Chemical Thermodynamics

Chemical Thermodynamics • Why Changes Take Place • Temperature, Thermal Energy and Heat • Law of Conservation of Energy • Energy Units • Heat Capacity and Specific Heat • Measurement of Thermal Energy Changes • Enthalpy • Hess’s Law • Entropy • Gibb’s Free Energy

Spontaneous Nonspontaneous Why changes take place Spontaneous process Takes place ‘naturally’ with no apparent cause or stimulus. Nonspontaneous process Requires that something be done in order for it to occur.

Spontaneous Processes Thermodynamics is concerned with the question: can a reaction occur? First Law of Thermodynamics: energy is conserved. Any process that occurs without outside intervention is spontaneous. Example: When two eggs are dropped they spontaneously break. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous.

Spontaneous Processes A process that is spontaneous in one direction is not spontaneous in the opposite direction. The direction of a spontaneous process can depend on temperature: Ice turning to water or water turning to ice depends on T< or > 0C.

Spontaneous Processes Reversible and Irreversible Processes A reversible process is one that can go back and forth between states along the same path. ice + 334 J/g  water When 1 mole of water is frozen at 1 atm at 0C to form 1 mole of ice, q = Hfusion of heat is removed. To reverse the process, q = Hfusion must be added to the 1 mole of ice at 0C and 1 atm to form 1 mole of water at 0C. Therefore, converting between 1 mole of ice and 1 mole of water at 0C is a reversible process.

Spontaneous Processes Chemical systems in equilibrium are reversible. In any spontaneous process, the path between reactants and products is irreversible. Exothermic reactions tend to be spontaneous…..systems in nature move towards a lower energy state. Reactions that increase the amount of disorder, or entropy, of the system also tend to be spontaneous. Thermodynamics gives us the direction of a process. It cannot predict the speed at which the process will occur.

When will a reactionbe spontaneous? • Spontaneity of a reaction can be determined by a study of thermodynamics. • Thermodynamics can be used to calculate the amount of useful work that is produced by some chemical reactions. • The two factors that determine spontaneity are enthalpy and entropy.

Energy • Energy - the ability to do work. • Work - when a force is applied to an object. • w = F xd • There are several types of energy: • Thermal - heat • Electrical • Radiant - including light • Chemical • Mechanical - like sound • Nuclear

Energy • Energy can be classified as: • Potential energy • Stored energy - ability to do work. • Kinetic energy • Energy of motion - actually doing work. • Energy can be transferred from one object to another. It can also change form. • Internal energy • the sum of all potential and kinetic energy in a system

Energy units • Earlier, kinetic energy was defined as: • kinetic energy = mv2 • m = mass and v = velocity. • Joule (J) - the energy required to move a 2 kg mass at a speed of 1 m/s. It is a derived SI unit. • J = kinetic energy = (2 kg) (1 m/s)2 • = 1 kg m2 s-2 1 2 1 2

Energy units • The SI unit of energy is the joule (J): • An older, non-SI unit is still in widespread use: the calorie (cal): • 1 cal = 4.184 J • Dietary Calorie • This is what you see listed on food products. • It is actually a kilocalorie.

Relating Heat and Work • A specific reaction under study is called “the system” and everything else is considered “the surroundings”. • For the system, ∆E = q + w For gases, the pressure-volume work,w = - P∆V

Practice Exercises • Exercise 1 Internal Energy • Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. • 17.0 kJ • Exercise 2 PV Work • Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. ‒270 L•atm

Practice Exercises (continued) • Exercise 3 Internal Energy, Heat, and Work • A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 × 106L to 4.50 × 106 L by the addition of 1.3 × 108 J of energy as heat. Assuming the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = 101.3 J.) 8.0 × 107 J

Heat capacity • Every material will contain thermal energy. • Identical masses of substances may contain different amounts of thermal energy even if at the same temperature. • Heat capacity. The quantity of thermal energy required to raise the temperature of an object by one degree. (Joules/ °C) • Specific heat, Cp. The amount of thermal energy required to raise the temperature of one gram of a substance by one degree. (J/g.oC) • q mDT Cp =

Substance Cp Al(s) 0.90 Br2 (l) 0.47 C (diamond) 0.51 C (graphite) 0.71 CH2CH2OH (l) 2.42 CH3(CH2)6CH3 (l) 2.23 Substance Cp Fe (s) 0.45 H2O (s) 2.09 H2O (l) 4.18 H2O (g) 1.86 N2 (g) 1.04 O2 (g) 0.92 Specific Heats at 25oC, 1 atm Cp = specific heat, J g-1oC-1

Heat capacity • Example. • How many joules must be added to a 50.0 g block of aluminum to heat it from 22oC to 85oC? • Heat required = mass x specific heat x DT • = 50.0 g x 0.90 J g-1oC-1 x (85-22)oC • = +2.8 kJ • This is an endothermic change; + sign.

Heat capacity • Molar heat capacity -- same as Cp but specific to one mole of substance (J/mol K or J/mol °C ) • Energy (q) released or gained at constant pressure: q = mCpΔT • q = quantity of heat (Joules or calories) • m = mass in grams • ΔT = Tf - Ti (final – initial) • Cp = specific heat capacity ( J/g°C) • Specific heat of water (liquid state) = 4.184 J/g°C (1.00 cal/g°C)

ENTHALPY ENTHALPY,ΔH Measure only the change in enthalpy, (the difference between the potential energies of the products and the reactants) ΔH is a state function ΔH = q at constant pressure (i.e. atmospheric pressure); (true most of the time for us and a very handy fact!)

ENTHALPY • Enthalpy can be calculated from several sources including: • Stoichiometry • Calorimetry • From tables of standard values • Hess’s Law • Bond energies

ENTHALPY • Stoichiometrically: • Sample Problem A: Upon adding solid potassium hydroxide pellets to water the following reaction • takes place: • KOH(s) → KOH(aq) + 43 kJ/mol • Answer the following questions regarding the addition of 14.0 g of KOH to water: • Does the beaker get warmer or colder? • Is the reaction endothermic or exothermic? • What is the enthalpy change for the dissolution of the 14.0 grams of KOH? • Answers: (a) warmer (b) exothermic (c) −10.7 kJ/mol

Measuring ENTHALPY changes with Calorimetry • Thermal energy cannot be directly measured. • We can only measure differences in energy. • To be able to observe energy changes, we must be able to isolate our system from the rest of the universe. • Calorimeter - a device that is used to measure thermal energy changes and provide isolation of our system. • Heat lost by substance =heat gained by water • (if this does not happen, calculate the heat capacity of the substance)

V Constant Calorimetry • A bomb calorimeter, for determining heat with fuel and food combustion, has • DH = -q • Q rxn = - (q bomb + q H2O) • Usually heat capacity for the calorimeter is given in J/oC (Heat is Evolved!)

Calorimetry • Q evolved at a Constant volume is not usually equal to Q evolved at a constant Pressure! (An open one, like a coffee cup!)

Conduct a reaction and look at the temperature change. Coffee cup calorimeter

Temperature, energy and heat • Temperature. An intensive property of a material. Kelvin temperature is proportional to KE (1/2mv2). • Thermal energy. Energy of motion of molecules, atoms or ions. All materials have this energy if at a temperature above 0 K. • Heat. Thermal energy transfer that results from a difference in temperature. Thermal energy flows from warm objects to cool ones.

Law of conservation of energy • “Energy cannot be created or destroyed in a chemical reaction.” • During a reaction, energy can change from one form to another. • Example. Combustion of natural gas. • Chemical bonds can be viewed as potential energy. So during the reaction: • 2CH4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l) + thermal energy + light • some potential energy is converted to thermal energy and light.

Calorimetry example • You are given the two solutions listed below. Each has an initial temperature of 20.0 oC. • 50 ml of 0.50 M NaOH • 50 ml of 0.50 M HCl • Both are rapidly added to a coffee cup calorimeter and stirred. The reaction takes place rapidly. The highest temperature is 23.3 oC. Solution density is 1.0 g/ml. Heat capacity of calorimeter = 75 J/oC • Determine the heat of reaction if the specific heat of the solution is 4.18 J/g•oC

Heat lost by substance =heat gained by water and calorimeter Calorimetry example • First, determine the energy given off. • = 100.0 g (4.18 J/g•oC) (23.3 - 20.0) oC • + (75 J/oC) (23.3 - 20.0) oC • = 1379 J + 248 J = 1627 J • = - 1.63 x103 J (use ‘-’ because heat is given off) • Next, determine the moles of HCl or NaOH involved in the reaction -- both are the same. • molHCl = (0.5 ) ( 0.05 L) = 0.025 mol HCl mol L

Calorimetry example • The heat of neutralization for the reaction: • HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l) is = -1.63 x103 J / 0.025 mol = 6.52 x 104 J/mol = 65 kJ/mol (final answer rounded to 2 SF)

Enthalpy • The energy gained or lost when a change takes place under constant pressure. • DH = Hfinal - Hinitial • Subscripts are used to show the type of change. • DHvap heat of vaporization • DHneut heat of neutralization • DHfusion heat of fusion • DHsol heat of solution • DHrxn heat of reaction

Energy and chemical bonds • During a chemical reaction • Old bonds break. • New bonds are formed. • Energy is either absorbed or released. • Exothermic Energy is released. • New bonds are more stable. • EndothermicEnergy is required. • New bonds are less stable.

Draw diagrams! Exothermic Reactants Energy Products Since excess energy is released, the products are more stable.

Endothermic Products Energy Reactants Additional energy is required because the products are less stable.

Practice Exercises • Sample Problem B: • In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOHand 100.0 mL of 1.0 M HCl are mixed. Both • solutions were originally at 24.6°C. After the Reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter. • ‒5.6 kJ/mol

Practice Exercises • Exercise 4 Enthalpy • When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8-g sample of methane is burned at constant pressure. • ΔH = heat flow = ‒320 kJ/mol

Practice Exercises • Exercise 5 Constant-Pressure Calorimetry • When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat • capacity of the solution is 4.18 J/°C⋅g, and that the density of the final solution is 1.0 g/mL, calculate • the enthalpy change per mole of BaSO4 formed. • ‒26 kJ/mol

Practice Exercises • Exercise 6 Constant-Volume Calorimetry • It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principally methane). To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/°C. When a 1.50g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C. When a 1.15-g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Calculate the energy of combustion (per gram) for hydrogen and methane. • Methane = 55 kJ/g • Hydrogen = 141 kJ/g

At 1 atm and 0ºC, an equilibrium formsH2O(s) H2O (l) • Above 0ºC • Energy+ H2O(s)H2O (l) • Below 0ºC • H2O(l)H2O(s) +Energy

Endothermicchanges of state • Sublimation • The direct conversion of a solid to a gas. • Example - dry ice (solid CO2) • Melting or fusion • The conversion of a solid to a liquid. • Example - melting of ice • Evaporation or vaporization • Converting a liquid to a gas. • Example - boiling water • Most materials first melt then vaporize as you raise the temperature.

EndothermicChanges of state Gas evaporation or vaporization sublimation Solid Liquid melting

Exothermicchanges of state • Condensation or liquifaction • The conversion of a gas to a liquid or solid. • Example - steam becoming water • Freezing or crystallization • When a liquid becomes a solid. • Examples - formation of ice from water • Substances usually first condense to liquids and then become solids.

Exothermicchanges of state Gas liquification or condensation deposition Solid Liquid freezing or crystallization

Chemical MW Polarity Mp Bp N2 28 Nonpolar -210 -196 Cl2 71 Nonpolar -102 -34 NH3 17 Polar -78 -33 H2O 18 Polar 0 100 NaCl 58 Ionic 801 1465 Changes in state andattractive forces • As the attractive forces between molecules become larger, more energy is needed to separate them. • Vapor pressures become smaller, boiling points and melting points become larger.

Heating/Cooling Curve Temperature 5 gas liquid & gas T2 4 …………………………….….……... 3 solid & liquid liquid ………..... T1 2 ……….…....... solid 1 ………. Heating Time

Heating/Cooling Curve • Position 1 = Solid phase • Position 2 = Solid/Liquid phase • Position 3 = Liquid phase • Position 4 = Liquid/Vapor phase • Position 5 = Vapor phase • (Plasma is above the gas phase!)

Heating and Cooling • Heat of fusion, DHfus • The amount of thermal energy necessary to melt one mole (or 1 gram) of a substance at its melting point. • Heat of vaporization, DHvap • The amount of thermal energy necessary to boil one mole (or 1 gram) of a substance at its boiling point.

Heating and Cooling • mp DHfusbpDHvap • Substance oC kJ/mol oC kJ/mol • Br2 -7.3 10.57 59.2 29.5 • CH3CH2OH -117.0 4.60 79.0 43.5 • CH3(CH2)6CH3 -56.8 20.65 125.7 38.6 • H2O 0.0 6.01 100.0 40.7 • Na 97.8 2.60 883 98.0

Substance J/g•oC Substance J/g•oC Aluminum, solid 0.90 Ice 2.06 Copper, solid 0.38 Water 4.18 Hydrogen, gas 14.3 Steam 2.02 Mercury, liquid 0.14 Specific heats • Each substance requires a different amount of energy to increase its temperature. • Specific heat - amount of energy needed to increase a substance’s temperature by 1oC. • It also depends on the state of the substance.

Heating and Cooling • Changes in state involve several steps. • (Could have a maximum of 5 Steps !) • Example. How much energy is needed to heat 250 g of ice at a temperature of -20oC to 150oC? • 1. Heat ice up to the melting point of 0 oC.q = m∆TCpice= (250g)(20ºC)(2.06J/gºC) • = 10300 J • 2. Convert the ice to water by melting.q = m∆Hfusion= (250g)(334 J/g) • = 83500 J

Chemical Thermodynamics