Energy Enthalpy Calorimetry Thermochemistry Energy is The ability

Energy, Enthalpy Calorimetry & Thermochemistry

Energy is. . . The ability to do work. Conserved. made of heat and work. a state function. independent of the path, or how you get from point A to B. • Work is a force acting over a distance. • Heat is energy transferred between objects because of temperature difference. • • •

The Universe is divided into two halves. the system and the surroundings. The system is the part you are concerned with. The surroundings are the rest. Exothermic reactions release energy to the surroundings. • Endothermic reactions absorb energy from the surroundings. • • •

Potential energy Heat

Potential energy Heat

Units of Heat • Calorie (cal) – The quantity of heat required to change the temperature of one gram of water by one degree Celsius. • Joule (J) – SI unit for heat 1 cal = 4. 184 J

Direction • 1. 2. 3. • • Every energy measurement has three parts. A unit ( Joules or calories). A number how many. and a sign to tell direction. negative - exothermic positive- endothermic

Surroundings System Energy E <0

Surroundings System Energy E >0

First Law of Thermodynamics • • • The energy of the universe is constant. Law of conservation of energy. q = heat Take the systems point of view to decide signs. q is negative when the system loses heat energy q is positive when the system absorbs heat energy

Enthalpy Heat content of a substance Depends on many things Symbol is “H” Can’t be directly measured Changes in enthalpy can be calculated, H = Hfinal - Hinitial • Changes in heat (q) can be used to find H • • •

Same rules for Enthalpy • Heat energy given off by the system is negative, H < 0 • Heat energy absorbed by the system is positive, H > 0

q and H q is the “heat flow for a system” It can be for any amount of mass or moles of substance H is “change in heat” It is for a specific chemical or physical change and it refers to that reaction or process.

Heat Capacity • The quantity of heat required to change the temperature of a system by one degree. • Three “types” with different “systems: – Specific Heat Capacity – Molar Heat Capacity – Heat Capacity

Heat Capacity Types – Specific heat capacity • • • System is one gram of substance Units are J/go. C q = heat change m = mass C = Specific Heat Capacity T = final temperature – initial temperature q = (m)(C)( T)

Heat Capacity Types – Molar heat capacity. • • • System is one mole of substance. Units are J/mol o. C q = heat change n = moles C = molar heat capacity T = final temperature – initial temperature q= n. C T

Heat Capacity Types – Heat capacity of an object • Specific for a given amount or unit so it is not mass dependent • Units are J/o. C • q = heat • C = heat capacity • T = final temperature – initial temperature q = C T

Heat Capacity Types • So how do you know which one to use when? • Look at the known information • Select the value with appropriate units to cancel

Example • 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.

Calorimetry • Measuring temperature changes to calculate heat changes. • Use a calorimeter. • Two kinds – Constant pressure calorimeter (called a coffee cup calorimeter) – Constant volume calorimeter (called a bomb calorimeter)

Calorimetry • A coffee cup calorimeter measures temperature and calculates q. • An insulated cup, full of water (constant pressure, variable volume) • Water is the surroundings in which the system changes • Calculate the heat change of water. • The specific heat of water is 4. 184 J /gºC • Heat of water q. H 2 O = CH 2 O x m. H 2 O x T

qsystem + qsurroundings = 0 • q. H 2 O + qrxn = 0 • q. H 2 O = - qrxn In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed. Law of Conservation of Energy

Coffee Cup Calorimeter • A simple calorimeter. – Well insulated and therefore isolated. – Measure temperature change. qrxn = -qcal

Determination of Specific Heat

Example Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. qlead = -qwater = mc T = (50. 0 g)(4. 184 J/g °C)(28. 8 - 22. 0)°C qwater = 1. 4 x 103 J qlead = -1. 4 x 103 J = mc T = (150. 0 g)(c)(28. 8 - 100. 0)°C clead = 0. 13 Jg-1°C-1

Example • When a piece of copper (5. 0 g) is heated for 2. 0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20. 0 ˚C to 71. 9 ˚C. • What is the specific heat of the copper?

Examples If 10. 0 g of Cu is heated for 2. 0 seconds from 20. 0 ˚C and 200 J of heat are absorbed, what is the final temperature of the block? a. 71. 9 ˚C b. 100 ˚C c. 46 ˚C d. 123. 9 ˚C e. 719 ˚C

Example Equal masses of liquid A, initially at 100 ˚C, and liquid B, initially at 50 ˚C, are combined in an insulated container. The final temperature of the mixture is 80 ˚C. Which has the larger specific heat capacity, A or B? • A • B • A and B have the same specific heat capacity.

Example • When 86. 7 grams of water at a temperature of 73. 0 ˚C is mixed with an unknown mass of water at a temperature of 22. 3 ˚C the final temperature of the resulting mixture is 61. 7 ˚C. What was the mass of the second sample of water? a. b. 24. 9 g 302 g c. 48. 2 g d. 419 g

States of Matter

General Heating Curve http: //cwx. prenhall. com/petrucci/medialib/med ia_portfolio/text_images/031_Changes. State. MOV

Changes of State of Water qphase change = DHphase change (mass or moles) Molar enthalpy of vaporization: H 2 O (l) → H 2 O(g) Hv = 40. 65 k. J/mol @ 373. 15 K Hv = (40. 65 k. J/mol)(1 mol/18. 02 g) = 2256 J/g Molar enthalpy of fusion (melting): H 2 O (s) → H 2 O(l) Hf = 6. 01 k. J/mol @273. 15 K Hf = (6. 01 k. J/mol)(1 mol/18. 02 g) = 333 J/g

Heating Curve for Water Total heat absorbed by the water as it is warmed can be determined by finding the heat involved in each “step” of the process. q 1+ q 2 + q 3 …. = qtotal Note: q= (m)(C)( T) Cice= Csteam = 2. 09 J/go. C qphase change = Hphase change (mass or moles)

Example Calculate q for the process in which 50. 0 g of water is converted from liquid at 10. 0°C to vapor at 125. 0°C.

Example • What is the heat of fusion of lead in J/g if 6. 30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a liquid? a. 0. 0250 J/g c. 1. 61 J/g b. 24. 7 J/g d. 40. 5 J/g

Example • What quantity of heat is required to heat 1. 00 g of lead from 25 ˚C to the melting point (327 ˚C) and melt all of it? (The specific heat capacity of lead is 0. 159 J/g • K and it requires 24. 7 J/g to convert lead from the solid to the liquid state. ) a. b. 2. 47 J 48. 0 J c. 39. 4 J d. 72. 7 J

Calorimetry • 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. • The heat capacity of the calorimeter is known and tested.

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

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

Bomb Calorimeter qrxn = -qcal = qbomb + qwater + qwires +… Define the heat capacity of the calorimeter: qcal = mici T = Ccal T heat

q and H revisited • If you know “q” for a process for a known amount of sample • And you have a balanced chemical equation • “q” can be converted to H

q to H If you know the heat per amount of substance, find it per gram then use the molar mass to find the heat per mole.

Thermochemical Equations • Balanced chemical equation • Includes a term which indicates the change in enthalpy, H • Enthalpy term can be embedded in the reaction or written after the reaction with H notation.

Examples The combustion of 1. 010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24. 92 to 28. 33°C. The heat capacity of the calorimeter assembly is 4. 90 k. J/°C. (a) What is the heat of combustion of sucrose, expressed in k. J/mol C 12 H 22 O 11 (a) Verify the claim of sugar producers that one teaspoon of sugar (about 4. 8 g) contains only 19 calories.

Endothermic Thermochemical Equations Al 2 O 3(s) 2 Al(s) + 3/2 O 2(g) Hrxn = 1, 676 k. J/mol rxn • THE EXACT SAME AS WRITING: • Al 2 O 3(s) + 1, 676 k. J 2 Al(s) + 3/2 O 2(g)

Exothermic Thermochemical Equations H 2(g) + ½ O 2(g) H 2 O(l) + 285. 8 k. J Same as: H 2(g) + ½ O 2(g) H 2 O(l) Hrxn = -285. 8 k. J/mol rxn

Stoichiometry and Thermochemical Equations • Just as the mole ratios can be used to determine reacting and produced amounts of chemicals • The ratios can be used to relate amounts of chemicals with energy

Example Al 2 O 3(s) 2 Al(s) + 3/2 O 2(g) Hrxn = 1, 676 k. J/mol rxn • How many grams of aluminum would be produced if the aluminum absorbed 3500 k. J of energy?

Hrxn is for ANY general chemical or physical reaction There are specific types of reactions which have a special designation for their H.

Some Important Types of Enthalpy Changes Heat of combustion ( Hcomb) C 4 H 10(l) + 13/2 O 2(g) 4 CO 2(g) + 5 H 2 O(g) Heat of formation ( Hf) K(s) + 1/2 Br 2(l) KBr(s) Heat of fusion ( Hfus) Na. Cl(l) Heat of vaporization ( Hvap) C 6 H 6(l) C 6 H 6(g)

Standard States and Standard Enthalpy Changes • Define a particular state as a standard state. • Standard enthalpy of reaction, H° – The enthalpy change of a reaction in which all reactants and products are in their standard states. • Standard State – The pure element or compound at a pressure of 1 bar and at the temperature of interest.

Indirect Determination of H: Hess’s Law • H is an extensive property. Enthalpy change is directly proportional to the amount of substance in a system. N 2(g) + O 2(g) → 2 NO(g) H = +180. 50 k. J ½N 2(g) + ½O 2(g) → NO(g) H = +90. 25 k. J o H changes sign when a process is reversed NO(g) → ½N 2(g) + ½O 2(g) H = -90. 25 k. J

Hess’s Law Hess’s law of constant heat summation n If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2(g) + ½O 2(g) → NO(g) H = +90. 25 k. J NO(g) + ½O 2(g) → NO 2(g) H = -57. 07 k. J ½N 2(g) + O 2(g) → NO 2(g) H = +33. 18 k. J

Example Find ΔH for 2 H 2(g) + O 2(g) 2 H 2 O(l) Given: CH 3 COOH(l) + 2 O 2(g) 2 CO 2(g) + 2 H 2 O(l) ΔH = -1, 959. 8 k. J C(graphite) + O 2(g) CO 2(g) ΔH = -886. 5 k. J CH 3 COOH(l) 2 C(graphite) + 2 H 2(g) + O 2(g) ΔH = 1, 100. 2 k. J

Example 11 • Find H for 2 NH 3(g) N 2 H 4 (l) + H 2(g) • Given: • N 2 H 4(l) + CH 4 O(l) CH 2 O(g) + N 2(g) + 3 H 2(g) ΔH = -92. 5 KJ N 2(g) + 3 H 2(g) 2 NH 3(g) ΔH = -115 KJ CH 2 O(g) + H 2(g) CH 4 O(l) ΔH = 162. 5 KJ

Standard Enthalpies of Formation H° f • The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. • The standard enthalpy of formation of a pure element in its reference state is 0.

Standard Enthalpies of Formation

Standard Enthalpies of Formation

Writing Formation Equations • Remember that f. Ho refers to forming ONE mole of a pure substance FROM ITS ELEMENTS in their standard state • Write thermochemical formation equations for the formation of: – Al 2 O 3(s) – Na. HCO 3(s)

Standard Enthalpies of Reaction H°rxn H°overall = -2 f. H°Na. HCO 3+ Hf°Na 2 CO 3+ Hf°CO 2 + Hf°H 2 O

Enthalpy of Reaction H°rxn = S Hf°products- S Hf°reactants

Enthalpies of Formation of Ions in Aqueous Solutions