"Hard" Equilibrium problem

Question

Given the initial concentrations below, calculate the concentration of CO₂ (g) at equilibrium. You know that the initial concentration of CH₄ is 0.5 M and the initial concentration of the O₂ is 0.4 M. Also, K for this reaction is 5.2 x 10⁵. 

CH₄ (g) +2O_{2 <=>} CO_{2 +}2H₂O (g)

 

Keep in mind simplifications are to be and should be made to answer this problem.

 

Answer 1

For this reaction, the forward direction is strongly favored and O₂  is the limiting reactant. 

After equilibrium is reached,   very little O₂ will remain, and the concentration of methane will be approximately .5 - ½.4 = .3,  the concentration of CO₂ will approximately .2 , and the concentration of water will be approximately .4 .

 

These values can be derived with the aid of a RICE table approach.   If we call the small concentration of O₂ δ,

 

the equilibrium formula is

 

    5.2 x 10⁶ =   .2 (.4)²  /( .3 δ² )  + terms of order 1/δ

 

Solving for  δ    gives   δ =  1.4 x 10^-4   + correction terms of order 10^-6

 

 RICE table approach shows that the concentration of CO₂ =  .2 -  δ/2