Question

QUESTION #1:

Carbon monoxide and chlorine gas react to form phosgene:
CO(g)+Cl2(g)⇌COCl2(g) Kp = 3.10 at 700 K

If a reaction mixture initially contains 321 torr of CO and 157 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached?

QUESTION #2:

Calculate the concentration of all species in a 0.150 M solution of H2CO3

[H2CO3], [HCO−3], [CO2−3], [H3O+], [OH−] = _____?_____

QUESTION #3:

Write the formula for the conjugate base of each of the following acids expressing the answer as a chemical formula: HNO2, H3PO4, HCHO2, HF

Answer 1

Question 1:

Kp = P(COCl₂) / P(CO) P(Cl₂)
In going to equilibrium, P(COCl₂) gains the stoichiometric partial pressure p from an initial pressure of 0, while both P(CO) and P(Cl₂) lose p pressure. The Kp equation becomes
3.10 = p / (321-p) (157-p)
which rearranges to the quadratic
3.1p² - 1482p + 156230.7 = 0.
Using the quadratic formula, we get two answers, p = 321.13 torr and p = 156.94 torr. Since Cl₂ can't lose any more than 157 torr, the former value is the one we want. At equilibrium then:
P(COCl₂) = p = 156.94 torr
P(CO) = 321-p = 164.06torr
P(Cl₂) = 157-p = 0.06 torr.
The equilibrium composition is checked by substituting these partial pressures back into the equilibrium equation to recalculate Kp.
The question asks for mole fraction of COCl₂ in the equilibrium composition; we can do this easily by remembering the ideal gas law
PV=nRT.
This tells us that, if V and T are constant, the number of moles of a gas is directly proportional to its partial pressure in the mixture. The mole fraction of COCl₂ is therefore:
X(COCl₂) = 156.94 / (156.94 + 164.06 + 0.06) = 0.489, or about 48% by mole.