PCl5 --> PCl3 + Cl2
A 0.318 mol sample of PCl5(g) is injected into an empty 4.15 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
Starting with 0.290mol SbCl3 and 0.150mol Cl2 , how many moles of SbCl5 , SbCl3 , and Cl2 are present when equilibrium is established at 248∘C in a 2.50−L flask? SbCl5(g)⇌SbCl3(g)+Cl2(g) KC=2.5×10−2at248∘C
0.318mol/4.15L = 0.0766M
.......PCl5 ==> PCl3 + Cl2
I...0.0766.......0.......0
C........-x......x.......x
E...0.0766-x.....x.......x
Substitute the E line into the Kc expression and solve for x and evaluate 0.0766-x/
To calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium, we will use the concept of equilibrium constants. The balanced equation for the reaction is:
PCl5(g) --> PCl3(g) + Cl2(g)
The equilibrium constant expression for this reaction can be written as:
Kc = [PCl3] * [Cl2] / [PCl5]
Where [PCl3], [Cl2], and [PCl5] represent the equilibrium concentrations of PCl3(g), Cl2(g), and PCl5(g), respectively.
Given that the initial moles of PCl5 is 0.318 mol and the reaction vessel volume is 4.15 L, we can calculate the initial concentration of PCl5 as follows:
Initial concentration of PCl5 = (moles of PCl5) / (volume of reaction vessel)
= 0.318 mol / 4.15 L
≈ 0.0767 M
At equilibrium, let's assume the concentration of PCl3 is x M and the concentration of Cl2 is also x M. Since one mole of PCl5 produces one mole of PCl3 and one mole of Cl2, the change in concentration for PCl5 will be -x M, and the change in concentration for PCl3 and Cl2 will be +x M.
Substituting these values into the equilibrium constant expression, we get:
Kc = (x * x) / (0.0767 - x)
The value of Kc for this reaction is not provided, so we cannot calculate the exact equilibrium concentrations without it. However, we can make an approximation assuming that Kc is large, which would indicate that the forward reaction is favored.
With this approximation, we can assume that x is much smaller than 0.0767. Therefore, we can neglect -x in the denominator of the equilibrium constant expression:
Kc ≈ (x * x) / (0.0767)
Since we have a starting amount of 0.318 mol of PCl5 and we know that some amount of it will be converted to PCl3 and Cl2, we can also write an equation for the initial amount of PCl5:
Initial amount of PCl5 = 0.318 mol - x
Now, let's solve the expression for x:
0.0767 = (x * x) / (0.318 - x)
Rearranging the equation, we get:
0.0767 * (0.318 - x) = x * x
0.0244 - 0.0767x + x^2 = x^2
Simplifying the equation, we find:
0.0767x = 0.0244
Dividing by 0.0767, we get:
x = 0.318 M
Therefore, the equilibrium concentration of PCl3 is approximately 0.318 M, and the equilibrium concentration of Cl2 is also approximately 0.318 M.
Please note that this calculation assumes that the approximation of neglecting -x in the denominator of the equilibrium constant expression is valid. If a value of Kc is provided, or if more accurate calculations are required, please supply that information to obtain a more precise answer.