a 6.0 L vessel contained 0.0222 mol of phosphorus trichloride, 0.0189 mol phosphorus pentachloride and 0.1044 mol of chlorine at 230 degrees celsius in an equilibrium mixture. calculate the equilibrium constant kc and kp for the reaction. PCl3 +Cl2-PCL5
Oh, chemistry, my favorite subject! Let's have some fun crunching those numbers, shall we?
To calculate the equilibrium constant, we need to use the equation:
Kc = ([PCL5]/[PCL3][Cl2])
First, let's find the concentrations of each component:
[PCL3] = 0.0222 mol / 6.0 L = 0.0037 M
[Cl2] = 0.1044 mol / 6.0 L = 0.0174 M
[PCL5] = 0.0189 mol / 6.0 L = 0.0031 M
Now we can plug in these values into the equilibrium constant equation:
Kc = (0.0031) / (0.0037 * 0.0174)
Kc ≈ 0.162
Now, to calculate the equilibrium constant in terms of partial pressures, we'll consider the ideal gas law:
PV = nRT
Let's use this equation to find the partial pressure of each component:
PPCL3 = (0.0037 M) * (0.0821 atm·L/mol·K) * (230 °C + 273.15) ≈ 2.20 atm
PCl2 = (0.0174 M) * (0.0821 atm·L/mol·K) * (230 °C + 273.15) ≈ 10.90 atm
PPCL5 = (0.0031 M) * (0.0821 atm·L/mol·K) * (230 °C + 273.15) ≈ 1.82 atm
Now we can calculate the equilibrium constant in terms of partial pressures:
Kp = (1.82)/(2.20 * 10.90)
Kp ≈ 0.077
So, the equilibrium constant Kc is around 0.162, and the equilibrium constant Kp is approximately 0.077. Voila!
To calculate the equilibrium constant Kc, we need to determine the molar concentrations of the reactants and products at equilibrium.
Given:
- Volume of the vessel (V) = 6.0 L
- Moles of PCl3 (initially) = 0.0222 mol
- Moles of PCl5 (initially) = 0.0189 mol
- Moles of Cl2 (initially) = 0.1044 mol
First, we need to determine the change in moles of the reactants and products at equilibrium. Since the stoichiometry of the reaction is:
PCl3 + Cl2 → PCl5
We can see that, at equilibrium, the moles of PCl3 will decrease by x, moles of Cl2 will decrease by x, and moles of PCl5 will increase by x.
Now, let's calculate the final moles of each species at equilibrium:
Moles of PCl3 (final) = 0.0222 mol - x
Moles of Cl2 (final) = 0.1044 mol - x
Moles of PCl5 (final) = 0.0189 mol + x
Since the total volume remains constant, we can calculate the molar concentrations of each species using the following equations:
Concentration (molar) = Moles / Volume
Concentration of PCl3 (final) = (0.0222 mol - x) / 6.0 L
Concentration of Cl2 (final) = (0.1044 mol - x) / 6.0 L
Concentration of PCl5 (final) = (0.0189 mol + x) / 6.0 L
The equilibrium constant Kc is then given by:
Kc = (Concentration of PCl5)^1 * (Concentration of Cl2)^1 / (Concentration of PCl3)^1
Kc = [(0.0189 mol + x) / 6.0 L]^1 * [(0.1044 mol - x) / 6.0 L]^1 / [(0.0222 mol - x) / 6.0 L]^1
To calculate the equilibrium constant Kp, we need to consider the stoichiometry of the reaction and the ideal gas law. Since there are no gases involved in the reaction, the expression for Kp is:
Kp = Kc
So, Kp = [(0.0189 mol + x) / 6.0 L]^1 * [(0.1044 mol - x) / 6.0 L]^1 / [(0.0222 mol - x) / 6.0 L]^1
Simplifying the expression would involve substituting the given values of moles and solving for x. Due to the complexity of the calculations, it would be best to use a calculator or software program to determine the equilibrium constant Kc and Kp.
To calculate the equilibrium constant, Kc, and equilibrium constant in terms of partial pressures, Kp, for the given reaction, you need to know the concentrations or partial pressures of the reactants and products at equilibrium.
Given:
- Volume of the vessel (V) = 6.0 L
- Moles of reactants and products at equilibrium:
- Moles of PCl3 (n1) = 0.0222 mol
- Moles of PCl5 (n2) = 0.0189 mol
- Moles of Cl2 (n3) = 0.1044 mol
Step 1: Calculate the total moles (n_total) at equilibrium:
n_total = n1 + n2 + n3
Step 2: Calculate the concentrations of reactants and products:
Concentration of PCl3 (C1) = n1 / V
Concentration of PCl5 (C2) = n2 / V
Concentration of Cl2 (C3) = n3 / V
Step 3: Calculate the equilibrium constant (Kc):
Kc = (C2 * C3) / C1
Step 4: Calculate the equilibrium constant (Kp):
Kp = (P2 * P3) / P1
Assuming the ideal gas behavior, we can use the ideal gas law to relate the partial pressures to the concentrations:
P = (n * R * T) / V
where:
P = partial pressure
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Step 5: Calculate the partial pressures:
Partial pressure of PCl3 (P1) = (n1 * R * T) / V
Partial pressure of PCl5 (P2) = (n2 * R * T) / V
Partial pressure of Cl2 (P3) = (n3 * R * T) / V
Step 6: Substitute the values into the equations for Kc and Kp to get the final answers.
Note: Remember to convert the temperature from Celsius to Kelvin by adding 273.15.
Let's calculate the equilibrium constants (Kc and Kp) using the above steps:
Step 1: n_total = 0.0222 + 0.0189 + 0.1044 = 0.1455 mol
Step 2:
C1 = 0.0222 / 6.0 = 0.0037 M
C2 = 0.0189 / 6.0 = 0.0031 M
C3 = 0.1044 / 6.0 = 0.0174 M
Step 3:
Kc = (0.0031 * 0.0174) / 0.0037
Step 4:
Kp = (P2 * P3) / P1
Step 5:
P1 = (0.0222 * 0.0821 * (230 + 273.15)) / 6.0
P2 = (0.0189 * 0.0821 * (230 + 273.15)) / 6.0
P3 = (0.1044 * 0.0821 * (230 + 273.15)) / 6.0
Step 6:
Substitute the values of P1, P2, and P3 into the equation for Kp.
After performing the calculations, you will have the equilibrium constants (Kc and Kp) for the given reaction.
(PCl3) = mols/L = 0.0222/6L = ? and I'll call this w.
(Cl2) = mols/L = ? = y
(PCl5) = mols/L = ? = z
........... PCl3 + Cl2 ---> PCL5
Eq....... w.........y.............z
Write the expression for the equilibrium constant, Kc, plug in the numbers and calculate Kc.
Then Kp = Kc(RT)^delta n
Post your work if you get stuck.