11 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 6 Energy Changes, Reaction Rates and Equilibrium Prepared by Andrea D. Leonard University of Louisiana at Lafayette

22 Introduction to Chemical Reactions A chemical change—a chemical reaction—converts one substance into another. Chemical reactions involve: breaking bonds in the reactants (starting materials) forming new bonds in the products CH 4 and O 2 CO 2 and H 2 O

33 Introduction to Chemical Reactions A chemical equation is an expression that uses chemical formulas and other symbols to illustrate what reactants constitute the starting materials in a reaction and what products are formed. The reactants are written on the left. The products are written on the right. Coefficients show the number of molecules of a given element or compound that react or are formed.

44 Introduction to Chemical Reactions

55 Balancing Chemical Equations HOW TO Balance a Chemical Equation Example Write a balanced chemical equation for the reaction of propane (C 3 H 8 ) with oxygen (O 2 ) to form carbon dioxide (CO 2 ) and water (H 2 O). Step [1] Write the equation with the correct formulas. C 3 H 8 + O 2 CO 2 + H 2 O The subscripts in a formula can never be changed to balance an equation, because changing a subscript changes the identity of a compound.

66 Balancing Chemical Equations Step [2] HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Finally, balance the O’s:

77 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Example Ethanol (C 2 H 6 O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C 2 H 4, molar mass 28.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene?

88 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product 14 g C 2 H 4 x 1 mol C 2 H g C 2 H 4 x 1 mol C 2 H 6 O 1 mol C 2 H 4 x 46.1 g C 2 H 6 O 1 mol C 2 H 6 O Grams of reactant Grams of reactant molar mass conversion factor molar mass conversion factor mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor =23 g C 2 H 6 O Grams of product Grams of product Grams C 2 H 4 cancel. Moles C 2 H 4 cancel. Moles C 2 H 6 O cancel.

99 Percent Yield The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation. The actual yield is the amount of product isolated from a reaction. Usually, however, the amount of product formed is less than the maximum amount of product predicted.

10 Percent Yield Calculating Percent Yield Percent yield = actual yield (g) theoretical yield (g) x 100% Sample Problem 5.14 If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed? = 15 g 23 g x 100% =65%

11 Energy Energy is the capacity to do work. Potential energy is stored energy; kinetic energy is the energy of motion. The law of conservation of energy states that the total energy in a system does not change. Energy cannot be created or destroyed. Chemical bonds store potential energy. Reactions that form products having lower potential energy than the reactants are favored. A compound with lower potential energy is more stable than a compound with higher potential energy.

12 Energy The Units of Energy A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 o C. A joule (J) is another unit of energy. 1 cal = J Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal). 1,000 J = 1 kJ1,000 cal = 1 kcal 1 kcal = kJ

13 Energy Energy Changes in Reactions When molecules come together and react, bonds are broken in the reactants and new bonds are formed in the products. Bond breaking always requires an input of energy. Bond formation always releases energy. Cl To cleave this bond, 58 kcal/mol must be added. To form this bond, 58 kcal/mol is released.

14 Energy Changes in Reactions  H is the energy absorbed or released in a reaction; it is called the heat of reaction or the enthalpy change. When energy is absorbed, the reaction is said to be endothermic and  H is positive (+). When energy is released, the reaction is said to be exothermic and  H is negative (−). Cl To form this bond,  H = −58 kcal/mol. To cleave this bond,  H = +58 kcal/mol.

15 Energy Changes in Reactions Bond Dissociation Energy The bond dissociation energy is the  H for breaking a covalent bond by equally dividing the e − between the two atoms. Bond dissociation energies are positive values, because bond breaking is endothermic (  H > 0). Bond formation always has negative values, because bond formation is exothermic (  H < 0). HH H + H HH  H = +104 kcal/mol  H = −104 kcal/mol

16 Energy Changes in Reactions Bond Dissociation Energy The stronger the bond, the higher its bond dissociation energy. In comparing bonds formed from elements in the same group, bond dissociation energies generally decrease going down the column.

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18 Energy Changes in Reactions Calculations Involving  H Values  H indicates the relative strength of the bonds broken and formed in a reaction. More energy is released in forming bonds than is needed to break the bonds. When  H is negative: The bonds formed in the products are stronger than the bonds broken in the reactants. CH 4 (g) + 2 O 2 (g)CO 2 (g) + 2 H 2 O(l)  H = −213 kcal/mol Heat is released.

19 Energy Changes in Reactions Calculations Involving  H Values More energy is needed to break bonds than is released in the formation of new bonds. When  H is positive: The bonds broken in the reactants are stronger than the bonds formed in the products. 6 CO 2 (g) + 6 H 2 O(l)C 6 H 12 O 6 (aq) + 6 O 2 (g) ΔH = +678 kcal/mol Heat is absorbed.

20 Energy Changes in Reactions Calculations Involving  H Values

21 Sample problem 6.5

22 Energy Diagrams For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds. The orientation of the two molecules must be correct as well.

23 Energy Diagrams E a, the energy of activation, is the difference in energy between the reactants and the transition state.

24 Energy Diagrams The E a is the minimum amount of energy that the reactants must possess for a reaction to occur. E a is called the energy barrier and the height of the barrier determines the reaction rate. When the E a is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. When the E a is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

25 Energy Diagrams The difference in energy between the reactants and the products is the  H. If  H is negative, the reaction is exothermic:

26 Energy Diagrams If  H is positive, the reaction is endothermic:

27 Reaction Rates How Concentration and Temperature Affect Reaction Rate Increasing the concentration of the reactants: increases the number of collisions increases the reaction rate Increasing the temperature: increases the kinetic energy of the molecules increases the reaction rate

28 Reaction Rates Catalysts A catalyst is a substance that speeds up the rate of a reaction. A catalyst is recovered unchanged in a reaction, and does not appear in the product. Catalysts accelerate a reaction by lowering E a without affecting  H. In the following reaction, Pd acts as a catalyst:

29 Reaction Rates Catalysts The uncatalyzed reaction (higher E a ) is slower. The catalyzed reaction (lower E a ) is faster.  H is the same for both reactions.

30 Reaction Rates Focus on the Human Body: Biological Catalysts Enzymes (usually protein molecules) are biological catalysts held together in a very specific three-dimensional shape. The enzyme lactase converts the carbohydrate lactose into the two sugars glucose and galactose. People who lack adequate amounts of lactase suffer from abdominal cramping and diarrhea because they cannot digest lactose when it is ingested. The active site binds a reactant, which then under- goes a very specific reaction with an enhanced rate.

31 Figure 6.3

32 Equilibrium A reversible reaction can occur in either direction, from reactants to products or from products to reactants. CO(g) + H 2 O(g)CO 2 (g) + H 2 (g) The forward reaction proceeds to the right. The reverse reaction proceeds to the left. The system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The net concentrations of reactants and products do not change at equilibrium.

33 Figure 6.5

34 Equilibrium The Equilibrium Constant The relationship between the concentration of the products and the concentration of the reactants is the equilibrium constant, K. Brackets, [ ], are used to symbolize concentration in moles per liter (mol/L). For the reaction: a A + b Bc C + d D equilibrium constant =K= [products] [reactants] = [C] c [D] d [A] a [B] b

35 Figure 6.4

36 Equilibrium The Equilibrium Constant For the following balanced chemical equation: N 2 (g) + O 2 (g) 2 NO(g) equilibrium constant =K= [N 2 ] [O 2 ] [NO] 2 The coefficient becomes the exponent.

37 Equilibrium The Magnitude of the Equilibrium Constant When K is much greater than 1 (K > 1): [products] [reactants] The numerator is larger. Equilibrium favors the products and lies to the left. When K is much less than 1 (K < 1): [products] [reactants] The denominator is larger. Equilibrium favors the reactants and lies to the right.

38 Equilibrium The Magnitude of the Equilibrium Constant When K is around 1 (0.01 < K < 100): [products] [reactants] Both are similar in magnitude. Both reactants and products are present. For the reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O(g)K = 2.9 x The product is favored because K > 1. The equilibrium lies to the right.

39 Equilibrium

40 Calculating the Equilibrium Constant HOW TO Calculate the Equilibrium Constant for a Reaction Example Calculate K for the reaction between the general reactants A 2 and B 2. The equilibrium concentrations are as follows: [A 2 ] = 0.25 M[B 2 ] = 0.25 M[AB] = 0.50 M A 2 + B 2 2 AB Step [1] Write the expression for the equilibrium constant from the balanced equation. [AB] 2 [A 2 ][B 2 ] K =

41 Calculating the Equilibrium Constant HOW TO Calculate the Equilibrium Constant for a Reaction Step [2] Substitute the given concentrations in the equilibrium expression and calculate K. [AB] 2 [A 2 ][B 2 ] K == [0.50] 2 [0.25][0.25] = = 4.0 Since the concentration is always reported in mol/L, these units are omitted during the calculation.

42 Question: Calculate K for the following equation: The equilibrium concentrations are as follows: Sample Problem 6.9 Solution