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If the value of K is greater than 1, the products in the reaction are favored. If the value of K is less than 1, the reactants in the reaction are favored. If K is equal to 1, neither reactants nor products are favored.
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Read More »In the previous section, you learned about reactions that can reach a state of equilibrium, in which the concentration of reactants and products aren't changing. If these amounts are changing, we should be able to make a relationship between the amount of product and reactant when a reaction reaches equilibrium. *Note that the water is the solvent, and thus has a value of 1. Dividing by 1 does not change the value of K. *Note that the only product is a solid, which is defined to have a value of 1. That leaves just 1 on top in the numerator. *Note that the solids have a value of 1, and multiplying or dividing by 1 does not change the value of K. [2 ce{TiCl_3} left( s ight) + 2 ce{HCl} left( g ight) ightleftharpoons 2 ce{TiCl_4} left( s ight) + ce{H_2} left( g ight)
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Read More », the concentration of the products is much greater than the concentration of the reactants. The reaction essentially "goes to completion"; all, or most of, the reactants are used up to form the products. If K eq is very small , the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent—most of the reactants remain unchanged, and there are few products produced. , the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent—most of the reactants remain unchanged, and there are few products produced. When K eq is not very large or very small (close to a value of 1) then roughly equal amounts of reactants and products are present at equilibrium.
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Read More »Example (PageIndex{4}) Determine the value of (K) for the reaction [ce{SO_2} left( g ight) + ce{NO_2} left( g ight) ightleftharpoons ce{SO_3} left( g ight) + ce{NO} left( g ight) onumber ] when the equilibrium concentrations are: (left[ ce{SO_2} ight] = 1.20 : ext{M}), (left[ ce{NO_2} ight] = 0.60 : ext{M}), (left[ ce{NO} ight] = 1.6 : ext{M}), and (left[ ce{SO_3} ight] = 2.2 : ext{M}). Solution Step 1: Write the equilibrium constant expression: [K = dfrac{left[ ce{SO_3} ight] left[ ce{NO} ight]}{left[ ce{SO_2} ight] left[ ce{NO_2} ight]} onumber ] Step 2: Substitute in given values and solve: [K = dfrac{left( 2.2 ight) left( 1.6 ight)}{left( 1.20 ight) left( 0.60 ight)} = 4.9
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