Chapter 8 Chemical and Physical Change: Energy, Rate, and Equilibrium Denniston Topping Caret 4 th Edition Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

8.1 Thermodynamics Thermodynamics - the study of energy, work, and heat. –applied to chemical change –applied to physical change The laws of thermodynamics help us to understand why some chemical reactions occur and others do not.

8.1 Thermodynamics The Chemical Reaction and Energy Important points to kinetic molecular theory –molecules and atoms in a reaction mixture are in constant, random motion; –these molecules and atoms frequently collide with each other; –only some collisions, those with sufficient energy, will break bonds in molecules; and –when reactant bonds are broken, new bonds may be formed and products result.

8.1 Thermodynamics As the bonds are broken and new bonds are formed, energy is required or released. We can measure the change in energy during these changes. System - the process under study –Usually the chemical reaction or physical change of interest. Surroundings - the rest of the universe.

We will be able to measure the change in energy in the form of heat as the temperature changes. 8.1 Thermodynamics

Exothermic and Endothermic Reactions 1 The first law of thermodynamics - energy of the universe is constant. Where does the energy come from that is released and where does the energy go when it is absorbed? The chemical bond is stored chemical energy. A-B + C-D  A-D + C-B

8.1 Thermodynamics A-B + C-D  A-D + C-B These bonds must be broken. This requires energy. These bonds are formed. This releases energy If the energy required to break the bonds is less then the energy released when the bonds are formed, there is a net release of energy…Exothermic reaction. If the energy required to break the bonds > the energy released when the bonds are formed, there will need to be an external supply of energy…Endothermic reaction.

8.1 Thermodynamics Exothermic Reaction Endothermic Reaction

8.1 Thermodynamics Combustion of organic compounds CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) kcal exothermic reaction Decomposition of ammonia 22 kcal + 2NH 3 (g)  N 2 (g) + 3H 2 (g) endothermic reaction

8.1 Thermodynamics Enthalpy Enthalpy - represents heat energy. Change in Enthalpy (  H o ) - energy difference between the products and reactants. Energy released (exothermic), enthalpy change is negative –In the combustion of CH 4,  H o = -211 kcal Energy absorbed (endothoermic), enthalpy change is positive. –In the decomposition of NH 3,  H o =+22 kcal 2

8.1 Thermodynamics Spontaneous and Nonspontaneous Reactions Spontaneous reaction - occurs without any external energy input. Often, but not always, exothermic reactions are spontaneous. Thermodynamics is used to help predict if a reaction will occur. Another factor is needed.

8.1 Thermodynamics Entropy The second law of thermodynamics - the universe spontaneously tends toward increasing disorder or randomness. Entropy (S o ) - a measure of the randomness of a chemical system. High entropy - highly disordered system Low entropy - well organized system No such thing as negative entropy. 2

8.1 Thermodynamics  S o of a reaction = S o (products) - S o (reactants) A positive  S o means an increase in disorder for the reaction. A negative  S o means a decrease in disorder for the reaction.

8.1 Thermodynamics  S o is positive  S o is negative

8.1 Thermodynamics All of the above have a positive  S o

8.1 Thermodynamics If exothermic and positive  S o … SPONTANEOUS If endothermic and negative  S o … NONSPONTANEOUS For any other situations, it depends on the relative size of  H o and  S o. 2

8.1 Thermodynamics Free Energy Free energy (  G o ) - represents the combined contribution of the enthalpy and entropy values for a chemical reaction. Predicts spontaneity Negative  G o …Spontaneous Positive  G o …Nonspontaneous  G o =  H o - T  S o T in Kelvins 2

8.2 Experimental Determination of Energy Change in Reactions Calorimetry - the measurement of heat energy changes in a chemical reaction. Calorimeter - devise which measures heat changes in calories The change in temperature is used to measure the heat loss or gain. 3

8.2 Calorimetry Specific heat (S.H.) - the number of calories of heat needed to raise the temperature of 1 g of the substance 1 o C. S.H. for water is 1.0 cal/g o C To determine heat released or absorbed, need: –specific heat –total number of grams of solution –temperature change (increase or decrease)

8.2 Calorimetry 1.If 0.10 mol of HCl is mixed with 0.10 mol of KOH in a “coffee cup” calorimeter, the temperature of 1.50 x 10 2 g of the solution increases from 25.0 o C to 29.4 o C. If the specific heat of the solution is 1.00 cal/g o C, calculate the quantity of energy evolved in the reaction. 2.Is the reaction endothermic or exothermic 3.What would be the energy evolved for each mole of HCl reacted?

3 8.2 Calorimetry Nutritional Calorie (large Calorie) = 1kilocalorie (1kcal) or 1000 calorie the fuel value of food Bomb Calorimeter is used to measure nutritional Calories

8.2 Calorimetry One gram of a certain carbohydrate was burned in a bomb calorimeter. The temperature of 1.25 x 10 3 g H 2 O was raised from 24.5 o C to 31.5 o C. Calculate the fuel value of the carbohydrate (in Kcal/g).

8.3 Kinetics Thermodynamics determines if a reaction will occur but tells us nothing about the time it will take Kinetics - the study of the rate of chemical reactions –Also gives the mechanism - step-by-step description of how reactants become products. 4

8.3 Kinetics We will look at: –disappearance of reactants and –appearance of products. for the reaction A  B

8.3 Kinetics

The Chemical Reaction Let’s consider the following reaction: CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O(g) kcal C-H and O=O bonds must be broken and C=O and O-H bonds must be formed Energy is required to break the bonds. –Comes from the collision of the molecules. –Effective collision - one that leads to a chemical reaction.

8.3 Kinetics Activation Energy and the Activated Complex Activation energy - the minimum amount of energy required to produce a chemical reaction. Activated complex - extremely unstable complex. Formation of activated complex requires energy. (E a ) Exothermic - net release of energy (  H o ) 5

8.3 Kinetics Potential energy diagram for an endothermic reaction. Net absorption of energy (  H o )

6 8.3 Kinetics Factors That Affect Reaction Rate structure of the reacting species, concentration of reactants, temperature of reactants,physical state of reactants, and presence of a catalyst

8.3 Kinetics Structure of Reacting Species Oppositely charged species react more rapidly Ions with the same charge do not react. Bond strength plays a role. Magnitude of the activation energy is related to bond strength Size and shape influence the rate. Large molecules may block the reactive part of the molecule.

8.3 Kinetics The Concentration of Reactants Rate will generally increase as concentration increases. Caused by a greater number of collisions The Temperature of Reactants Rate increases as the temperature increases. Higher temp. means higher K.E. Higher K.E. means higher percentage of these collisions will result in product formation.

8.3 Kinetics The Physical State of Reactants Solid state: atoms, ions or cpds. are close but restrictive in motion. Gaseous state: particles are free to move but are far apart causing collisions to be relatively infrequent. Liquid state: particles are free to move and are close together. According to the above statement, what would be the typical order of rate? Liquid > gas> solid

8.3 Kinetics The Presence of a Catalyst Catalyst - a substance that increases the reaction rate. Catalyst interacts with the reactants to create an alternative pathway for product production

8.3 Kinetics Haber Process - N 2 +3H 2  2NH 3

8.3 Kinetics Enzyme - a biological catalyst that controls and speeds up thousands of essential biochemical reactions. Again what are the factors that affect rate? structure concentration temperature physical state catalyst Hold those factors in yellow constant and observe how concentration affects rate...

8.3 Kinetics Mathematical Representation of Reaction Rate 7 Consider the following decomposition: When holding all factors except concentration constant, find: or k is called the rate constant

8.3 Kinetics For a reaction A  products the equation will be: rate = k[A] n this is called the rate equation (or rate law) n is the order of the reaction n=1, first order n=2, second order etc. n must be determined experimentally

8.3 Kinetics For the equation A + B  products the rate equation is: rate = k[A] n [B] m What would be the general form of the rate equation for the reaction: CH 4 +2O 2  CO 2 +2H 2 O Rate = k[CH 4 ] n [O 2 ] m Knowing the rate equation and the order helps industrial chemists determine the optimum conditions for preparing a product.

8.4 Equilibrium Rate and Reversibility of Reactions Equilibrium reactions - chemical reactions that do not go to completion. –Incomplete reactions After no further obvious change, measurable quantities of reactants and products remain. 8

8.4 Equilibrium Physical Equilibrium Reversible reaction - a process that can occur in both directions –Use the double arrow symbol Dynamic equilibrium - the rate of the forward process is exactly balanced by the rate of the reverse process Example: Sugar in Water –If put 2-3 g of sugar in 100 mL water, all will dissolve. –sugar (s)  sugar (aq)

8.4 Equilibrium If dissolving 100 g in 100 mL of water, not all of it will dissolve. –Over time, you observe no further change in the amount of dissolved sugar. –Individual sugar molecules are constantly going into and out of solution. –Happens at the same rate. The double arrow serves as –an indicator of a reversible process –an indicator of an equilibrium process, and –a reminder of the dynamic nature of the process. sugar(s) sugar(aq)

8.4 Equilibrium rate f = forward rate rate r = reverse rate at equilibrium: rate f =rate r We learned from Kinetics that rate f =k f [sugar(s)] rate r =k r [sugar(aq)] so k f [sugar(s)]=k r [sugar(aq)] sugar(s) sugar(aq)

8.4 Equilibrium k f [sugar(s)]=k r [sugar(aq)] Equilibrium constant (K eq )- ratio of the two rate constants

8.4 Equilibrium Chemical Equilibrium Example: The Reaction of N 2 and H 2 N 2 (g) + 3H 2 (g) qe 2NH 3 (g) Forward and reverse reaction rates are equal.

8.4 Equilibrium N 2 (g) + 3H 2 (g) qe 2NH 3 (g) subdivides into: forward rxn: N 2 (g) + 3H 2 (g)  2NH 3 (g) reverse rxn: 2NH 3 (g)  N 2 (g) + 3H 2 (g) and rate f = k f [N 2 ] n [H 2 ] m rate r = k r [NH 3 ] p rate f =rate r

8.4 Equilibrium The exponents in the rate expression are numerically equal to the coefficients. So, K eq is a constant at constant temperature.

8.4 Equilibrium The Generalized Equilibrium-Constant Expression for a Chemical Reaction 9 aA + bB cC + dD

8.4 Equilibrium Writing Equilibrium-Constant Expressions Each chemical reaction has a unique equilibrium constant value at a specified temperature. The brackets represent molar concentration. All equilibrium constants are shown as unitless. Only the concentration of gases and substances in solution are shown. concentration for pure liquids and solids are not shown.

8.4 Equilibrium Write equilibrium expressions for the following equilibria: 1.2HI(aq) H 2 (g) + I 2 (g) 2.BaCl 2 (s) Ba 2+ (aq) + 2Cl - (aq) 3.CaCO 3 (s) CaCO(s) + CO 2 (g)

8.4 Equilibrium Interpreting Equilibrium Constants The numerical value of the equilibrium constant tells us the extent to which reactants have converted to products. 1.K eq greater than 1 x Large value of K eq : numerator > denominator. At equilibrium mostly product present.

2.K eq less than 1 x Which is larger, numerator or denominator? Ans: Denominator. At equilibrium which is present in the larger quantity? Reactants or Products? Ans: Reactants. 3.K eq between 1 x and 1 x 10 2 Equilibrium mixture contains significant concentration of both reactants and products. 8.4 Equilibrium

Calculating Equilibrium Constants A reversible reaction is allowed to proceed until the system reaches equilibrium Amount of reactants and products no longer changes. 2NO 2 (g) N 2 O 4 (g)

Reaction mixture is analyzed to determine the molar concentrations of each of the product and reactants. Concentrations are substituted into the expression. What would the equilibrium expression be for the reaction: 8.4 Equilibrium 2NO 2 (g) N 2 O 4 (g)

8.4 Equilibrium At equilibrium, [NO 2 ]= M and [N 2 O 4 ]=0.643 M. What is the value for K eq ? 215 Would the value for K eq be the same if the equation were written: N 2 O 4 (g) 2NO 2 (g)? NO

Equilibrium LeChateleir’s Principle LeChateleir’s Principle - if a stress is placed on a system at equilibrium, the system will respond by altering the equilibrium composition in such a way as to minimize the stress.

Equilibrium LeChateleir’s Principle We will examine the following “stresses.” 1. Effect of Concentration 2. Effect of Heat 3. Effect of Pressure 4. Effect of Catalyst

8.4 Equilibrium 1. Effect of Concentration Adding or removing reactants and products at a fixed volume. Addition of N 2 or H 2. To minimize the stress, which way will the reaction shift? To the right. Forming more products. If NH 3 is put in the reaction vessel? Equilibrium shifts to the left, forming more reactants. N 2 (g) + 3H 2 (g) 2NH 3 (g)

8.4 Equilibrium 2. Effect of Heat Addition of heat is similar to increasing the amount of product. If heat is added by raising the temperature, which way will the equilibrium shift? To the left. Exothermic reactions: treat heat as a product N 2 (g) + 3H 2 (g) 2NH 3 (g) + 22 kcal

8.4 Equilibrium Which way will this reaction shift if heat is added by increasing the temperature? To the right. Endothermic Reaction - treat heat as a reactant. 39 kcal + 2N 2 (g) + O 2 (g) 2NH 3 (g)

8.4 Equilibrium 3. Effect of Pressure Pressure affects the equilibrium only if one or more substances in the reaction are gases Relative number of gas moles on reactant and product side must differ. When pressure goes up…shift to side with less moles of gas When pressure goes downs…shifts to side with more moles of gas

8.4 Equilibrium If increase pressure, which way will the equilibrium shift? Right. Producing more NH 3 N 2 (g) + 3H 2 (g) 2NH 3 (g) If increase the pressure, which way will the equilibrium shift? No change! 2HI(g) H 2 (g) + I 2 (g)

8.4 Equilibrium 4. Effect of a Catalyst A catalyst has no effect on the equilibrium composition. It increases the rate of both the forward and reverse reaction to the same extent.

The End Chapter 8