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How to predict the extent of a reaction. |
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Answer» SOLUTION :The magnitude of the equilibrium constant tells us how far a REACTION has proceeded by the time equilibrium has been reached. That is, the value of K of a reversible reaction indicates the extent of the reaction. Consider a reaction `A+BhArrC+D` and its equilibrium constant `K_(c)`. `K_(c)=([C][D])/([A][B])` It can be seen that the value of `K_(c)` is directly proportional to the concentration of the products, and is inversely proportional to the concentration of the reactants. Therefore, a high value of `K_(c)` indicates a high concentration of products and vice versa. Case I : `K_(c)gtgt1` The reaction is strongly product-favoured. Because, equilibrium concentartion of products and much greater than equilibrium concentration of reactions. A large value of `K_(c)` suggests that the reaction proceeds nearly to completion. Examples : (1) The ozone depletion reaction in the stratosphere. `NO(g)+O_(3)(g)hArrNO_(2)(g)+O_(2)(g), K_(c)=6XX10^(34)" at "25^(@)C` Case II : `K_(c)ltlt1` : The reaction is strongly reactant favoured. Because, equilibrium concentrations of reactants are greater than that of products. A small value of `K_(c)` suggests that the reaction proceeds only to a negligible extent. Examples : (1) Ozonisation of oxygen. `3O_(2)(g)hArr2O_(3)(g),""K_(c)=6.25xx10^(-58)" at "25^(@)C` Such low values of `K_(c)` in the above two examples indicate that the reactions do not take PLATE at `25^(@)C` Case III : `K_(c)=1` : Equilibrium mixture contains significant concentrations of reactants and products. In this case, `K_(c)` is neither extremely large nor extremely small. Therefore neither the forward reaction nor the backward reaction goes to completion. Examples: 1. Dissociation of `N_(2)O_(4)` has neither a very large nor a very small equilibrium constant. `N_(2)O_(4)(g)hArr2NO_(2)(g),""K_(c)=4.64xx10^(-3)" at "25^(@)C,andK_(c)=1.00" at "118^(@)C` |
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