Phosphorous pentachloride is a fine grained powder, ranging in colour from colourless to pale yellow. It has a pungent odour. In the gaseous state it shows the equilibrium shown in the equation below. Initially, a sample of phosphorus pentachloride is introduced to an empty vessel. If the equilibrium constant for the above reaction is 4.0 × 10–4 and the equilibrium concentration of [PCl5(g)] is 0.090 mol/L, the equilibrium concentration of [PCl3(g)] is _______ mol/L.
SHOW ALL YOUR WORK.
You didn't show the equation but I believe I have written it below.
...................PCl5(g) ==> PCl3(g) + Cl2(g)
I....................y...................0..............0
C..................-x...................x..............x
E.................y-x...................x..............x
Kc = 4E-4 = (PCl3)(Cl2)/(PCl5)
So y-x = 0.090 M. Substitute 0.090, x, and x into the Kc expression above and solve for x =(PCl3) = (Cl2)
To find the equilibrium concentration of PCl3(g), we need to use the equilibrium constant and the given equilibrium concentration of PCl5(g).
Let's represent the equilibrium concentrations as follows:
[PCl5(g)] = x mol/L
[PCl3(g)] = y mol/L
[Cl2(g)] = z mol/L
The given equilibrium constant (Kc) is 4.0 × 10^-4.
The equilibrium equation is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
According to the equation, the stoichiometry is 1:1:1. Therefore, the equilibrium concentrations can be expressed as:
[PCl5(g)] = x
[PCl3(g)] = y
[Cl2(g)] = z
At equilibrium, the molar ratio of PCl5(g), PCl3(g), and Cl2(g) is 1:1:1. So, we can write:
Kc = [PCl3(g)][Cl2(g)] / [PCl5(g)]
Substituting the given values into the equation:
4.0 × 10^-4 = y * z / x
4.0 × 10^-4 = y * z / 0.090
Now, we can solve for y.
Rearranging the equation:
y * z = 4.0 × 10^-4 * 0.090
y * z = 3.6 × 10^-5
Since the molar ratio between y and z is 1:1, we can assume y = z = a (a is a constant).
a * a = 3.6 × 10^-5
a^2 = 3.6 × 10^-5
a = √(3.6 × 10^-5)
a ≈ 0.006
So, the equilibrium concentration of [PCl3(g)] is approximately 0.006 mol/L.
The given equilibrium reaction is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Let's assume the initial equilibrium concentration of PCl5(g), PCl3(g), and Cl2(g) as [PCl5], [PCl3], and [Cl2], respectively.
At equilibrium, the equilibrium concentrations can be expressed in terms of the initial concentration and the change in concentration using the equilibrium constant (K).
The equation for the equilibrium constant is:
K = ([PCl3] * [Cl2]) / [PCl5]
Given:
K = 4.0 × 10–4
[PCl5] = 0.090 mol/L
We need to find the equilibrium concentration of [PCl3].
Let's assume the change in concentration for PCl5 is -x (as it decreases), and the change in concentration for PCl3 and Cl2 is +x (as they increase).
Using the equilibrium constant expression, we can express the equilibrium concentrations of PCl5, PCl3, and Cl2 as follows:
[PCl5] = 0.090 - x
[PCl3] = x
[Cl2] = x
Substituting the values in the equilibrium constant expression:
K = ([PCl3] * [Cl2]) / [PCl5]
4.0 × 10–4 = (x * x) / (0.090 - x)
Cross-multiplying and rearranging the equation:
4.0 × 10–4 * (0.090 - x) = x^2
0.00004 * (0.090 - x) = x^2
0.0000036 - 0.00004x = x^2
Rearranging the equation and setting it equal to zero:
x^2 + 0.00004x - 0.0000036 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using the values:
a = 1
b = 0.00004
c = -0.0000036
x = (-0.00004 ± √(0.00004^2 - 4 * 1 * -0.0000036)) / (2 * 1)
Simplifying the equation:
x = (-0.00004 ± √(0.0000016 + 0.0000144)) / 0.00008
x = (-0.00004 ± √0.000016) / 0.00008
x = (-0.00004 ± 0.004) / 0.00008
Using the positive root:
x = (-0.00004 + 0.004) / 0.00008
x = 0.00396 / 0.00008
x = 49.5 mol/L
Therefore, the equilibrium concentration of [PCl3] is 49.5 mol/L.