Thermodynamics. Heat Capacity Calorimetry Enthalpy Thermodynamic cycles Adiabatic processes. NC State University

Thermodynamics Heat Capacity Calorimetry Enthalpy Thermodynamic cycles Adiabatic processes NC State University

Motivation The enthalpy change DH is the change in energy at constant pressure. When a change takes place in a system that is open to the atmosphere, the volume of the system changes, but the pressure remains constant. In any chemical reactions that involve the creation or consumption of molecules in the vapor or gas phase there is a work term associated with the creation or consumption of the gas.

Molar Enthalpy Enthalpy can be expressed as a molar quantity: H m = H n We can also express the relationship between enthalpy and internal energy in terms of molar quantities: H m = U m + PV m For an ideal or perfect gas this becomes: H m = U m + RT Usually when we write DH for a chemical or physical change we refer to a molar quantity for which the units are kj/mol.

Enthalpy for reactions involving gases If equivalents of gas are produced or consumed in a chemical reaction, the result is a change in pressure-volume work. This is reflected in the enthalpy as follows. DH = DU + PDV which can be rewritten for an ideal gas: DH = DU + DnRT at const. T and P at const. T and P The number of moles n is the number of moles created or absorbed during the chemical reaction. For example, CH 2 =CH 2 (g) + H 2 (g) CH 3 CH 3 (g) Dn = -1 We arrive at this value from the formula Dn = n products - n reactants = 1-2 = -1

The measurement of heat We must carefully distinguish between heat and temperature. When we add heat to the system its temperature increases. We can use measurement of the temperature to determine how much heat has been added. However, we need to know the heat capacity of the system in order to do this. Heat capacity = Heat supplied Temperature rise The heat capacity is called C. If we perform a heat exchange at constant volume then we designate the heat capacity as C V. If the process occurs at constant pressure we call the heat capacity C P. C V,P = q V,P DT

Molar and Specific Heat Capacities We use molar heat capacities for pure substances. As the name implies the units are J/mol-K for the molar heat capacity. We write the molar heat capacity at constant volume as C V,m. For mixtures we cannot use a molar heat capacity and so we use the specific heat capacity, which is the heat capacity per gram of material with units of J/g-K.

Heat Capacity for a Diatomic Molecule For a diatomic molecule there is contribution from rotations as well as translations. This means that as heat is added to the system the rotational levels can be populated in addition to an increase in molecular speed. The kinetic theory of gases considers only the speed. An approximate rule is that we obtain a contribution to the heat capacity, C V of 1/2nR for each degree of freedom. We saw that for a monatomic gas the heat capacity was C V = 3/2nR. A diatomic gas has two rotational degrees of freedom and so the heat capacity is approximately C V = 5/2nR. What does this say about C P? Well, the relationship between C P and C V holds for all gases so C P = 7/2nR for a diatomic ideal gas.

The temperature dependence of the enthalpy change Based on the discussion the heat capacity from the last lecture we can write the temperature dependence of the enthalpy change as: DH = C P DT Note that we can use tabulated values of enthalpy at 298 K and calculate the value of the enthalpy at any temperature of interest. We will see how to use this when we consider the enthalpy change of chemical reactions (the standard enthalpy change). The basic physics of all temperature dependence is contained in the above equation or more frequently in the equation below as molar quantity: DH m = C P,m DT

Another view of the heat capacity At this point it is worth noting that the expressions for the heat capacity at constant volume and constant pressure can be related to the temperature dependence of U and H, respectively. DH = C P DT DU = C V DT C P = DH DT = H T P C V = DU DT = U T V The heat capacity is the rate of change of the energy with temperature. The partial derivative is formal way of saying this.

Variation in Reaction Enthalpy with Temperature Since standard enthalpies are tabulated at 298 K we need to determine the value of the entropy at the temperature of the reaction using heat capacity data. Although we have seen this procedure in the general case the calculation for chemical reactions is easier if you start by calculating the heat capacity difference between reactants and products: D r C P = C P (products) C P (reactants) and then substitute this into the expression: D r H (T 2 ) = D r H (T 1 ) + D r C P dt If the heat capacities are all constant of the temperature range then: D r H (T 2 ) = D r H (T 1 ) + D r C P DT T 2 T 1

Calorimetry The science of heat measurement is called calorimetry. A calorimeter consists of a container in a heat bath. A physical or chemical process occurs in the container and heat is added or removed from the heat bath. The temperature increases or decreases as result. By knowing the heat capacity of the bath we can measure the amount of heat that has been added or removed from the system. Energy in the form of heat flows into the bath.

Calorimetry In the studies of biological systems there are two important types of calorimetry. 1. Differential scanning calorimetry (DSC) 2. Isothermal titration calorimetry (ITC) In DSC the temperature is increased at a constant heating rate and the heat capacity is measured. DSC is used for determining the parameters associated with phase transitions e.g. protein unfolding, denaturation, DNA hybridization etc. In ITC the temperature is held constant while one component is added to another. The heat of interaction (e.g. binding) is measured using this method. ITC is widely used to determine the enthalpy of binding, e.g. for protein-protein and protein-drug interactions among other types of biological applications.

A note on using current flow in calorimetry measurements Energy is measured in Joules. Electrical energy is used to delivery heat in calorimetry applications. In electrostatics the units of energy are: Joules = Coulombs * Volts J = CV Coulomb is a unit of charge and volt is a unit of potential. A charge moving through a potential is a little like a waterfall. The problem here is that we need a dynamic description since moving charge is not static. C V O

A note on using current flow in calorimetry measurements As charge moves through the potential energy (heat) is Released. Power = Amperes * Volts W = IV Power has units of energy per unit time. So as the current flows through a wire at a certain rate, heat energy is added to the system at that rate. C V O

A note on using current flow in calorimetry measurements Over a period of time, t, the total energy added to the system is: Energy = Amperes * Volts * time E = IVt This equation is used in the text to describe a number of calorimetry applications. C V O

The standard state The standard state of a substance is its pure form a 1 bar of pressure or 1 molar concentration. The standard state can have any temperature, but the temperature should be specified. For example, the standard enthalpy of vaporization of water Is the enthalpy of vaporization at 373 K (the boiling point) and 1 bar of pressure. If we lower the temperature below 373 K (for example to 372 K) then water no longer boils. We can say that the reaction is defined at the standard state of 1 bar (or 1 atm in practice) of pressure. A reaction can take place at concentrations other than the standard state, but there will be a dependence of the enthalpy on pressure. Vaporization is a special case since the atmosphere is always at 1 atm at sea level.

Enthalpy of physical change A physical change is when one state of matter changes into another state of matter of the same substance. The difference between physical and chemical changes is not always clear, however, phase transitions are obviously physical changes. Fusion Freezing DH fus DH vap Solid Liquid Gas DH freeze DH cond Vaporization Condensation Sublimation DH sub Solid Vapor Deposition DH vap dep Gas

Properties of Enthalpy as a State Function The fact that enthalpy is a state function is useful for the additivity of enthalpies. Clearly the enthalpy of forward and reverse processes must be related by: D forw ard H = D reverse H so that the phase changes are related by: D fus H = D freeze H D vap H = D cond H D sub H = D vap dep H Moreover, it should not matter how the system is transformed from the solid phase to the gas phase. The two processes of fusion (melting) and vaporization have the same net enthalpy as sublimation.

Addivity of Enthalpies Because the enthalpy is a state function the same magnitude must be obtained for direct conversion from solid to gas as for the indirect conversion solid to liquid and then liquid to gas. D sub H = D fus H + D vap H Of course, these enthalpies must be measured at the same temperature. Otherwise an appropriate correction would need to be applied as described in the section on the temperature dependence of the enthalpy.

Chemical Change In a chemical change the identity of substances is altered during the course of a reaction. One example is the hydrogenation of ethene: CH 2 =CH 2 (g) + H 2 (g) CH 3 CH 3 (g) DH = -137 kj/mol The negative value of DH signifies that the enthalpy of the system decreases by 137 kj/mol and, if the reaction takes place at constant pressure, 137 kj/mol of heat is released into the surroundings, when 1 mol of CH 2 =CH 2 combines with 1 mol of H 2 at 25 o C.

Standard Enthalpies of Formation The standard enthalpy of formation D f H o is the enthalpy for formation of a substance from its elements in their standard states. The reference state of an element is its most stable form at the temperature of interest. The enthalpy of formation of the elements is zero. For example, let s examine the formation of water. H 2 (g) + 1/2 O 2 (g) H 2 O(l) DH o = -286 kj Therefore, we say that D f H o (H 2 O, l) = -286 kj/mol. Although D f H o for elements in their reference states is zero, D f H o is not zero for formation of an element in a different phase: C(s, graphite) C(s, diamond) D f H o = + 1.895 kj/mol

Standard Enthalpy Changes The reaction enthalpy depends on conditions (e.g. T and P). It is convenient to report and tabulate information under a standard set of conditions. Corrections can be made using heat capacity for variations in the temperature. Corrections can also be made for variations in the pressure. When we write DH in a thermochemical equation, we always mean the change in enthalpy that occurs when the reactants change into the products in their respective standard states.

Standard Reaction Enthalpy The standard reaction enthalpy, D r H o, is the difference between the standard molar enthalpies of the reactants and products, with each term weighted by the stoichiometric coefficient,, The standard state is for reactants and products at 1 bar of pressure or 1 molar concentration. The unit of energy used is kj/mol. The temperature is not part of the standard state and it is possible to speak of the standard state of oxygen gas at 100 K, 200 K etc. It is conventional to report values at 298 K and unless otherwise specified all data will be reported at that temperature.

Hess s Law We often need a value of DH that is not in the thermochemical tables. We can use the fact that DH is a state function to advantage by using sums and differences of known quantities to obtain the unknown. We have already seen a simple example of this using the sum of DH of fusion and DH of vaporization to obtain DH of sublimation. Hess s law is a formal statement of this property. The standard enthalpy of a reaction is the sum of the standard enthalpies of the reactions into which the overall reaction may be divided.

Application of Hess s Law We can use the property known as Hess s law to obtain a standard enthalpy of combustion for propene from the two reactions: C 3 H 6 (g) + H 2 (g) C 3 H 8 (g) DH = -124 kj C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) DH = -2220 kj If we add these two reactions we get: C 3 H 6 (g) + H 2 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) DH = -2344 kj and now we can subtract: H 2 (g) + 1/2O 2 (g) H 2 O(l) DH = -286 kj to obtain: C 3 H 6 (g) + 9/2O 2 (g) 3CO 2 (g) + 3H 2 O(l) DH = -2058 kj

Understanding the additivity of enthalpies Calculate D rxn H o for the rxn : 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) given the following : a) S(s) + O 2 (g) SO 2 (g) D f H o = -297 kj b) S(s) + 3 / 2 O 2 (g) SO 3 (g) D f H o = -396 kj

Enthalpy of solution Note the trends in the heats of solution for these common compounds. SOLUTE hydrochloric acid -74.84 ΔH o in kj/mol ammonium nitrate +25.69 ammonia -30.50 potassium hydroxide -57.61 caesium hydroxide -71.55 sodium chloride +3.88 potassium chlorate +41.38 acetic acid -1.51 sodium hydroxide -44.51

Tabulated bond enthalpies Bond pm kj/mol Bond pm kj/mol H H 74 436 C C 154 348 H C 109 413 C=C 134 614 C C 154 348 C N 147 308 H N 101 391 CºC 120 839 N N 145 170 C O 143 360 H O 96 366 C S 182 272 O O 148 145 O O 148 145 H F 92 568 C F 135 488 F F 142 158 O=O 121 498 H Cl 127 432 C Cl 177 330 Cl-Cl 199 243 C Br 194 288 H Br 141 366 N N 145 170 Br-Br 228 193 C I 214 216 H I 161 298 NºN 110 945 I I 267 151

Thermodynamic cycles The energies and enthalpies of ionic solids are dominated by Coulombic interactions. The lattice enthalpy can be calculated from Coulombic interactions. Using the calculated lattice enthalpy and experimental data one can obtain the enthalpy of formation of an ionic solid by means of the Born-Haber cycle. This is illustrated on the next slide for a generic salt of a monovalent ion M + X -. The reactions needed to obtain M + X - in the solid phase to M + + X - in the vapor phase are given below. The overall process of separating the ions in an ionic solid into the constituent ions is written as M + X - (s) M + (g) + X - (g) Lattice enthalpy

Enthalpies of Ionization The molar enthalpy of ionization is the enthalpy that accompanies the removal of an electron from a gas phase atom or ion: H(g) H + (g) + e - (g) DH o = +1312 kj/mol For ions that are in higher charge states we must consider successive ionizations to reach that charge state. For example, for Mg we have: Mg(g) Mg + (g) + e - (g) DH o = +738 kj/mol Mg + (g) Mg 2+ (g) + e - (g) DH o = +1451 kj/mol We shall show that these are additive so that the overall enthlalpy change is 2189 kj/mol for the reaction: Mg(g) Mg 2+ (g) + 2e - (g) DH o = +2189 kj/mol

Electron Gain Enthalpy The reverse of ionization is electron gain. The corresponding enthalpy is called the electron gain enthalpy. For example: Cl(g) + e - (g) Cl - (g) DH = -349 kj/mol The sign can vary for electron gain. Sometimes, electron gain is endothermic. The combination of ionization and electron gain enthalpy can be used to determine the enthalpy of formation of salts. Other types of processes that are related include molecular dissociation reactions.

The Born-Haber cycle The thermodynamic cycle is illustrated. The scheme shows that if one knows all of the energies except one (difficult to measure) quantity, one can calculate it using Hess law. M + X - Lattice Enthalpy M + + X - Enthalpy of Formation 1/2 X 2 (g) + M (s) Ionization Potential Enthalpy of Vaporization M (s) Electron Affinity Bond Dissociation X (g)

Using enthalpies of formation Calculate the enthalpy change for the following rxn: NH 3 (g) + O 2 (g) NO(g) + H 2 O(g)

Using enthalpies of formation Calculate the enthalpy change for the following rxn: 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g) Solution: Look up the enthalpies of formation of each molecule in the thermodynamic tables. Remember that the enthalpy of formation of O 2 is zero.

Enthalpies of Combustion Standard enthalpies of combustion refer to the complete combination with oxygen to carbon dioxide and water. For example, for methane we have: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) D comb H o = -890 kj/mol Enthalpies of combustion are commonly measured in a bomb calorimeter (a constant volume device). Thus, DU m is measured. To convert from DU m to DH m we need to use the relationship: DH m = DU m + D gas RT The quantity D gas is the change in the stoichiometric coefficients of the gas phase species. We see in the above express that D gas = -2. Note that H 2 O is a liquid.

Enthalpy of formation from combustion The enthalpy of combustion D comb H o for propane, C 3 H 8, is -2220 kj. What is D f H o of C 3 H 8?

Enthalpy of formation from combustion The enthalpy of combustion D comb H o for propane, C 3 H 8, is -2220 kj. What is D f H o of C 3 H 8? Solution: Write down the balanced equation Look up the enthalpies of formation of the CO 2 and H 2 O. Remember that the enthalpy of formation of O 2 is zero. The tabulated values of enthalpy are given at 298 K, Keep in mind that H 2 O is in the liquid phase.

Enthalpy of formation from combustion The enthalpy of combustion, D comb H o for propane, C 3 H 8, is -2220 kj. What is DH o f of C 3 H 8? Solution: Write down the balanced equation

Enthalpy of formation What is the enthalpy of combustion for propane, C 3 H 8 assuming that water is in the vapor phase? Solution: Write down the balanced equation Look up the enthalpies of formation of the CO 2 and H 2 O. Remember that the enthalpy of formation of O 2 is zero.

Expansion following combustion What volume of gas is produced if 50 grams of C 3 H 8 is combusted at 373 K at 1 atm? You may assume all molecules are in the vapor phase.

Expansion following combustion What volume of gas is produced if 50 grams of C 3 H 8 is combusted? You may assume all molecules are in the vapor phase (T = 373 K). Solution: Write down the balanced equation Calculate the change in number of moles of gas, Dn gas. Then calculate DV.

Expansion following combustion What volume of gas is produced if 50 grams of C 3 H 8 is combusted? You may assume all molecules are in the vapor phase. Solution: Write down the balanced equation Calculate the change in number of moles of gas, Dn gas. Then calculate DV. Dn gas = S n products S n reactants Dn gas = 3 + 4 1 5 = 1

Work done following combustion How much work can be extracted for an expansion against a constant pressure of 10 atm if 50 grams of C 3 H 8 is combusted? You may assume all molecules are in the vapor phase at 373 K.A

Comparing fuels on a mass basis We can consider the heat released upon combustion as the energy available to do work in an engine. If we consider hydrocarbon fuels (methane, ethane, propane, butane, and octane) as well as ethanol, we can compare them both on a per mole and per mass basis. The mole basis is obtained using the enthalpy of combustion.

Comparing fuels on a mass basis Fuel D comb H o CH 4-882 C 2 H 6-1561 C 3 H 8-2219 C 4 H 10-2878 C 8 H 18-5430 C 2 H 6 O -1370

Comparing fuels on a mass basis Fuel D comb H o M m CH 4-882 16 C 2 H 6-1561 30 C 3 H 8-2219 44 C 4 H 10-2878 58 C 8 H 18-5430 114 C 2 H 6 O -1370 46

Comparing fuels on a mass basis Fuel D comb H o M m q (kj/gram) CH 4-882 16 55.1 C 2 H 6-1561 30 52.0 C 3 H 8-2219 44 50.4 C 4 H 10-2878 58 49.6 C 8 H 18-5430 114 47.6 C 2 H 6 O -1370 46 29.8

Methane vs. Methanol as a fuel On a per mole basis methane and methanol have similar enthalpies of combustion. On a mass basis methanol provides 22 kj/g while methane produces 55 kj/g. Methanol is convenient because it is a liquid.

Atmospheric methane: a greenhouse gas

Methane Capture: An important ecological problem On a per mole basis methane is converted to methanol with a heat of reaction of -164.5 kj/ mol. The direct conversion shown above does not work well because it is slow unless the temperature is so high that one risks complete oxidation to CO 2.

Methane Conversion via Syngas The conversion of CH 4 to syngas is the most common industrial method for producing CH 3 OH from CH 4. The first step or production of syngas is called cracking. The second step requires a Cu or Pd catalyst. The overall process is only about 10% efficient.

Photochemical Methane Conversion An alternative approach is to use light to activate water in order to make a hydroxyl radical, which then reacts with methane. This process requires sufficient light energy to cleave the HO-H bond in H 2 O.

Methane Monooxygenase

Heating of a fuel in an engine The combustion of a fuel leads to heating of the gas produced in the reaction. The heating can be calculated using the heat capacity (or specific heat). For octane, c p = 255.7 J/mol/K. What is the s (the specific heat)?

Heating of a fuel in an engine Consider the fact that after combustion the octane fuel has been converted into CO 2 and H 2 O in the vapor phase. For H 2 O vapor, c p = 33 J/mol/K. What is the final temperature if 12 microliters of octane are combusted?

Focus on energy The work done in the internal combustion engine is called pressure volume work. For a simple irreversible stroke the work is: work = P DV In a 3.0 L 6 cylinder engine DV = 0.5 L. Assuming the initial volume is 80 microliters, what is P? We can estimate the pressure from the amount of octane is injected and combusted

From combustion to useful work A typical fuel injector will inject 12 microliters of octane fuel per stroke. What is the pressure of the gas created by combustion in a volume of 80 microliters? In our first examination of this problem we will ignore the heating and simply calculate the conversion from the volume of liquid fuel to the volume of H 2 O and CO 2 according to the reaction stoichiometry.

Calculating pressure First convert from volume to moles: The change in the number of moles is: Therefore the pressure is: The pressure is ~100 atm.

Solar Heating

Equilibration In an equilibration two objects that are initially at different temperatures come to equilibrium at a final temperature. For example, in a solar water heater we can imagine that a slab of black stone acts as the absorber. If the stone has a mass of 20 kg and a temperature of 80 o C. What is the final temperature of 1 L of water that is initially at 10 o C? (for H 2 O s 1 = 4.18 J/g- o C and stone s 2 = 0.5 J/g- o C )

Biological implications Obviously, the food we eat release heat in our bodies. This heat is used both to maintain body temperature and for processes that build up our bodies (anabolic processes). We often talk about calories in the food we eat. The calorie on a cereal box is equal to 1000 calories in the chemical nomenclature. 1 calorie = 4.184 Joules. Therefore, the Intake required for an average man of 12 MJ/day is about 3000 calories per day (in the sense of diet).

Differential relationships for enthalpy We have defined a relationship between the enthalpy and Internal energy H = U + PV The infinitesimal change in the state function H results in H + dh = U + du + (P + dp)(v + dv) Therefore dh = du + PdV + VdP Now we substitute du = dq + dw into this expression dh = dq + dw + PdV + VdP Since dw = - PdV dh = dq + VdP At constant pressure, dp = 0, and we have dh = dq P

Path Functions We have seen that work and heat are path functions. The magnitude of the work and heat depends not just on the final values of the T and P, but also on the path taken. We can summarize the paths and their implications in the table below. Path Condition Result Isothermal DT = 0 w = -q Constant V DV = 0 w = 0, DU = C v DT Constant P DP = 0 w = -PDV, q P = C p DT Adiabatic q = 0 DU = w

State Functions At present we have introduced two state functions: Internal Energy DU Enthalpy DH State functions do not depend on the path, only on the value of the variables. We can make the analogy with elevation. The potential energy at an elevation h, which we call V(h) does not depend on how we got to that elevation. If we compare V(h 1 ) in Raleigh to V(h 2 ) on Mt. Mitchell the difference V(h 2 ) - V(h 1 ) is the same regardless of whether we drive to Mt. Mitchell through Statesville or Asheville. The work we do to get (i.e. how much gas we use in a car!) is a path function.