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Value of equilibrium constant `(K_P)` for a reaction A(g) 2B(g) is 0.0005 atm. What does it indicate ? (I) Products are more stable than reactants. (II) Reactants are very less dissociated. (III) Reactants are more stable than products. (IV) Value of `K_C` will be less than `K_P` at 300 K.A. II and IIIB. III and IVC. II, III and IVD. I, II, III and IV |
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Answer» Correct Answer - C Very small value of `K_P` indicates that reactants are more stable and very less dissociation. Unit of `K_P` = atm indicates that `Deltan_g` = 1 and `K_P` is greater than `K_C` |
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