A sample of PCl5 weighing 2.69 grams is placed in a 1.000 liter flask and vaporized completely at 250 o C.

The final pressure observed at this temperature is 1.000 atmospheres. Note that some, but not all of the PCl5 decomposes according to the equation: PCl5 (g) ---------> PCl3 (g) + Cl2 (g)
What are the pressures ( in atm) of PCl5 , PCl3 and Cl2 once the reaction is complete and at equilibrium ?

I assume you don't need a Kp but if you have one listed, please let me know. I believe this is the way to solve the problem although it seems the long way around to me.

moles PCl5 initially = 2.69/molar mass PCl5 = about 0.0129 but you need to confirm ALL of these numbers.
Total moles AT EQUILIBRIUM from PV = nRT. Plug in P, V, R, and T, and I get something like 0.0233 for total moles AT EQUILIBRIUM (not starting.

...........PCl5 ==> PCl3 + Cl2
initial...0.0129......0.....0
change......-x........x.....x
equil....0.0129-x.....x.....x

total moles at equilibrium = 0.0129 - x + x + x = 0.0233
x = 0.0104 and 0.0129-x = 0.0025.

Now solve PV = nRT three times, once for PPCl5, once for PPCl3 and once for PCl2 using the value for T, n (x), R, and total P of 1 atm above. The values you get will be the partial pressure of each of the gases at equilibrium and at the conditions listed. Check my thinking.

Well, it seems like PCl5 is feeling the pressure to decompose! Let's see how the pressures of PCl5, PCl3, and Cl2 change once the reaction is complete and at equilibrium.

First, let's represent the initial pressure of PCl5 as P(PCl5). Since the entire sample of PCl5 is vaporized, we can assume that the initial pressure of PCl5 is equal to 1 atm.

Now, according to the balanced equation, 1 mole of PCl5 decomposes to form 1 mole of PCl3 and 1 mole of Cl2. So, at equilibrium, the mole ratio of PCl5:PCl3:Cl2 is 1:1:1.

Since the initial pressure of PCl5 is 1 atm, the pressures of PCl3 and Cl2 at equilibrium must also be 1 atm.

So, once the reaction is complete and at equilibrium, the pressures of PCl5, PCl3, and Cl2 are all 1 atm.

Looks like the pressures are perfectly balanced, just like a clown juggling multiple balls!

To determine the pressures of PCl5, PCl3, and Cl2 once the reaction is complete and at equilibrium, we need to use the ideal gas law and the concept of partial pressure.

1. First, let's calculate the number of moles of PCl5 using its molar mass. The molar mass of PCl5 is 208.25 g/mol.

moles of PCl5 = mass / molar mass
moles of PCl5 = 2.69 g / 208.25 g/mol
moles of PCl5 = 0.0129 mol

2. Since the reaction is balanced as follows:

PCl5 (g) -> PCl3 (g) + Cl2 (g)

We can see that 1 mole of PCl5 will produce 1 mole of PCl3 and 1 mole of Cl2.

3. At equilibrium, the total number of moles of the reactant and products gases will remain constant. The total number of moles in the flask will be:

total moles = moles of PCl5 + moles of PCl3 + moles of Cl2
total moles = 0.0129 mol + 0.0129 mol + 0.0129 mol
total moles = 0.0387 mol

4. Since the flask has a volume of 1.000 L, we can calculate the total pressure at equilibrium using the ideal gas law equation:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

total pressure = (total moles * R * T) / V
total pressure = (0.0387 mol * 0.0821 L·atm/(mol·K) * 250 °C) / 273.15 K
total pressure = 0.0897 atm

5. The partial pressure of each gas is determined by the ratio of its moles to the total moles. Hence,

partial pressure of PCl5 = (moles of PCl5 / total moles) * total pressure
partial pressure of PCl5 = (0.0129 mol / 0.0387 mol) * 0.0897 atm
partial pressure of PCl5 = 0.03 atm

partial pressure of PCl3 = (moles of PCl3 / total moles) * total pressure
partial pressure of PCl3 = (0.0129 mol / 0.0387 mol) * 0.0897 atm
partial pressure of PCl3 = 0.03 atm

partial pressure of Cl2 = (moles of Cl2 / total moles) * total pressure
partial pressure of Cl2 = (0.0129 mol / 0.0387 mol) * 0.0897 atm
partial pressure of Cl2 = 0.03 atm

Therefore, at equilibrium, the pressures of PCl5, PCl3, and Cl2 are all 0.03 atm.

To find the pressures of PCl5, PCl3, and Cl2 at equilibrium, we need to consider the balanced chemical equation and use the ideal gas law.

The balanced chemical equation for the decomposition of PCl5 is:
PCl5 (g) => PCl3 (g) + Cl2 (g)

Let's assign variables to the unknown pressures:
Let x = pressure of PCl5 (g)
Let y = pressure of PCl3 (g)
Let z = pressure of Cl2 (g)

Based on the information given, the initial pressure of PCl5 is 1.000 atmospheres. Since some PCl5 decomposes to form PCl3 and Cl2, the total pressure at equilibrium will be 1.000 atmospheres.

According to the ideal gas law:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin

We can rearrange the ideal gas law equation to solve for the number of moles:

n = (PV) / (RT)

Since the volume (V) and temperature (T) remain constant, we can simplify the equation as:

n ∝ P

Now let's consider the balanced equation and the moles of each gas. From the equation, we can see that:
1 mole of PCl5 (g) => 1 mole of PCl3 (g) + 1 mole of Cl2 (g)

Using the molar ratio, we can express the moles of PCl5, PCl3, and Cl2 in terms of x, y, and z:

Moles of PCl5 = x
Moles of PCl3 = y
Moles of Cl2 = z

Since the molar ratio is 1:1:1, the moles of each gas are equal.

Therefore:
x = y = z

Now, since the total pressure at equilibrium is 1.000 atm, we can express the pressures in terms of x:

Pressure of PCl5 (g) = x atm
Pressure of PCl3 (g) = x atm
Pressure of Cl2 (g) = x atm

To get the values of x, we need additional information. One crucial information we need is the number of moles of PCl5 initially present in the flask. Without the number of moles, we cannot calculate the value of x and consequently the pressures of PCl5, PCl3, and Cl2 at equilibrium.

Once we know the number of moles of PCl5, we can divide it by the total volume of the flask (1.000 L) to get the concentration in moles per liter. Then, we can use the ideal gas law to calculate the value of x.