Water decomposes at 2300 K. H_2O_{(g)} right | vedclass

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(A) The reaction is $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$.Let the degree of dissociation be $\alpha$. At equilibrium,the par

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$2$

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(A) The reaction is $H_2O_{(g)} ightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$.Let the degree of dissociation be $\alpha$. At equilibrium,the partial pressures are $P_{H_2O} = P(1-\alpha)$,$P_{H_2} = P\alpha$,and $P_{O_2} = P(\frac{\alpha}{2})$,where $P = 1 \ bar$.The total pressure is $P_{total} = P(1-\alpha) + P\alpha + P\frac{\alpha}{2} = P(1 + \frac{\alpha}{2}) = 1 \ bar$.Since $\alpha$ is very small,$1 + \frac{\alpha}{2} \approx 1$,so $P \approx 1 \ bar$.The equilibrium constant $K_p$ is given by $K_p = \frac{P_{H_2} \cdot (P_{O_2})^{1/2}}{P_{H_2O}} = \frac{(P\alpha) \cdot (P\alpha/2)^{1/2}}{P(1-\alpha)} = 2 	imes 10^{-3}$.Assuming $\alpha \ll 1$,we have $1-\alpha \approx 1$,so $\frac{\alpha \cdot (\alpha/2)^{1/2}}{1} = 2 	imes 10^{-3}$.$\frac{\alpha^{3/2}}{\sqrt{2}} = 2 	imes 10^{-3} \implies \alpha^{3/2} = 2 	imes 10^{-3} 	imes \sqrt{2} = 2^{1.5} 	imes 10^{-3}$.$\alpha = (2^{1.5} 	imes 10^{-3})^{2/3} = 2 	imes 10^{-2} = 0.02$.The percentage of decomposition is $\alpha 	imes 100 = 0.02 	imes 100 = 2 \%$.

Water decomposes at $2300 \ K$. $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2} O_{2(g)}$. 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 $2300 \ K$.

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