Thermochemistry

A. Phase Changes

A. Phase Changes Evaporation molecules at the surface gain enough energy to overcome IMF Volatility measure of evaporation rate depends on temp & IMF

A. Phase Changes Equilibrium trapped molecules reach a balance between evaporation & condensation

directly related to volatility A. Phase Changes Vapor Pressure pressure of vapor above a liquid at equilibrium v.p. depends on temp & IMF directly related to volatility temp

A. Phase Changes Boiling Point temp at which v.p. of liquid equals external pressure depends on Patm & IMF Normal B.P. - b.p. at 1 atm

A. Phase Changes Melting Point equal to freezing point Which has a higher m.p.? polar or nonpolar? covalent or ionic? polar ionic

EX: dry ice, mothballs, solid air fresheners A. Phase Changes Sublimation solid  gas v.p. of solid equals external pressure EX: dry ice, mothballs, solid air fresheners

B. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE 

energy required to raise the temp of 1 gram of a substance by 1°C B. Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Heat Capacity energy required to raise the temp of 1 gram of a substance by 1°C

energy required to melt 1 gram of a substance at its m.p. B. Heating Curves Phase Change change in PE (molecular arrangement) temp remains constant Heat of Fusion (Hfus) energy required to melt 1 gram of a substance at its m.p.

EX: sweating, steam burns, the drinking bird B. Heating Curves Heat of Vaporization (Hvap) energy required to boil 1 gram of a substance at its b.p. usually larger than Hfus…why? EX: sweating, steam burns, the drinking bird

C. Phase Diagrams Show the phases of a substance at different temps and pressures.

Thermal Energy A. Temperature & Heat 1. Temperature is related to the average kinetic energy of the particles in a substance.

2. SI unit for temp. is the Kelvin a. K = C + 273 (10C = 283K) b. C = K – 273 (10K = -263C) 3. Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance.

4. Thermal energy relationships a. As temperature increases, so does thermal energy (because the kinetic energy of the particles increased). b. Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy).

Cup gets cooler while hand gets warmer 5. Heat a. The flow of thermal energy from one object to another. b. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler

a. Some things heat up or cool down faster than others. 6. Specific Heat a. Some things heat up or cool down faster than others. Land heats up and cools down faster than water

b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). 1) C water = 4184 J / kg C 2) C sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

Why does water have such a high specific heat? water metal Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

How to calculate changes in thermal energy q = m x c x  T q = change in thermal energy m = mass of substance T = change in temperature (Tf – Ti) c = specific heat of substance Heat loss (released) = heat gained (absorbed) -q = + q

First, mass and temperature of water are measured c. A calorimeter is used to help measure the specific heat of a substance. First, mass and temperature of water are measured Knowing its Q value, its mass, and its T, its Cp can be calculated Then heated sample is put inside and heat flows into water This gives the heat lost by the substance T is measured for water to help get its heat gain

Project Description and Outline DUE FRIDAY! MAY 2nd . HAPPY Monday! PRE-AP Take out your notebooks!! Turn in any late work/check your boxes for returned quiz and returned redox homework. Heat Capacity Homework DUE WEDNESDAY REGULAR Take out your Enthalpy Homework – we are going to go over some of the problems. Enthalpy Hmk DUE TOMORROW. Pre-TEST TOMORROW!! TEST ON THURSDAY REMINDERS Project Description and Outline DUE FRIDAY! MAY 2nd .

STUDY FOR PRE-TEST LeChatelier’s Principle (Disney ride analogy!) Add too much to L, move to right Take away from L, move to left Phase Changes Heating Curve Phase Change Diagram Exothermic/Endothermic Vocab for phase changes (sublimation etc)

3. Basic Temperature Vocabulary Kinetic Energy and Potential Energy Thermal Energy Intermolecular forces

4. q and heat capacity problems 1. Q = amount of heat transferred (neg/pos) 2. C = specific heat 3. Units come from the C value q = mc T

5. Enthalpy Problems 1. H = (Products added) – (Reactants added) 2. Multiply by the coefficents 3. Elements in basic form are 0.

TWO Trends in Nature Order  Disorder   High energy  Low energy 

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g) 6.2

DH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 and is positive 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ 6.4

Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 and is negative 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ 6.4

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ/mol ΔH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 mol x 6.01 kJ/mol = 12.0 kJ 6.4

Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DHreaction = -3013 kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.4

Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f 6.6

6.6

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn = S DH0 (products) f - S DH0 (reactants) f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) 6.6

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn DH0 (products) f = S DH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ -6535 kJ 2 mol = - 3267 kJ/mol C6H6 6.6

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6

Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J

DHsoln = Hsoln - Hcomponents The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.7

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

Energy Diagrams Exothermic Endothermic Activation energy (Ea) for the forward reaction Activation energy (Ea) for the reverse reaction (c) Delta H 50 kJ/mol 300 kJ/mol 150 kJ/mol 100 kJ/mol -100 kJ/mol +200 kJ/mol

Entropy (S) is a measure of the randomness or disorder of a system. If the change from initial to final results in an increase in randomness DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0 18.3

First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 18.4

Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) DS0 rxn S0(products) = S S0(reactants) - What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol 18.4

Entropy Changes in the System (DSsys) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) The total number of gas molecules goes down, DS is negative. 18.4

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0C and ice melts above 0 0C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

For a constant-temperature process: Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium. 18.5

DG = DH - TDS 18.5

DG0 of any element in its stable form is zero. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) DG0 rxn DG0 (products) f = S DG0 (reactants) - Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f DG0 of any element in its stable form is zero. f 18.5

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) What is the standard free-energy change for the following reaction at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DG0 rxn DG0 (products) f = S DG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous 18.5

Recap: Signs of Thermodynamic Values Negative Positive Enthalpy (ΔH) Exothermic Endothermic Entropy (ΔS) Less disorder More disorder Gibbs Free Energy (ΔG) Spontaneous Not spontaneous

Heat (q) absorbed or released: The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.5

Dt = tfinal – tinitial = 50C – 940C = -890C How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = 0.444 J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J 6.5

Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.5

Sample Problem How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 1: Heat the ice Q=mcΔT Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ Step 2: Convert the solid to liquid ΔH fusion Q = 2.0 mol x 6.01 kJ/mol = 12 kJ Step 3: Heat the liquid Q=mcΔT Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

Now, add all the steps together Sample Problem How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C? Step 4: Convert the liquid to gas ΔH vaporization Q = 2.0 mol x 44.01 kJ/mol = 88 kJ Step 5: Heat the gas Q=mcΔT Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ Now, add all the steps together 0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ