Chapter 16: Energy and Chemical Change

1. Chapter 16: Energy and Chemical Change Objectives: The student will be able to: evaluate energy changes is chemical and physical processes; apply the concept of enthalpy to solving energy changes in problems; compare and contrast energy changes in physical and chemical processes; apply Hess’s law to solving multi-step chemical reaction problems; apply heat of formation information to chemical reactions; apply the concept of entropy to solving energy changes in problems; combine enthalpy and entropy concepts to understand free energy. CH: 5 The student will investigate and understand that the phases of matter are explained by kinetic theory and forces of attraction between particles. 5d: molar heats of fusion and vaporization; 5e: specific heat capacity.

2. Chapter 16 Assignments Assignment 1: End of chapter problems: 75, 78, 80, 82, 86 Assignment 2: End of chapter problems: 88, 89, 90, 91, 92, 93, 95, 97, 104, 107 Assignment 3: Chapter 16 Mastery worksheet. Assignment 4: Standardized test practice on page 527-all 9 questions. SOL Practice: choose all strands and increase the number of questions to 40.

3. Law of conservation of energy: In a chemical reaction or physical process, energy can neither be created nor destroyed. Therefore, energy can only be converted from one form to another. Chemical Potential Energy: Energy stored in compounds because of the bonds present in the compound. When a chemical reaction occurs, some bonds are broken and some bonds are made. In many reactions, energy changes are felt as heat.

4. Units for Heat: 1 calorie = the heat required to raise the temperature of 1 g of pure water by 1oC. Nutritional calories are 1000 times larger than a chemical calorie and the units are Cal-notice the capital “C”. 1 Cal = 1000 cal = 1 kcal SI unit for heat is the joule (J). 1 cal = 4.184 J N is Newton and is a unit of force. Force = mass*acceleration = (kg)(m/s2) Force times distance is energy, so a joule is a Newton of force applied over a meter of distance. J = (N)(m) = (kgm/s2)(m) = (kg)(m2)/(s2)

5. Convert 245 Cal into cal Convert 245 Cal into kcal Convert 245 Cal into J Complete practice problems 1-3 on page 492.

6. Specific Heat (c) is the amount of heat (J) required to raise the temperature of 1 gram (g) of a substance by 1degree Celsius (oC) so the units are J/(goC) Since the calorie is defined by this same concept, the specific heat of water is 4.184 J/(goC) Notice that specific heat is made up of three units-it is a derived unit. The specific heats for several substances is given in table 16-2 on page 492. q = (m)(c)(DT) q is the heat absorbed (+) or released (-) during a change in temperature m is the mass of the substance c is the specific heat of the substance DT is the change in temperature (DT = Tfinal – Tinitial)

7. How much heat is required to raise the temperature of 50.0 g of water from 25.0oC to 35.0 oC? What will the final temperature of a 20.0 g piece of iron be if it absorbs 450.0 J of heat when it was initially 25.0 oC? Complete practice problems 4 – 6 on page 495.

8. A simple laboratory calorimeter measures changes in temperature in an insulated environment. A calorimeter is based on the principle that the heat gained by one substance must equal the heat lost by another. For example, the temperature change for water can be measured when a hot piece of metal is placed into a calorimeter filled with water. Since all of the heat required to change the temperature of the water came from the metal, the specific heat of the metal can be determined.

9. A calorimeter contains 125.0 g of water at a temperature of 25.6oC. A 50.0 g sample of an unknown metal has a temperature of 115.0oC when it is placed into the calorimeter. The final temperature of the system is 29.3oC. What is the specific heat of the unknown metal? qH2O = (m)(cH2O)(DT) qH2O = (125.0 g)(4.184 J/goC)(29.3oC – 25.6oC) = 1935.1 J = 1900 J qgained = - qlost qH2O = - qmetal cmetal = (qmetal)/(mDT) cmetal = (-1935.1 J)/(50.0 g)(29.3oC – 115.0oC) = 0.451598599 J/goC = 0.45 J/goC

10. Complete practice problems 12 – 13 on page 498.

11. Since energy can not be created or destroyed in a chemical process, the energy of the universe must be constant. So if a system you are studying loses energy, what must happen to the surroundings if the energy of the universe must remain constant. Universe = System + Surroundings The system is what you are studying (the chemicals and the solvent), the surroundings are everything else in the universe. Enthalpy (H): the heat content of a system at constant pressure. DH is the change in enthalpy DHrxn is the change in enthalpy of a reaction DHrxn = DHproducts-DHreactants

12. If the system gives off energy, the final state if the system is lower in energy than it started. Since DE is Efinal – Einitial, DE will be negative when energy is given off by the system. If the system takes in energy, the final state if the system is higher in energy than it started. Since DE is Efinal – Einitial, DE will be positive when energy is taken in by the system.

13. An exothermic reaction DH is like DE DHrxn = DHproducts-DHreactants DHrxn = negative

14. An endothermic reaction DH is like DE DHrxn = DHproducts-DHreactants DHrxn = positive

15. Breaking bonds always requires energy (endothermic process) Making bonds always releases energy (exothermic process) Since a chemical reaction always involves bond breaking and bond making, the energy change for the reaction can be determined by comparing the total energy of all the bonds broken to the total energy of all the bonds being made. If the energy released during bond making is greater than the energy required for bond breaking, a reaction will be exothermic. If the energy released during bond making is less than the energy required for bond breaking, a reaction will be endothermic. Another way of saying this is: if the energy of the products is less than the energy of the reactants, the reaction will be exothermic. If the energy of the products is more than the energy of the reactants, the reaction will be endothermic.

16. Thermochemical Equations A chemical equation that includes the heat of reaction is a thermochemical equation. The heat of reaction may be included in the equation as a reactant or a product, or the heat of reaction can be included as a DHrxn value given along with the equation. If heat is a product, the reaction is exothermic. If heat is a reactant, the reaction is endothermic. 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) + 1625 kJ 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) DH = -1625 kJ NH4NO3 (s) + 27 kJ → NH4+ (aq) + NO3- (aq) NH4NO3 (s) → NH4+ (aq) + NO3- (aq) DH = 27 kJ

17. The heat of reaction can also be used in stoichiometry problems. 4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s) + 1625 kJ How many moles of Fe were reacted if 455.7 kJ of heat were released during the above reaction? How many grams of Fe would be required to produce 945.0 kJ of heat using the above reaction?

18. DHrxn for a combustion reaction is renamed DHcomb or the heat of combustion. The heats of combustion for a variety of fuels are given in Table 16-5 on page 501. Which fuel produces the most heat per mole of substance? Which fuel produces the most heat per gram of substance?

19. Heat changes for phase changes are also given new names. Molar Heat of Vaporization (DHvap): heat required to vaporize 1 mol of a liquid Heat is required to make this phase change occur, so the process is endothermic and values for heats of vaporization are always positive. Molar Heat of Fusion (DHfus): heat required to melt 1 mol of a solid Heat is required to make this phase change occur, so the processes is endothermic and values for heats of fusion are always positive. Condensation is the opposite process to vaporization, so a heat of condensation (DHcond) and a heat vaporization have opposite signs but are of equal magnitude. Solidification is the opposite process to fusion, so a heat of solidification (DHsolid) and a heat fusion have opposite signs but are of equal magnitude.

20. DHvap = -DHcond DHfus = -DHsolid Notice that the energy required to turn liquid water into gaseous water is almost 7 times larger than the energy required to turn ice into liquid water. Why might this be the case? Is the heat of vaporization for other substances likely to be larger than the heat of fusion for those substances as well?

21. See page 503: Heat is added to solid water at a constant rate, and the temperature changes are measured over time. What is happening in the places where the graph is flat?

22. How much heat is required to convert a 100.0 g block of ice at -25oC to steam at 115oC? qtotal = q1 + q2 + q3 + q4 + q5 but q2 = (n)(DHfus) and q4 = (n)(DHvap) q1 = (m)(cs)(DT) Solid phase Liquid phase q3 = (m)(cl)(DT) Gaseous phase q5 = (m)(cg)(DT) For water: cs = 2.03 J/goC cl = 4.184 J/goC cg = 2.01 J/goC n = moles of substance

23. Complete practice problems 20 – 22 on page 504.

24. Hess’s Law: if you add two or more thermochemical equations to produce a final equation, the sum of the enthalpy changes of individual thermochemical equations is equal to the enthalpy change for the overall reaction. Reaction 1: S (s) + O2 (g) → SO2 (g) + 297 kJ Reaction 2: 2 S (s) + 3 O2 (g) → 2 SO3 (g) + 792 kJ 2 SO2 (g) + O2 (g) → 2 SO3 (g) DH = ? How can we use Hess’s Law to find the DHrxn for this reaction? Since SO2 is a reactant in the final equation, reaction 1 must be reversed. Since there are 2 SO2 in the final equation, reaction 1 must be multiplied by 2. Then add reaction 1 to reaction 2.

25. Target Equation: 2 SO2 (g) + O2 (g) → 2 SO3 (g) DH = ? Reverse this equation S (s) + O2 (g) → SO2 (g) + 297 kJ SO2 (g) + 297 kJ → S (s) + O2 (g) Multiply by 2 Notice that reversing the equation changes the sign of DH. 2 SO2 (g) + 594 kJ → 2 S (s) + 2 O2 (g) Notice that multiplying the equation by 2 doubles DH. Now add the two equations together. 2 SO2 (g) + 594 kJ → 2 S (s) + 2 O2 (g) 2 S (s) + 3 O2 (g) → 2 SO3 (g) + 792 kJ 2 SO2 (g) + O2 (g) → 2 SO3 (g) + 792 kJ – 594 kJ 198 kJ DH = - 198 kJ

26. S (s) + O2 (g) → SO2 (g) DH = - 297 kJ SO2 (g) → S (s) + O2 (g) DH = 297 kJ 2SO2 (g) → 2S (s) + 2O2 (g) DH = 594 kJ Rules for using Hess’s Law: Reversing an equation changes the sign of DH (like multiplying by -1) Multiplying an equation by a number requires you to multiply the DH by the same number. 2 S (s) + 3 O2 (g) → 2 SO3 (g) DH = - 792 kJ DH = - 792 kJ + 594 kJ 2 SO2 (g) + O2 (g) → 2 SO3 (g) DH = - 198 kJ

27. Complete practice problems 28 and 29 on page 508.

28. Standard Enthalpy (Heat) of Formation: DHof is the change in enthalpy that occurs when one mole of a substance is produced in its standard state from it elements. The small circle indicates “standard state” which is the normal physical state at 1 atm and 25oC. Even though most compounds can not actually be made directly from their elements, this method allows us to determine the change in enthalpy for reactions using Hess’s Law. DHrxn = S(DHof (products)) -S(DHof (reactants)) The heat of formation of any element in its standard state is 0 kJ.

29. What is the heat of reaction for the following reaction? H2S (g) + 4 F2 (g) → 2 HF (g) + SF6 (g) Example: the heat of formation for HF (g) = -273 kJ for SF6 (g) = -1220. kJ for H2S (g) = -21 kJ DHrxn = S(DHof (products)) -S(DHof (reactants)) S(DHof (products)) = (2(-273 kJ) + (-1220 kJ)) = -1766 kJ S(DHof (reactants)) = ((-21 kJ) + 4(0 kJ)) = -21 kJ DHorxn = (-(1766 kJ) - (-21 kJ)) = -1745 kJ Notice that the coefficients of the balanced equation must be included!

30. Complete practice problem 31 on page 512.

31. Spontaneous Processes: any process that, once started, will continue on its own without any outside intervention. Example: A ball on the side of a hill will roll to the bottom once it starts rolling. Example: A block of ice will melt if left in a room that is at 25oC. Example: Steam will condense if left in a room that is at 25oC. Example: A cup of gasoline will continue to burn in air once it is lit. Non-spontaneous Process: any process that, once started, will not continue on its own without any outside intervention. Example: A cup of water will not freeze if left in a room at 25oC. Example: A ball will not roll up a hill on its own. Example: A room will not get neater on its own.

32. Entropy (S): a measure of the order (or randomness) of a system. Entropy is an energy term, so its units are joules, but it is not like heat. DSsystem = Sproducts-Sreactants If entropy increases in a reaction, Sproducts > Sreactants, so DS > 0 or DS is positive. What does it mean for entropy to be greater in one state than another? Greater entropy means that the system is more disordered than it was before. Compare a new deck of cards to a deck of cards that has been thrown on the floor and then quickly pushed together into a stack. Which has greater entropy?

33. Anything that allows for particles to be arranged in more ways is an indication of greater entropy. Examples: A flask with a pure substance compared to a flask with a mixture. A flask with solid water compared to a flask with liquid water. A flask with 2 moles of chemicals compared to a flask with 4 moles of chemicals. A flask with a gas in it compared to a flask with a solid in it. Complete practice problem 38 on page 516.

34. Second Law of thermodynamics: The entropy of the universe must increase for any spontaneous process. DSuniverse > 0 DSuniverse = Ssystem+Ssurroundings Therefore, if the system becomes more ordered (DSsystem is negative), the surroundings must become disordered (DSsurroundings is positive) to an even larger extent. Free energy (G) is the concept chemists use to determine whether a process is spontaneous or not. If DG< 0, a process is spontaneous. If DG> 0, a process is not spontaneous. If DG= 0, a process is at equilibrium-no longer changing in any direction.

35. DGsystem = DHsystem- TDSsystem DGosystem = DHosystem- TDSosystem If DGosystem = 0, DHosystem- TDSosystem = 0 and T = (DHosystem)/(DSosystem) Note: S is in J/mol·K while H and G are in kJ/mol In order to complete the math correctly, S must be changed to kJ/mol·K. Other important Thermodynamic Equations: DGorxn = S(DGof (products)) -S(DGof (reactants)) DHorxn = S(DHof (products)) -S(DHof (reactants)) DSorxn = S(DSof (products)) -S(DSof (reactants))