Phosphorus pentachloride decomposes according to the chemical equation
PCl5<-->PCl3 +Cl2
A 0.260 mol sample of PCl5(g) is injected into an empty 2.75 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
Please help I'm not sure where to go after you solved for x (Using the quadratic formula)
I got x=0.0921
is this correct? and how do i get the PCl5 and PCl3 from that?
Sorry Kc is 1.8 at 250 degree C
You can check these things by substituting your value of x into the Kc expression and see if you come out with 1.8. I did this with 0.0921 and it doesn't pan out. When I solve the quadratic I obtained 0.09 for x and that checks out.
Phosphorus pentachloride gas, PCl5(g), will decompose to phosphorus trichloride,
PCl3(g), and chlorine, Cl2(g), at 160⁰C. In a sealed vessel, the reaction will proceed
to equilibrium:
PCl5(g) ↔ PCl3(g) + Cl2(g)
A chemist places 3.00 mol of phosphorus pentachloride gas into a sealed 1.50 L
flask at 160⁰C. At equilibrium, he observes there is 0.300 mol of phosphorus
trichloride gas and some chlorine gas. Calculate the equilibrium concentrations of
gaseous phosphorus pentachloride and chlorine gas. (6 points)
PCl5(g) ↔ PCl3(g) + Cl2(g)
A chemist places 3.00 mol of phosphorus pentachloride gas into a sealed 1.50 L
flask at 160⁰C. At equilibrium, he observes there is 0.300 mol of phosphorus
trichloride gas and some chlorine gas. Calculate the equilibrium concentrations of
gaseous phosphorus pentachloride and chlorine gas. (6 points)
To determine the concentrations of PCl5(g) and PCl3(g) at equilibrium, you need to start by setting up an ICE (Initial, Change, Equilibrium) table and use the given information.
Step 1: Initial conditions (I)
The initial moles of PCl5(g) is 0.260 mol, and initially, there are no moles of PCl3(g) and Cl2(g). Therefore, the initial concentrations are as follows:
[PCl5] = 0.260 mol / 2.75 L
[PCl3] = 0 mol / 2.75 L
[Cl2] = 0 mol / 2.75 L
Step 2: Change (C)
Since the reaction involves the decomposition of PCl5, let's assume that x mol of PCl5 decomposes at equilibrium. It will then produce x mol of PCl3 and x mol of Cl2. Thus, the change in concentration is as follows:
[PCl5] = -x mol / 2.75 L
[PCl3] = +x mol / 2.75 L
[Cl2] = +x mol / 2.75 L
Step 3: Equilibrium (E)
To determine the equilibrium concentrations, you need to consider the initial concentrations plus the changes. Therefore, at equilibrium:
[PCl5] = (0.260 - x) mol / 2.75 L
[PCl3] = x mol / 2.75 L
[Cl2] = x mol / 2.75 L
Now, you have set up the equilibrium concentrations in terms of x.
To find the value of x, you can substitute these expressions into the equilibrium expression of the given reaction:
Kc = [PCl3][Cl2] / [PCl5] = (x/2.75)^2 / (0.260 - x)
In this case, Kc is the equilibrium constant, which you can look up or be given in the question.
Once you have this equation, you can solve it either by substituting the given temperature (250 °C) to get Kc and using the quadratic formula, or you can use an online solver to get the value of x.
Based on your previous calculations, if you found x = 0.0921, plug that value back into the expressions for [PCl5] and [PCl3] at equilibrium to calculate their concentrations:
[PCl5] = (0.260 - 0.0921) mol / 2.75 L
[PCl3] = 0.0921 mol / 2.75 L
Simplify these expressions to get the final concentrations.
I can't check this without knowing Kc.
If your value of x is correct that is the value of PCl3 and Cl2 because you probably did it this way.
........PCl5 ==> PCl3 + Cl2
I.......0.0945.....0.....0
C.........-x.......x......x
E.......0.0945-x...x......x
You can see x = (PCl3) = (Cl2) while (PCl5) = 0.0945-x