Question
Phosgene (COCl2) is an extremely poisonous gas that was used extensively in World War I as a chemical warfare agent. At 300oC, phosgene decomposes to carbon monoxide and chlorine with a Kp = 6.8 x 10-2. Calculate the pressure in the vessel at equilibrium after placing 5.00 grams of pure phosgene in a 1.00 litre reaction vessel at 300oC.
COCl2(g)---> CO(g) + Cl2(g).
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COCl2(g)---> CO(g) + Cl2(g).
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Answers
First get the number of moles of COCl2 and the total pressure before dissociation starts. 5 g is 5/99 mol. That would occupy (5/99)(673/273)22.4 = 2.79 liter at 300C=673 K and 1 atm. Since is is confined to 1.00 l, the pressure is 2.79 atm initially. When a fraction x dissociates, You will have x moles of CO, x moles of Cl2, and 1-x moles of COCl2.
Now write the Kp relationship
(2.79 x)^2/[2.79(1-x)] = Kp = 6.8*10^-2
2.79 x^2/(1-x) = 6.8*10^-2
Calculate x and then realize you have (5/99)(1+x) total moles. Use that and the gas law to get the new pressure at equilibrium.
Now write the Kp relationship
(2.79 x)^2/[2.79(1-x)] = Kp = 6.8*10^-2
2.79 x^2/(1-x) = 6.8*10^-2
Calculate x and then realize you have (5/99)(1+x) total moles. Use that and the gas law to get the new pressure at equilibrium.
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