Carbon dioxide decomposes into \({\bf{CO}}\) and \({{\bf{O}}_{\bf{2}}}\)at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at \({\bf{100}}{{\bf{0}}^{\bf{^\circ }}}{\bf{C}}\) for which the initial pressure of \({\bf{C}}{{\bf{O}}_{\bf{2}}}\)was \({\bf{1}}{\bf{.15\;atm}}\)?

Each gas in a gas mixture contributes to the total pressure of the mixture. The partial pressure is the result of this contribution. The partial pressure of a gas is the pressure it would have if it were in the same volume and temperature as itself.

According to Dalton's law, the total pressure of an ideal gas mixture is the sum of the partial pressures of each individual gas.

The reaction in the problem is:

\(2{\rm{C}}{{\rm{O}}_2}(g) \to 2{\rm{CO}}(g) + {{\rm{O}}_2}(g)\)

We can calculate the equilibrium constant for the reaction at this temperature using the formula

\(\Delta G = - RT\ln K\)

That is:

\(K = {e^{ - \frac{{\Delta G}}{{RT}}}}\)

We can calculate the \(\Delta G\)using the formula:

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

We can look up the standard formation free energy changes for each species in the Appendix \(G\)in the book.

\(\Delta G = 2{G_f}(CO,g) + {G_f}\left( {{O_2},g} \right) - 2{G_f}\left( {C{O_2},g} \right)\)

\(\Delta G = (2 \cdot ( - 137.15) + 0 - 2 \cdot ( - 394.36))\frac{{kJ}}{{\;{\rm{mol}}}}\)

\(\Delta G = 514.42\frac{{kJ}}{{\;{\rm{mol}}}}\)

\(\Delta G = 514420\frac{{\rm{J}}}{{{\rm{mol}}}}\)

We can convert the energy change from \(\frac{J}{{mol}}\) to \(\frac{J}{{mol}}\) by multiplying it by\(1000(1kJ = 1000J)\). Therefore the partial equilibrium pressure of \({\rm{C}}{{\rm{O}}_2}\) at room temperature is \(4.7 \times {10^{ - 15}}\;{\rm{atm}}\).

We can convert the temperature in degrees Celsius to Kelvins by adding 273 to the temperature in degrees

Celsius. Now we can calculate the equilibrium constant:

\(K = {e^{ - \frac{{\Delta G}}{{RT}}}}\)

\(K = {e^{ - \frac{{514420{\rm{Jmo}}{{\rm{l}}^{ - 1}}}}{{8.314{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}.1273\;{\rm{K}}}}}}\)

\(K = 7.78 \times {10^{ - 22}}\)

Now that we know the equilibrium constant, we can calculate the equilibrium partial pressure of oxygen. We calculate the equilibrium constant as a ratio of the product of equilibrium concentrations/partial pressures of products to the power of the reaction coefficients and the product of concentrations/partial pressures of reactants to the power of their reaction coefficients. Here it would be:

\(K = \frac{{{p^2}(CO) \cdot p\left( {{O_2}} \right)}}{{{p^2}\left( {CO{O_2}} \right)}}\)

At the beginning of the reaction, we already have initial partial pressures. The initial partial pressure of \({\rm{C}}{{\rm{O}}_2}\) is \(1.15\;{\rm{atm}}\)and those of \({\rm{CO}}\)and \({{\rm{O}}_2}\)are \(0\;{\rm{ atm}}\). When the reaction occurs, the partial pressure of \({\rm{C}}{{\rm{O}}_2}\)drops by \(2x\)(because of the reaction coefficient), the pressure of oxygen rises for \(x\)and the pressure of \(CO\)rises by\(2x\). Now at the equilibrium, the partial pressures are:

\(p\left( {C{O_2}} \right) = 1.15 - 2x\)

\(p\left( {C{O_{}}} \right) = 2x\)

\(p\left( {{O_2}} \right) = x\)

If we insert the partial pressures written as above in the equilibrium constant equation, we get:

\(K = \frac{{{{(2x)}^2} \cdot x}}{{{{(1.15 - 2x)}^2}}}\)

\(K = \frac{{4{x^3}}}{{1.32 - 4.6x - 4{x^2}}}\)

Here we have a cube equation which we can solve via a program on the computer. The result for \(x\)equals \(0.23\;{\rm{atm}}\), which is the equilibrium partial pressure of oxygen.

\(p\left( {{O_2}} \right) = 0.23\;{\rm{atm}}\)

The equilibrium partial pressure of oxygen in a sample at 1000⁰C for which the initial pressure of \({\rm{C}}{{\rm{O}}_2}\)was \(1.15\;{\rm{ atm}}\) is \(0.23\;{\rm{atm}}\).