Carbonyl fluoride, COF_2, is an important intermediate used in the production of
ID: 634004 • Letter: C
Question
Carbonyl fluoride, COF_2, 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) + CF_4(g) Kc=8.40
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?/
I'm having the most trouble at the end of this problem. I'm getting to x=5.79-5.79x but don't know where to go from there. I know 0<x<.5 right?
I need help walking through the last few steps...step by step.
Explanation / Answer
Let's take it from the beginning, and we'll catch up with your equation to make sure everything checks out:
The equilibrium constant is: Kc = [CF4][CO2]/[COF2]^2 = 8.40
Recall that we need the exponent of 2 since COF2 has a coefficient of 2 in our equation.
So, call the amount of COF2 that reacts "x"
Then, we have 0.5 x each of CO2 and CF4 (because 2 moles of COF2 are needed for each mole of CO2 and CF4 that we generate). The remaining concentration of COF2 is then 2 - x.
Now it's time for some algebra:
[0.5 x][0.5x]/[(2-x)^2 = 8.40
[0.25x^2]/[4 - 4x + x^2] = 8.40
.25 x ^ 2 = 8.40 (x^2 - 4 x + 4)
Or, rearranging and expanding:
8.15x^2 - 33.6 x + 33.6 = 0. So you should end up with a quadratic equation, which is typical for these sort of "one-becomes-two" reactions.
The two roots of this equation are 2.417 and 1.706. the root at 2.42 is impossible, since it would lead to a final concentration of -0.417 M for COF2.
So, the last possibility if that 1.706 M of COF2 was consumed, leading to 0.853 M each CO2 and CF4, and 0.294 M COF2 remains.
Plugging this into our equation for Kc:
Kc = [CF4][CO2]/[COF2^2]
Kc = (0.853)*(0.853)/(0.294)^2 = 8.418, which is slightly different from the given value of 8.40 due to rounding.
Thus, at equilibrium the final concentrations are: [CO2] = [CF4] = 0.853 M; [COF2] = 0.294 M.
I never reached an equation like the one you posted. If you'd like, I could follow through your work and see if I can spot what's different.