At certain temperature a 1.0 L flask initially contained 0.298mol PCl3(g) and 8.7 x 10-3 mol PCl5(g). After the system had reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5(g) decomposes according to the reaction:
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium concentration of all species and the value of K.
The reaction will proceed in the forward direction
Cl2]0 = 0
[PCl3]0 = 0.298 mol/ 1 L = 0.298 M
[PCl5]0 = 8.70x10-3 mol/ 1L = 8.70x10-3M
Then determine the change
Since no Cl2(g) was present before but 2.00x10-3 M developed that means that 2.00x10-3 M of PCl5 had to have decomposed. This also means since Cl2(g) and PCl3 have a 1:1 ratio that 2.00x10-3M PCl3 also had to have formed.
Now apply the change to the initial
[Cl2] = 0 + 2.00x10-3mol/ 1L = 2.00x10-3M
[PCl3] = 0.298 M + 2.00x10-3mol/ 1L = 0.30M
[PCl5] = 8.70x10-3M – 2.00x10-3mol/1L = 6.70X10-3 M
Plug these values into the equilibrium expression and solve for K
K = [Cl2][PCl3] = (2.00x10-3)(0.3) = 8.96x10-2
[PCl5] 6.7x10-3
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