The equilibrium constant for the decompositio | vedclass

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(D) The reaction is $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$.At $t = 0$,we have $1 \ mole$ of $H_2O$.At equilibrium,the moles a

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$5$

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(D) The reaction is $H_2O_{(g)} ightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$.At $t = 0$,we have $1 \ mole$ of $H_2O$.At equilibrium,the moles are: $H_2O = 1-\alpha$,$H_2 = \alpha$,$O_2 = \frac{\alpha}{2}$.Total moles $n_T = 1 + \frac{\alpha}{2} \approx 1$ (since $\alpha \ll 1$).The equilibrium constant $K_p$ is given by:$K_p = \frac{P_{H_2} \cdot P_{O_2}^{1/2}}{P_{H_2O}} = \frac{(\alpha \cdot P) \cdot (\frac{\alpha}{2} \cdot P)^{1/2}}{(1-\alpha) \cdot P}$.Given $P = 1 \ bar$,$K_p = \frac{\alpha \cdot (\alpha/2)^{1/2}}{1} = \frac{\alpha^{3/2}}{\sqrt{2}}$.$8.0 	imes 10^{-3} = \frac{\alpha^{3/2}}{\sqrt{2}}$.$\alpha^{3/2} = 8.0 	imes 10^{-3} 	imes \sqrt{2} \approx 11.31 	imes 10^{-3}$.$\alpha = (11.31 	imes 10^{-3})^{2/3} \approx 5.04 	imes 10^{-2}$.The nearest integer value is $5$.

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