Chapter 18, Problem 127CP

Interpretation Introduction

Interpretation:

The value of standard change in enthalpy, change in Gibbs free energy and equilibrium constant needs to be determined for the production of ozone from oxygen.

Concept Introduction: The relation between change in Gibbs free energy equilibrium constant can be represented as follows:

  $\Delta G=-RT\ln K_P$

Here, R is Universal gas constant and T is temperature.

Explanation of Solution

The formation of ozone gas can be represented as follows:

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

The change in enthalpy can be calculated as follows:

  $\Delta H=\Delta H_P-\Delta H_R$

For oxygen molecule, the value is zero and that for ozone gas is 143 kJ/mol.

Thus, for the reaction:

  $\Delta H_{rxn}^o=2\Delta H^o{}_{{\text{O}}_{\text{3}}\left(g\right)}-3\Delta H^o{}_{{\text{O}}_{\text{2}}\left(g\right)}$

Putting the values,

  $\Delta H_{rxn}^o=2\left(143\text{ kJ/mol}\right)-3\left(0\right)=286\text{ kJ/mol}$

Thus, the standard enthalpy of the reaction is 286 kJ/mol.

The standard Gibbs free energy for the reaction can be calculated as follows:

  $\begin{array}{l}\Delta G_{rxn}^o=2\Delta G^o{}_{{\text{O}}_{\text{3}}\left(g\right)}-3\Delta G^o{}_{{\text{O}}_{\text{2}}\left(g\right)}\\ =2\left(163\text{ kJ/mol}\right)-0\\ =326\text{ kJ/mol}\end{array}$

Thus, the change in Gibbs free energy for the reaction is 326 kJ/mol.

The relation between change in Gibbs free energy equilibrium constant can be represented as follows:

  $\Delta G=-RT\ln K_P$

Putting all the values,

  $326\times {10}^3\text{ J/mol}=-\left(8.314\text{ J/K mol}\right)\left(298\text{ K}\right)\ln K_P$

On calculating,

  $K_P=7.22\times {10}^{-58}$

For temperature 230 K, the value of equilibrium constant can be calculated as follows:

  $\begin{array}{l}\ln K_2-\ln K_1=\left(\frac{-\Delta H^o}{R{T}_2}+\frac{\Delta S^o}{R}\right)-\left(\frac{-\Delta H^o}{R{T}_1}+\frac{\Delta S^o}{R}\right)\\ =\frac{-\Delta H^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\end{array}$

Putting all the values,

  $\ln \left(7.22\times {10}^{-58}\right)-\ln K_1=\frac{-286\times {10}^3}{8.314}\left(\frac{1}{298}-\frac{1}{230}\right)$

The value of $K_1$ can be calculated as follows:

  $\begin{array}{l}\ln \frac{7.22\times {10}^{-58}}{K_1}=\frac{-286\times {10}^3}{8.314}\left(\frac{298-230}{230\times 298}\right)\\ \ln \frac{7.22\times {10}^{-58}}{K_1}=34.13\end{array}$

On rearranging,

  $K_1=1.1\times {10}^{-72}$

The equilibrium at 230 K is related to the partial pressure of oxygen and ozone as follows:

  $K=\frac{P_{O_3}{}^2}{P_{O_2}{}^3}$

Putting the values,

  $1.1\times {10}^{-72}=\frac{P_{O_3}{}^2}{{\left(1.0\times {10}^{-3}\right)}^3}$

Or,

  $P_{O_2}=3.3\times {10}^{-41}\text{ atm}$

Using the ideal gas equation, the volume can be calculated as follows:

  $V=\frac{nRT}{P}$

For 1 molecule, the number of moles will be $\frac{1}{6.022\times {10}^{23}}\text{mol}$

Putting the values,

  $\begin{array}{l}V=\frac{\left(\frac{1}{6.022\times {10}^{23}}\text{mol}\right)\left(0.082\text{ L atm/K mol}\right)\left(230\text{ K}\right)}{\left(3.3\times {10}^{-41}\text{ atm}\right)}\\ =9.5\times {10}^{17}\text{ L}\end{array}$

Since, the calculated volume can be occupied by only 2 mol of ozone thus, equilibrium between oxygen and ozone is not maintained.

Here, the value of reaction quotient is greater than equilibrium constant thus, reaction shifts in backward direction.

Here, the volume is very large thus, the gas molecules cannot collide. The concentration of ozone is thus, not sufficient to maintain the equilibrium.