Asked by Alex
At a certain temperature, .5 mol of PCl5 was placed in a .25 L vessel and permitted to react as shown
PCl5 (g) --> PCl3 (g)+ Cl2 (g)
At equilibrium, the container held .1 mol of PCl5. What is the value of "K"?
PCl5 (g) --> PCl3 (g)+ Cl2 (g)
At equilibrium, the container held .1 mol of PCl5. What is the value of "K"?
Answers
Answered by
DrBob222
........PCl5 ==> PCl3 + Cl2
I.......0.5.........0.....0
C........-x........x.......x
E........0.5-x.......x.......x
Therefore, mol PCl5 = 0.5-x = 0.1 and x = 0.4
PCl3 = 0.4 mol
Cl2 = 0.4 mol
Convert mols to M = mols/L and substitute into K expression; solve.
I.......0.5.........0.....0
C........-x........x.......x
E........0.5-x.......x.......x
Therefore, mol PCl5 = 0.5-x = 0.1 and x = 0.4
PCl3 = 0.4 mol
Cl2 = 0.4 mol
Convert mols to M = mols/L and substitute into K expression; solve.
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