The equilibrium constant for decomposition of $$H_2O(g)$$
$$H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)$$ ($$\Delta G^\circ = 92.34$$ kJ mol$$^{-1}$$) is $$8.0 \times 10^{-3}$$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($$\alpha$$) of water is _____ $$\times 10^{-2}$$. (nearest integer value)
[Assume $$\alpha$$ is negligible with respect to 1]

Let us start with $$1$$ mole of $$H_2O(g)$$ and let the degree of dissociation at equilibrium be $$\alpha$$.

For the reaction $$H_2O(g) \rightleftharpoons H_2(g) + \frac12 O_2(g)$$ the mole table is:

Initial moles: $$H_2O = 1$$, $$H_2 = 0$$, $$O_2 = 0$$
Change in moles: $$H_2O = -\alpha$$, $$H_2 = +\alpha$$, $$O_2 = +\dfrac{\alpha}{2}$$
Equilibrium moles: $$H_2O = 1-\alpha$$, $$H_2 = \alpha$$, $$O_2 = \dfrac{\alpha}{2}$$

Total moles at equilibrium: $$n_{\text{tot}} = 1 - \alpha + \alpha + \dfrac{\alpha}{2} = 1 + \dfrac{\alpha}{2}$$ $$-(1)$$

The total pressure is given as $$P_{\text{tot}} = 1\text{ bar}$$, therefore the partial pressures are

$$P_{H_2O} = \dfrac{1-\alpha}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$,
$$P_{H_2} = \dfrac{\alpha}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$,
$$P_{O_2} = \dfrac{\dfrac{\alpha}{2}}{1+\dfrac{\alpha}{2}}\;(1\text{ bar})$$

The equilibrium constant for the gaseous reaction is

$$K_p = \dfrac{P_{H_2}\,(P_{O_2})^{1/2}}{P_{H_2O}}$$

Substituting the expressions of partial pressures:

$$K_p = \dfrac{\dfrac{\alpha}{1+\dfrac{\alpha}{2}}\left(\dfrac{\dfrac{\alpha}{2}}{1+\dfrac{\alpha}{2}}\right)^{1/2}}{\dfrac{1-\alpha}{1+\dfrac{\alpha}{2}}}$$

Simplifying (the common factor $$1+\dfrac{\alpha}{2}$$ cancels once):

$$K_p = \dfrac{\alpha\,\sqrt{\dfrac{\alpha}{2}}}{(1-\alpha)\sqrt{1+\dfrac{\alpha}{2}}}$$ $$-(2)$$

The data give $$K_p = 8.0 \times 10^{-3}$$. Since dissociation of steam at 2300 K is small, we first make the approximation $$\alpha \ll 1$$ so that $$1-\alpha \approx 1$$ and $$1+\dfrac{\alpha}{2}\approx 1$$. With this, equation $$(2)$$ reduces to

$$K_p \approx \dfrac{\alpha^{3/2}}{\sqrt{2}}$$

Hence

$$\alpha^{3/2} \approx K_p\,\sqrt{2} = (8.0\times 10^{-3})(1.414) = 1.131\times 10^{-2}$$

$$\alpha \approx \left(1.131\times 10^{-2}\right)^{2/3} \approx 5.0\times 10^{-2}$$

To improve accuracy, put $$\alpha = 0.05$$ back into the exact expression $$(2)$$:

Numerator: $$\alpha\,\sqrt{\dfrac{\alpha}{2}} = 0.05\sqrt{\dfrac{0.05}{2}} = 0.05(0.1581)=0.00791$$
Denominator: $$(1-\alpha)\sqrt{1+\dfrac{\alpha}{2}} = 0.95\sqrt{1.025}=0.95(1.012)=0.961$$
$$K_p = \dfrac{0.00791}{0.961}=8.2\times10^{-3}\approx 8.0\times10^{-3}$$

The calculated value matches the given $$K_p$$, confirming $$\alpha \approx 0.05$$.

Expressing $$\alpha$$ as $$\alpha \times 10^{-2}$$ gives $$5.0 \times 10^{-2}$$, whose nearest integer value is

5.

Therefore, the degree of dissociation of water under the stated conditions is $$\boxed{5\times 10^{-2}}$$ (nearest integer 5).