Question

Question: At 1200 K in the presence of solid carbon, an equilibrium mixture of CO and${\mathbf{CO}}_{\mathbf{2}}$(called “producer gas”) contains 98.3 mol percent CO and 1.69 mol percent of $\mathit{C}{\mathit{O}}_{\mathbf{2}}$when the total pressure is 1 atm. The reaction is $\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathit{C}\left(graphite\right)\mathbf{\rightleftarrows }\mathbf{2}\mathit{C}\mathit{O}\left(g\right)$

(a)Calculate${\mathbf{P}}_{\mathbf{CO}}$and${\mathbf{P}}_{{\mathbf{CO}}_{\mathbf{2}}}$.

(b)Calculate the equilibrium constant.

(c)Calculate$\mathit{\Delta }\mathit{G}\mathbf{°}$for this reaction.

Short answer

a)The partial pressure of CO is 0.983 and ${\mathrm{CO}}_2$is 0.016.

b) The equilibrium constant for the given reaction is 60.37.

c) The change in Gibb’s free energy is$-40901{\mathrm{Jmol}}^{-1}$.

Step-by-step solution

Subpart a)

The Given reaction is, ${\mathrm{CO}}_2\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{graphite}\right)\rightleftarrows 2\mathrm{CO}\left(\mathrm{g}\right)$.

For the calculation of ${\mathrm{P}}_{\mathrm{CO}}$and ${\mathrm{P}}_{{\mathrm{CO}}_2}$we have to calculate the mole fractions of CO and ${\mathrm{CO}}_2$so,

$$\begin{gathered} \mathrm{Mole}\mathrm{percentage}\mathrm{of}\mathrm{CO}={\mathrm{X}}_{\mathrm{CO}}\times 100\% \\ 98.3={\mathrm{X}}_{\mathrm{CO}}\times 100\% \\ {\mathrm{X}}_{\mathrm{CO}}=\frac{98.3}{100} \\ {\mathrm{X}}_{\mathrm{CO}}=0.983 \end{gathered}$$

$$\begin{gathered} \mathrm{Mole}\mathrm{percentage}\mathrm{of}{\mathrm{CO}}_2={\mathrm{X}}_{{\mathrm{CO}}_2}\times 100\% \\ 1.69={\mathrm{X}}_{{\mathrm{CO}}_2}\times 100\% \\ {\mathrm{X}}_{{\mathrm{CO}}_2}=\frac{1.69}{100} \\ {\mathrm{X}}_{{\mathrm{CO}}_2}=0.016 \end{gathered}$$

According to the partial pressure expression,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{CO}}={\mathrm{X}}_{\mathrm{CO}}\times {\mathrm{P}}_{\mathrm{T}} \\ =0.983\times 1\mathrm{atm} \\ =0.983 \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{CO}}_2}={\mathrm{X}}_{{\mathrm{CO}}_2}\times {\mathrm{P}}_{\mathrm{T}} \\ =0.016\times 1\mathrm{atm} \\ =0.016 \end{gathered}$$

Hence, the partial pressure of CO is 0.983 and ${\mathrm{CO}}_2$is 0.016.

Subpart b)

The equilibrium constant for the given reaction is as follows,

$$\begin{gathered} \mathrm{K}=\frac{{\left({\mathrm{P}}_{\mathrm{CO}}\right)}^2}{\left({\mathrm{P}}_{{\mathrm{CO}}_2}\right)} \\ =\frac{{\left(0.983\right)}^2}{0.016} \\ =\frac{0.966}{0.016} \\ =60.37 \end{gathered}$$

Therefore, the equilibrium constant for the given reaction is 60.37.

Subpart c)

The change in Gibb’s free energy for the reaction is as follows,

$$\begin{gathered} \mathrm{\Delta G}°=-\mathrm{RTlnK} \\ =-\left(8.314{\mathrm{Jk}}^{-1}{\mathrm{mol}}^{-1}\right)\left(1200\mathrm{K}\right)\ln \left(60.37\right) \\ =-9976{\mathrm{Jmol}}^{-1}\times 4.100 \\ =-40901{\mathrm{Jmol}}^{-1} \end{gathered}$$

Hence, the change in Gibb’s free energy is $-40901{\mathrm{Jmol}}^{-1}$.