Question

Suppose 1.50 atm of CH₄(g), 2.50 atm of C₂H₆(g), and 15.00 atm of O₂(g) are placed in a flask at a given temperature. The reactions are

$\begin{array}{rcl}\mathit{C}{\mathit{H}}_{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{O}}_{\mathbf{2}}& \mathbf{\rightleftharpoons }& \mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\\ \mathbf{2}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{+}\mathbf{7}{\mathit{O}}_{\mathbf{2}}& \mathbf{\rightleftharpoons }& \mathbf{4}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\end{array}$

Calculate the equilibrium pressures of all gases.

Short answer

The equilibrium pressures of all gases will be as follows

$\begin{array}{rcl}{P}_{C{H}_4}& =& 6.8\times {10}^{-3}atm\\ P_{C_2{H}_6}& =& 7.9\times {10}^{-2}atm\\ P_{O_2}& =& 3.53atm\\ P_{C{O}_2}& =& 6.342atm\\ P_{H_{2}O}& =& 10.263atm\end{array}$

Step-by-step solution

Assumption

As the K_P values are much higher then it can be said that equilibrium lies to the right or product side. Let us assume that each reaction go to completion initially. It can be solved for the back equilibrium problems.

Again, assume that both the reactions go to completion. From the above assumption, we have 3.25 atm O₂, 6.50 atm CO₂, and 10.50 atm H₂O. Now we need to solve the back equilibrium problems in order to determine the equilibrium concentrations.

Determine the equilibrium pressure

$\begin{array}{rcl}{K}_p& =& 1.00\times {10}^4\\ K_p& =& \frac{P_{C{O}_2}\times {P_{H_{2}O}}^2}{P_{C{H}_4}\times {P_{O_2}}^2}\\ \frac{P_{C{O}_2}\times {P_{H_{2}O}}^2}{P_{C{H}_4}\times {P_{O_2}}^2}& =& 1.00\times {10}^4\\ \frac{\left(6.50-x\right){\left(10.50-2x\right)}^2}{x{\left(3.25+2x\right)}^2}& =& 1.00\times {10}^4\\ x& =& 6.8\times {10}^{-3}atm\\ P_{C{H}_4}& =& 6.8\times {10}^{-3}atm\end{array}$

$\begin{array}{rcl}{K}_p& =& 1.00\times {10}^8\\ K_p& =& \frac{{P_{C{O}_2}}^2\times {P_{H_{2}O}}^6}{{P_{C_2{H}_6}}^2\times {P_{O_2}}^7}\\ \frac{{P_{C{O}_2}}^2\times {P_{H_{2}O}}^6}{{P_{C_2{H}_6}}^2\times {P_{O_2}}^7}& =& 1.00\times {10}^8\\ \frac{{\left(6.50-4x\right)}^2{\left(10.50-6x\right)}^6}{{\left(2x\right)}^2{\left(3.25+7x\right)}^7}& =& 1.00\times {10}^8\\ x& =& 3.95\times {10}^{-2}atm\\ P_{C{H}_4}& =& 6.8\times {10}^{-3}atm\end{array}$

$\begin{array}{rcl}{P}_{C_2{H}_6}& =& 2x=7.9\times {10}^{-2}atm\\ P_{O_2}& =& 3.25+7x=3.53atm\\ P_{C{O}_2}& =& 6.50-4x=6.342atm\\ P_{H_{2}O}& =& 10.50-6x=10.263atm\end{array}$

The equilibrium pressures of all gases will be

$\begin{array}{rcl}{P}_{C{H}_4}& =& 6.8\times {10}^{-3}\\ P_{C_2{H}_6}& =& 7.9\times {10}^{-2}atm\\ P_{O_2}& =& 3.53atm\\ P_{C{O}_2}& =& 6.342atm\\ P_{H_{2}O}& =& 10.263atm\end{array}$