Principles of Chemistry: A Molecular Approach , 1st Ed. Nivaldo Tro

1. Principles of Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro Chapter 6 Thermochemistry

2. Nature of Energy • Even though Chemistry is the study of matter, energy affects matter. • Energy is anything that has the capacity to do work. • Work is a force acting over a distance. • energy = work = force × distance • Energy can be exchanged between objects through contact. • collisions

3. Energy, Heat, and Work • You can think of energy as a quantity an object can possess. • or collection of objects • You can think of heat and work as the two different ways that an object can exchange energy with other objects. • either out of it, or into it

4. Classification of Energy • Kinetic energy is energy of motion or energy that is being transferred. • Thermal energy is kinetic.

5. Classification of Energy • Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object. • Energy stored in the structure of a compound is potential.

6. Exchanging Energy • Energy cannot be created or destroyed. • Energy can be exchanged between objects. • Energy can also be transformed from one form to another. • heat → light → sound

7. Some Forms of Energy • electrical • kinetic energy associated with the flow of electrical charge • heat or thermal energy • kinetic energy associated with molecular motion • light or radiant energy • kinetic energy associated with energy transitions in an atom • nuclear • potential energy in the nucleus of atoms • chemical • potential energy due to the structure of the atoms, the attachment between atoms, the atoms’ relative position to each other in the molecule, or the molecules’ relative positions in the structure

8. Surroundings Surroundings System System System and Surroundings • We define the system as the material or process that contains the energy changes we are studying. • We define the surroundings as everything else in the universe. • What we study is the exchange of energy between the system and the surroundings.

9. When the mass is in kg and velocity in m/s, the unit for kinetic energy is . • One joule of energy is the amount of energy needed to move a 1-kg mass at a speed of 1 m/s. • 1 J = 1 Units of Energy • The amount of kinetic energy an object has is directly proportional to its mass and velocity. • KE = ½mv2

10. Units of Energy • One joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter. • 1 J = 1 N∙m = 1 kg∙m2/s2 • One calorie(cal) is the amount of energy needed to raise one gram of water by 1 °C. • kcal = energy needed to raise 1000 g of water 1 °C • food Calories = kcals

11. Energy Use

12. The First Law of Thermodynamics—Law of Conservation of Energy • When energy is exchanged between objects or transformed into another form, all the energy is still there. • The total amount of energy in the universe before the change has to be equal to the total amount of energy in the universe after. • You can, therefore, never design a system that will continue to produce energy without some source of energy.

13. Energy Flow and Conservation of Energy • Conservation of energy requires that the total energy change in the system and the surroundings must be zero. • DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings • D is the symbol that is used to mean change. • final amount – initial amount

14. Internal Energy • The internal energy is the total amount of kinetic and potential energy a system possesses. • The change in the internal energy of a system depends only on the amount of energy in the system at the beginning and end. • A state function is a mathematical function whose result depends only on the initial and final conditions, not on the process used. • DE = Efinal – Einitial • DEreaction = Eproducts – Ereactants

15. State Function You can travel either of two trails to reach the top of the mountain. One is long and winding; the other is shorter but steeper. Regardless of which trail you take, when you reach the top you will be 10,000 ft. above the base. The height of the mountain is a state function. It depends only on the distance from the base to the peak, not on how you arrive there!

16. final energy added DE = + Internal Energy initial initial energy removed DE = – Internal Energy final Energy Diagrams • Energy diagrams are a “graphical” way of showing the direction of energy flow during a process. • If the final condition has a • larger amount of internal • energy than the initial • condition, the change in the • internal energy will be +. • If the final condition has a • smaller amount of internal • energy than the initial • condition, the change in the • internal energy will be –.

17. Surroundings DE + System DE – Energy Flow • When energy flows out of a system, it must all flow into the surroundings. • When energy flows out of a system, DEsystem is –. • When energy flows into the surroundings, DEsurroundings is +. • therefore: –DEsystem= DEsurroundings

18. Surroundings DE – System DE + Energy Flow • When energy flows into a system, it must all come from the surroundings. • When energy flows into a system, DEsystem is +. • When energy flows out of the surroundings, DEsurroundings is –. • therefore: DEsystem= –DEsurroundings

19. energy absorbed DErxn = + C(s), O2(g) Internal Energy CO2(g) energy released DErxn = – Energy Flow in a Chemical Reaction • The total amount of internal energy in 1 mol of C(s) and 1 mole of O2(g) is greater than the internal energy in 1 mole of CO2(g). • at the same temperature and pressure • In the reaction C(s) + O2(g) → CO2(g), there will be a net release of energy into the surroundings. • −DEreaction = DEsurroundings • In the reaction CO2(g) → C(s) + O2(g), there will be an absorption of energy from the surroundings into the reaction. • DEreaction = −DEsurroundings Surroundings System CO2 → C + O2 System C + O2 → CO2

20. Energy Exchange • Energy is exchanged between the system and surroundings through heat and work. • q = heat (thermal) energy • w = work energy • q and w are NOT state functions; their value depends on the process. DE = q + w

21. Energy Exchange • Energy is exchanged between the system and surroundings through either heat exchange or work being done.

22. Heat and Work • On a smooth table, most of the kinetic energy is transferred from the first ball to the second—with a small amount lost through friction.

23. Heat and Work • On a rough table, most of the kinetic energy of the first ball is lost through friction—less than half is transferred to the second.

24. Heat, Work, and Internal Energy • In the previous billiard ball example, the DE of the white ball is the same for both cases, but q and w are not. • On the rougher table, the heat loss, q, is greater. • q is a more negative number. • But on the rougher table, less kinetic energy is transferred to the purple ball, so the work done by the white ball, w, is less. • w is a less negative number. • The DE is a state function and depends only on the velocity of the white ball before and after the collision. • In both cases, it started with 5.0 kJ of kinetic energy and ended with 0 kJ because it stopped. • q + w is the same for both tables, even though the values of q and w are different.

25. q, w q, w potato q, w fuel DE Example 6.1 If burning fuel in a potato cannon gives 855 J of kinetic energy to the potato and releases 1422 J of heat from the cannon, what is the change in internal energy of the fuel? Given: Find: qpotato = 855 J, wpotato = 1422 J DEfuel,J Concept Plan: Relationships: qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w Solution: Check: The unit and sign are correct.

26. Practice—Reacting 50 mL of H2(g) with 50 mL of C2H4(g) produces 50 mL of C2H6(g) at 1.5 atm. If the reaction produces 3.1 × 102 J of heat and the decrease in volume requires the surroundings to do 7.6 J of work on the system, what is the change in internal energy of the system?

27. q, w w surrounding DE w reaction If the reaction produces 3.1 × 102 J of heat and the decrease in volume requires the surroundings to do 7.6 J of work on the system, what is the change in internal energy of the system? Given: Find: qreaction = −310 J, wsurroundings = −7.6 J DEfuel,J Concept Plan: Relationships: qsystem = −qsurroundings, wsystem = −wsurroundings, DE = q + w Solution: Check: The unit and sign are correct.

28. Heat Exchange • Heat is the exchange of thermal energy between the system and surroundings. • It occurs when the system and surroundings have a difference in temperature. • Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature. • thermal equilibrium

29. Quantity of Heat Energy AbsorbedHeat Capacity • When a system absorbs heat, its temperature increases. • The increase in temperature is directly proportional to the amount of heat absorbed. • The proportionality constant is called the heat capacity, C. • Units of C are J/°C or J/K. q = C × DT • The heat capacity of an object depends on its mass. • 200 g of water requires twice as much heat to raise its temperature by 1 °C as does 100 g of water. • The heat capacity of an object depends on the type of material. • 1000 J of heat energy will raise the temperature of 100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C.

30. Specific Heat Capacity • measure of a substance’s intrinsic ability to absorb heat • The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1 °C. • Cs • units are J/(g∙°C) • The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1 °C.

31. Specific Heat of Water • The rather high specific heat of water allows water to absorb a lot of heat energy without a large increase in its temperature. • The large amount of water absorbing heat from the air keeps beaches cool in the summer. • Without water, the Earth’s temperature would be about the same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F). • Water is commonly used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating. • or even melting • It can also be used to transfer the heat to something else because it is a fluid.

32. Quantifying Heat Energy • The heat capacity of an object is proportional to its mass and the specific heat of the material. • So we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object. heat = (mass) × (specific heat capacity) × (temp. change) q = (m) × (Cs) × (DT)

33. Cs m, DT q Example 6.2 How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from –8.0 °C to 37.0 °C? Sortinformation Given: Find: T1 = –8.0 °C, T2 = 37.0 °C, m = 3.10 g q,J q = m ∙ Cs ∙ DT Cs = 0.385 J/g (Table 6.4) Strategize Conceptual Plan: Relationships: Follow the conceptual plan to solve the problem. Solution: Check Check: The unit and sign are correct.

34. Practice—Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C.

35. Cs m, DT q Practice—Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C. Sort information Given: Find: T1 = 49 °C, T2 = 29 °C, m = 7.40 g q,J q = m ∙ Cs ∙ DT Cs = 4.18 J/gC (Table 6.4) Strategize Conceptual Plan: Relationships: Follow the conceptual plan to solve the problem Solution: Check Check: The unit and sign are correct.

36. Pressure–Volume Work • PV work is work that is the result of a volume change against an external pressure. • When gases expand, DV is +, but the system is doing work on the surroundings, so wgas is –. • As long as the external pressure is kept constant, –work = external pressure × change in volume w = –PDV • To convert the units to joules, use 101.3 J = 1 atm∙L.

37. P, DV w Example 6.3 If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done? Given: Find: V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm w,J Conceptual Plan: Relationships: 101.3 J = 1 atm∙L Solution: Check: The unit and sign are correct.

38. Practice—A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas?

39. A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? Given: Find: DE = −68.3 kJ, q = +15.8 kcal, V1 = 10.0 L, P = 1.00 atm V2, L w Conceptual Plan: Relationships: q, DE V2 P, V1 DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L Solution: Check:

40. A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? Given: Find: DE = −6.83 × 104 J, q = +6.604 × 104 J, V1 = 10.0 L, P = 1.00 atm V2, L w Conceptual Plan: Relationships: q, DE V2 P, V1 DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L Solution: Check: 40

41. A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? Given: Find: DE = −6.83 × 104 J, q = +6.604 × 104 J, V1 = 10.0 L, P = 1.00 atm V2, L w Conceptual Plan: Relationships: q, DE V2 P, V1 DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L Solution: Check: 41

42. A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is absorbed by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? Given: Find: DE = −6.83 × 104 J, q = +6.604 × 104 J, V1 = 10.0 L, P = 1.00 atm V2, L w Conceptual Plan: Relationships: q, DE V2 P, V1 DE = q + w, w = –PDV, 1 kJ = 1000 J, 1 cal = 4.18 J, 101.3 J = 1 atm∙L Solution: Check: Since DE is − and q is +, w must be −; and it is. When w is − the system is expanding, so V2 should be greater than V1; and it is. 42

43. Exchanging Energy betweenSystem and Surroundings • exchange of heat energy q = mass × specific heat × DTemperature • exchange of work w = –Pressure × DVolume

44. Measuring DE, Calorimetry at Constant Volume • Since DE = q + w, we can determine DE by measuring q and w. • In practice, it is easiest to do a process in such a way that there is no change in volume, w = 0. • at constant volume, DEsystem = qsystem • In practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction. So instead, we use an insulated, controlled surroundings and measure the temperature change in it. • The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water. qsurroundings = qcalorimeter = –qsystem –DEreaction=qcal = Ccal× DT

45. used to measure DE because it is a constant volume system Bomb Calorimeter

46. qrxn, mol DE Example 6.4 When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If Ccal = 4.90 kJ/°C, find DE for burning 1 mole. Given: Find: 1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C, Ccal = 4.90 kJ/°C DErxn,kJ/mol 2.9506 × 10−3 mol C12H22O11, DT = 3.41 °C, Ccal = 4.90 kJ/°C DErxn,kJ/mol Conceptual Plan: Relationships: qcal Ccal, DT qrxn qcal = Ccal × DT = –qrxn MM C12H22O11 = 342.3 g/mol Solution: The units and sign are correct. Check:

47. Practice—When 1.22 g of HC7H5O2 (MM 122.12) is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C. If DE for the reaction is −3.23 × 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C?

48. qcal, DT Ccal When 1.22 g of HC7H5O2 is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C. If DE for the reaction is −3.23 × 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? Given: Find: 1.22 g HC7H5O2, T1 = 20.27 °C, T2 = 22.67 °C, DE = –3.23 × 103 kJ/mol Crxn,kJ/°C 9.990 × 10−3 mol HC7H5O2, DT = 2.40 °C, DE = –3.23 × 103 kJ/mol Ccal,kJ/°C Conceptual Plan: Relationships: qrxn mol, DE qcal qcal = Ccal x DT = –qrxn MM HC7H5O2 = 122.12 g/mol Solution: The units and sign are correct. Check:

49. Enthalpy • The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume (work done). • H is a state function. H = E + PV • The enthalpy change,DH, of a reaction is the heat evolved in a reaction at constant pressure. DHreaction = qreaction at constant pressure • Usually DH and DE are similar in value; the difference is largest for reactions that produce or use large quantities of gas.

50. Endothermic and Exothermic Reactions • When DH is –, heat is being released by the system. • Reactions that release heat are called exothermic reactions. • When DH is +, heat is being absorbed by the system. • Reactions that absorb heat are called endothermic reactions. • Chemical heat packs contain iron filings that are oxidized in an exothermic reaction. Your hands get warm because the released heat of the reaction is absorbed by your hands. • Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process. Your hands get cold because they are giving away your heat to the reaction.