Water decomposes at 2300 K H_2O(g) -> H_2(g)+(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 x 10^-3 at 2300 K.

H_2 O -> H_2+(1 / 2) O_2 K_ p=frac P_ T^(3 / 2)(1+(alpha / 2)) alpha^(3 / 2)2^(1 / 2) P_ T(1+(alpha / 2))^(3 / 2)(1-alpha) \ 2 x 10^-3=fracalpha^(3 / 2)2^(1 / 2)[ as alpha

Water decomposes at 2300 K H2Og H2gfrac 12O2g The perce... | null (Chemistry)

\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}

\begin{aligned} & \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{T}}^{\frac{3}{2}}\left(1+\frac{\alpha}{2}\right) \alpha^{\frac{3}{2}}}{2^{\frac{1}{2}} \mathrm{P}_{\mathrm{T}}\left(1+\frac{\alpha}{2}\right)^{\frac{3}{2}}(1-\alpha)} \\\\ & 2 \times 10^{-3}=\frac{\alpha^{\frac{3}{2}}}{2^{\frac{1}{2}}}\left[\text { as } \alpha<<1 \text { and let } \mathrm{P}_{\mathrm{T}}=1\right] \end{aligned}

\begin{aligned} & \alpha^{\frac{3}{2}}=2^{\frac{3}{2}} \times 10^{-3} \\\\ & \approx 2 \times 10^{-2} \end{aligned}

\therefore \%

of water decomposition

=2 \%

Explanation

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