At 760◦C, Kc= 33.3 for the reaction PCl5(g)⇀↽PCl3(g) + Cl2(g) If a mixture that consists of 0.31 mol PCl3 and 0.22 mol Cl2 is placed in an 8 L reaction vessel and heated to 760◦C, what is the equilibrium composition of PCl5?Answer in units of mol/L.
To find the equilibrium composition of PCl5, we need to use the given equilibrium constant (Kc) and the initial amounts of PCl3 and Cl2.
The balanced equation for the reaction is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Given:
Kc = 33.3
Initial amount of PCl3 = 0.31 mol
Initial amount of Cl2 = 0.22 mol
Volume of reaction vessel = 8 L
Let's assume the equilibrium amount of PCl5 in the reaction vessel is x mol.
The equilibrium amounts of PCl3 and Cl2 can be calculated using the initial amounts and the changes in moles during the reaction:
PCl3 = 0.31 mol - x mol
Cl2 = 0.22 mol - x mol
Now we can use the equilibrium constant expression to set up an equation:
Kc = [PCl3][Cl2] / [PCl5]
33.3 = (0.31 - x)(0.22 - x) / x
Expanding the equation:
33.3x = (0.31 - x)(0.22 - x)
Solving for x:
33.3x = 0.0682 - 0.53x + x^2
x^2 - 0.53x + 0.0682 - 33.3x = 0
Rearranging the equation:
x^2 - 33.83x + 0.0682 = 0
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 1, b = -33.83, c = 0.0682
Calculating the values:
x = (33.83 ± sqrt((-33.83)^2 - 4(1)(0.0682))) / (2(1))
x ≈ 33.75 mol/L (approximately)
Therefore, the equilibrium composition of PCl5 is approximately 33.75 mol/L.
To find the equilibrium composition of PCl5, we need to use the equilibrium constant (Kc) and the initial concentrations of PCl3 and Cl2.
Let's start by writing the balanced equation for the reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
The equilibrium constant expression for this reaction is:
Kc = [PCl3] * [Cl2] / [PCl5]
Given values:
Kc = 33.3
[PCl3] = 0.31 mol
[Cl2] = 0.22 mol
Volume (V) = 8 L
Now we can plug in these values into the equilibrium constant expression and solve for [PCl5]:
Kc = [PCl3] * [Cl2] / [PCl5]
Rearranging the equation to solve for [PCl5]:
[PCl5] = [PCl3] * [Cl2] / Kc
Now let's substitute the given values into the equation and solve for [PCl5]:
[PCl5] = (0.31 mol) * (0.22 mol) / 33.3
Calculating:
[PCl5] = 0.002052 moles
To convert from moles to mol/L, we divide the moles by the volume (in L):
[PCl5] = 0.002052 mol / 8 L
Calculating:
[PCl5] = 0.0002565 mol/L
Therefore, the equilibrium composition of PCl5 is approximately 0.0002565 mol/L.
0.31/8 = estimated 0.04 M for PCl3 but you need a better number for all of the calculations that follow.
0.22/8 = about 0.03 M for Cl2.
................PCl5(g)⇀↽PCl3(g) + Cl2(g)
I...................0...............0.04..........0.03
C.................+x............ -x...............-x
E..................+x............0.04-x......0.03-x
Plug the E line into Kc expression and solve for x M = (PCl5)
Kc = 33.3