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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), K=7.80 at 850 K If only COF2 is present initially at a pressure of 240 kPa, what is the partial pressure of COF2 at equilibrium? You can convert the units of pressure from kPa to bar, using the relation 1 kPa=0.01 bar.

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Answer:

pCOF₂ = 101.8 kPa = 1.018 bar

Explanation:

What we need is to determine the equilibrium pressure for COF₂ .

Now the equilibrum is

                           2COF2(g)⇌   CO2(g)+  CF4(g)  with Kp = 7.80

initially kPa          240                 0               0

react                      -2x                + x            + x

equilibrium        240-2x              +x            +x

Kp = pCO₂ x pCF₄ / p COF₂

7.80 = x² /( 240-2x )²

taking square roots to both sides of the equation

√7.80 = √(x² /( 240-2x )²)

2.8 = x / ( 240 - 2x )

672 - 5.6x = x

672 = 6.6x

672//6.6 = x = 101.8

pCOF₂ = 101.8 kPa = 1.018 bar

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