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Chemical Equilibrium Part 2

Section: Chemistry

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Question : 15 of 22

Marks: +1, -0

The equilibrium constant for decomposition of

H _2 O ( g )

H_2 O(g) ⇌ H_2(g)+\;{1}/{2} O_2(g) (∆ G^{∘}=92.34 kJ mol^{-1})

8.0 × 10^{-3}

at 2300 K and total pressure at equilibrium is 1 bar.
Under this condition, the degree of dissociation (

α

\_\_\_\_

× 10^{-} { }^2

(nearest integer value)
[Assume

α

is negligible with respect to 1 ]

[8 Apr 2025 Shift 2]

Your Answer:

Solution:

{\H}_2 {\O}( {\g}) ⇌ {\H}_2( {\g})+\;{1}/{2} {\O}_2( {\g})

\;\text " Initial mole "\; \;\;\;\;\;\; 1\;-\;-

\;\text " Moles at equil. "\;\;1-α\;α\;\;{α}/{2}

\;\text " Equilibrium pressure "\;=1 \;\text " bar "\;\;

K_p=\;{ ({2 α}/{2+α}) ({α}/{2+α}) ^{{1}/{2}}}/{2 (\;{1-α}/{2+α}) }=8.0 × 10^{-3}

\;\;{α^{{3}/{2}}}/{{√{2}}}=8.0 × 10^{-3}

\;α= (8 {√{2}} × 10^{-3}) ^{\;{2}/{3}}

\;= (2^{\;{7}/{2}} × 10^{-3}) ^{\;{2}/{3}}=2^{\;{7}/{3}} × 10^{-2}=5 × 10^{-2}

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