We are going to consider the gas$-$phase chemical reaction $CO(g)+\frac{1}{2}{O}_2(g)C{O}_2(g),$ treating the gases as ideal gases at room temperature. We start with $1$ mole of $CO,$ and $0\mathrm{.}5$ moles of $O_2.$

$($a$)$ Assume that all of the $CO$ and $O_2$ are converted to $C{O}_2.$ Using your knowledge of ideal gases, and the tables on pages $403-404(428-9$ online$)$ of Schroeder, and assuming that the initial and final temperature and pressure are $298K$ and $1$ atmosphere, calculate the change in: $($i$)$ the volume; $($ii$)$ the Gibbs free energy; $($iv$)$ the enthalpy; and $($v$)$ the entropy.

$($b$)$ What does the sign of $G$ tell you about the direction of this reaction, if it is allowed to proceed spontaneously?

In principle, if the reaction is allowed to proceed spontaneously, not quite all of the CO and $O_2$ are used up in the reaction. We will calculate the equilibrium concentrations.

$($c$)$ Minimize the Gibbs free energy $dG={\displaystyle \underset{i}{\overset{?}{}}}{}_{i}d{N}_i$ for the reaction we are considering,

$1$

to show that in equilibrium

${}_{CO}+\frac{1}{2}{}_{O_2}-{}_{C{O}_2}=0.$

$($Hint: for this and the next part, it may be helpful to see Tutorial $6,$ Q$1.)$

In Lecture $16,$ slide five, we showed that for an ideal gas

$(P)=(P)+kTln(\frac{P}{P}),$

where the temperature is held fixed as the pressure changes, and $P$ is a reference pressure, taken to be atmospheric pressure by chemists, as in the Tables in Schroeder.

$($d$)$ Use this, plus the result of part $($c$),$ to show that

$P_{CO}^{1\mathrm{.}5}=\sqrt[\phantom{2}]{2}{P}_{C{O}_2}(1$atm${)}^{0\mathrm{.}5}{e}^{-\frac{G_{CO}+\frac{1}{2}{G}_{O_2}-G_{C{O}_2}}{R}T}.$

Note: on Thursday, $14$ March, the previous equation was corrected from $(1$atm${)}^{1\mathrm{.}5}$ to atm