A vessel of 10 L was filled with 6 mole of Sb 2 S 3 and 6 mole of H 2 to attain the equilibrium at 44 0 ∘ C as : S b 2 S 3 ( s ) + 3 H 2 ( g ) ⇌ 2 S b ( s ) + 3 H 2 S ( g ) After equilib

Question

A vessel of 10 L was filled with 6 mole of $Sb_{2} S_{3}$ and 6 mole of $H_{2}$ to attain the equilibrium at $44 0^{\circ} C$ as :

$S b_{2} S_{3} (s) + 3 H_{2} (g) ⇌ 2 S b (s) + 3 H_{2} S (g)$

After equilibrium, the $H_{2} S$ formed was analyzed by dissolving it in water and treating it with excess $P b^{2 +}$ to give 708 g of $P b S$ as a precipitate. What is the value of $K_{c}$ for the reaction at $440℃$ ? (Atomic weight of Pb = 206).

TestHub · Accepted Answer

708 g P b S = 206 + 32 708 ​ = 238 708 ​ = 2.97 ≈ 3 m o lP b S ⇒ 3 m o l H 2 ​ S f or m e d . From S b 2 ​ S 3 ​ + 3 H 2 ​ ⇌ 2 S b + 3 H 2 ​ S , H 2 ​ consumed = 3 m o l , so H 2 ​ left = 6 − 3 = 3 m o l . In a 10 L vessel: [ H 2 ​ ] = [ H 2 ​ S ] = 10 3 ​ = 0.3 M . Since solids are omitted, K c ​ = [ H 2 ​ ] 3 [ H 2 ​ S ] 3 ​ = ( 0.3 ) 3 ( 0.3 ) 3 ​ = 1. Answer: K c ​ = 1 (Option A).

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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