A vessel of `250` litre was filled with `0.01` mole of `Sb_(2)S_(3)` and `0.01` mole of `H_(2)` to attain the equilibrium at `440^(@)C` as
`Sb_(2)S_(3)(s)3H_(2)(g)hArr2Sb(s)+3H_(2)D(g)` After equilibrium, the `H_(2)S` formed was analysed was analysed by dissloved it in water and treating with execedd of `Pb^(20+)` to give `1.19` g of PbS as precipitate. What is the value of `K_(c)` at `440^(@)C` ?

A vessel of 2.50 litre was filled with 0.01 mole of Sb_(2)S_(3) and 0.01 mole of H_(2) to attain the equilibrium at 440^(@)C as: Sb_(2)S_(3(s))+3H_(2(g))hArr2Sb_((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 1.029g of PbS as precipitate. What is value of K_(c) of the reaction at 440^(@)C ? (At weight of Pb=206 )

A vessel of 250 L was filled with 0.01 mole of Sb_(2)S_(3) and 0.01 mole of H_(2) to attain the equilibrium at 440^(@)C as Sb_(2)S_(3(s))+3H_(2(g))hArr2Sb_(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 1.19 g of PbS as precipitate. The value of K_(C) at 440^(@)C is (Pb=20S)

Sb_(2)S_(3) is -

Write the equilibrium constant of the reaction C(s)+H_(2)O(g)hArrCO(g)+H_(2)(g)

What is the equilibrium expression K_c , for the reaction : 2S(s)+3O_(2)(g)hArr 3SO_3(g) ?

The equilibrium constant for the reaction CaSO_(4).5H_(2)H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g) is equal to

" The equilibrium constant "Kp" for the reaction NH_(4)HS_((s))to NH_(3(g))+H_(2)S_((g)) is