**The equilibrium composition for the reaction PCl3 + Cl2 –> PCl5 at 298 K is given below:[PCl _{3}]_{eq} = 0.2mol/L ,[Cl_{2}]_{eq} = 0.1 mol/L , [PCl_{5}]_{eq} = 0.40 mol/L.If 0.2 mol of Cl_{2} is added at the same temperature, the equilibrium concentrations of PCl_{5} is _____× 10^{-2} mol/LGiven: K_{C} for the reaction at 298 K is 20Answer: 49Explanation: **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:

0.24x

^{2}– 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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