Question

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 1

Answer

0.1861

Explanation

CO(g) + Cl2(g) <-------> COCl2(g)

K_p = P_COCl2 /(P_CO × P_Cl2) = 3.10

Initial partial pressure

P_CO = 468

P_Cl2 = 399

P_COCl2 = 0

change in concentration

P_CO = -x

P_Cl2 = - x

P_COCl2 = + x

change in concentration

P_CO = 468 - x

P_Cl2 = 399 - x

P_COCl2 = 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