Question

Suppose 1.50 atm of CH₄(g), 2.50 atm of C₂H₆(g), and15.00 atm of O₂(g) are placed in a flask at a given temperature. The reactions are

$\begin{array}{c}\mathbf{C}{\mathbf{H}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\\ \mathbf{2}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{7}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{4}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\end{array}$

Calculate the equilibrium pressures of all gases.

Short answer

The equilibrium pressures of all gases will be as follows:

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

Step-by-step solution

Step 1:Define assumptions

If the K_P values are much higher,then it can be said thatthe equilibrium lies to the right or product side. Let us assume that each reaction goes tocompletion initially.It can be solved for the back equilibrium problems.

$\begin{array}{l}C{H}_4+2O_2\to C{O}_2+2H_{2}O\\ Before:1.5015.0000\\ After:012.001.503.00\\ 2C_2{H}_4+7O_2\to 4C{O}_2+6H_{2}O\\ Before:2.5012.001.503.00\\ After:03.256.5010.50\end{array}$

Again, assume that both the reactions go to completion. From the above assumption we have 3.25 atm O₂, 6.50 atm CO₂, and10.50 atm H₂O. Now we need to solve the back equilibrium problems in order to determine the equilibrium concentrations.

Determine the equilibrium pressures

$\begin{array}{l}C{H}_4+2O_2\rightleftharpoons C{O}_2+2H_{2}O\\ Initial:03.256.5010.50\\ Change:+x+2x-x-2x\\ Equillibrium:x3.25-2x6.50-x10.50-2x\end{array}$

$\begin{array}{c}{K}_P=1.0\times {10}^4\\ K_P=\frac{P_{C{O}_2}\times P_{H_{2}O}^2}{P_{C{H}_4}\times P_{O_2}^2}\end{array}$

$\begin{array}{c}\frac{P_{C{O}_2}\times P_{H_{2}O}^2}{P_{C{H}_4}\times P_{O_2}^2}=1.0\times {10}^4\\ \frac{(6.50-x){(10.50-2x)}^2}{x{(3.25+2x)}^2}=1.0\times {10}^4\\ x=6.8\times {10}^{-3}atm\\ P_{C{H}_4}=6.8\times {10}^{-3}atm\end{array}$

$\begin{array}{l}2C{H}_4\left(g\right)+7O_2\left(g\right)\rightleftharpoons 4C{O}_2\left(g\right)+6H_{2}O\left(g\right)\\ Initial:03.256.5010.50\\ Change:+2x+7x-4x-6x\\ Equillibrium:2x3.25+7x6.50-4x10.50-6x\end{array}$

$\begin{array}{c}{K}_P=1.0\times {10}^8\\ K_P=\frac{P_{C{O}_2}^2\times P_{H_{2}O}^6}{P_{C_2{H}_6}^2\times P_{O_2}^7}\end{array}$

$\begin{array}{c}\frac{P_{C{O}_2}^2\times P_{H_{2}O}^6}{P_{C_2{H}_6}^2\times P_{O_2}^7}=1.0\times {10}^8\\ \frac{{(6.50-4x)}^4{(10.50-6x)}^6}{{\left(2x\right)}^2{(3.25+7x)}^7}=1.0\times {10}^4\\ x=3.95\times {10}^{-2}atm\\ P_{C_2{H}_6}=2x=7.9\times {10}^{-2}atm\\ P_{O_2}=3.25+7x=3.53atm\\ P_{C{O}_2}=6.50-4x=6.342atm\\ P_{H_{2}O}=10.50-6x=10.263atm\end{array}$

The equilibrium pressures of all gases will be

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