Answer :
Answer:
[tex]\boxed{\text{0.50 mol/L}}[/tex]
Explanation:
The balanced equation is
2COF₂ ⇌ CO₂+CF₄; Kc = 9.00
1. Set up an ICE table
[tex]\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}[/tex]
2. Solve for x
[tex]K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}[/tex]
3. Calculate the equilibrium concentration of COF₂
c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol
[tex]\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}[/tex]
OK.