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AP Notes Chapter 5 Principles of Reactivity: Energy and Chemical Reactions

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study of energy transfer or heat flow Energy Kinetic Energy Thermal - Heat Mechanical Electrical Sound Potential Energy Chemical Gravitational Electrostatic Thermodynamics

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Energy is... The ability to do work. The ability to do work. Conserved. Conserved. made of heat and work. made of heat and work. Work is a force acting over a distance. Work is a force acting over a distance. W=F x d Power is work done over time. Power is work done over time. P=W t Heat is energy transferred between objects because of temperature difference. Heat is energy transferred between objects because of temperature difference.

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First Law of Thermodynamics Law of conservation of energy Law of conservation of energy Total energy of the universe is constant Total energy of the universe is constant Temperature and heat Temperature and heat Heat is not temperature Heat is not temperature Thermal energy = particle motion Thermal energy = particle motion Total thermal energy is sum of all a materials individual energies Total thermal energy is sum of all a materials individual energies

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The Universe is divided into two halves. is divided into two halves. the system and the surroundings. the system and the surroundings. The system is the part you are concerned with. The system is the part you are concerned with. The surroundings are the rest. The surroundings are the rest. q into system = -q from surroundings

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System: region of space where process occurs Surroundings: region of space around system

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Thermodynamic Properties 1. State functions -depend on nature of process NOT method -path independent -denoted by capital letters E state function

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2. Path functions -depends on how process is carried out -path dependent -denoted by lower case Thermodynamic Properties q & w path functions

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Work w w = - (PV) w = - P V - V P Constant P w P = - P V Constant V w V = - V P

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Work w w = - P V assume ideal piston (i.e. no heat loss) w = E E = - P V in isothermics w = -q

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PV = nRT R=0.0821L atm/mol K In isothermic conditions -q = w = -nRT R=8.314 J/mol K

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U = q + w In isothermics U = 0 so q=-w U = q - P V In constant V or constant P w=0 so U = q

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SI unit Joule 1 cal = J 1000cal = 1Kcal=4.184KJ Calorie - Amount of heat needed to raise the temperature of 1 gram H 2 0, 1 degree centigrade Nutritional calorie [Calorie]= 1 Kcal

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Direction of energy flow Every energy measurement has three parts. Every energy measurement has three parts. 1. A unit ( Joules or calories). 2. A number how many. 3. and a sign to tell direction. negative - exothermic negative - exothermic positive- endothermic positive- endothermic No change in energy - isothermic No change in energy - isothermic

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Factors Determining Amount of Heat Amount of material Temperature change Heat capacity

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Specific Heat Amount of heat needed to raise the temperature of 1 gram of material 1 degree centigrade Heat = q q = m C p T where C p = Heat capacity T = T 2 – T 1

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C P (J/mol.o C) H 2 O(s) = H 2 O(l) = H 2 O(g) = 33.60

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What amount of heat is needed to just boil away 100. mL of water in a typical lab?

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System Surroundings Energy E <0

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Exothermic reactions release energy to the surroundings. q < 0 (-) Exothermic

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Potential energy Heat

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System Surroundings Energy E >0

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Endothermic reactions absorb energy from the surroundings. q > 0 (+) Endothermic

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Potential energy Heat

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1. How much heat is absorbed when 10.0 g of H 2 0 are heated from 20 0 C to 25 0 C?

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2. How much heat is needed to completely vaporize 10.0 g of H 2 O at its boiling point?

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Heat of Vaporization H vap (H 2 O) = kJ/mol Heat of Fusion H fus (H 2 O) = 6.01 kJ/mol

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Energy processes: Within a phase q=mHv or mHf Between phases q=mCp T

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3. How much total heat is used to raise the temperature of g H 2 O from-15 o to 125 o C?

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Some rules for heat and work Heat given off is negative. Heat given off is negative. Heat absorbed is positive. Heat absorbed is positive. Work done by system on surroundings is positive. Work done by system on surroundings is positive. Work done on system by surroundings is negative. Work done on system by surroundings is negative. Thermodynamics- The study of energy and the changes it undergoes. Thermodynamics- The study of energy and the changes it undergoes.

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First Law of Thermodynamics The energy of the universe is constant. The energy of the universe is constant. Law of conservation of energy. Law of conservation of energy. q = heat q = heat w = work w = work E = q + w E = q + w Take the systems point of view to decide signs. Take the systems point of view to decide signs.

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What is work? Work is a force acting over a distance. Work is a force acting over a distance. w= F x d w= F x d P = F/ area P = F/ area d = V/area d = V/area w= (P x area) x (V/area)= P V w= (P x area) x (V/area)= P V Work can be calculated by multiplying pressure by the change in volume at constant pressure. Work can be calculated by multiplying pressure by the change in volume at constant pressure. units of liter - atm L-atm units of liter - atm L-atm

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Work needs a sign If the volume of a gas increases, the system has done work on the surroundings. If the volume of a gas increases, the system has done work on the surroundings. work is negative work is negative w = - P V w = - P V Expanding work is negative. Expanding work is negative. Contracting, surroundings do work on the system w is positive. Contracting, surroundings do work on the system w is positive. 1 L atm = J 1 L atm = J

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Examples What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? If 2.36 J of heat are absorbed by the gas above. what is the change in energy? If 2.36 J of heat are absorbed by the gas above. what is the change in energy? How much heat would it take to change the gas without changing the internal energy of the gas? How much heat would it take to change the gas without changing the internal energy of the gas?

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Calorimetry Measuring heat. Measuring heat. Use a calorimeter. Use a calorimeter. Two kinds Two kinds Constant pressure calorimeter (called a coffee cup calorimeter) Constant pressure calorimeter (called a coffee cup calorimeter) heat capacity for a material, C is calculated heat capacity for a material, C is calculated C= heat absorbed/ T = H/ T C= heat absorbed/ T = H/ T specific heat capacity = C/mass specific heat capacity = C/mass

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Calorimetry molar heat capacity = C/moles molar heat capacity = C/moles heat = specific heat x m x T heat = specific heat x m x T heat = molar heat x moles x T heat = molar heat x moles x T Make the units work and youve done the problem right. Make the units work and youve done the problem right. A coffee cup calorimeter measures H. A coffee cup calorimeter measures H. An insulated cup, full of water. An insulated cup, full of water. The specific heat of water is 1 cal/gºC The specific heat of water is 1 cal/gºC Heat of reaction= H = sh x mass x T Heat of reaction= H = sh x mass x T

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Bomb Calorimeter

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Calorimeter Constant heat capacity that is constant over a temperature range where q(cal) = CC x T C (CC = calorimeter constant)

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4. Determine the calorimeter constant for a bomb calorimeter by mixing 50.0 mL of water at 25 o C with 50.0 mL of water at 60.0 o C. The final temp. attained is 40 o C.

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5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?

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6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.

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7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of C. After equilibrium was reached, the final temperature was C.

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Examples The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper? A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

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Calorimetry Constant volume calorimeter is called a bomb calorimeter. Constant volume calorimeter is called a bomb calorimeter. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. The heat capacity of the calorimeter is known and tested. The heat capacity of the calorimeter is known and tested. Since V = 0, P V = 0, E = q Since V = 0, P V = 0, E = q

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Bomb Calorimeter thermometer thermometer stirrer stirrer full of water full of water ignition wire ignition wire Steel bomb Steel bomb sample sample

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Properties intensive properties not related to the amount of substance. intensive properties not related to the amount of substance. density, specific heat, temperature. density, specific heat, temperature. Extensive property - does depend on the amount of stuff. Extensive property - does depend on the amount of stuff. Heat capacity, mass, heat from a reaction. Heat capacity, mass, heat from a reaction.

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Enthalpy abbreviated H abbreviated H H = E + PV (thats the definition) PV = w H = E + PV (thats the definition) PV = w at constant pressure. at constant pressure. H = E + P V H = E + P V the heat at constant pressure q p can be calculated from the heat at constant pressure q p can be calculated from E = q p + w = q p - P V E = q p + w = q p - P V q p = E + P V = H q p = E + P V = H

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Hesss Law Enthalpy is a state function. Enthalpy is a state function. It is independent of the path. It is independent of the path. We can add equations to come up with the desired final product, and add the H We can add equations to come up with the desired final product, and add the H Two rules Two rules If the reaction is reversed the sign of H is changed If the reaction is reversed the sign of H is changed If the reaction is multiplied, so is H If the reaction is multiplied, so is H

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Enthalpy and H f N 2 + 2O 2 2NO 2 N 2 + 2O KJ/mol 2NO 2 Which side is the heat written on? Why? Look on your Heat Sheets! Why is it only 34KJ/mol?

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N2N2 + 2O 2 68 kJ 2NO 2 H (kJ)

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N2N2 + 2O 2 O2O2 + NO 2 2NO kJ -112 kJ H (kJ)

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N2N2 + 2O 2 O2O2 + NO 2 68 kJ 2NO kJ -112 kJ H (kJ)

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Hess Law Reactants Products The enthalpy change is the same whether the reaction occurs in one step or in a series of steps

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Molar heat capacity

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8. Calculate the heat needed to decompose one mole of calcium carbonate.

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Standard Enthalpy The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) The enthalpy change for a reaction at standard conditions (25ºC, 1 atm, 1 M solutions) Symbol Hº Symbol Hº When using Hesss Law, work by adding the equations up to make it look like the answer. When using Hesss Law, work by adding the equations up to make it look like the answer. The other parts will cancel out. The other parts will cancel out.

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Hº= -394 kJ Hº= -286 kJ Example Given calculate Hº for this reaction Hº= kJ

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394 kJ 286 kJ Example calculate Hº for this reaction kJ Given

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394 kJ 286 kJ Example calculate Hº for this reaction kJ Given 1300.KJ + 2CO 2 (g) + H 2 0(l) C 2 H 2 (g) + 5/2 O 2 (g)

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394 kJ) 286 kJ Example calculate Hº for this reaction Given 1300.KJ + 2CO 2 (g) + H 2 0(l) C 2 H 2 (g) + 5/2 O 2 (g) 222

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kJ 286 kJ Example calculate Hº for this reaction Given 1300.KJ + 2CO 2 (g) + H 2 0(l) C 2 H 2 (g) + 5/2 O 2 (g) 422 2

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kJ 286 kJ Example calculate Hº for this reaction Given 1300.KJ + 2CO 2 (g) + H 2 0(l) C 2 H 2 (g) + 5/2 O 2 (g) 422 2

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Example calculate Hº for this reaction Given 1300.KJ C 2 H 2 (g) 2C(s) 788KJ H 2 (g) 286KJ 1300.KJ 788kJ + 286KJ 1300.KJ 1074KJ

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Example calculate Hº for this reaction Given 1300.KJ 1074KJ 226KJ H f = 226KJ (+sign shows Endothermic) 226KJ + H f = 1300KJ – 1074KJ

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Example Given Calculate Hº for this reaction Hº= +77.9kJ Hº= +495 kJ Hº= kJ Now, You try it!

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Standard Enthalpies of Formation Hesss Law is much more useful if you know lots of reactions. Hesss Law is much more useful if you know lots of reactions. Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. Standard states are 1 atm, 1M and 25ºC Standard states are 1 atm, 1M and 25ºC For an element it is 0 For an element it is 0 There is a table in Appendix 4 (pg A22) There is a table in Appendix 4 (pg A22)

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Standard Enthalpies of Formation Need to be able to write the equations. Need to be able to write the equations. What is the equation for the formation of NO 2 ? What is the equation for the formation of NO 2 ? ½N 2 (g) + O 2 (g) NO 2 (g) ½N 2 (g) + O 2 (g) NO 2 (g) Have to make one mole to meet the definition. Have to make one mole to meet the definition. Write the equation for the formation of methanol CH 3 OH. Write the equation for the formation of methanol CH 3 OH.

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Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O which leads us to this rule. which leads us to this rule.

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Since we can manipulate the equations We can use heats of formation to figure out the heat of reaction. We can use heats of formation to figure out the heat of reaction. Lets do it with this equation. Lets do it with this equation. C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O C 2 H 5 OH +3O 2 (g) 2CO 2 + 3H 2 O which leads us to this rule. which leads us to this rule.

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9. Calculate H f of SO 3 (g) from the data given. Forms of Equations: 2SO 2(g) + O 2(g) 2SO 3(g) H = kj or 2SO 2(g) + O 2(g) 2SO 3(g) kj

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10. If the metabolism of glucose is combustion of C 6 H 12 O 6 (s), how much heat is produced by the metabolism of 1.00 g of glucose?

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H = Enthalpy H = E + PV Additivity of Heats of Reaction

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at constant pressure: E = q P + w or E = q P - P V

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now, at constant P: V P = 0 thus, H = E + P V & if E = q P - P V then H = q P

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H = Enthalpy a state function cannot be measured measure H H = H products - H reactants

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Standard Enthalpy of Formation Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states

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reference point- for ELEMENTS in standard state Thermodynamic Data Appendix L of text p. A31 Thermodynamic Standard State T = 25 o C and P = 1 atm

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11. Combustion of Methane H C = ?

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Write equations and find ΔH f 0 for methane, carbon dioxide, & water

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H C = ?

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