# Chemical Thermodynamics - Berkeley City College

THERMODYNAMICS Spontaneity Entropy: Criteria for Spontaneity Second Law of Thermodynamics

Third Law of Thermodynamics Free Energy Coupled Reactions Thermodynamics The chemistry that deals with the flow of energy during a chemical or physical process; Determine whether a reaction is endothermic or exothermic; Provide criteria to predict whether a reaction is spontaneous or non-spontaneous;

Provide relationship between free energy and work; Thermodynamics vs. Kinetics Kinetics Domain Reaction rates depend on the path from reactants to products. Thermodynamics domain: tells us in which direction energy flows during

changes, and whether a reaction is spontaneous or non-spontaneous; First Law of Thermodynamics This is the law of conservation of energy Energy cannot be created or destroyed; The energy content of a thermodynamic universe is constant. For a system that absorbs q amount of heat and does

W amount of work, the change in its internal energy (DE) is: DE = q + W For work that only involves gas expansion or compression, W = -pDV; (p = pressure exerted on the system) Enthalpy versus Energy DE = energy change for a process at constant constant volume;

DE = q + W = qv; (at constant volume, W = 0) DH = enthalpy or energy change for a process at constant pressure P; DH = qp = DE W = DE + PDV; q > 0: system absorbs heat; q < 0: system loses heat W < 0: work done by system; W > 0: work done on system; Predicting Spontaneous Processes Thermodynamics allows us to predict whether a process will take occur spontaneously, but it does not tell us how fast it will occur;

A spontaneous process is one that, once it starts, will continue to occur without outside intervention; like a ball rolling down a slope; A nonspontaneous process one that requires a continuous external intervention to occur; like pushing a bicycle uphill; A spontaneous but very slow process The conversion of carbon from the diamond to graphite is predicted to be spontaneous at ambient pressure, but its rate is immeasurably slow at low to

moderate temperatures. The rate becomes measurable only at temperatures in the 10002000 K range. (credit diamond photo: modification of work by Fancy Diamonds/Flickr; credit graphite photo: modificaton of work by images-of-elements.com/carbon.php) Expansion of Gas into a Vacuum (Spontaneous and relatively fast) An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly

distributed between the flasks. The Flow of Heat When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object. Scientists responsible for introducing criteria for spontaneity (a) Nicholas Lonard Sadi Carnots extensive research on the efficiency of steampowered machinery led to the concept of heat flow and spontaneous processwes;

(b) Rudolf Clausiuss study on Carnots work resulted in the introduction of Entropy as thermodynamic function for predicting spontaneous processes. Spontaneous Processes Some examples of spontaneous processes/reactions: 1. The corrosion of iron is spontaneous, but the reverse is not: 4Fe(s) + 3 O2(g) 2Fe2O3(s) 2. Ice melts spontaneously at room temperature, but water will not freeze at room temperature: H2O(s) H2O(l)

3. Heat spontaneously flows from a hot object to a cold object if the objects are in contact. Entropy A thermodynamic function related to the distribution of heat: A process that is accompanied by an increase in the heat distribution or dispersion implies an increase in entropy, and is a spontaneous process; Ludwig Boltzmann described entropy as the number

of microstates of energy in which a system can exist; A system that can exist in many different microstates of energy has a very high entropy. Entropy is related to the degree of energy states a thermodynamic universe can exist. What is the significance of entropy? Nature spontaneously progress into a situation that has the highest probability of energy states or highest entropy;

Entropy change is used to predict whether a given process/reaction is thermodynamically possible; Microstates for 4 Particles in 2 Boxes The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are grouped into five distributions: (a), (b), (c), (d), and (e), based on the numbers of particles in each box. Microstates for Heat Flow

This is a microstate model describing the flow of heat from a hot object to a cold object. (a) Before the heat flow occurs, the object comprised of particles A and B contains both units of energy and as represented by a distribution of three microstates. (b) If the heat flow results in an even dispersal of energy (one energy unit transferred), a distribution of four microstates results. (c) If both energy units are transferred, the resulting distribution has three microstates. Physical States and Entropy Entropy can be related to the degree of freedom a system can exist in its microstate.

The more degree of freedom the microstate of a system can exist, the higher its entropy. Sgas >> Sliquid > Ssolid Changes in Entropy with Phase Changes The entropy of a substance increases (S > 0) as it transforms from a relatively ordered solid, to a less-ordered liquid, and then to a still less-ordered gas. The entropy decreases (S < 0) as the substance

transforms from a gas to a liquid and then to a solid. Entropy as a Function of Temperature Entropy increases as the temperature of a substance is raised, which corresponds to the greater spread of kinetic energies. When a substance melts or vaporizes, its entropy increases significantly. Relative Entropy of Substances Entropy: 1. increases from solid-to-liquid-to-vapor/gas;

2. increases when temperature is increased; 3. increases as gas volume is increased at constant temperature; 4. increases when gases are mixed. 5. increases from top to bottom down a group in the periodic table; 6. increases as the structure of a compound becomes more complex. Concept Check Consider 1.0 moles of a gas contained in a 1.0-L

bulb at a constant temperature of 25C. This bulb is connected by a valve to an evacuated 5.0-L bulb. At constant temperature, a) What should happen to the gas when the valve is opened? b) What are the values of DH, DE, q, and w for this process? c) Base on the answers given to parts (a) and (b), what is the driving force for the process? Concept Check

Predict the sign of DS for each of the following, and explain: a) The evaporation of alcohol; b) Snowing; c) Compressing an ideal gas at constant temperature; d) Mixing of gases; e) Dissolving NaCl in water. o

Standard Entropy Values, S This is the entropy of a substance in its most stable state at 1 atm and 25oC. For ionic species in solution, it is the entropy of ions in 1 M solution at 25oC relative to that of H+(aq), which is assigned zero entropy value. Standard Entropy Change (DS) The standard entropy change (DSo) for a reaction is calculated using the following

equation: DSreaction = SnpSproducts SnrSreactants Entropy Change in Chemical Reactions At constant temperature and pressure, DSorxn = SnpSoproducts SnrSoreactants In general, DSorxn > 0 if Snp > Snr Example-1: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g),

(Snp > Snr) DSorxn = {(3 x SoCO ) + (4 x SoH O)} {(SoC H ) + (5 x SoO )} = {(3 x 214) + (4 x 189)}J/K {270 + (5 x 205)}J/K = (642 + 756) J/K (270 + 1025) J/K = 103 J/K 2 2

3 8 2 Entropy Change in Chemical Reactions Example-2: CO(g) + 2H2(g) CH3OH(g), (Snp < Snr) DSorxn = (SoCH3OH) {(SoCO) + (2 x (SoH2)}

= 240 J/K {198 J/K + 2 x (131 J/K)} = 240 J/K 460 J/K = -220 J/K DS o rxn

< 0 (has a negative value) if Snp < Snr Second Law of Thermodynamics A spontaneous process is always accompanied by an increase in the entropy of the universe. The entropy of universe always increases. DSuniverse = DSsystem + DSsurroundings Second Law of Thermodynamics Energy flows from a concentrated energy state to

a diffused energy state; Diffusion of energy state is a spontaneous process. Diffusion of energy state represents increase in entropy; According to SLoT, a process is spontaneous if the entropy of the universe increases. Third Law of Thermodynamics A perfect crystalline solid at 0 K has no kinetic energy; A perfect crystalline solid at 0 K with no kinetic

energy has only one microstate (W = 1) According to Boltzmann equation, entropy S = k ln W = k ln (1) = 0 The entropy of a pure, perfect crystalline solid at 0 K is zero. Dssys versus DSsurr The algebraic sign of DSsys (+ or -) depends on relative energy distribution before and after the change; The algebraic sign of DSsurr (+ or -) depends on the

direction of the heat flow; Heat flows from system to surrounding, DSsurr > 0; Heat flows from surrounding to system, DSsurr < 0; Magnitude of DSsurr also depends on the temperature at which the exchange of heat occurs. Magnitude & Algebraic Sign of DSsurr DSsurr DSsurr =

For exothermic reactions, DHsys < 0, DSsurr > 0; For endothermic reactions, DHsys > 0, DSsurr < 0; Conditions for Spontaneous Process Predicting spontaneity of a process using DSuniv: DSuniv = DSsys + DSsurr > 0; spontaneous; DSuniv = DSsys + DSsurr < 0; nonspontaneous; DSuniv = DSsys + DSsurr = 0; state of equilibrium; DSuniv = DSsys + DSsurr = DSsys Interplay of DSsys and DSsurr in

Determining the Sign of DSuniv Gibbs Free Energy, DG DSuniv = DSsys + DSsurr DSuniv = DSsys > 0; spontaneous TDSuniv = TDSsys DHsys DG = TDSuniv DG = DHsys TDSsys DG < 0; a spontaneous process Effect of DH and DS on Spontaneity

DH DS Result + spontaneous at all temps +

+ spontaneous at high temps spontaneous at low temps

+ not spontaneous at any temp Concept Check Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: (a) Exothermic and becomes more random

Spontaneous (b) Exothermic and becomes less random Cannot tell (c) Endothermic and becomes more random Cannot tell (d) Endothermic and becomes less random Not spontaneous Explain how temperature would influence your answers. Free Energy (G)

TDSuniv = - DG (at constnt T and P) At constant temperature and pressure, a process is spontaneous if DG < 0; A process progress spontaneous in the direction that decreases the free energy, G, until it react equilibrium. The Meaning of G for a Chemical Reaction A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.

Algebraic Signs of DS, DH and DG There are four possibilities regarding the signs of enthalpy and entropy changes. Temperature Dependence of DG These plots show the variation in G with temperature for the four possible combinations of arithmetic sign for H and S. Effect of Temperature on DG and Spontaneity

DH DS T DG Comments Examples

- + high spontaneous at 2H2O2(l) 2H2O(l) + O2(g) or low all temperature + + high spontaneous at

CaCO3(s) CaO(s) + CO2(g) high temperature - low spontaneous at N2(g) + 3H2(g) 2NH3(g) low temperature + high + nonspontaneous at 2H2O(l) + O2(g) 2H2O2(l)

or low all temperature ________________________________________________________ Solar Energy is Free What Free Energy? Free Energy, G It is the maximum amount of chemical energy from

a spontaneous reaction that can be utilized to do work or to drive a nonspontaneous process; Wmax = G It is the minimum amount of energy that must be supplied to make a nonspontaneous reaction occur. Free Energy and Work Maximum work from free energy of a spontaneous process is only achievable in a hypothetical slow reversible process. In a real process occurring at normal, measurable

rate, only a fraction of free energy can be turned into work; most of it is lost as heat, which is a waste form of energy. Standard Free Energy (G) Under standard condition, G = H TS For a reaction under standard condition, Greaction = npGproducts nrGreactants General Expression for DG

For a reaction under any condition: DG = DGo + RTlnQ For a reaction at equilibrium: DG = DGo + RTlnK = 0; lnK = lnK = - () + (R = 8.314 J/K.mol) DG and Pressure for Phase Equilibrium For liquid-vapor equilibrium at any temperature, DG = DGo + RTlnP = 0; (P = vapor pressure in atm)

lnP = - () + (T is in Kelvin) (In this case, DHo = DHvap = enthalpy of vaporization) If lnP is plotted against 1/T, the slope = - ; the intercept = Estimating boiling point from DHvap and DSvap For phase equilibrium between liquid and vapor at 1 atm, the equilibrium temperature is also the boiling

point of the liquid. At phase equilibrium: DGo = DHo TDSo = 0; T= ; For liquid-vapor equilibrium, Tb = ; (DDHovap and DSovap are standard enthalpy and standard entropy of vaporization, and Tb = boiling point.) Estimating Boiling Point Consider the process: C2H5OH(l) C2H5OH(g)

at 1 atm (standard pressure condition) Given: DHof[C2H5OH(l)] = -277.6 kJ/mol; DHof[C2H5OH(g)] = -234.8 kJ/mol; So[C2H5OH(l)] = 160.7 J.mol-1.K-1; So[C2H5OH(g)] = 281.6 J.mol-1.K-1; Calculate DHovap and DSovap, and estimate the boiling point of ethanol, C2H5OH. Estimating Boiling Point (Solution

to previous problem): For the process: C2H5OH(l) C2H5OH(g) at 1 atm (standard pressure condition) DHovap = (-234.8) (-277.6) = 42.8 kJ.mol-1; DSovap = (281.6 160.7) = 120.9 J.mol-1.K-1; Tb = = = 354 K ~ 81 oC (This is an estimate, because we use DHovap and DSovap values at 25 oC) Concept Check

A liquid is vaporized at its boiling point. Predict the signs of: w + q + DH + DSsys

DSsurr 0 DG Explain your answers. Transition Temperature T = is also the transition temperature when a reaction changes from spontaneous to nonspontaneous, and vice versa; At transition temperature, Tr, DGo = 0;

DGo = DHo TrDSo = 0; Tr = Estimating Transition Temperature Consider the reaction: N2(g) + 3H2(g) 2NH3(g)

where DHo = -91.8 kJ and DSo = -198.1 J/K = -0.1981 kJ/ K At 298, TDSo = 298 K x (-0.199 J/K) = -59.3 kJ DGo = DHo - TDSo = -92 kJ (-59.3 kJ) = -33 kJ; Reaction is spontaneous at 298 K under standard condition; Transition temperature, Tr = = 463 K ~ 190 oC This reaction will become nonspontaneous above 463 K if carried out under standard condition, (such that Qp = 1)

Estimating Transition Temperature Consider the reaction: CH4(g) + H2O(g) CO(g) + 3H2(g), where DHo = 205.9 kJ and DSo = 214.7 J/K = 0.2147 kJ/K At 298 K, TDSo = 298 K x (0.2147 J/K) = 64.0 kJ DGo = DHo - TDSo = 205.9 kJ (64.0 kJ) = 141.9 kJ; Reaction is not spontaneous at 298 K (under standard condition) Transition temperature, Tr = = 959.0 K ~ 686 oC This reaction will become spontaneous above 959 K if carried

out under standard condition, (such that Qp = 1) Reaction spontaneous at low temperature but not at high temperature (DH < 0; DS < 0) Consider the following reaction: N2(g) + 3H2(g) 2NH3(g), DHo = -91.8 kJ and DSo = -198.1 J/K = -0.1981 kJ/K At 298 K, TDSo = 298 K x (-0.1981 J/K) = -59.0 kJ DGo = DHo - TDSo = -91.8 kJ (-59.0 kJ) = -32.8 kJ; DGo < 0 reaction is spontaneous at 298 K;

At 500 K, TDSo = 500 K x (-0.1981 J/K) = -99.0 kJ; DGo = DHo - TDSo = -91.8 kJ (-99.0 kJ) = 7.2 kJ; DGo > 0 reaction is nonspontaneous at 500 K Reaction spontaneous at high temperature but not at low temperature (DH > 0; DS > 0) Consider the following reaction: CH4(g) + H2O(g) CO(g) + 3H2(g), where DHo = 205.9 kJ and DSo = 214.7 J/K = 0.2147 kJ/K

At 298 K, TDSo = 298 K x (0.2147 J/K) = 64.0 kJ DGo = DHo - TDSo = 205.9 kJ 64.0 kJ = 141.9 kJ; DGo > 0 reaction is nonspontaneous at 298 K. At 1000 K, TDSo = 1000 K x (0.2147 J/K) = 214.7 kJ; DGo = DHo - TDSo = 205.9 kJ 214.7 kJ) = -8.8 kJ; DGo > 0 reaction is spontaneous at 1200 K DG under Nonstandard Conditions Free energy also depends on concentration and partial pressure;

DG = DGo + RTlnQp, Under nonstandard conditions (Qp 1), Consider the reaction: N2(g) + 3H2(g) 2NH3(g) at 500 K, where PN2 = 40. atm; PH2 = 120. atm, and PNH3 = 20. atm Qp = = = 5.8 x 10-6 = 7.2 kJ + (0.008314 kJ/(mol.K))(500 K)ln(5.8 x 10-6) = 7.2 kJ + (-50. kJ) ~ 43 kJ DG < 0 reaction is spontaneous under above pressure conditions; At 500 K, DG Free Energy and Equilibrium

Equilibrium occurs when reaction/process reaches the lowest possible value of free energy available to the system. G = G + RT lnK = 0; G = RT lnK; lnK = Qualitative Relationship Between DGo and K for Reaction Calculating K from DG

o Consider the reaction: N2(g) + 3H2(g) 2NH3(g), At 298 K, DGo = -32.8 kJ lnK = = 13.2 K = e13.2 = 5.40 x 105 (large K at low temperature) At 500 K, DGo = 7.2 kJ; lnK = = -1.7

K = e-1.7 = 0.18 (small K at high temperature) For exothermic reactions, K values decreases as the temperature is increased. Coupling Reactions A nonspontaneous reaction can be coupled to a spontaneous one to make it happen. Example: Fe2O3(s) 2Fe(s) + 3/2 O2(g); DGo = 742.2 kJ (1) 3CO(g) + 3/2O2(g) 3CO2(g); DGo = -771.6 kJ (2) Coupling reactions (1) and (2) yields a net reaction that is

spontaneous: Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g); DGo = -29.4 kJ Coupling Reactions in Biological System The formation of ATP from ADP and H2PO4- is nonspontaneous, but it can be coupled to the hydrolysis of creatine-phosphate that has a negative DGo. ADP + H2PO4-

ATP + H2O; DGo = +30 kJ Creatine-phosphate creatine + phosphate; DGo = -43 kJ Coupling the two reactions yields a spontaneous overall reaction: Creatine-phosphate + ADP Creatine + ATP; DGo = -13 kJ

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