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1 CHEMICAL EQUILIBRIUM Chapter 16. “Systems”: two reactions that differ only in direction Any reversible reaction H 2 + I 2 ↔ 2HI noted by the double.

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Presentation on theme: "1 CHEMICAL EQUILIBRIUM Chapter 16. “Systems”: two reactions that differ only in direction Any reversible reaction H 2 + I 2 ↔ 2HI noted by the double."— Presentation transcript:

1 1 CHEMICAL EQUILIBRIUM Chapter 16

2 “Systems”: two reactions that differ only in direction Any reversible reaction H 2 + I 2 ↔ 2HI noted by the double arrow; ↔

3 TWO REACTIONS only difference is the Direction H 2 + I 2 ↔ 2HI 2HI ↔ H 2 + I 2 reactants products LeftRight

4 Reversible Reactions H 2 + I 2 ↔ 2HI the products may react back to original reactants. “closed system”: ONLY if all reactant are present If one piece is completely gone it has ”gone to competition” and no longer reversible

5 Examples: Reversible Reactions. Unopened Soda Breathing Rechargeable batteries Color changing shirt

6 Equilibrium The state in which a chemical reaction and its reverse reaction occur at the same rate.

7 Equilibrium = No change in amount over time

8 8 Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Pink to blue 2 H 2 O Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2

9 Equilibrium Rates = 0 No change in the amounts

10 10 Reversible Reactions Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium Equilibrium achieved

11 11 Chemical Equilibrium At Equilibrium: RATES ARE EQUAL the concentrations of reactants and products are constant.  [ ]’s = 0 The forward and reverse reactions continue after equilibrium is attained. Fe 3+ + SCN -  FeSCN 2+

12 12 Reaction Quotient At any point in the reaction ---> H 2 + I 2 ---> 2 HI At any point in the reaction ---> H 2 + I 2 ---> 2 HI

13 13 The Reacton Quotient, Q In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. When the system is at equilibrium, Q = K

14 14 Equilibrium Constant Equilibrium achieved In the equilibrium region

15

16 16 THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B ---> c C + d D the following is a CONSTANT (at a given T). If K is known, then we can predict concs. of products or reactants.

17 17 Writing and Manipulating K Expressions Solids NEVER appear in equilibrium expressions. S(s) + O 2 (g) ---> SO 2 (g)

18 18 Writing and Manipulating K Expressions Liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq) ---> NH 4 + (aq) + OH - (aq)

19 19 Writing Equilibrium Expressions

20 20 The Determination of K K comes from thermodynamics. See Chapter 19, page 812-813 ∆G˚ negative: product favored ∆G˚ positive: reactant-favored If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive

21 21 Product- or Reactant Favored Product-favoredReactant-favored

22 22 For: N 2 (g) + 3 H 2 (g) ---> 2 NH 3 (g) Using K: Is the reaction product-favored or reactant-favored? When K is much greater than 1 the reaction is strongly product-favored.

23 23 For AgCl(s)  Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 If K is much less than 1 The reaction is strongly reactant-favored. Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq)  AgCl(s) is product-favored. Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq)  AgCl(s) is product-favored.

24 24 Using K: Can determine if the reaction is at equilibrium.

25 25 If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction “shift” to approach equilibrium?

26 26 REACTION QUOTIENT, Q Characterize all chemical systems Q (2.33) < K (2.5) Reaction is NOT at equilibrium, _. [iso] must ________ and [n] must ____________.. If Q = K, then system is at equilibrium.

27 27 Experimental Determination of Equilibrium Constant, K 2 NOCl(g) ---> 2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Set of an “ICE” table of concentrations [NOCl][NO][Cl 2 ] Initial2.0000 Change Equilibrium0.66

28 28 Determining K 2 NOCl(g) ---> 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

29 29 2 NOCl(g) ---> 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33


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