Question

A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions

Answer 1

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. In this case, we can use the ideal gas law to calculate the partial pressures of PCl5, PCl3, and Cl2. The partial pressure of PCl5 is equal to the observed pressure of 1.00 atm, while the partial pressures of PCl3 and Cl2 can be calculated using the ideal gas law with the given temperature, volume, and number of moles of each gas.

Step-by-step explanation:

Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. In this case, we have PCl5, PCl3, and Cl2 gases. Since PCl5 dissociates into PCl3 and Cl2, the partial pressure of PCl5 would be equal to the observed pressure of 1.00 atm. The partial pressure of PCl3 would be equal to the partial pressure of Cl2, and we can calculate them using the ideal gas law with the given temperature, volume, and number of moles of each gas. Let's calculate:

First, let's find the number of moles of PCl5 using its molar mass:

1. 
   
3. Calculate the number of moles of PCl5:
   
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   • Molar mass of PCl5 = 208.24 g/mol
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   • Number of moles of PCl5 = mass / molar mass = 2.69 g / 208.24 g/mol
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Next, let's calculate the partial pressure of PCl5, which is equal to the observed pressure of 1.00 atm:

1. 
   
3. Calculate the partial pressure of PCl5:
   
   • 
     
   • Partial pressure of PCl5 = observed pressure = 1.00 atm
   • 
     
   
   

Now, let's calculate the number of moles of each product (PCl3 and Cl2) formed from the dissociation of PCl5:

1. 
   
3. Calculate the number of moles of PCl3 and Cl2:

Since the stoichiometry of the dissociation is 1:1:1, the number of moles of PCl3 and Cl2 formed will be the same.

1. 
   
3. Calculate the partial pressure of PCl3 and Cl2 using the ideal gas law:

Using the ideal gas law equation PV = nRT, we can rearrange it to solve for the partial pressure (P):

P = (n/V) * R * T

Where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant, and T is the temperature in Kelvin. Let's calculate:

1. 
   
3. Calculate the partial pressure of PCl3 and Cl2 using the ideal gas law equation:

Partial pressure of PCl3 = (number of moles of PCl3 / total moles) * total pressure

Partial pressure of Cl2 = (number of moles of Cl2 / total moles) * total pressure

With these calculations, you can determine the partial pressures of PCl5, PCl3, and Cl2 under the given experimental conditions.