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.

Heat Potential energy

Heat Potential energy

Units of Heat Calorie (cal) Joule (J) 1 cal = 4.184 J 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 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 DE <0

Surroundings System Energy DE >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, DH = Hfinal - Hinitial Changes in heat (q) can be used to find DH

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

q and DH q is the “heat flow for a system” It can be for any amount of mass or moles of substance DH 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 q = (m)(C)(T) Specific heat capacity System is one gram of substance Units are J/goC q = heat change m = mass C = Specific Heat Capacity DT = final temperature – initial temperature q = (m)(C)(T)

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

Heat Capacity Types q = CT Heat capacity of an object Specific for a given amount or unit so it is not mass dependent Units are J/oC q = heat C = heat capacity DT = 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 qH2O = CH2O x mH2O x DT

qsystem + qsurroundings = 0 qH2O + qrxn = 0 qH2O = - 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 qlead = -qwater Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. qlead = -qwater qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C qwater = 1.4x103 J qlead = -1.4x103 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 d. 123.9 ˚C b. 100 ˚C e. 719 ˚C c. 46 ˚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. 24.9 g c. 48.2 g b. 302 g d.419 g

States of Matter

General Heating Curve http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_ChangesState.MOV

Changes of State of Water qphase change = DHphase change (mass or moles) Molar enthalpy of vaporization: H2O (l) → H2O(g) Hv = 40.65 kJ/mol @ 373.15 K Hv = (40.65 kJ/mol)(1 mol/18.02 g) = 2256 J/g Molar enthalpy of fusion (melting): H2O (s) → H2O(l) Hf = 6.01 kJ/mol @273.15 K Hf = (6.01 kJ/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. q1+ q2 + q3 …. = qtotal Note: Cice= Csteam = 2.09 J/goC q = (m)(C)(T) qphase change = DHphase 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. 2.47 J c. 39.4 J b. 48.0 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 qcal = qbomb + qwater + qwires +… Define the heat capacity of the calorimeter: qcal = miciT = CcalT heat System includes everything inside the double walled container. All the water, wires, stirrer, thermometer, and the reaction chamber. Bomb is filled with sample and assembled. Then pressurized witih O2. A short pulse of electric current heats the sample and ignites it. The final temperature of the calorimeter assembly is determined after the combustion reaction. Because the reaction is carried out at a fixed volume we say that this is at constant volume.

q and DH 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 DH

q to DH 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, DH Enthalpy term can be embedded in the reaction or written after the reaction with DH 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 kJ/°C. What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11 Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

Endothermic Thermochemical Equations Al2O3(s)  2Al(s) + 3/2 O2(g) DHrxn = 1,676kJ/mol rxn THE EXACT SAME AS WRITING: Al2O3(s) + 1,676 kJ  2Al(s) + 3/2 O2(g)

Exothermic Thermochemical Equations H2(g) + ½ O2(g)  H2O(l) + 285.8kJ Same as: H2(g) + ½ O2(g)  H2O(l) DHrxn = -285.8 kJ/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 DHrxn = 1,676 kJ/mol rxn Al2O3(s)  2Al(s) + 3/2 O2(g) DHrxn = 1,676 kJ/mol rxn How many grams of aluminum would be produced if the aluminum absorbed 3500 kJ of energy?

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

Some Important Types of Enthalpy Changes Heat of combustion (DHcomb) C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g) Heat of formation (DHf) K(s) + 1/2Br2(l) KBr(s) Heat of fusion (DHfus) NaCl(s) NaCl(l) Heat of vaporization (DHvap) C6H6(l) C6H6(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. Temperature must be specified because H varies with temperature.

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. N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ H changes sign when a process is reversed NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ

Hess’s law of constant heat summation 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. ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ ½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

Example Find ΔH for 2H2(g) + O2(g)  2H2O(l) Given: CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l) ΔH = -1,959.8 kJ C(graphite) + O2(g)  CO2(g) ΔH = -886.5 kJ CH3COOH(l)  2C(graphite) + 2H2(g) + O2(g) ΔH = 1,100.2 kJ

Example 11 Find DH for 2NH3(g) N2H4 (l) + H2(g) Given: N2H4(l) + CH4O(l)  CH2O(g) + N2(g) + 3H2(g) ΔH = -92.5KJ N2(g) + 3H2(g)  2NH3(g) ΔH = -115KJ CH2O(g) + H2(g)  CH4O(l) ΔH = 162.5KJ

Standard Enthalpies of Formation fH° 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. Absolute enthalpy cannot be determined. H is a state function so changes in enthalpy, H, have unique values. Reference forms of the elements in their standard states are the most stable form of the element at one bar and the given temperature. The superscript degree symbol denotes that the enthalpy change is a standard enthalpy change and The subscript “f” signifies that the reaction is one in which a substance is formed from its elements.

Standard Enthalpies of Formation

Standard Enthalpies of Formation

Writing Formation Equations Remember that DfHo refers to forming ONE mole of a pure substance FROM ITS ELEMENTS in their standard state Write thermochemical formation equations for the formation of: Al2O3(s) NaHCO3(s)

Standard Enthalpies of Reaction H°rxn H°overall = -2fH°NaHCO3+ Hf°Na2CO3+ Hf°CO2 + Hf°H2O

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

Enthalpies of Formation of Ions in Aqueous Solutions