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Chapter 14 Chemical Equilibrium This chemical engineer is testing a process for the formation of new liquid fuels from coal and petroleum. Such methods.

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Presentation on theme: "Chapter 14 Chemical Equilibrium This chemical engineer is testing a process for the formation of new liquid fuels from coal and petroleum. Such methods."— Presentation transcript:

1 Chapter 14 Chemical Equilibrium This chemical engineer is testing a process for the formation of new liquid fuels from coal and petroleum. Such methods may transform the generation of energy worldwide. Dynamic equilibrium Physical processes: solid-liquid vapor-liquid…. Chemical reactions: reactants  products Dynamic equilibria are responsive to changes in the conditions. Watch animation…chemical reaction reaching equilibrium equilibrium being dynamic

2 Assignment for Chapter ,

3 Figure 14.1 In the synthesis of ammonia, the molar concentrations of N 2, H 2, and NH 3 change with time until there is no further net change and the concentrations settle into values corresponding to a mixture in which all three substances are present. N 2 (g)+3H 2 (g)  2NH 3 (g) N 2 (g)+3H 2 (g)  2NH 3 (g) No chemical reaction is ‘complete’, No chemical reaction is ‘impossible’.

4 Figure 14.2 When we plot the rates of the forward and reverse reactions for the formation of ammonia on one graph, we can see that as the forward rate decreases, the reverse rate increases, until they are equal. At this point, the reaction is at equilibrium and the rates remain constant. Reactants  Products

5 Figure 14.3 In an experiment showing that equilibrium is dynamic, a reaction mixture in which N 2 (pairs of blue spheres), D 2 (pairs of yellow spheres), and ND 3 have reached equilibrium is mixed with one with the same concentrations of N 2, H 2 (pairs of gray spheres), and NH 3. After some time, the concentrations of nitrogen, hydrogen, and ammonia are found to be the same, but the D atoms are distributed among the hydrogen and ammonia molecules. N 2 (g)+3D 2 (g)  2ND 3 (g) N 2 (g)+3H 2 (g)  2NH 3 (g) N 2 (g)+2HD (g)+D 2 (g)  2NHD 2 (g) N 2 (g)+H 2 (g)+2HD(g)  2NH 2 Dg) Chemical reaction is dynamic, forward and reverse reactions taking place all the time.

6 Try Yourself to Define the Equilibrium Constant? aA+bB+cC+…  pP+qQ+rR+… Kc=[P]+[Q]+[R]+…-[A]-[B]-[C]-…? Kc=p[P]+q[Q]+r[R]+…-a[A]-b[B]-c[C]-…? Kc=[P][Q][R]…/[A][B][C]…? Kc=pqr…[P][Q][R]…/abc…[A][B][C]…?

7 Equilibrium Constant aA+bB+cC+…  pP+qQ+rR+… Cato Guldberg & Peter Waage (1864)

8 ‘Justification’ of the Definition of Equilibrium Constant aA+bB+cC+…  pP+qQ+rR+… A+A+…B+B+…+C+C+…  P+P+..+Q+Q+…+R+R+…

9 Is the same for all experiment no matter what the initial compositions are. Example

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12 Classroom Exercise Write the equilibrium constant for the following reactions:

13 Multiply a reaction: Reverse a reaction: Using smallest possible stoichiometric coefficients to write equilibrium constants

14 Classroom Exercise Write down the equilibrium constant of the following reaction: H 2 +D 2  2HD Suppose the equilibrium constant of above reaction at 500 K is 3.6, what is the equilibrium constant of its reverse reaction? K c =[HD] 2 /[H 2 ][D 2 ] Reverse reaction: K c -1 =0.28

15 Multi-Step Reactions

16 Equilibrium constant depends on reaction conditions

17 Reaction Rate and Equilibrium

18 Figure 14.4 The equilibrium constant for a reaction is equal to the ratio of the rate constants for the forward and reverse elementary reactions that continue in a state of dynamic equilibrium. (a) A relatively large forward rate constant means that the forward rate can match the reverse rate even though only a small amount of reactants is present. (b) Conversely, if the reverse rate constant is relatively large, then the forward and reverse rates are equal when only small amounts of products are present.

19 Multi-Step Reactions

20 Homogeneous Equilibria: All products and reactants are of the same phase. Heterogeneous Equilibria: Reacting systems with more than one phase.

21 Molar concentration of pure solid and liquid is a constant, independent of the amount present. It is ignored in the calculation of equilibrium constant. Ignored! Another way of understanding: the concentration of pure solid/liquid Is always 100%  1. Since 1 a =1, it does not affect K c.

22 Ignored! Classroom Exercise

23 Gaseous Equilibria

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26 N 2 (g)+3H 2 (g)  2NH 3 (g) T=500 K, equilibrium concentrations: [NH 3 ]=0.796 mol/L, [N 2 ]=0.305 mol/L, [H 2 ]=0.324 mol/L.

27 Classroom Exercise 2S(s)+3O 2 (g)  2SO 3 (g) T=300K, equilibrium concentrations: [O 2 ]=0.25 mol/L, [SO 3 ]=0.3 mol/L. calculate the equilibrium constant Kc and Kp.

28 Gaseous Equilibria Equilibrium constants for gaseous reactions can be written by using either molar concentrations or partial pressures.

29 The equilibrium constant is the ratio of the concentrations or partial pressures of the products to those of the reactants, each concentration raised to a power equal to its stoichiometric coefficient in the balanced equation. Summary of Equilibrium Constants

30 Using Equilibrium Constants A.The Extent of Reaction B.The Direction of Reaction C.Equilibrium Tables

31 The Extent of Reaction K>1000, product dominant; 0.001

32 Figure 14.6 The size of the equilibrium constant indicates whether the reactants (blue squares) or the products (yellow squares) are favored. Note that reactants are favored when K c is small (left), products are favored when K c is large (right), and reactants and products are in almost equal abundance when K c is close to 1 (middle). Here, for simplicity, we compare reactions with K c  10  2 and K c  10 2.

33 Calculating equilibrium concentration H 2 (g)+Cl 2 (g)  2HCl(g)

34 Classroom Exercise Suppose that the equilibrium molar concentrations of H 2 and Cl 2 at 300 K are both 1.0× mol/L. What is the equilibrium molar concentration of HCl, given K c = 4.0×10 31 ? H 2 (g)+Cl 2 (g)  2HCl(g)

35 Using K to determine a partial pressure

36 Classroom Exercise: Using K to determine a partial pressure

37 The Direction of Reaction Reaction quotient:

38 Figure 14.7 The relative sizes of the reaction quotient Q and the equilibrium constant K indicate the direction in which a reaction mixture tends to change. The arrows point from reactants to products when Q,  K (left) or from products to reactants when Q  K (right). There is no tendency to change once the reaction quotient has become equal to the equilibrium constant.

39 Quiz At certain stage, it is measured that Will the reaction be moving to formation of more product or not? The reaction will be moving to formation of more product.

40 A table that shows the initial concentrations, the changes needed to reach equilibrium, and the final equilibrium compositions. Equilibrium Tables

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42 Classroom Exercise

43 Le Chatelier’s Principle When a stress is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the effect of the stress. Henry Louis LE CHATELIER ( )

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45 Use of Le Chatelier Principle (I)  Adding a reactant or removing a product  reaction tends to form products. Adding a product or removing a reactant  more reactant tends to form.

46 Use of Le Chatelier Principle (II)  Compression of a reaction mixture  the reaction that reduces the number of gas-phase molecules. Increasing by introducing an inert gas  no effect on the equilibrium!

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48 Commercial ammonia synthesis vessel

49 Introducing inert gas (yellow) has no effect on the equilibrium composition. More generally, introducing anything that does not react with any product and reactant of a reaction would not change the equilibrium composition of the reaction. (A catalyst can increase reaction rate but not equilibrium constant.)

50 Cheating equilibrium (1)

51 Cheating equilibrium (2)

52 Use of Le Chatelier Principle (III)  Raising the temperature of an exothermic reaction  reaction tends to form more reactants. Raising the temperature of an endothermic reaction  reaction tends to form more products.

53 (a) Endothermic reaction (b) Exothermic reaction

54 Radar image of Venus: high partial pressure of carbon dioxide

55 Catalysts A catalyst can increase the rate of a chemical reaction. A catalyst has no effect on the equilibrium composition of a reaction mixture.

56 Haber’s Achievements Fritz Haber ( , Nobel Prize for Chemistry, 1916) N 2 (g)+3H 2 (g)  2NH 3 (g)

57 Assignment for Chapter ,


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