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September 21, 2014

September 21, 2014

Posted by **JP** on Sunday, April 15, 2007 at 4:55pm.

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 = pCOF2

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.

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