Category: Chemical Engineering Math
"Published in Vacaville, California, USA"
PCl5(g) dissociates into PCl3(g) and Cl2(g) when heated at 250°C and 1 atm. If the density of the gas mixture at equilibrium is 4.4 g/L, what is the fraction of PCl5(g) dissociated?
Solution:
Consider the chemical reactions above as follows
At equilibrium, there are PCl5(g), PCl3(g) and Cl2(g) gas mixture in 1 liter. If the volume of the reaction at equilibrium is 1 liter, then the number of moles of gas mixture at equilibrium is
At the start of the reaction, the volume is also 1 liter. The number of moles of PCl5(g) which is also the number of moles of gas mixture at the start of the reaction is
Hence, the amount of PCl5(g) dissociated or converted into PCl3(g) and Cl2(g) is
Therefore, the fraction of PCl5(g) dissociated is

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Sunday, June 21, 2015
Saturday, June 20, 2015
Dalton's Law of Partial Pressure, 4
Category: Chemical Engineering Math
"Published in Newark, California, USA"
A solvent-water mixture is to be distilled at 95°C. The vapor pressure of the solvent at this temperature is 130 mm Hg and that of water is 640 mm Hg. The solvent is immiscible in water and has a molecular weight of 150. What is the weight of the solvent in kilograms that will be carried over in the distillate with 200 kg of water?
Solution:
From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors which are solvent and water. In this problem, the molecular weight of a solvent is unknown but the partial pressure of solvent and water vapor are given. From the given partial pressures, we can solve for the mole fraction of a solvent vapor as follows
If W is the weight of a solvent and its molecular weight is 150 kg/kgmole, therefore the weight of a solvent that will be carried over in the distillate with 200 kg water is
"Published in Newark, California, USA"
A solvent-water mixture is to be distilled at 95°C. The vapor pressure of the solvent at this temperature is 130 mm Hg and that of water is 640 mm Hg. The solvent is immiscible in water and has a molecular weight of 150. What is the weight of the solvent in kilograms that will be carried over in the distillate with 200 kg of water?
Solution:
From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors which are solvent and water. In this problem, the molecular weight of a solvent is unknown but the partial pressure of solvent and water vapor are given. From the given partial pressures, we can solve for the mole fraction of a solvent vapor as follows
If W is the weight of a solvent and its molecular weight is 150 kg/kgmole, therefore the weight of a solvent that will be carried over in the distillate with 200 kg water is
Friday, June 19, 2015
Dalton's Law of Partial Pressure, 3
Category: Chemical Engineering Math
"Published in Newark, California, USA"
If ethanol and methanol are mixed 50/50 by weight at 60°C, and the solution assumed ideal, what is the composition of the vapor above the solution?
Solution:
From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors. Ethanol and methanol are mixed in equal proportion or amount.
Basis: 100 grams of the mixture
Let's consider first the liquid phase. The number of moles of ethanol and methanol in the solution are
Next, consider the gas or vapor phase. From Saturated Vapor Pressure Data, the vapor pressure of pure ethanol at 60°C is 0.461395 atm and for the pure methanol is 0.832997 atm. Hence, the partial pressure of ethanol is
and the partial pressure of methanol is
Therefore, the % ethanol in the vapor is
and the % methanol in the vapor is
"Published in Newark, California, USA"
If ethanol and methanol are mixed 50/50 by weight at 60°C, and the solution assumed ideal, what is the composition of the vapor above the solution?
Solution:
From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors. Ethanol and methanol are mixed in equal proportion or amount.
Basis: 100 grams of the mixture
Let's consider first the liquid phase. The number of moles of ethanol and methanol in the solution are
Next, consider the gas or vapor phase. From Saturated Vapor Pressure Data, the vapor pressure of pure ethanol at 60°C is 0.461395 atm and for the pure methanol is 0.832997 atm. Hence, the partial pressure of ethanol is
and the partial pressure of methanol is
Therefore, the % ethanol in the vapor is
and the % methanol in the vapor is
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