Carbonyl fluoride, COF2, is an important intermediate for organic fluorine compounds. It can be prepared by the following reaction:
CO2(g) + CF4(g) <==> 2COF2(g)
At 1000*C, K for this reaction is .50. What are the partial pressures of all the gases at equilibrium when the initial partial pressures of CO2 and CF4 are .713 atm?
I'm sort of confused by this problem, but this is what I have so far:
K= (.713)^2/(.713)(.713)= 1.00
.50= x^2/x^2
...which doesn't make sense. Any help would be appreciated.
K = pCOF22/(pCO2*pCF4) = 0.50.
If we have 0.713 atm for CO2 and 0.713 for CF4 initially, and I assume COF2 is 0 initially.
How much will it change? COF2 will change to 2x. CO2 and CF4 will diminish by x so the final partial pressures at equilibrium will be
pCO2 = 0.713-x
pCF4 = 0.713-x
pCOF2 = 2x
Plug in K and solve for x (Don't forget to square 2x.
Post your work if you get stuck.
The problem is that in the denominator the pressure of each gas is NOT .713 That is what it started the reaction was used up. If the partial pressure of 2CO is x, then the partial pressure of each reactant is .714-x
K= (2x)^2/(.713-x)(.713-x)= 1.00
Now solve for x.
To solve for x, we can rearrange the equation as follows:
1.00 = (2x)^2 / (.713 - x)(.713 - x)
Now, let's simplify the equation by multiplying both sides by (.713 - x)(.713 - x):
1.00 * (.713 - x)(.713 - x) = (2x)^2
(.713 - x)(.713 - x) = 4x^2
Expanding both sides:
(.713^2 - x * .713 - x * .713 + x^2) = 4x^2
.510169 - 1.426x + 2x^2 = 4x^2
Rearranging the equation and simplifying further:
3x^2 + 1.426x - .510169 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 1.426, and c = -0.510169. Plugging these values into the quadratic formula:
x = (-1.426 ± √(1.426^2 - 4 * 3 * -0.510169)) / (2 * 3)
Simplifying further:
x = (-1.426 ± √(2.035876 - (-6.1220672))) / 6
x = (-1.426 ± √(8.1579432)) / 6
x = (-1.426 ± 2.8573537) / 6
Now, we can calculate the two possible values of x:
x1 = (-1.426 + 2.8573537) / 6 = 0.2385589
x2 = (-1.426 - 2.8573537) / 6 = -0.7135589
Since the pressure cannot be negative, we discard the negative value of x.
So, x = 0.2385589.
Now, we can calculate the partial pressures of all the gases at equilibrium:
pCO2 = 0.713 - x = 0.713 - 0.2385589 = 0.4744411 atm
pCF4 = 0.713 - x = 0.713 - 0.2385589 = 0.4744411 atm
pCOF2 = 2x = 2 * 0.2385589 = 0.4771178 atm
Therefore, at equilibrium, the partial pressures of CO2, CF4, and COF2 are approximately 0.4744411 atm, 0.4744411 atm, and 0.4771178 atm, respectively.