Thermodynamics: Enthalpy & Entropy

Thermodynamics thermo = heat (energy) dynamics = movement, motion

Some thermodynamic terms chemists use: System: the portion of the universe that we are considering open system: energy & matter can transfer closed system: energy transfers only isolated system: no transfers

Surroundings: everything else besides the system Isothermal: a system that is kept at a constant temperature by adding or subtracting heat from the surroundings.

Heat Capacity: the amount of heat energy required to raise the temperature of a substance by 1°C (or 1 K). (little c in equations)

Heat Capacity: Specific Heat Capacity: 1 g by 1°C Capital “C” Molar Heat Capacity: 1 mole by 1°C “Cmol”

c C Cmolar Symbol Name formula units heat capacity q =c*m*∆T J/g*K specific heat capacity q =C*m*∆T Cmolar Molar heat capacity amount of heat to raise 1 mole of a substance 1K q =Cmol*m*∆T J/mol*Ks

Calorie: the amount of heat required to raise the temperature of 1g of water by 1°C. specific heat of water = 1 cal/g °C 1 calorie = 4.18 joules

Substance Specific Heat (J/°C•g) Molar Heat (J/°C•mol) Al 0.90 24.3 Cu 0.38 24.4 Fe 0.45 25.1 CaCO3 0.84 83.8 Ethanol 2.43 112.0 Water 4.18 75.3 Air 1.00 ~ 29 important to: Engineers Chemist

Heat Energy Heat energy = (specific heat)(mass)(ΔT) q =(C)(m)(Tf-Ti) Note that ΔT can be ºC or K, but NOT ºF. When just T is being used in a scientific formula it will usually be kelvin (K).)

EXAMPLE: How many joules of energy are needed to raise the temperature of an iron nail (7.0 g) from 25°C to 125°C? The specific heat of iron is 0.45 J/°C•g. Heat energy = (0.45 J/°C•g)(7.0 g)(125-25°C) = 315 J

Problem: How much energy does it take to raise the body temperature 2.5ºC (a fever of just over 103ºF) for someone who weighs 110 pounds (50. kg). Assume an average body specific heat capacity of 3.0 J/ºC•g. 380,000 J

-30oC to OoC – kinetic energy increases 0 o C to 0oC – phase change – potential energy change – solid to liquid 0 o C to 100oC – kinetic energy change 100 o C to 100oC – phase change – potential energy change – liquid to gas 100 oC to 125oC – kinetic energy change

Phase Change Calculations How much energy and time is needed to take 40.0g of ice from -30oC to 125oC Given Cice = The specific heat capacity of ice = 0.50 cal/g oC Lf = The Latent heat of fusion of ice or energy needed to melt = 80 cal/g Cwater = The specific heat capacity of water = 1.0 cal/g oC Lv = The latent heat of vaporization of water = 540 cal/g Csteam = The specific heat capacity of steam = 0.48 cal/g oC Rate of heating = 12 cal/ s

Phase Change Calculations How much energy and time is needed to take 40.0g of ice from -30oC to 125 Q=mcT= 40.0g (0.50 cal/g oC) ( 0oC – (-30.0 oC)) = 600 cal Time = 600 cal | 12 cal = 50s 1 s Q= mLf = 40.0g (80 cal )= 3200 cal g Time = 3200 cal | 1 s = 267 s 12 cal

Phase Change Calculations How much energy and time is needed to take 40.0g of ice from -30oC to 125oC Q=m  Ttc= 40.0g (1.0 cal/g oC) ( 100oC – 0 oC) = 4000 cal Time = 4000 cal / 12 cal / s = 333s Q= mLv = 40.0g (540 cal /g ) = 21,600 cal Time = 21,600 cal / 12 cal / s = 1800s Q=m c T= 40.0g (0.48 cal/g oC) ( 125oC – 100 oC) = 480 cal Time = 480 cal / 12 cal / s = 40s

Phase Change Calculations How much energy and time is needed to take 40.0g of ice from -30oC to 125oC Qprocess = 600cal + 3200 cal + 4000 cal + 480 cal +21,600 cal = 29880cal S process = 50 sec + 267 sec + 333 s + 1800s + 40 s = 2490 s It takes 29880 cal and will heat in 2490 seconds

A calorimeter A calorimeter is used to measure the transfer of heat during a physical or chemical change.  The formula for heat transfer calculations is: q = m(ΔT)Cp Amount of heat transferred = mass x change in temperature x specific heat Where q = heat transferred, ΔT = the change in temperature and Cp = the specific heat.  Know the C H2O = 4.18J/(g*K)  

Heat loss by metal = heat gained by water - q = q - mCpT = mCpT If a mass of cadmium of 65.6g and a temperature of 100oC is dropped into 25.0 cm3 of water at a temperature of 23.0oC what will the final temperature be of the water? Cp = 0.2311 J/oC*g Cadmium Heat loss by metal = heat gained by water - q = q - mCpT = mCpT Tf= 33.07oC

Phileas Fogg, the character who went around the world in 80 days, was very fussy about his bathwater temperature. It had to be exactly 38.0o C. You are his butler, and one morning while checking his bath temperature, you notice that it’s 42.0oC. You plan to cool the 113.2 kg of water to the desired temperature by adding an aluminum -duckie originally at freezer temperature (-24.0oC). Of what mass should the Al-duckie be? [Specific heat of Al = 0.900 J/(goC); density of water =1 .00 g/ml]. Assume that no heat is lost to the air. 33.89 Kg for duck (Don’t forget) –q=q

Specific Heat Capacity of Copper Calorimeters are designed to be well-insulated, so no heat is gained from or lost to the surroundings. If no heating element is used to introduce heat in the system, the total heat transferred (q) for the entire calorimeter system must equal zero. The total heat can be split into heats for each component in the system. Imagine an experiment in which a hot copper ball is dropped into a calorimeter containing water at room temperature. The copper ball will lose heat, which will be absorbed by the calorimeter and water. Because no heat enters or leaves the system the heat balance for this experiment is 0 = q = qCu + qcal + qw

Specific Heat Capacity of Copper In this case qCu < 0, because the copper ball will lose heat to the calorimeter and water. Similarly qcal > 0 and qw > 0, because both the calorimeter and the water will gain heat. In chemistry the thermodynamic sign convention is things entering the system a positive in value, and things leaving the system are negative in value.

Specific Heat Capacity of Copper The basic strategy in calorimetry is to use a temperature change and a heat capacity to determine a heat flow. In this experiment, all substances have the same final temperature (Tf), but not all substances have the same initial temperature. The copper ball is initially at temperature TCu while the calorimeter and water are initially at temperature Ti. qCu = mCu sCu ( Tf - TCu ) qcal = Ccal ( Tf - Ti ) qw = mw sw ( Tf - Ti )

Specific Heat Capacity of Copper The heat capacity of the calorimeter must be obtained from a separate calibration experiment (for example, a heating element can be used to introduce a known amount of heat). The specific heat capacity of water is known (4.184 J oC-1), and the temperatures TCu, Ti, and Tf can be measured experimentally. The masses of the copper and water (mCu and mCu) can also be measured experimentally. The only unknown property in the above equations is the specific heat capacity of the copper. sCu = - (Ccal + mw sw) (Tf - Ti) mCu (Tf - TCu)

The temperature of a piece of copper with a mass of 95 The temperature of a piece of copper with a mass of 95.4g increases from 25oC to 48.0oC when the metal absorbs 849J of heat. What is the specific heat of copper? What do we know? C = q m x  T mcu =  T = q= Don’t know: Ccu = ? J . g x oC To solve Ccu = 95.4 g 48.0- 25 oC 849 J 849 J = 0.387 J/g·oC 95.4 g * 23oC

Energy: "The capacity to do work and/or transfer heat”

Forms of Energy: Kinetic (Ekinetic = ½mv2) Potential Heat Light (Electromagnetic) Electricity Chemical Nuclear Matter (E = mc2)

Matter (E = mc2) WORK

First Law of Thermodynamics The total amount of energy (and mass) in the universe is constant. In any process energy can be changed from one form to another; but it can never be created nor destroyed.

"You can't get something for nothing"

Enthalpy (Heats) of Reaction The amount of heat released or absorbed by a chemical reaction at constant pressure (as one would do in a laboratory) is called the enthalpy or heat of reaction. We use the symbol H to indicate enthalpy.

Sign notation (EXTREMELY IMPORTANT!!): +ΔH indicates that heat is being absorbed in the reaction (it gets cold) endothermic −ΔH indicates that heat is being given off in the reaction (it gets hot) exothermic

Pressure 1 atm (760 torr) Concentration 1.0 M Standard Enthalpy = ΔH° (° is called a “not”) Occurring under Standard Conditions: Pressure 1 atm (760 torr) Concentration 1.0 M

Temperature is not defined or part of Standard Conditions, but is often measured at 298 K (25°C).

Standard Enthalpy of Formation - ΔHf°

Standard Enthalpy of Formation The amount of heat absorbed (endothermic) or released (exothermic) in a reaction in which one mole of a substance is formed from its elements in their standard states, usually at 298 K (25°C). Also called heat of formation.

Also called heat of formation. ΔHf° = 0 for any element in its standard state (the natural elemental form at 1 atm or 1 M) at 298 K.

C (graphite,s) + O2 (g) CO2 (g) ΔHrxn °= 0 kJ/mol + 0 kJ/mol - 393 C (graphite,s) + O2 (g) CO2 (g) ΔHrxn °= 0 kJ/mol + 0 kJ/mol - 393.5 kJ/mol elements in their product standard state (one mole) negative sign heat released - exothermic rxn ΔHf ° (CO2 ) = - 393.5 kJ/mol

2H2(g) + O2 (g)  2H2O (g) ΔHrxn °= 0 kJ/mol + 0 kJ/mol - 483 2H2(g) + O2 (g)  2H2O (g) ΔHrxn °= 0 kJ/mol + 0 kJ/mol - 483.6 kJ/2mol elements in their product standard state (TWO mole) negative sign heat released - exothermic rxn Divide by 2 to get to per mole level ΔHf ° (H2O) = - 241.8 kJ/mol

Hess's Law -- Adding Reactions The overall heat of reaction (ΔHrxn) is equal to the sum of the ΔHf (products) minus the sum of the ΔHf (reactants): ∆Horxn = ∑ΔHof (products) -Σ ΔHof (reactants)

So, by knowing ΔHf of the reactants and products, we can determine the ΔHrxn for any reaction that involves these reactants and products.

2Mg(s) + CO2 (g)  2MgO(s) + C(s) CO2 is used in certain kinds of fire extinguishers to put out simple fires. It works by smothering the fire with "denser" CO2 that replaces oxygen needed to maintain a fire. CO2 is not good, however, for more exotic electrical and chemical fires. 2Mg(s) + CO2 (g)  2MgO(s) + C(s) ΔH°f = 0 kJ/mol -393 kJ/mol -602 kJ/mol + 0 kJ/mol    ΔH°rxn = Σ (#mol)ΔH°f (products) - (#mol)ΔH°f(reactants) ΔH°rxn =Σ(2)(-602 kJ/mol)+(1)(0 kJ/mol) - Σ(2)(0kJ/mol) + (1)(-393 kJ/mol)

ΔH°rxn = (-1204 kJ/mol) - (-393 kJ/mol) ΔH°rxn = -811 kJ/mol } highly exothermic rxn !! So, Mg will "burn" CO2 !

Calculate ΔHrxn for the following reaction: You can also add two reactions together to get the ΔHrxn for another new reaction: Calculate ΔHrxn for the following reaction: C2H4 (g) + H2O(l)  C2H5OH(l) ΔHrxn = ??  Given these two reactions and thermodynamic data: 2CO2 (g) + 3 H2O(l)  C2H5OH(l) + 3O2(g) ΔH°rxn = -1367 kJ/mol 2CO2 (g) + 3 H2O(l)  C2H5OH(l) + 3O2(g) ΔH°rxn = +1367 kJ/mol C2H5OH(l) + 3O2(g)  2CO2 (g) + 3 H2O(l) ΔH°rxn = -1367 kJ/mol C2H4(g) + 3O2(g)  2CO2 (g) + 2 H2O(l) ΔH°rxn = -1411 kJ/mo C2H5OH is on the product side of the first reaction -- so we want to switch equation a) around to get C2H5OH on the product side: Note that when we reverse the reaction, ΔH°rxn changes sign!!!

C2H4 (g) + H2O(l)  C2H5OH(l) ΔHrxn = ?? Now we can add the two reactions together to give us the desired net reaction: C2H4 (g) + H2O(l)  C2H5OH(l) ΔHrxn = ?? 1 2CO2 (g) + 3 H2O(l)  C2H5OH(l) + 3O2(g) ΔH°rxn = +1367 kJ/mol C2H4(g) + 3O2(g)  2CO2 (g) + 2 H2O(l) ΔH°rxn = -1411 kJ/mol C2H4(g) + H2O(l) C2H5OH(l) ΔH°rxn = +1367 kJ/mol + -1411 kJ/mol ΔH°rxn = - 44 kJ/mol 46

Calculate the standard enthalpy change, ΔHo, for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements. Sr (s) + C (graphite) + 3/2 O2 (g)  SrCO3 (s) The information available is: Sr (s) + 1/2 O2 (g)  SrO (s) ∆Ho = -592 kJ SrO (s) + CO2 (g)  SrCO3 (s) ∆Ho = -234 kJ C (graphite) + O2  CO2 (g) ∆Ho = -394 kJ ∆Ho = -1220 kJ

2C(s) graphite + 2H2O (g)  CH4 (g) + CO2 (g) The combination of cake and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. The equation for coal gas is; 2C(s) graphite + 2H2O (g)  CH4 (g) + CO2 (g) Determine the standard enthalpy change for this reaction from the following information C(s) graphite + 2H2O (g)  2H2 (g) + CO2 (g) ∆Ho = 131.3 kJ CO (g) + H2O (g)  H2 (g) + CO2 (g) ∆Ho = -41.2 kJ CH4 (g) + H2O (g)  3H2 (g) + CO (g) ∆Ho = 206.1 kJ ∆Ho = 15.3 kJ

If we have to multiply one (or more) of the reactions by some constant to get them to add correctly, then we also would have to multiply ΔHrxn for that reaction by the same amount.

Problem: Calculate ΔHrxn for the following reactions given the following ΔHf ° values: ΔHf ° (SO2, g) = −297 kJ/mol ΔHf ° (SO3, g) = −396 kJ/mol ΔHf ° (H2SO4, l) = −814 kJ/mol ΔHf ° (H2SO4, aq) = −908 kJ/mol ΔHf ° (H2O, l) = −286 kJ/mol ΔHf ° (H2S, g) = −20 kJ/mol S(s) + O2(g)  SO2(g) 2SO2(g) + O2(g) 2SO3(g) SO3 (g) + H2O(l) H2SO4(l) 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)

Calculate Enthalpy of Reaction from Bond Energies Because of the principle of conservation of energy the total energy before and after the reaction must not change. So, the energy of a reaction released or absorbed in the reaction must come from the difference in bond energies of the products and the reactants.

Calculate Enthalpy of Reaction from Bond Energies Example The bond energy (kJ) for H2, F2, and HF are 436, 158 and 568 kJ (assume one mole) respectively. Calculate the enthalpy (energy) of the reaction, H2(g) + F2(g)  2 HF

H2(g) + F2(g)  2 HF Solution Based on the bond energies given, we have H2  2H                 D = 436 kJ/mol F2  2F                  D = 158 kJ/mol 2H + 2F  2HF     H = -568 kJ/mol * 2 = -1136 Adding all three equations and energies H2(g) + F2(g) = 2 HF     ΔH = -542 kJ/equation Note that D represent bond dissociation energy, and H the enthalpy of the reaction as written. We use ΔH in the last equation to denote enthalpy of change of the overall reaction.