# Question #faa2e

Apr 28, 2015

The equilibrium concentration of carbonyl fluoride will be 0.340 M.

Before doing any calculations, try to predict what's going to happen when you place the carbonyl fluoride in the vessel.

Since you start with no products, the reaction will proceed to the right and favor the formation of carbon dioxide and carbon tetrafluoride. Because the equilibrium constant is greater than 1, the equilibrium will lie to the right.

Because products will be favored at equilibrium, you can expect the concentration of the carbonyl fluoride to be the smallest of the three.

Set up an ICE table to determine the concentration you need

$\text{ } \textcolor{red}{2} C O {F}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s C {O}_{2 \left(g\right)} + C {F}_{4 \left(g\right)}$

I......2.00....................0..................0
C....(-$\textcolor{red}{2}$x)..................(+x)..................(+x)
E....2-2x.......................x..................x

At equilibrium, you'll get

${K}_{c} = \frac{\left[C {O}_{2}\right] \cdot \left[C {F}_{4}\right]}{{\left[C O {F}_{2}\right]}^{\textcolor{red}{2}}} = \frac{x \cdot x}{2 - 2 x} ^ 2 = {x}^{2} / {\left(2 - 2 x\right)}^{2} = 6.00$

${x}^{2} = 6 \cdot \left(4 - 8 x + 4 {x}^{2}\right) \implies 23 {x}^{2} - 48 x + 24 = 0$

Solving for $x$ will produce 2 values, 1.256 and 0.830. Since you need the concentration of carbonyl fluoride to be positive, the first solution is eliminated. As a result,

$x = \left[C {O}_{2}\right] = \left[C {F}_{4}\right] = \text{0.830 M}$

Therefore,

$\left[C O {F}_{2}\right] = 2 - 2 \cdot 0.830 = \textcolor{g r e e n}{\text{0.340 M}}$

Notice that the initial prediction is correct, carbonyl fluoride has the smallest concentration at equilibrium.