Question

"Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=9.00 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Answer 1

Answer:

$\boxed{\text{0.50 mol/L}}$

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

$\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}$

2. Solve for x

$K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}$

3. Calculate the equilibrium concentration of COF₂

c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol

$\text{The equilibrium concentration of COF _{2} at equilibrium is \boxed{\textbf{0.50 mol/L}} }$

Check:

$\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}$

OK.