For the reaction
A + B $$\rightleftharpoons$$ 2C
the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is $$x \times 10^{-1}$$ M. The value of x is ___. (Nearest integer)

For the homogeneous reaction $$\mathrm{A + B \rightleftharpoons 2C}$$ at 298 K the equilibrium constant is given to be $$K_c = 100$$.

Initially all three species have the same concentration, $$[\mathrm A]_0 = [\mathrm B]_0 = [\mathrm C]_0 = 1\;\text{M}$$.

Let the extent of the forward reaction at equilibrium be $$y$$ M. For every mole of A that reacts, one mole of B also reacts and two moles of C are produced. Therefore:

$$ \begin{aligned} [\mathrm A]_{\text{eq}} &= 1 - y,\\[3pt] [\mathrm B]_{\text{eq}} &= 1 - y,\\[3pt] [\mathrm C]_{\text{eq}} &= 1 + 2y. \end{aligned} $$

The equilibrium constant expression for the reaction $$\mathrm{A + B \rightleftharpoons 2C}$$ is

$$ K_c \;=\; \frac{[\mathrm C]_{\text{eq}}^{\,2}}{[\mathrm A]_{\text{eq}}\,[\mathrm B]_{\text{eq}}}. $$

Substituting the equilibrium concentrations gives

$$ 100 \;=\; \frac{(1 + 2y)^{2}}{(1 - y)(1 - y)} \;=\; \left(\frac{1 + 2y}{1 - y}\right)^{2}. $$

Taking the positive square root (all concentrations are positive) we obtain

$$ \frac{1 + 2y}{1 - y} \;=\; 10. $$

Now we solve for $$y$$:

$$ 1 + 2y \;=\; 10(1 - y) \quad\Longrightarrow\quad 1 + 2y \;=\; 10 - 10y. $$

Collecting the $$y$$ terms on one side:

$$ 2y + 10y \;=\; 10 - 1 \quad\Longrightarrow\quad 12y \;=\; 9. $$

Hence

$$ y \;=\; \frac{9}{12} \;=\; \frac34 \;=\; 0.75. $$

Substituting this value back to find the equilibrium concentration of C:

$$ [\mathrm C]_{\text{eq}} \;=\; 1 + 2y \;=\; 1 + 2(0.75) \;=\; 1 + 1.5 \;=\; 2.5\;\text{M}. $$

This concentration is written in the required form as

$$ 2.5\;\text{M} \;=\; 25 \times 10^{-1}\;\text{M}, $$

so the value of $$x$$ is $$25$$.

So, the answer is $$25$$.