Final answer:
To find the required partial pressure of C2H4 gas to maintain the given solubility, we convert the solubility to molarity, then use Henry's law to determine the pressure, which is 0.366 atm or 278.16 mm Hg.
Step-by-step explanation:
The question asks what partial pressure of ethylene gas (C2H4) is needed to maintain a certain solubility in water based on the given Henry's law constant. First, we need to convert the solubility from grams per liter to moles per liter (molarity), since the Henry's law constant is given in mol/L·atm. The molar mass of ethylene (C2H4) is approximately 28.05 g/mol.
Let's calculate the molarity (M) of the C2H4 solution using the known solubility:
Solubility = 4.92 × 10-2 g/L
Molar mass of C2H4 = 28.05 g/mol
Molarity (M) = Solubility / Molar mass
M = (4.92 × 10-2 g/L) / (28.05 g/mol) = 1.75 × 10-3 mol/L
Henry's law states that the solubility of a gas (Cg) in a liquid at a constant temperature is directly proportional to the partial pressure of that gas (Pg) above the liquid:
Cg = kH × Pg
Given the Henry's law constant (kH) for C2H4 is 4.78 × 10-3 mol/L·atm, we can rearrange the formula to solve for the partial pressure (Pg):
Pg = Cg / kH
Pg = (1.75 × 10-3 mol/L) / (4.78 × 10-3 mol/L·atm)
Pg = 0.366 atm
To convert this to mm Hg, we use the conversion factor that 1 atm = 760 mm Hg:
Pg (in mm Hg) = Pg (in atm) × 760 mm Hg/atm
Pg (in mm Hg) = 0.366 atm × 760 mm Hg/atm = 278.16 mm Hg
Therefore, the partial pressure of C2H4 gas required to maintain the given solubility is 278.16 mm Hg.