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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Well, well... phosgene, you sure know how to make an impression! Let's calculate the pressure in that reaction vessel and see what we're dealing with.
First, we need to convert the mass of phosgene into moles. The molar mass of COCl2 is about 98.92 g/mol. So, 5.00 grams of phosgene gives us:
5.00 g COCl2 * (1 mol COCl2 / 98.92 g COCl2) = 0.0505 mol COCl2
Now, we have to use the ideal gas law to determine the pressure. The equation is P = nRT/V, where P is the pressure, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume.
Let's plug in the given values:
n = 0.0505 mol
R = 0.0821 L*atm/(K*mol)
T = 300 oC = 573 K
V = 1.00 L
P = (0.0505 mol * 0.0821 L*atm/(K*mol) * 573 K) / 1.00 L
P = 2.34 atm
So, the pressure in the vessel at equilibrium is 2.34 atmospheres. Keep that phosgene contained, because we don't want any clowning around with poisonous gases!
Remember, safety first!
To calculate the pressure in the vessel at equilibrium, we need to use the ideal gas law and the equilibrium constant (Kp) to determine the partial pressures of carbon monoxide (CO), chlorine (Cl2), and phosgene (COCl2) at equilibrium.
Step 1: Convert the mass of phosgene (COCl2) to moles.
To do this, we need to calculate the molar mass of phosgene.
The molar mass of COCl2 = (atomic mass of carbon) + 2 * (atomic mass of oxygen) + 2 * (atomic mass of chlorine)
The atomic masses are:
Carbon (C) = 12.01 g/mol
Oxygen (O) = 16.00 g/mol
Chlorine (Cl) = 35.45 g/mol
Molar mass of COCl2 = 12.01 g/mol + 2 * 16.00 g/mol + 2 * 35.45 g/mol = 99.00 g/mol
Now, we can calculate the number of moles of COCl2:
Number of moles = Mass of COCl2 / Molar mass of COCl2
Number of moles = 5.00 g / 99.00 g/mol
Step 2: Calculate the initial pressure of COCl2.
Since the volume of the reaction vessel is given as 1.00 liter, we can assume that the initial pressure is equal to the partial pressure of COCl2.
Partial pressure of COCl2 = (Number of moles of COCl2 * R * Temperature) / Volume
Where:
R = Ideal gas constant = 0.0821 L.atm/mol.K
Temperature = 300 oC = 300 + 273.15 K
Partial pressure of COCl2 = (Number of moles * 0.0821 L.atm/mol.K * 573.15 K) / 1.00 L
Step 3: Calculate the equilibrium partial pressures of CO and Cl2 using the equilibrium constant (Kp).
Since the stoichiometric coefficients of CO and Cl2 in the balanced equation for the decomposition reaction of phosgene are both 1, the equilibrium partial pressure of both gases will be equal to each other.
At equilibrium, let the partial pressure of CO and Cl2 be P.
According to the balanced equation,
Kp = (P_CO * P_Cl2) / P_COCl2
Given that Kp = 6.8 x 10^-2 and we know the initial pressure of COCl2, we can determine the partial pressure of CO and Cl2, which is equal to P.
Step 4: Calculate the pressure in the vessel at equilibrium.
Since the partial pressure of CO and Cl2 at equilibrium is equal to P, the total pressure at equilibrium will be the sum of these partial pressures.
Pressure at equilibrium = P + P
Now, let's plug in the values and calculate.
Note: Make sure to convert temperature from °C to K.
Mass of COCl2 = 5.00 g
Molar mass of COCl2 = 99.00 g/mol
Number of moles of COCl2 = 5.00 g / 99.00 g/mol
R = 0.0821 L.atm/mol.K
Temperature = (300 + 273.15) K
Partial pressure of COCl2 = (Number of moles * R * Temperature) / Volume
Equilibrium constant Kp = 6.8 x 10^-2
Pressure at equilibrium = P + P
By plugging in the values into the equations, you can find the pressure in the vessel at equilibrium after placing 5.00 grams of pure phosgene in a 1.00-litre reaction vessel at 300°C.
To calculate the pressure at equilibrium, we can use the expression for Kp, which is the ratio of the product of the partial pressures of the products and the product of the partial pressures of the reactants, each raised to the power of their stoichiometric coefficients.
The balanced chemical equation is:
COCl2(g) ⇌ CO(g) + Cl2(g)
According to the stoichiometry of the reaction, the molar ratio between COCl2, CO, and Cl2 is 1:1:1.
Now, let's solve the problem step by step:
1. Convert the mass of phosgene (COCl2) to moles:
To convert grams to moles, we need to know the molar mass of COCl2, which is calculated as:
(12.01 g/mol (C) + 16.00 g/mol (O) + 35.45 g/mol (Cl)) x 2 = 98.96 g/mol
Mass of COCl2 = 5.00 g
Moles of COCl2 = 5.00 g / 98.96 g/mol = 0.0505 mol
2. Convert the volume of the reaction vessel to moles:
Since the coefficients of the balanced equation are all 1, the moles of COCl2 will be equal to the moles of CO and Cl2.
Volume of the reaction vessel = 1.00 L
3. Calculate the initial partial pressure of COCl2:
Partial pressure is given by the ideal gas law: P = n/V, where n is the number of moles and V is the volume.
Initial partial pressure of COCl2 = mole fraction of COCl2 x total pressure
The mole fraction of COCl2 is calculated as:
Mole fraction of COCl2 = moles of COCl2 / total moles
Total moles = moles of COCl2 = 0.0505 mol
Therefore, the mole fraction of COCl2 = 0.0505 mol / 0.0505 mol = 1
Assuming an ideal gas, the total pressure is unknown, so we'll use the variable P.
Initial partial pressure of COCl2 = 1 x P = P
4. Calculate the equilibrium concentration of CO and Cl2:
Since the stoichiometric coefficient is 1 for all compounds, the concentration of CO and Cl2 at equilibrium will also be equal to the moles of COCl2.
Equilibrium partial pressure of CO = moles of COCl2 = 0.0505 mol
Equilibrium partial pressure of Cl2 = moles of COCl2 = 0.0505 mol
5. Calculate the equilibrium constant (Kp) using the given information:
Kp = (P_CO * P_Cl2) / P_COCl2
Substituting the given value of Kp = 6.8 x 10^-2 and the equilibrium partial pressure values:
6.8 x 10^-2 = (0.0505 * 0.0505) / P_COCl2
Simplifying this equation, we find:
P_COCl2 = (0.0505^2) / (6.8 x 10^-2)
P_COCl2 = 0.003771 mol
6. Calculate the equilibrium pressure:
Total pressure at equilibrium = P_COCl2 + P_CO + P_Cl2
Total pressure at equilibrium = 0.003771 + 0.0505 + 0.0505 = 0.1048 atm
Therefore, the pressure in the vessel at equilibrium after placing 5.00 grams of pure phosgene in a 1.00-liter reaction vessel at 300°C is approximately 0.1048 atm.
0.125
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.