Unit 4 How do we model chemical change? The central goal of this unit is to help you understand and apply ways of thinking that can be used to model chemical change in a system. M1. Understanding Proportions Determining the amount of substance formed or consumed. Every Course Unit includes a brief description of the central learning objectives of each Module in the Unit. M2. Tracking Energy Predicting the amount of energy absorbed or released. M3. Analyzing Rate and Extent Identifying the factors that affect chemical transformations.

How do we model chemical change? Module 2: Tracking Energy Unit 4 How do we model chemical change? Module 2: Tracking Energy Central goal: To make qualitative and quantitative predictions about the amount of energy absorbed or released during a chemical reactions based on the nature of the chemical bonds in the molecules of reactants and products. Each Module begins with a description of its central learning objective.

Modeling How do I explain it? The Challenge Modeling How do I explain it? We would like to generate models that allow us to answer questions such as this: What amounts of reactants and products are involved? How much energy will be needed or produced? “The Challenge” describes the types of problems or questions that we would like to be able to answer by the end of the Module. Questions in green are rhetorical questions. They are included to invite students to think and brainstorm initial ideas. Images: Ultrafast laser pulses was used to polymerize a photoresist resin (bull image) Organic LED How fast will the process go and how can I control it? To what extent will the reactants be changed into products?

Focus on Energy CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) T Let’s go back to our combustion problem, nut now paying attention to energy issues: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) T Why does the temperature increase? Energy is released during the process. Methane CH4

The system releases energy The system absorbs energy Energy Transfer A common signature of a chemical change is the release or absorption of energy. Surroundings System heat Surroundings System heat Tsurr Tsurr q < 0 The system releases energy Heat q > 0 The system absorbs energy EXOTHERMIC ENDOTHERMIC

Exothermic or Endothermic How can we explain the differences Combustion reactions are prototypical examples of exothermic processes: CH4 + 2 O2  CO2 + 2 H2O + ENERGY The decomposition of stable compounds, such as H2O, tends to be endothermic: 2 H2O  2 H2 + O2 + ENERGY “Let’s think” designed for students to brainstorm ideas. How can we explain the differences

Do we need energy to form bonds or do we get energy from bond forming? A Chemical Model The amount of energy released or absorbed in a chemical reaction is related to the nature and numbers of bonds broken or formed in the process. During a chemical reaction some bonds are broken and some new bonds are formed. These questions are included to give a better sense of the goal of the Module. Do we need energy to break bonds or do we get energy from bond breaking? Do we need energy to form bonds or do we get energy from bond forming? Let′s think!

By convention, Ep = 0 when the atoms are infinitely separated. Bond Energy Energy is needed to separate a pair of bonded atoms (“break” the bond). The energy provided is transformed into potential energy of the separated atoms. Ep By convention, Ep = 0 when the atoms are infinitely separated.

Bond Energy Bond formation corresponds to a minimum in the potential energy (electrostatic forces are balanced). Potential energy is transformed into kinetic energy when a bond is formed (energy is released). There are two movies embedded in this page. On the upper right corner, click on and drag the atom in the right hand-side to displace them and observe the formation of a bond . Click on the dynamic molecule to open a more accurate represenattion. Dissociation Energy

Go to: http://www.chem.arizona.edu/chemt/C21/sim (Bond Energy) Let’s Explore Go to: http://www.chem.arizona.edu/chemt/C21/sim (Bond Energy) We recommend that students work in groups or pairs on this simulation. Ideally, each student will have a computer where they can access the simulation. Click on image to open the simulation in the Molecular Workbench (java applet). Analyze and compare energy transformations for systems with different bond energies. Discuss the questions posted on the Web page.

Bond Strength (Dissociation Energy) Bond Comparison Ep Ep Ep 414 kJ/mol 464 kJ/mol 799 kJ/mol r r r Bond Strength (Dissociation Energy) End day one Energy released when the bond is formed Energy needed to break the bond Let’s Think Construct these PE diagrams for C-C, C=C and C≡C bonds.

Our Model Let’s analyze this reaction: What is net energy is released or absorbed? Let’s analyze this reaction: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Ep CH4 + 2 O2  CO2 + 2 H2O 4 C + 4 H Energy needed to break bonds 4 O 2 C=O = - 2 x 799 = -1598 kJ Energy released when forming bonds 4 O-H = - 4 x 464 = -1856 kJ 2 O=O = - 2 x 498 = +996 kJ 4 C-H = 4 x 414 = +1656 kJ

Net Energy Released = -802 kJ = qrxn Heat of Reaction Ep C + H + H + H + H + O + O + O + O - 3454 = -1598 -1856 Formation + 2652 Energy Profile 1656 + 996 = Dissociation CH4 + 2 O2 -2652 kJ CO2 + 2 H2O -3454 kJ Introduce the term DHrxn as a label for heat of reaction Net Energy Released = -802 kJ = qrxn CH4 + 2 O2  CO2 + 2 H2O

Energy Conservation As with mass, the total energy of a closed system undergoing a chemical change is conserved. If so, where are the -802 kJ in this reaction coming from? Ep The main contributors are the molecules with the highest potential energy. The actual fuel for the reaction is O2, the molecule with the weakest bonds (higher potential energy). Ep is transformed into kinetic energy or vise versa. It is energy that was stored as Ep in chemical bonds (reactants).

Measurements The actual heat of reaction can be measured experimentally using Calorimetry. Calorimetry: Heat transfer is indirectly measured by quantifying changes in temperature. These measurements are commonly done using substances in their standard state at 1 atm and 25 oC. This heat of reaction is commonly called standard enthalpy change DHorxn.

Let’s Think Given the information provided, build the “Energy Profile” for the decomposition of water. How much energy is released or absorbed? 2 H-O-H  2 H-H + O=O Ep Bond Dissoc. Energy (KJ/mol) O-H 464 H-H 436 O=O 498 Or 486/2 = 243 kJ/mol of H2O 2 H2O 4 x 464 2 H2 + O2 -2 x 436 -498 “Let’s think” designed for students to apply what they have learned. DHrxn = 486 kJ What happens to the energy we put in? It is stored as Ep in the products

Let’s Think Compare these two sets of bond energies: Energy (kJ/mol) H-H 436 C-H 414 C-C 347 O-H 464 O-O 142 C-O 360 N-N 163 N-H 389 Which types of bonds are easier to break? In general, A-A bonds are weaker than A-B bonds. “Let’s thin” designed for students to identify patters in bond dissociation energies. Based on this trend, predict which of these reactions is likely to be exothermic or endothermic: A-A + B-B  2 A-B 2 A-B-C  2 A + 2 B + C-C Exothermic Endothermic

Let’s Think A + B  X + Y Based on the Energy Profiles for these 3 reactions, what can you say about the relative strength of the bonds in compounds A and B, versus C and D, versus E and F? Are the reactions endo or exothermic? Which reaction releases more energy? C + D  X + Y Ep E + F  X + Y C + D A + B X + Y E + F

DHcombustion in kJ/mol (when 6.02 x 1023 molecules react ) Let’s Think Different substances produce different amounts of energy during combustion. Octane Methane Methanol CH4 CH4O C8H18 X + O2  CO2 + H2O It is important to point out that this is a graph of the energy of combustion of each substance, not of the actual chemical potential energy of each substance. “Let’s think” designed for students to apply what they have learned. This type of information is crucial when making decisions about fuel use. DHcombustion in kJ/mol (when 6.02 x 1023 molecules react )

All of them produce H2O and CO2 when burned. Let’s Think Give a possible explanation to the difference in the energy released when one mole of each of these substances reacts with enough O2. less Methanol CH4O Remember: All of them produce H2O and CO2 when burned. Bond Energy (KJ/mol) C-C 347 C-O 360 C-H 414 O-H 464 O=O 498 C=O 799 Methane CH4 Octane C8H18 more DHrxn(-)

There are always costs and benefits! Important Trends The O=O is a relatively weak bond, while the C=O and O-H bonds are strong. ½ [ ] The more oxygen is used during combustion of burned fuel, the more energy will be released. In that sense, “oxygenated” fuels will produce less energy per mole. However, they normally generate less CO and other pollutants. There are always costs and benefits!

Heat of Reaction Ep CH4 + 2 O2 2652 kJ needed to break bonds CH4 + 2 O2 2652 kJ needed to break bonds CO2 + 2 H2O -3454 kJ released in forming bonds CH4O + 1.5 O2 2811 kJ needed to break bonds -2652 kJ -4x414 -2x498 = -2652 Potential energy -3x414 -1x358 -1x464-1.5x498= -2811 -2811 kJ End day 2 ½ [ ] Less oxygen involved -3454 kJ -2x799 -4x464= -3454

Let′s apply! Assess what you know

Predict Let′s apply! Most of the energy our body needs is supplied by the combustion of carbohydrates and lipids. The most common carbohydrate is glucose. Most lipids result from the combination of fatty acids, such as oleic acid. C6H12O6 C18H34O2 “Let’s apply” are designed to have students apply what they have learned in a module. We recommend to have students work in pairs or groups to complete the task and to collect their work. Students may need some hints and guidance to complete the activity. Predict which combustion reaction will produce more energy per mole of each substance.

Estimate Let′s apply! Verify your prediction by estimating the amount of energy produced by the combustion of one mole of each substance. C6H12O6 Bond Energy (KJ/mol) C-C 347 C=C 611 C-H 414 O-H 464 O=O 498 C-O 360 C=O 799 Have half a class doing one and the other half the other. Show the following slide to simplify counting. C18H34O2

Estimate Let′s apply! C6H12O6 + 6 O2  6 CO2 + 6 H2O C-C C-H C=C C-O O-H C=O O=O B 5 7 1 6 F 12 Bond Energy (KJ/mol) C-C 347 C=C 611 C-H 414 O-H 464 O=O 498 C-O 360 C=O 799 C18H34O2 + 51/2 O2  18 CO2 + 17 H2O C-C C-H C=C C-O O-H C=O O=O B 16 33 1 25.5 F 34 36 Remember, energy is needed to break bonds; energy is released when bonds form.

Estimate Let′s apply! C6H12O6 + 6 O2  6 CO2 + 6 H2O C-C C-H C=C C-O O-H C=O O=O B 5x347 7x414 5x360 5x464 1x799 6x498 12540 F -12x464 -12x799 -15156 DHmodel = -2.62 x 103 kJ/mol DHexp = -2.80 x 103 kJ/mol C18H34O2 + 51/2 O2  18 CO2 + 17 H2O C-C C-H C=C C-O O-H C=O O=O B 16x347 33x414 1x611 1x360 1x464 1x799 25.5x498 34147 F -34x464 -36x799 -44540 DHmodel = -1.04 x104 kJ/mol DHexp = -1.12 x 104 kJ/mol

Compare Let′s apply! It is common to express the energy produced during fuel combustion in amount of energy per unit mass (energy density; kJ/g). Glucose C6H12O6 DHexp = -2.80 x 103 kJ/mol Oleic Acid C18H34O2 DHexp = -1.12 x 104 kJ/mol Determine and compare the energy density, in kcal/g = Cal/g , for these two substances. Discuss why this quantity may be more useful for purposes of fuel analysis and evaluation than its equivalent in kJ/mol. 1 calorie = 1 cal = Energy required to raise the temperature of 1.00 g of H2O by 1.0 oC = 4.184 J “Let’s apply” are designed to have students apply what they have learned in a module. We recommend to have students work in pairs or groups to complete the task and to collect their work. Students may need some hints and guidance to complete the activity.

DHexp = -2.8 x 103 kJ/mol M = 180.2 g/mol Compare Let′s apply! Glucose C6H12O6 DHexp = -2.8 x 103 kJ/mol M = 180.2 g/mol DHexp = -11180 kJ/mol M = 282.5 g/mol C18H34O2 Oleic Acid Lipids (fats) generate close to 2.5 more energy than carbohydrates. They are normally called a concentrated source of energy.

Analyze Our body uses the excess energy from food to synthesize “fat.” Let′s apply! Our body uses the excess energy from food to synthesize “fat.” How much would your body weight change if this energy was used to synthesize glucose (glycogen)? Description Women Men Athletes 16–20% 6–13% Fitness 21–24% 14–17% Acceptable 25–31% 18–25% Obese 32%+ 25%+ Discuss why the accumulation of fat provides an evolutionary advantage over the accumulation of glycogen or starch? % Body Fat American Council on Exercise

Analyze Let′s apply! If my weight is 175 lb, for example, and my %body fat is 20%, then: 175 x 0.20 = 35 lb of my weight comes from fat. If my body used glucose to store the same amount of energy, I would need 2.5 times this mass or: 2.5 x 35 = 87.5 lb My weight then would be: 175 – 35 + 87.5 = 227.5 lb, a weight increase of 30%. We would be slower animals, more energy would be needed to move around, etc.

Discuss how would the world be different if we did not know what you learned in this module.

Tracking Energy Summary A common signature of a chemical reaction is the absorption (endothermic) or release (exothermic) of energy. The transfer of energy during chemical reactions can be explained by assuming that energy is needed to break bonds and energy is released when new bonds are formed.

Energy is stored in the form of potential energy in chemical bonds. Tracking Energy Summary Energy is stored in the form of potential energy in chemical bonds. We can make predictions about the amount of energy transferred during a chemical reaction by paying attention to the nature of the bonds that are broken and formed. Bond Energy (kJ/mol) H-H 436 C-H 414 C-C 347 O-H 464 O-O 142 C-O 360 N-N 163 N-H 389

For next class, Investigate what factors determine how fast a reaction goes and whether it will reach completion or not. What is the difference between being a fast and being a product-favored reaction?