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CGP EDU Academic Team
Published on: August 14, 2026
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 440 o C as:
Sb 2 S 3 (s) + 3H 2 (g) 
2Sb(s) + 3H 2 S(g)
After equilibrium the H 2 S formed was analysed by dissolving it in water and treating with excess of Pb 2+ to give 708 g of PbS as precipitate. What is value of K c of the reaction at 440ºC? (At. weight of Pb = 206).
Text Solution
Verified by ExpertsThe correct answer is:
A
Mole of PbS = 708 / 236 = 3 mole = mole of H 2 S
Sb 2 S 3 (s) + 3H 2 (g) 
2Sb(s) + 3H 2 S(g)
Initial 6 6 0 0
at eq. 5 3 2 3
K C = 
= 
= 0.08
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