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.
2 answers
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/
.......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/
1 answer