For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5.
PCl5(g) = PCl3(g) + Cl2(g)
Suppose that 2.510 g of PCl5 is placed in an evacuated 470. mL bulb, which is then heated to 600. K.
(a) What would be the pressure of PCl5 if it did not dissociate?
___atm
(b) What is the partial pressure of PCl5 at equilibrium?
___ atm
(c) What is the total pressure in the bulb at equilibrium?
___atm
(d) What is the degree of dissociation of PCl5 at equilibrium?
___%
i got a to be 1.263 but i cnat find b or c and i didn't get to d yet
im having a lot of trouble! help!!
Your answer to a is correct IF the unit is atmospheres.
b. Set up at ICE chart.
Initial: pressure PCl5 = 1.263 atm
partial p PCl3 = 0
partial p Cl2 = 0
change: partial p PCl3 = +y
partial p Cl2 = y
partial pressure PCl5 = -y
equilibrium pressures:
add the columns to obtain
partial p PCl3 = y
partial p Cl2 = y
partial pressure PCl5 = 1.263-y
Now plug all that into the equilibrium constant expression; Kp = ------
and solve for y, the only unknown in the equation. I expect you will get a quadratic equation. That will give you y and 1.263-y (1.263-y will be the partial pressure of PCl5).
c. Add partial pressure PCl5 + partial pressure PCl3 + partial pressure Cl2.
d. The degree of dissociation is
partial p PCl5 at equilibrium/partial p PCl5 initially. By the way, if you then multiply that by 100 you will have percent dissociation. The degree of dissociation IS NOT expressed in percent terms.
part c.
To solve this problem, we can use the ideal gas law, along with the given equilibrium constant, to calculate the partial pressures and other quantities.
(a) The pressure of PCl5 if it did not dissociate can be calculated using the ideal gas law formula: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
Mass of PCl5 = 2.510 g
Volume = 470 mL = 0.47 L
Temperature = 600 K
Molar mass of PCl5 = 208.25 g/mol (obtained from periodic table)
First, calculate the number of moles of PCl5:
n = mass / molar mass = 2.510 g / 208.25 g/mol = 0.01206 mol
Now, substitute the values into the ideal gas law equation:
PV = nRT
P * 0.47 L = 0.01206 mol * 0.0821 L*atm/mol*K * 600 K
P = (0.01206 mol * 0.0821 L*atm/mol*K * 600 K) / 0.47 L
P ≈ 0.299 atm
Therefore, the pressure of PCl5 if it did not dissociate would be approximately 0.299 atm.
(b) To find the partial pressure of PCl5 at equilibrium, we need to first determine the initial moles of PCl5. Since there is no dissociation initially, the initial moles of PCl5 are equal to the moles placed in the bulb.
n_initial = 0.01206 mol
The equilibrium concentration of PCl5 can be calculated using the equilibrium constant expression:
Kp = (PCl3)(Cl2) / (PCl5)
Understanding that PCl3 and Cl2 are produced in equimolar amounts, we can call their concentration at equilibrium as x and consider PCl5 = n_initial - x.
Kp = (x)(x) / (n_initial - x)
11.5 = x^2 / (0.01206 - x)
By solving this equation, we find that x ≈ 0.1097 mol.
The partial pressure of PCl5 at equilibrium can now be calculated:
P_PCl5 = (moles of PCl5) / volume
P_PCl5 = (0.01206 - 0.1097) mol / 0.47 L
P_PCl5 ≈ 0.225 atm
Therefore, the partial pressure of PCl5 at equilibrium is approximately 0.225 atm.
(c) The total pressure in the bulb at equilibrium is equal to the sum of the partial pressures of all the gases present at equilibrium.
Total Pressure = P_PCl5 + P_PCl3 + P_Cl2
From the balanced equation, we know that the moles of PCl3 and Cl2 produced are equal to the moles of PCl5 dissociated. Therefore, we can use the value of x (0.1097 mol) as the moles of PCl3 and Cl2.
P_PCl3 = P_Cl2 = 0.1097 mol / 0.47 L
P_PCl3 = P_Cl2 ≈ 0.233 atm
Total Pressure = 0.225 atm + 0.233 atm + 0.233 atm
Total Pressure ≈ 0.691 atm
Therefore, the total pressure in the bulb at equilibrium is approximately 0.691 atm.
(d) The degree of dissociation (α) of PCl5 at equilibrium can be calculated using the formula:
Degree of dissociation (α) = (moles dissociated) / (initial moles)
Moles dissociated = 0.1097 mol
Initial moles = 0.01206 mol
Degree of dissociation (α) = 0.1097 mol / 0.01206 mol
Degree of dissociation (α) ≈ 9.1
To express it as a percentage, multiply α by 100:
Degree of dissociation (α) ≈ 9.1 %
Therefore, the degree of dissociation of PCl5 at equilibrium is approximately 9.1%.
To solve this problem, you will need to use the information given about the equilibrium constant and the initial amount of PCl5 to find the partial pressures and degree of dissociation at equilibrium. Let's go step-by-step through the problem to find the answers:
(a) To find the pressure of PCl5 if it did not dissociate, you can assume that all of the PCl5 remains as PCl5. The pressure is directly proportional to the number of moles and inversely proportional to the volume.
First, we need to find the number of moles of PCl5:
n(PCl5) = mass / molar mass = 2.510 g / 208.2 g/mol ≈ 0.01206 mol
Next, we need to find the pressure:
PV = nRT
P = nRT / V
P = (0.01206 mol) (0.0821 atm/mol K) (600 K) / (0.470 L)
P ≈ 1.222 atm
So, the pressure of PCl5 if it did not dissociate would be approximately 1.222 atm.
(b) To find the partial pressure of PCl5 at equilibrium, we need to calculate the partial pressure by using the equilibrium constant.
Kp = (P_PCl3 * P_Cl2) / P_PCl5
Since we are assuming that PCl3 and Cl2 come from the dissociation of PCl5, their initial pressures are zero. Therefore, we only need to consider the partial pressure of PCl5 at equilibrium.
Let x be the partial pressure of PCl5 at equilibrium.
Kp = (x * 0) / (1.222 - x)
Simplifying the equation gives:
11.5 = 0 / (1.222 - x)
Since 0 divided by any non-zero number is equal to 0, we have:
0 = 0
This means that any value of x that satisfies the equation will be the partial pressure of PCl5 at equilibrium. Therefore, the partial pressure of PCl5 can be any value between 0 and 1.222 atm.
(c) To find the total pressure in the bulb at equilibrium, we need to consider the partial pressures of PCl5, PCl3, and Cl2. Since PCl3 and Cl2 come from the dissociation of PCl5, their initial pressures are also zero.
Total pressure = P_PCl5 + P_PCl3 + P_Cl2
Total pressure = x + 0 + 0
Total pressure = x atm
Therefore, the total pressure in the bulb at equilibrium is equal to the partial pressure of PCl5, which can be any value between 0 and 1.222 atm.
(d) The degree of dissociation of PCl5 at equilibrium can be calculated using the formula:
Degree of dissociation = (initial moles of PCl5 - moles of PCl5 at equilibrium) / initial moles of PCl5
The initial moles of PCl5 can be calculated using the given mass and molar mass:
Initial moles of PCl5 = 2.510 g / 208.2 g/mol ≈ 0.01206 mol
The moles of PCl5 at equilibrium can be calculated using the partial pressure of PCl5 at equilibrium and the ideal gas equation:
PV = nRT
0.470 L * x atm = moles at equilibrium * 0.0821 atm/mol K * 600 K
Simplifying the equation gives:
x = moles at equilibrium / 0.02832 mol/L
Substituting this into the degree of dissociation formula gives:
Degree of dissociation = (0.01206 mol - moles at equilibrium) / 0.01206 mol
The value of moles at equilibrium can vary, depending on the partial pressure of PCl5. Therefore, you need to find the value of moles at equilibrium using the equation derived in part (c) above.
I hope this helps you to solve the remaining parts of the problem!
I didn't check a)
But to get b,c you have to write the equilibrium expression
and solve for the moles of the products at equialbrium. I assume you know how to do that.
Then, you can calculate b, c given the moles of each product, and the moles left of the reactant.