Carbon monoxide and chlorine gas react to form phosgene: CO(g)+C l 2 (g)⇌COC l 2 (g) K p = 3.10 at 700 K Part A If a reaction mixture initially contains 468 torr of CO and 399 torr of Cl 2 , what is the mole fraction of COCl 2 when equilibrium is reached?
Answer
0.1861
Explanation
CO(g) + Cl2(g) <-------> COCl2(g)
Kp = PCOCl2 /(PCO × PCl2) = 3.10
Initial partial pressure
PCO = 468
PCl2 = 399
PCOCl2 = 0
change in concentration
PCO = -x
PCl2 = - x
PCOCl2 = + x
change in concentration
PCO = 468 - x
PCl2 = 399 - x
PCOCl2 = x
so,
x/(( 0.468 -x) ( 0.399 - x)) = 3.10
solving for x
x = 0.1861
at equillibrium
[CO] = 0.468 - 0.1861= 0.2819torr
[Cl2] = 0.399 - 0.1861 = 0.2129torr
[COCl2] = 0.1861torr
Total pressure = 0.2819torr + 0.2129torr + 0.1861torr = 0.6809torr
mole fraction = partial pressure /Total pressure
mole fraction of COCl2 at equillibrium = 0.1861torr/ 0.6809torr = 0.2733
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