He places 4.20 g of chlorine, Cl 2 , and an equal molar amount of carbon monoxid

Question

He places 4.20 g of chlorine, Cl2, and an equal molar amount of carbon monoxide, CO, into a 10.00 L reaction vessel at 395 C. After the reaction comes to equilibrium, he adds another 4.20 g of chlorine to the vessel in order to push the reaction to the right to get more product. What is the partial pressure of phosgene when the reaction again comes to equilibrium? K? = 1.23E+3

A chemist wants to prepare phosgene, COCl2, by the following reaction: CO(g) + Cl2(g) COCl2(g) He places 4.20 g of chlorine, Cl2, and an equal molar amount of carbon monoxide, CO, into a 10.00 L reaction vessel at 395 ½C. After the reaction comes to equilibrium, he adds another 4.20 g of chlorine to the vessel in order to push the reaction to the right to get more product. What is the partial pressure of phosgene when the reaction again comes to equilibrium? K? = 1.23E+3

Explanation / Answer

Concentration of chlorine gas = (4.2/70.9)*(1/10) = 0.592M

[CO] = 0.592M

[COCl2] = 0.529^2*Kc

Kc = Kp(RT)^delta n ==> Kp/(8.314*668) = 0.2215

[COCl2] at equilibrium = 0.529^2*0.2215 = 0.062 M

after adding 4.2 g of Cl2 more,

Kc = (0.062+x)/(0.467-x)(0.996-x)

solving for x, x = 0.041

therefore, [COCl2] = 0.103 M

moles of COCl2 = 1.03, CO =( 0.467-0.041)*10=4.26

Cl2=0.955*10=9.55

total moles = 14.84. Total pressure = nRT/V = 8.241x10^6 Pa

partial pressure of COCl2 = (1.03/14.84)*(P) = 5.72x10^5 Pa

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