The equilibrium constant for decomposition of H 2 O ( g ) H 2 O ( g ) ⇌ H 2 ( g ) + 2 1 O 2 ( g ) ( Δ G ∘ = 92.34 kJ mol − 1 ) is 8.0 × 1 0 − 3 at 2300 K and total pressure at equilibrium is

Question

The equilibrium constant for decomposition of $H_{2} O (g)$
$H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) (Δ G^{\circ} = 92.34 kJ mol^{- 1})$
is $8.0 \times 1 0^{- 3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation $(α)$ of water is ________ $\times 1 0^{- 2}$ (nearest integer value).
[Assume $α$ is negligible with respect to 1 ]

TestHub · Accepted Answer

​ H 2 ​ O ( g ) ⇌ H 2 ( g ) ​ + 2 1 ​ O 2 ( g ) ​ t = 0 1 mole t = t eq ​ 1 − α α 2 α ​ n T ​ = 1 + 2 α ​ ≃ 1 ( α ≪ 1 ) k P ​ = P H 2 ​ O ​ P H 2 ​ ​ ⋅ P O 2 ​ 1/2 ​ ​ = ( 1 − α ) P ( α ⋅ P ) ( 2 α ​ P ) 2 1 ​ ​ 8 × 1 0 − 3 = 2 ​ α 3/2 ​ α 3/2 = 8 2 ​ × 1 0 − 3 α 3 = 128 × 1 0 − 6 α = 3 128 ​ × 1 0 − 2 = 5.03 × 1 0 − 2 ​

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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