The equilibrium constant (K_p$K_p$) for the thermal decomposition of water vapor: textH_2textO(g) rightleftharpoons textH_2text(g) + frac12textO_2text(g) quad (Delta G^circ = 92.34 text kJ mol^-1)$$\text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad (\Delta G^\circ = 92.34 \text{ kJ mol}^{-1})$$ is evaluated as 8.0 times 10^-3$8.0 \times 10^{-3}$ at 2300 text K$2300 \text{ K}$ under a total pressure of 1 text bar$1 \text{ bar}$. Under these specific conditions, the degree of dissociation (alpha$\alpha$) of water is _________ times 10^-2$\times 10^{-2}$ (as the nearest integer value). [Assume alpha ll 1$\alpha \ll 1$].

### Related Formula Gas phase dissociation equilibrium setup: textH_2textO(g) rightleftharpoons textH_2text(g) + frac12textO_2text(g)$$\text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}$$ Partial pressure equilibrium expression: K_p = fracP_textH_2 cdot (P_textO_2)^1/2P_textH_2textO$$K_p = \frac{P_{\text{H}_2} \cdot (P_{\text{O}_2})^{1/2}}{P_{\text{H}_2\text{O}}}$$ ### Execution Step 1: Set up the mole distribution table at equilibrium assuming 1 initial mole: * textH_2textO = 1 - alpha$\text{H}_2\text{O} = 1 - \alpha$ * textH_2 = alpha$\text{H}_2 = \alpha$ * textO_2 = fracalpha2$\text{O}_2 = \frac{\alpha}{2}$ Step 2: Calculate the total moles (n_T$n_T$) at equilibrium: n_T = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2$$n_T = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$$ Given alpha ll 1$\alpha \ll 1$, we can approximate n_T approx 1$n_T \approx 1$. Step 3: Express the partial pressures using total pressure P = 1 text bar$P = 1 \text{ bar}$: P_textH_2textO = frac1-alpha1 cdot P approx 1 cdot 1 = 1$$P_{\text{H}_2\text{O}} = \frac{1-\alpha}{1} \cdot P \approx 1 \cdot 1 = 1$$ P_textH_2 = alpha cdot P = alpha$$P_{\text{H}_2} = \alpha \cdot P = \alpha$$ P_textO_2 = fracalpha2 cdot P = fracalpha2$$P_{\text{O}_2} = \frac{\alpha}{2} \cdot P = \frac{\alpha}{2}$$ Step 4: Substitute these partial pressures into the K_p$K_p$ expression: K_p = fracalpha cdot left(fracalpha2right)^1/21 = fracalpha^3/2sqrt2$$K_p = \frac{\alpha \cdot \left(\frac{\alpha}{2}\right)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$$ Step 5: Equate to the given value of K_p = 8.0 times 10^-3$K_p = 8.0 \times 10^{-3}$ and solve for alpha$\alpha$: 8.0 times 10^-3 = fracalpha^3/2sqrt2 implies alpha^3/2 = 8sqrt2 times 10^-3$$8.0 \times 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}} \implies \alpha^{3/2} = 8\sqrt{2} \times 10^{-3}$$ Cube both sides to clear fractional exponents: alpha^3 = left(8sqrt2 times 10^-3right)^2 = 128 times 10^-6$$\alpha^3 = \left(8\sqrt{2} \times 10^{-3}\right)^2 = 128 \times 10^{-6}$$ alpha = sqrt[3]128 times 10^-2 approx 5.03 times 10^-2$$\alpha = \sqrt[3]{128} \times 10^{-2} \approx 5.03 \times 10^{-2}$$ Matching the target template alpha = 5.03 times 10^-2$\alpha = 5.03 \times 10^{-2}$, the integer value is **5**. ### Pattern Recognition When alpha ll 1$\alpha \ll 1$, the total mole expression simplifies to 1, and the denominator (1-alpha)$(1-\alpha)$ drops out. This simplifies the expression to K_p propto alpha^1 + Delta n_g$K_p \propto \alpha^{1 + \Delta n_g}$, allowing you to quickly isolate alpha$\alpha$ via standard powers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Equilibrium Class 11 Chemistry: Chemical Thermodynamics