At 1990 K and 1 atm pressure, there are equal number of $$Cl_2$$ molecules and Cl atoms in the reaction mixture. The value of $$K_p$$ for the reaction $$Cl_{2g} = 2Cl_g$$ under the above conditions is $$x \times 10^{-1}$$. The value of $$x$$ is ______ (Rounded off to the nearest integer)

We are given that at 1990 K and 1 atm pressure, there are equal number of $$Cl_2$$ molecules and $$Cl$$ atoms in the equilibrium: $$Cl_{2}(g) \rightleftharpoons 2Cl(g)$$.

Since the number of $$Cl_2$$ molecules equals the number of $$Cl$$ atoms, their moles are also equal. Let $$n_{Cl_2} = n_{Cl} = n$$.

The total number of moles is $$n + n = 2n$$.

The mole fraction of $$Cl_2$$ is $$\frac{n}{2n} = \frac{1}{2}$$ and the mole fraction of $$Cl$$ is $$\frac{n}{2n} = \frac{1}{2}$$.

Since the total pressure is 1 atm, the partial pressures are:

$$P_{Cl_2} = \frac{1}{2} \times 1 = 0.5 \text{ atm}$$ $$P_{Cl} = \frac{1}{2} \times 1 = 0.5 \text{ atm}$$

Now, $$K_p$$ for the reaction $$Cl_2(g) \rightleftharpoons 2Cl(g)$$ is:

$$K_p = \frac{(P_{Cl})^2}{P_{Cl_2}} = \frac{(0.5)^2}{0.5} = \frac{0.25}{0.5} = 0.5$$

We are told $$K_p = x \times 10^{-1}$$. So:

$$0.5 = x \times 10^{-1}$$ $$x = 5$$

So, the answer is $$5$$.