An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

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