The two common chlorides of Phosphorus, PCl3 and PCl5, both important in the production of other phosphorous compounds, coexist in equilibrium as shown in the balanced chemical reaction below:
PCl3 (g) + Cl2 (g) <-----> PCl5 (g)
At 250 C, an equilibrium mixture in a 2.50 L flask contains 0.305 g PCl5, 0.440 g PCl3, and 4.24 g Cl2.
A. Show the equilibrium Constant formula.
B. What is the value of Kc for this reaction? Hint: Use molarity in the Kc calculations.
My answers:
A.
Kc = [PCl5]/([Cl2][PCl3])
B.
1.826
A is OK.
I don't get that for B
What concns do you have for each?
I got 0.23887323 for Cl2, 0.001281584 for PCl3, and 0.000586862 for PCl5.
Check that Cl2. 4.24/70.9.2.5 = 0.0239
A. The equilibrium constant (Kc) formula for this reaction is:
Kc = [PCl5]/([Cl2][PCl3])
In this formula, [PCl5], [Cl2], and [PCl3] represent the molar concentrations of PCl5, Cl2, and PCl3, respectively.
B. To calculate the value of Kc for this reaction, you need to determine the molar concentrations of PCl5, Cl2, and PCl3 at equilibrium.
First, calculate the number of moles of each component present in the equilibrium mixture.
Moles of PCl5 = Mass of PCl5 / Molar mass of PCl5
= 0.305 g / (30.973761 + (5 * 35.453))
= 0.005 mol
Moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 4.24 g / (2 * 35.453)
= 0.119 mol
Moles of PCl3 = Mass of PCl3 / Molar mass of PCl3
= 0.440 g / (30.973761 + (3 * 35.453))
= 0.006 mol
Next, calculate the molar concentrations of each component by dividing the number of moles by the volume of the flask.
Molar concentration of PCl5 = Moles of PCl5 / Volume of flask
= 0.005 mol / 2.50 L
= 0.002 M
Molar concentration of Cl2 = Moles of Cl2 / Volume of flask
= 0.119 mol / 2.50 L
= 0.048 M
Molar concentration of PCl3 = Moles of PCl3 / Volume of flask
= 0.006 mol / 2.50 L
= 0.0024 M
Now, substitute these values into the equilibrium constant formula to calculate Kc.
Kc = [PCl5]/([Cl2][PCl3])
= (0.002 M) / ((0.048 M)(0.0024 M))
= 1.826
Therefore, the value of Kc for this reaction is 1.826.