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 ==> 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.
To determine the value of the equilibrium constant "K" for the given reaction, we need to know the concentrations of the reactants and products at equilibrium.
In this case, we are given the initial amount of PCl5 (0.5 mol) and the amount at equilibrium (0.1 mol). The initial volume of the vessel is 0.25 L.
To calculate the concentrations, we divide the amount of each substance by the volume of the container:
[PCl5] = 0.1 mol / 0.25 L = 0.4 M
[PCl3] = 0 mol / 0.25 L = 0 M (because it is formed from PCl5)
[Cl2] = 0 mol / 0.25 L = 0 M (because it is formed from PCl5)
Now we can write the expression for the equilibrium constant "K" using these concentrations:
K = [PCl3] * [Cl2] / [PCl5]
= 0 * 0 / 0.4
Since both [PCl3] and [Cl2] are zero, the numerator becomes zero and the value of "K" is also zero.
Therefore, the equilibrium constant "K" for this reaction at the given temperature is zero.