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Quick Equilibrium review. The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed,

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Presentation on theme: "Quick Equilibrium review. The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed,"— Presentation transcript:

1 Quick Equilibrium review

2 The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 The double arrow implies the process is dynamic.

3 The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

4 A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

5 The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

6 The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b

7 Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant)

8 What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

9 Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g)

10 Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction = K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)

11 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

12 The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

13 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

14 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

15 Le Châtelier’s Principle Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

16 Le Châtelier’s Principle Effects of Volume and Pressure The system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

17 Le Châtelier’s Principle Effect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: –if  H > 0, cooling favors the reverse reaction, –if  H < 0, cooling favors the forward reaction. Consider for which  H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

18 CATALYST—EQUILIBRIUM is achieved faster, but the equilibrium composition remains unaltered.

19 Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

20 EQUILIBRIUM INVOLVING THE SOLUBILITY AND PRECIPITATION OF COMPOUNDS Equilibrium and Solubility

21 Saturated solutions A saturated solution is a solution that is in equilibrium with undissolved solute Example: BaSO 4 (s)   Ba +2 (aq)+ SO 4 -2 (aq)

22 Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2- ] where the equilibrium constant, K sp, is called the solubility product.

23 Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2- ] where the equilibrium constant, K sp, is called the solubility product.

24 Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

25 Will a Precipitate Form? In a solution,  If Q = K sp, the system is at equilibrium and the solution is saturated.  If Q < K sp, more solid will dissolve until Q = K sp.  If Q > K sp, the salt will precipitate until Q = K sp.

26 Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

27 Common Ion Effect If a solution containing two dissolved substances share a common ion, then the solubility of the salt is more difficult to determine Adding “common ion” will cause the solubility to be less in the presence of the common ion Causes less of the substance with the smaller K sp will dissolve.


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