The equilibrium composition for the reaction
$$PCl_3 + Cl_2 \rightleftharpoons PCl_5$$ at 298 K is given below:
[PCl$$_3$$]$$_{eq}$$ = 0.2 mol L$$^{-1}$$, [Cl$$_2$$]$$_{eq}$$ = 0.1 mol L$$^{-1}$$, [PCl$$_5$$]$$_{eq}$$ = 0.40 mol L$$^{-1}$$
If 0.2 mol of Cl$$_2$$ is added at the same temperature, the equilibrium concentrations of PCl$$_5$$ is ______ $$\times 10^{-2}$$ mol L$$^{-1}$$
Given: K$$_c$$ for the reaction at 298 K is 20

Given the equilibrium reaction:

$$PCl_3 + Cl_2 \rightleftharpoons PCl_5$$

Initial equilibrium concentrations: $$[PCl_3] = 0.2$$ mol/L, $$[Cl_2] = 0.1$$ mol/L, $$[PCl_5] = 0.40$$ mol/L, and $$K_c = 20$$.

After adding 0.2 mol of $$Cl_2$$ (assuming 1 L volume), the new initial concentrations before re-equilibrium are:

$$[PCl_3] = 0.2$$ mol/L, $$[Cl_2] = 0.1 + 0.2 = 0.3$$ mol/L, $$[PCl_5] = 0.40$$ mol/L

Let $$x$$ mol/L shift to the right to restore equilibrium:

$$[PCl_3] = 0.2 - x$$, $$[Cl_2] = 0.3 - x$$, $$[PCl_5] = 0.40 + x$$

Applying the equilibrium expression:

$$K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40 + x}{(0.2 - x)(0.3 - x)} = 20$$

Expanding:

$$0.40 + x = 20(0.06 - 0.5x + x^2) = 1.2 - 10x + 20x^2$$

$$20x^2 - 11x + 0.8 = 0$$

Using the quadratic formula:

$$x = \frac{11 \pm \sqrt{121 - 64}}{40} = \frac{11 \pm \sqrt{57}}{40}$$

Since $$x < 0.2$$, we take the smaller root:

$$x = \frac{11 - \sqrt{57}}{40} = \frac{11 - 7.55}{40} \approx 0.0863$$

The equilibrium concentration of $$PCl_5$$:

$$[PCl_5] = 0.40 + 0.0863 = 0.4863 \text{ mol/L} \approx 49 \times 10^{-2} \text{ mol/L}$$

Therefore, the answer is $$\mathbf{49}$$.