{{\text{RT}}_{\text{2}}}+\frac{\text{ΔS°}}{\text{R}} \end{gathered}$$

Subtracting the first equation from the second. and substituting the following,

${\text{K}}_{\text{1}}=7.22\times {10}^{-58}$

Now, the calculation for ${\text{K}}_{\text{2}}$, (${\text{K}}_{\text{p}}$at $T=\text{230K}$).

$$\begin{gathered} T_1=298\text{K} \\ {T}_2=230\text{K} \end{gathered}$$

$\begin{array}{rcl}\text{ΔH}°& =& \text{286kJ}\\ & =& 286\times {10}^3\text{J}\end{array}$

The value of ${\text{ΔS}}^{\text{o}}$is not needed since then it will be cancelled in the derived equation.

$$\begin{gathered} {\text{lnK}}_{\text{2}}-{\text{lnK}}_{\text{1}}=\left(\frac{-\text{ΔH}°}{{\text{RT}}_{\text{2}}}+\frac{\text{ΔS}}{\text{R}}\right)-\left(\frac{-\text{ΔH}°}{{\text{RT}}_{\text{1}}}+\frac{\text{ΔS}}{\text{R}}\right) \\ {\text{lnK}}_{\text{2}}={\text{lnK}}_{\text{1}}\text{+}\left(\frac{-\Delta \text{H}°}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{{\text{T}}_{\text{1}}}\right) \\ {\text{K}}_{\text{2}}=\text{exp}\left[{\text{lnK}}_{\text{1}}+\left(\frac{-{\text{ΔH}}^{\text{°}}}{\text{R}}\right)\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\right] \end{gathered}$$

On further calculation,

$\begin{array}{rcl}{\text{K}}_{\text{2}}& =& \text{exp}\left[\text{ln}(7.22\times {10}^{-58})+\left(\frac{-286\times {10}^3}{8.3145}\right)\left(\frac{1}{230}-\frac{1}{298})\right)\right]\\ & =& {\text{e}}^{-165.700}\\ & =& 1.1\times {10}^{-72}\end{array}$

Calculate Partial Pressure and volume occupied by a molecule of ozone:

To solve for the partial pressure of ozone in equilibrium with oxygen at above the earth's surface, it is given as,

$\begin{array}{rcl}{\text{K}}_{\text{2}}& =& {\text{K}}_{\text{r-200}}\\ & =& 1.1\times {10}^{-72}\\ & =& \frac{{\text{P}}_{{\text{O}}_{\text{3}}}^{\text{2}}}{{\text{P}}_{{\text{O}}_{\text{2}}}^{\text{3}}}\end{array}$

$$\begin{gathered} 1.1\times {10}^{-72}=\frac{{\text{P}}_{{\text{O}}_{\text{s}}}^{\text{2}}}{{(\text{1.0}\times {19}^{-3}\text{atm})}^{\text{3}}} \\ {\text{P}}_{{\text{O}}_{\text{3}}}^{\text{2}}=3.3\times {10}^{-41}\text{atm} \end{gathered}$$

The volume occupied by one molecule of ozone is,

$\begin{array}{rcl}{\text{V}}_{{\text{O}}_{\text{3}}}& =& \frac{n_{o_3}RT}{P}\\ & =& \frac{\left(\text{1molecule}\times {\displaystyle \frac{\text{1mol}}{\text{6.022}\times {\text{10}}^{\text{23}}\text{molecule}}}\right)\left(\text{0.08206}{\displaystyle \raisebox{1ex}{$\text{L}\cdot \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.}\right)\left(\text{230K}\right)}{\text{3}.3\times {10}^{-4\text{1}}\text{atm}}\\ & =& 9.5\times {10}^{17}\text{L}\end{array}$

Hence, the values for ${\text{ΔH}}^{\text{o}}{\text{,ΔG}}^{\text{o}}$, and${\text{K}}_{\text{p}}$ are $\text{286kJ}$,$\text{326kJ}$ and$7.22\times {10}^{-58}$ respectively. The value for partial pressure is $3.3\times {10}^{-41}\text{atm}$. So, under these conditions the equilibrium is not maintained.