Answer:
The concentration of COF₂ at equilibrium is 0.296 M.
Step-by-step explanation:
To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the concentrations or changes in concentration in that stage. For this reaction:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00 - 2x x x
Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".
![Kc=8.30=([CO_(2)] * [CF_(4)] )/([COF_(2)]^(2) ) =(x^(2) )/((2.00-2x)^(2) ) \\8.30=((x)/(2.00-2x) )^(2) \\√(8.30) =(x)/(2.00-2x)\\5.76-5.76x=x\\x=0.852](https://img.qammunity.org/2020/formulas/chemistry/college/dkufc9wdpsfjyffeq3lnfuncg584wzvvx8.png)
The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M