The equilibrium composition for the reaction PCl3 + Cl2 –> PCl5 at 298 K is given below:
[PCl3]eq =  0.2mol/L ,[Cl2]eq = 0.1 mol/L , [PCl5]eq = 0.40 mol/L.
If 0.2 mol of Cl2 is added at the same temperature, the equilibrium concentrations of PCl5 is _____× 10-2 mol/L
Given: KC for the reaction at 298 K is 20

Answer: 49
Explanation:
The equilibrium concentrations of the reactants and products are:
[PCl3]eq = 0.2 mol/L
[Cl2]eq = 0.1 mol/L
[PCl5]eq = 0.40 mol/L

When 0.2 mol of Cl2 is added, the new concentration of Cl2 is:
[Cl2]eq = 0.1 mol/L + 0.2 mol/L = 0.3 mol/L
Let x be the change in concentration of PCl3 and Cl2, and the change in concentration of PCl5 is +x. Then, the new concentrations at equilibrium are:
[PCl3]eq = 0.2 mol/L – x
[Cl2]eq = 0.3 mol/L – x
[PCl5]eq = 0.40 mol/L + x

The equilibrium constant expression for the reaction is:
Kc = [PCl5]eq / ([PCl3]eq * [Cl2]eq)

Substituting the new equilibrium concentrations, we get:
Kc = (0.40 mol/L + x) / ((0.2 mol/L – x) * (0.3 mol/L – x))

We know that Kc = 20, so we can set up the equation:
20 = (0.40 mol/L + x) / ((0.2 mol/L – x) * (0.3 mol/L – x))

Multiplying both sides by ((0.2 mol/L – x) * (0.3 mol/L – x)), we get:
4(0.2 mol/L – x)(0.3 mol/L – x) = 0.40 mol/L + x

Expanding and simplifying, we get:
0.24x2 – 0.14x + 0.028 = 0

Solving this quadratic equation, we get: x = 0.086 mol/L

Therefore, the equilibrium concentration of PCl5 is:
[PCl5]eq = 0.40 mol/L + x = 0.486 mol/L

Converting to scientific notation, we get:
[PCl5]eq = 4.86 × 10-1 mol/L

So the answer is 49.

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