Water decomposes at 2300 K
$$\text{H}_2\text{O}(g) \to \text{H}_2(g) + \frac{1}{2}\text{O}_2(g)$$
The percent of water decomposing at 2300 K and 1 bar is ______ (Nearest integer). Equilibrium constant for the reaction is $$2 \times 10^{-3}$$ at 2300 K

The decomposition reaction is:

$$H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2}O_2(g)$$

with $$K_p = 2 \times 10^{-3}$$ at 2300 K and total pressure = 1 bar.

Let $$\alpha$$ be the fraction of water that decomposes. Starting with 1 mole of H$$_2$$O:

At equilibrium: $$(1-\alpha)$$ mol H$$_2$$O, $$\alpha$$ mol H$$_2$$, $$\frac{\alpha}{2}$$ mol O$$_2$$.

Total moles = $$1 - \alpha + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$$.

Partial pressures (with $$P_{total} = 1$$ bar):

$$P_{H_2} = \frac{\alpha}{1 + \alpha/2}, \quad P_{O_2} = \frac{\alpha/2}{1 + \alpha/2}, \quad P_{H_2O} = \frac{1-\alpha}{1 + \alpha/2}$$

The equilibrium expression is:

$$K_p = \frac{P_{H_2} \cdot P_{O_2}^{1/2}}{P_{H_2O}} = \frac{\alpha \cdot (\alpha/2)^{1/2}}{(1-\alpha)(1+\alpha/2)^{1/2}}$$

Since $$K_p = 2 \times 10^{-3}$$ is very small, $$\alpha \ll 1$$. Using the approximation $$1 - \alpha \approx 1$$ and $$1 + \alpha/2 \approx 1$$:

$$K_p \approx \alpha \cdot \left(\frac{\alpha}{2}\right)^{1/2} = \frac{\alpha^{3/2}}{\sqrt{2}}$$

$$\alpha^{3/2} = K_p \sqrt{2} = 2 \times 10^{-3} \times 1.414 = 2.828 \times 10^{-3}$$

$$\alpha = (2.828 \times 10^{-3})^{2/3}$$

Computing: $$(2.828 \times 10^{-3})^{2/3} = (2.828)^{2/3} \times 10^{-2}$$. Since $$(2.828)^{2/3} \approx 2.0$$:

$$\alpha \approx 0.02 = 2\%$$

The percentage of water decomposing is $$\boxed{2}$$ (nearest integer).