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5.2 - EQUILIBRIUM CONSTANT: K EQ Unit 5: Equilibrium.

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Presentation on theme: "5.2 - EQUILIBRIUM CONSTANT: K EQ Unit 5: Equilibrium."— Presentation transcript:

1 5.2 - EQUILIBRIUM CONSTANT: K EQ Unit 5: Equilibrium

2 EQUILIBRIUM CONSTANT  For a general reaction...  aA + bB ↔ cC + dD  where a,b,c,d are balancing coefficients and A,B,C,D are substances,  an equilibrium constant (K eq ) expression can be written as:  K eq = [C] c x [D] d OR [A] a x [B] b  Keq = [Products] coefficient [Reactants] coefficient  This mathematical relationship is true for all equilibrium systems.

3 USING THE K EQ EXPRESSION  For the equilibrium reaction... H 2 (g) + I 2 (g) ↔ 2 HI (g)  find the equilibrium constant (k eq ) if  [H 2 ] = 0.022 M,  [I 2 ] = 0.022 M, and  [HI] = 0.156 M.

4 FACTS ABOUT K EQ VALUES  Since Keq is a constant for a reaction, it does not change unless the temperature of the system changes.  It does not matter on the initial concentrations used to reach equilibrium, just the concentrations at equilibrium.

5 FACTS ABOUT K EQ VALUES  For example, the following data was taken during an experiment with the equation H 2 + I 2 ↔ 2 HI at equilibrium:  Note that for this reaction, Keq is always the same (ignoring experimental error). Trial[HI][ H 2 ][ I 2 ]K eq 10.1560.0220 50.3 20.7500.106 50.1 31.000.8200.024250.4 41.000.02420.82050.4 51.560.220 50.3

6 FACTS ABOUT K EQ VALUES  When calculating Keq for a given reaction, we DO NOT INCLUDE substances in the liquid or solid phase.  This is because the concentrations of substances in these phases do not change, but are constant.  So, only include gaseous and aqueous states when calculating Keq.

7 FACTS ABOUT K EQ VALUES  Write the equilibrium expression for CaO(s) + CO 2 (g)  CaCO 3 (s)

8 SITUATION 1: IF K EQ IS VERY LARGE:  The concentration of the products are much greater than the concentration of the reactants.  This means that the reaction essentially ‘goes to completion’. That is, all or most of the reactants are used up to form the products.  We will call a number greater than 10 10 very large.  For example he decomposition of ozone, O 3 : 2 O 3 (g)↔ 3 O 2 (g) K eq = 2.0 × 10 57

9 SITUATION 2: IF K EQ IS VERY SMALL:  The concentration of the products are much smaller than the concentration of the reactants.  This means the reaction does not occur to a great extent. That is, most of the reactants remain unchanged because only a few products are formed.  We will call a value less than 10 -10 very small.  For example, The production of nitrogen monoxide: N 2 (g) + O 2 (g)↔2 NO(g) K eq = 1.0 × 10 -25

10 SITUATION 3: IF K EQ IS NEITHER VERY LARGE OR VERY SMALL:  This means that there significant amounts of both products and reactants formed at equilibrium.  We call values between 10 -10 and 10 10 neither very large or very small.  For example, the reaction of carbon monoxide and water:  CO(g) + H 2 O(g)↔CO 2 (g) + H 2 (g)  K eq = 5.09


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