phsophorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to
PCl3 (g) +Cl2 (g)->PCl5 (g) Kc = 2.01 at 500k
If a 1.000L reaction vessel is charged with .300 mol of PCl5(g) and allowed to equilibrate at this temperature, what would be the partical pressure of PCl5(g)
please help need this to study for a test tomorrow. I know I have to use the ice table in there somewhere but that is about it.
First, it is phosphorus trichloride and pentachloride. Second you probably meant partial pressure.
..........PCl3 + Cl2 ==> PCl5
I..........0......0.......0.300
c..........x......x........-x
E..........x......x.......0.0300-x
Substitute the E line into the kc expression ad solve for x and evaluate 0.300-x.
Well, I must say, this chemistry problem is no laughing matter! But don't worry, I'll do my best to help you understand it.
To solve this problem, we can start by setting up an ICE (Initial, Change, Equilibrium) table. Let's say the initial number of moles of PCl5 is x.
PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)
Initial: 0 0 x
Change: -x -x +x
Equilibrium: 0 0 x + 0.300
Now, we can use the equilibrium constant expression, Kc, to solve for x:
Kc = [PCl5] / [PCl3]^1 * [Cl2]^1
Given that Kc = 2.01, we can substitute the equilibrium concentrations (moles/liters) into the equation:
2.01 = (x + 0.300) / [0] * [0]
Since the initial concentrations of PCl3 and Cl2 are both zero, we can ignore them. Now we can solve for x:
2.01 = (x + 0.300) / 0
Hmm, that doesn't look right. It seems that we have run into a problem. It appears that the equilibrium concentrations of PCl3 and Cl2 are zero, which means they completely dissociated. As a result, we can say that the reaction goes to completion, and all the PCl5 dissociates.
Therefore, the number of moles of PCl5 is 0.300. Since you know the volume of the reaction vessel (1.000 L), you can calculate the partial pressure of PCl5 by using the ideal gas law:
PV = nRT
P * 1.000 L = 0.300 mol * R * 500 K
Solving for P:
P = (0.300 mol * R * 500 K) / 1.000 L
Now, we need to know the value of the ideal gas constant, R, which is 0.0821 L * atm / (mol * K). Substituting the values:
P = (0.300 mol * 0.0821 L * atm / (mol * K) * 500 K) / 1.000 L
P = 12.315 atm
Therefore, the partial pressure of PCl5 would be approximately 12.315 atm.
To solve this problem, we will use the concept of the equilibrium constant and the ice table.
First, let's set up the balanced equation:
PCl3(g) + Cl2(g) ⇌ PCl5(g)
According to the equilibrium equation:
Kc = [PCl5(g)] / ([PCl3(g)] * [Cl2(g)])
We are given Kc = 2.01. We'll assume the initial concentration of PCl3 and Cl2 is 0.
To set up the ice table, let's assign x as the change in concentration for PCl3 and Cl2. Since PCl5 has a concentration of 0.300 mol, the initial concentration of PCl3 and Cl2 is equal to x.
The equilibrium concentrations will be x for PCl3, x for Cl2, and 0.300 - x for PCl5.
Substituting these values into the equilibrium equation:
2.01 = (0.300 - x) / (x * x)
Now, let's solve for x:
2.01 = (0.300 - x) / (x^2)
2.01 * x^2 = 0.300 - x
2.01x^2 + x - 0.300 = 0
We can solve this quadratic equation for x using the quadratic formula:
x = [-1 ± √(1 + 4*2.01*0.300)] / (2*2.01)
x = [-1 ± √(1.2406)] / 4.02
Since we are dealing with concentration values, we can ignore the negative root:
x ≈ 0.0629 M
Now that we have obtained the value of x, we can calculate the partial pressure of PCl5.
PCl5(g) is the only component in the reaction that has a concentration of 0.300 - x = 0.300 - 0.0629 = 0.2371 M.
To convert the concentration to partial pressure, we can use the ideal gas law:
P = nRT/V
where P is pressure, n is the number of moles, R is the gas constant, T is the temperature, and V is the volume.
In this case, we have:
n = 0.2371 moles
T = 500 K (provided in the question)
V = 1.000 L (provided in the question)
R = 0.0821 L·atm/(mol·K) (gas constant)
P = (0.2371 mol) * (0.0821 L·atm/(mol·K)) * (500 K) / (1.000 L)
P ≈ 9.74 atm
Therefore, the partial pressure of PCl5(g) would be approximately 9.74 atm.
To solve this problem, you'll need to set up an ICE table and use the equilibrium constant expression. Here's a step-by-step guide:
Step 1: Write the balanced chemical equation:
PCl3 (g) + Cl2 (g) ⇌ PCl5 (g)
Step 2: Set up the ICE table:
Initial: PCl3(g) = 0 mol
Cl2(g) = 0 mol
PCl5(g) = 0.300 mol
Change: -x -x +x
Equilibrium: 0-x 0-x 0.300+x
Step 3: Write the equilibrium constant expression:
Kc = [PCl5] / ([PCl3] * [Cl2])
Step 4: Plug in the values from the ICE table into the equilibrium constant expression:
2.01 = (0.300+x) / (0-x * 0-x)
Step 5: Solve for x:
2.01 = (0.300+x) / (x^2)
Step 6: Rearrange the equation and solve the quadratic equation:
2.01x^2 = 0.300+x
2.01x^2 - x - 0.300 = 0
Using the quadratic formula, you would find the value of x. However, in this case, you can assume that the value of x is small compared to 0.300. This allows us to simplify the equation:
2.01x^2 ≈ x
2.01x ≈ 1
x ≈ 1 / 2.01
x ≈ 0.498 mol
Step 7: Calculate the partial pressure of PCl5 (g):
The initial amount of PCl5 (g) was 0.300 mol, and at equilibrium, the change in moles was 0.498 mol. Therefore, the equilibrium amount of PCl5 (g) is:
Equilibrium amount of PCl5 (g) = 0.300 + 0.498 = 0.798 mol
Remember that the reaction vessel has a volume of 1.000 L. We can calculate the partial pressure using the ideal gas law:
PV = nRT
Partial pressure of PCl5 (g) = (0.798 mol * 0.0821 L·atm/(mol·K) * 500 K) / 1.000 L
Partial pressure of PCl5 (g) ≈ 32.887 atm
Therefore, the approximate partial pressure of PCl5 (g) at equilibrium is 32.887 atm.