Energy can change form and flow, but it is always conserved. Section 1: Energy Energy can change form and flow, but it is always conserved. K What I Know W What I Want to Find Out L What I Learned

The Nature of Energy Energy is the ability to do work or produce heat. Energy exists in two basic forms: potential energy and kinetic energy. Potential energy is energy due to composition or position. Kinetic energy is energy of motion. Copyright © McGraw-Hill Education Energy

The Nature of Energy The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed—also known as the first law of thermodynamics. Chemical potential energy is energy stored in a substance because of its composition. Chemical potential energy is important in chemical reactions. Heat is energy that is in the process of flowing from a warmer object to a cooler object. q is used to symbolize heat. Copyright © McGraw-Hill Education Energy

Review What is the ability to do work or produce heat? What two basic forms does energy exist in? What is energy due to composition or position? What is energy of motion?

Review What states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed? What is energy stored in a substance because of its composition? What is energy that is in the process of flowing from a warmer object to a cooler object? What is used to symbolize heat? The law of conservation of mass is also known as what?

Measuring Heat A calorie is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius. The energy content of food is measured in Calories, or 1000 calories (kilocalorie). A joule is the SI unit of heat and energy, equivalent to 0.2390 calories. Copyright © McGraw-Hill Education Energy

Review What is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius? What is the SI unit of heat and energy?

CONVERT ENERGY UNITS Problem Response SOLVE FOR THE UNKNOWN Convert nutritional Calories to calories. Apply the relationship 1 Calorie = 1000 cal. 230 Calories × 1000 cal 1 Calorie = 2.3 × 105 cal Convert calories to joules. Apply the relationship 1 cal = 4.184 J. 2.3 × 105 cal × 4.184 J 1 cal = 9.6 × 105 J Use with Example Problem 1. Problem A breakfast of cereal, orange juice, and milk might contain 230 nutritional Calories. Express this energy in joules. Response ANALYZE THE PROBLEM You are given an amount of energy in nutritional Calories. You must convert nutritional Calories to calories and then convert calories to joules. EVALUATE THE ANSWER The minimum number of significant figures used in the conversion is two, and the answer correctly has two digits. A value of the order of 105 or 106 is expected because the given number of kilocalories is of the order of 102 and it must be multiplied by 103 to convert it to calories. Then, the calories must be multiplied by a factor of approximately 4. Therefore, the answer is reasonable. KNOWN amount of energy = 230 Calories UNKNOWN amount of energy = ? J Copyright © McGraw-Hill Education Energy

Specific Heat The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius. Some objects require more heat than others to raise their temperatures. Copyright © McGraw-Hill Education Energy

Specific Heat Calculating heat absorbed and released Energy Copyright © McGraw-Hill Education Energy

CALCULATE SPECIFIC HEAT KNOWN energy released = 114 J Ti = 50.4°C mass of iron = 10.0 g Fe Tf = 25.0°C UNKNOWN specific heat of iron, c = ? J/(g•°C) Use with Example Problem 2. Problem In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114J. What is the specific heat of iron? SOLVE FOR THE UNKNOWN Calculate ΔT. ΔT = 50.4°C – 25.0°C = 25.4°C State the equation for calculating heat. q = c × m × ΔT Solve for c. c × m × ΔT m × ΔT = q m × ΔT c = q m × ΔT Response ANALYZE THE PROBLEM You are given the mass of the sample, the initial and final temperatures, and the quantity of heat released. You can calculate the specific heat of iron by rearranging the equation that relates these variables to solve for c. Energy Copyright © McGraw-Hill Education

CALCULATE SPECIFIC HEAT EVALUATE THE ANSWER The values used in the calculation have three significant figures, so the answer is correctly stated with three digits. The value of the denominator of the equation is approximately two times the value of the numerator, so the final result, which is approximately 0.5, is reasonable. The calculated value is the same as that recorded for iron in Table 2. SOLVE FOR THE UNKNOWN (continued) c = q m × ΔT Substitute q = 114 J, m = 10.0 g, and ΔT = 25.4°C. c = 114 J (10.0 g)(25.4°C) Multiply and divide numbers and units. c = 0.449 J/(g•°C) Energy Copyright © McGraw-Hill Education

Section 2: Heat The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants. K What I Know W What I Want to Find Out L What I Learned

Calorimetry A calorimeter is an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process. Copyright © McGraw-Hill Education Heat

USING SPECIFIC HEAT Problem Response KNOWN mass of metal, m = 4.68 g Use with Example Problem 3. Problem A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C. What is the specific heat of the metal? Could the metal be one of the alkaline earth metals listed in Table 2 (on the next slide)? KNOWN mass of metal, m = 4.68 g quantity of heat absorbed, q = 256 J ΔT = 182°C UNKNOWN specific heat, c = ? J/(g•°C) Response ANALYZE THE PROBLEM You are given the mass of the metal, the amount of heat it absorbs, and the temperature change. You must calculate the specific heat. Use the equation for q, the quantity of heat, but solve for specific heat, c. SOLVE FOR THE UNKNOWN State the equation for the quantity of heat, q. q = c × m × ΔT Solve for c. c = q m × ΔT Heat Copyright © McGraw-Hill Education

USING SPECIFIC HEAT SOLVE FOR THE UNKNOWN (continued) Substitute q = 256 J, m = 4.68 g, and ΔT = 182°C. c = 256 J (4.68 g)(182°C) = 0.301 J/(g•°C) Table 2 indicates that the metal could be strontium. EVALUATE THE ANSWER The three quantities used in the calculation have three significant figures, and the answer is correctly stated with three digits. The calculations are correct and yield the expected unit. Heat Copyright © McGraw-Hill Education

Chemical Energy and the Universe Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. The system is the specific part of the universe that contains the reaction or process you wish to study. The surroundings are everything else other than the system in the universe. The universe is defined as the system plus the surroundings. Copyright © McGraw-Hill Education Heat

Chemical Energy and the Universe Chemists are interested in changes in energy during reactions. Enthalpy is the heat content of a system at constant pressure. Enthalpy (heat) of reaction is the change in enthalpy during a reaction symbolized as ΔHrxn. ΔHrxn = Hfinal – Hinitial ΔHrxn = Hproducts – Hreactants Enthalpy changes for exothermic reactions are always negative. Enthalpy changes for endothermic reactions are always positive. Copyright © McGraw-Hill Education Heat

Chemical Energy and the Universe Copyright © McGraw-Hill Education Heat

Chemical Energy and the Universe Copyright © McGraw-Hill Education Heat

Section 3: Thermochemical Equations Thermochemical equations express the amount of heat released or absorbed by chemical reactions. K What I Know W What I Want to Find Out L What I Learned

Writing Thermochemical Equations A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and energy change. The enthalpy (heat) of combustion of a substance is the enthalpy change for the complete burning of one mole of the substance. Copyright © McGraw-Hill Education Thermochemical Equations

Changes of State Molar enthalpy (heat) of vaporization refers to the heat required to vaporize one mole of a liquid substance. Molar enthalpy (heat) of fusion is the amount of heat required to melt one mole of a solid substance. Copyright © McGraw-Hill Education Thermochemical Equations

THE ENERGY RELEASED IN A REACTION Response ANALYZE THE PROBLEM You are given a mass of glucose, the equation for the combustion of glucose, and ΔHcomb. You must convert grams of glucose to moles of glucose. Because the molar mass of glucose is more than three times the mass of glucose burned, you can predict that the energy evolved will be less than one-third ΔHcomb. Use with Example Problem 4. Problem A bomb calorimeter is useful for measuring the energy released in combustion reactions. The reaction is carried out in a constant-volume bomb with a high pressure of oxygen. How much heat is evolved when 54.0 g glucose (C6H12O6) is burned according to this equation? C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l) ΔHcomb = -2808 kJ SOLVE FOR THE UNKNOWN Convert grams of C6H12O6 to moles of C6H12O6. Multiply by the inverse of molar mass, 1 mol 180.18 g . 54.0 g C6H12O6 × 1 mol C6H12O6 180.18 g C6H12O6 = 0.300 mol C6H12O6. KNOWN mass of glucose = 54.0 g C6H12O6 ΔHcomb = -2808 kJ UNKNOWN q = ? kJ Copyright © McGraw-Hill Education Thermochemical Equations

THE ENERGY RELEASED IN A REACTION EVALUATE THE ANSWER All values in the calculation have at least three significant figures, so the answer is correctly stated with three digits. As predicted, the released energy is less than one-third ΔHcomb. SOLVE FOR THE UNKNOWN (continued) Multiply moles of C6H12O6 by the enthalpy of combustion, ΔHcomb. Multiply moles of glucose by 2808 kJ 1 mol C6H12O6 . 0.300 mol C6H12O6 × 2808 kJ 1 mol C6H12O6 = 842 kJ Thermochemical Equations Copyright © McGraw-Hill Education

Combustion Reactions Combustion is the reaction of a fuel with oxygen. Food is the fuel in combustion reactions in biological systems. Copyright © McGraw-Hill Education Thermochemical Equations

Review What is a balanced chemical equation that includes the physical states of all reactants and products, and energy change? What of a substance is the enthalpy change for the complete burning of one mole of the substance? What refers to the heat required to vaporize one mole of a liquid substance? What is the amount of heat required to melt one mole of a solid substance? What is the reaction of a fuel with oxygen?

Section 4: Calculating Enthalpy Change The enthalpy change for a reaction can be calculated using Hess’s law. K What I Know W What I Want to Find Out L What I Learned

Hess's Law Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction. Copyright © McGraw-Hill Education Calculating Enthalpy Change

H2O2 (aq) → H2 (g) + O2 (g) ΔH = 188 kJ HESS’S LAW KNOWN a. 2H2(g) + O2(g) → 2 H2O(l) ΔH = -572 kJ b. H2(g) + O2(g) → H2O2(l) ΔH = -188 kJ UNKNOWN ΔH = ? kJ Use with Example Problem 5. Problem Use thermochemical Equations a and b below to determine ΔH for the decomposition of hydrogen peroxide (H2O2), a compound that has many uses ranging from bleaching hair to powering rocket engines. 2H2O2(l) → 2H2O(l) + O2(g) 2H2(g) + O2(g) → 2 H2O(l) ΔH = -572 kJ b. H2(g) + O2(g) → H2O2(l) ΔH = -188 kJ SOLVE FOR THE UNKNOWN H2O2 is a reactant. Reverse Equation b and change the sign of ΔH. H2O2 (aq) → H2 (g) + O2 (g) ΔH = 188 kJ Two moles of H2O2 are needed. Multiply the reversed Equation b by two to obtain Equation c. c. 2 H2O2 (aq) → 2 H2 (g) + 2 O2 (g) Response ANALYZE THE PROBLEM You have been given two chemical equations and their enthalpy changes. These two equations contain all the substances found in the desired equation. Calculating Enthalpy Change Copyright © McGraw-Hill Education

HESS’S LAW EVALUATE THE ANSWER The two equations produce the desired equation. All values are accurate to the ones place, so ΔH is correctly stated. SOLVE FOR THE UNKNOWN (continued) Multipy 188 kJ by two to obtain ΔH for Equation c. ΔH for Equation c = (188 kJ)(2) = 376 kJ Write Equation c and ΔH. c. 2H2O2(aq) → 2H2(g) + 2O2(g), ΔH = 376 kJ Add Equations a and c, canceling any terms common to both sides of the combined equation. Add the enthalpies of Equations a and c. Write Equation a. a. 2H2(g) + O2(g) → 2H2O(l), ΔH = -572 kJ Write Equation c. c. 2H2O2(l) → 2H2(g) + 2O2(g), ΔH = 376 kJ Add Equations a and c. Add the enthalpies. 2H2O2 (l) → 2H2O(l) + O2(g), ΔH = -196 kJ Calculating Enthalpy Change Copyright © McGraw-Hill Education

Standard Enthalpy (Heat) of Formation The standard enthalpy (heat) of formation is defined as the change in enthalpy that accompanies the formation of one mole of the compound in its standard state from its elements in their standard states. Elements in their standard states have a The formation of compounds are placed above or below elements in their standard states. Copyright © McGraw-Hill Education Calculating Enthalpy Change

Standard Enthalpy (Heat) of Formation Standard enthalpies of formation can be used to calculate the enthalpies for many reactions under standard conditions by using Hess’s law. The summation equation: Copyright © McGraw-Hill Education Calculating Enthalpy Change

ENTHALPY CHANGE FROM STANDARD ENTHALPIES OF FORMATION KNOWN UNKNOWN ΔH f ₒ (CO2) = –394 kJ ΔH rxn ₒ = ? kJ ΔH f ₒ (H2O) = –286 kJ ΔH f ₒ (CH4) = –75 kJ ΔH f ₒ °(O2) = 0.0 kJ Use with Example Problem 6. Problem Use standard enthalpies of formation to calculate ΔH rxn ₒ for the combustion of methane. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) SOLVE FOR THE UNKNOWN Use the formula ΔH rxn ₒ =Σ ΔH rxn ₒ (products) – Σ ΔH f ₒ (reactants). Expand the formula to include a term for each reactant and product. Multiply each term by the coefficient of the substance in the balanced chemical equation. Substitute CO2 and H2O for the products, CH4 and O2 for the reactants. Multiply H2O and O2 by two. ΔH rxn ₒ = [ ΔH f ₒ (CO2) + (2) ΔH f ₒ (H2O)] – [ ΔH f ₒ (CH4) + (2) ΔH f ₒ (O2)] Response ANALYZE THE PROBLEM You are given an equation and asked to calculate the change in enthalpy. The formula ΔH rxn ₒ = Σ ΔH rxn ₒ (products) – Σ ΔH f ₒ (reactants) can be used with data from Table R-11 (on the next slide). Calculating Enthalpy Change Copyright © McGraw-Hill Education

Calculating Enthalpy Change Copyright © McGraw-Hill Education Calculating Enthalpy Change

ENTHALPY CHANGE FROM STANDARD ENTHALPIES OF FORMATION SOLVE FOR THE UNKNOWN (continued) Substitute ΔH f ₒ (CO2) = –394 kJ, ΔH f ₒ (H2O) = –286 kJ, ΔH f ₒ (CH4) = –75 kJ, and ΔH f ₒ (O2) = 0.0 kJ into the equation. ΔH rxn ₒ = [(–394 kJ) + (2)(–286 kJ)] – [(–75 kJ) + (2)(0.0kJ)] ΔH rxn ₒ = [– 966 kJ] – [–75 kJ] = –966 kJ+75 kJ = –891 kJ The combustion of 1 mol CH4 releases 891 kJ. EVALUATE THE ANSWER All values are accurate to the ones place. Therefore, the answer is correct as stated. The calculated value is the same as that given in Table 3 (on the next slide). You can check your answer by using the stepwise procedure shown previously. Calculating Enthalpy Change Copyright © McGraw-Hill Education

ENTHALPY CHANGE FROM STANDARD ENTHALPIES OF FORMATION Copyright © McGraw-Hill Education Calculating Enthalpy Change