Chapter 19 – Principles of Reactivity: Entropy and Free Energy Objectives: Describe terms: entropy and spontaneity. Predict whether a process will be spontaneous. Describe: free energy. Describe the relationship between DG, K, and product favorability.
Thermodynamics • Thermodynamics is _______________ ____________________. • First Law of Thermodynamics • The law of conservation of energy: ______ ________________________________. DE = q + w • The change in internal energy of a system is the sum of the heat transferred to or from the system and the work done on or by the system.
Spontaneous Change • Chemical changes, physical changes • Spontaneous change: occurs _____________________. It leads to ____________. • Example: heat transfers spontaneously from a hotter object to a cooler object. • ____________ is reached in product-favored and in reactant-favored processes.
Spontaneous Chemical Reactions 2 H2 + O2 2 H2O CH4 + 2 O2 CO2 + 2 H2O 2 Na + Cl2 2 NaCl HCl + NaOH NaCl + H2O • Common feature: _____________ • But many processes are ____________ and spontaneous. • H2 + I2 2 HI (g) _______________ can be approached from either direction.
Spontaneous Processes • Dissolving NH4NO3 in water: DH = +25.7 KJ/mol • Expansion of a gas into a vacuum: energy neutral, heat is neither evolved nor required. • Phase changes: melting of ice requires ~ 6 kJ/mol; but only occurs if T > 0oC. • ______________ determines whether a process is spontaneous. • Heat transfer: The T of a cold substance in a warm environment will rise until the substance reaches the ambient T. • The required heat comes from the _____________.
Entropy • To predict whether a process will be spontaneous. • Entropy, S is a thermodynamic function • State function: a quantity whose value is determined only by the initial and final states of a system. • Second Law of Thermodynamics • ____________________________________________________________________________. • ____________________________________________________________________________.
Dispersal of Energy • By statistical analysis: • Energy is distributed of a number of particles • Most often case is when energy is distributed over all particles and to a large number of states. • As the number of particles and the number of energy levels grows, one arrangement turns out to be vastly more probable than all others.
Dispersal of Energy • Dispersal of __________ often contributes to energy dispersal.
Boltzmann Equation • Ludwig Boltzmann (1844-1906) • Look at the distribution of energy over different energy states as a way to calculate ____________. S = k log W • K – Boltzmann constant • W – represent the number of different ways that the energy can be distributed over the available energy levels. • A maximum entropy will be achieved at _________________ , a state in which W has the maximum value.
Summary: Matter and Energy Dispersal • A final state of a system can be more probable than the initial state if: • The atoms and molecules can be more ____________ and/or • ___________ can be dispersed over a greater number of atoms and molecules. • If energy and matter are both dispersed in a process, it is _______________. • If only matter is dispersed, then quantitative information is needed to decide whether the process is spontaneous. • If energy is not dispersed after a process occurs, then that process will ____________ _____________________.
Entropy • Entropy is used to __________________ ___________ resulting from dispersal of energy and matter. The greater the _______ in a system, the greater the value of S. • Third Law of Thermodynamics • There is no disorder in a perfect crystal at 0K, S=0. • The entropy of a substance at any T can be obtained by measuring the heat required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts).
Entropy • The entropy of a substance at any T can be obtained by measuring the ________ required to raise the T from 0K, where the conversion must be carried by a reversible process (very slow addition of heat in small amounts). • The entropy added by each incremental change is: DS = • Adding the entropy changes gives the total entropy. • All substances have ___________ entropy values at temperatures above 0K.
Thermodynamics • First Law: The total energy of the universe is a constant. • Second Law: The total entropy of the universe is always increasing. • Third Law: The entropy of a pure, perfectly formed crystalline substance at 0K is zero. - A local decrease in entropy (the assembly of large molecules) is offset by an increase in entropy in the rest of the universe -.
Standard Entropy • So, is the entropy gained by converting it from a perfect crystal at 0K to standard state conditions (1 bar, 1 molal solution). • Units: J/Kmol • Entropies of gases are ____________than those for liquids, entropies of liquids are ____________ than those for solids. • Larger molecules have a _________ entropy than smaller molecules, molecules with more complex structures have ________entropies than simpler molecules.
Entropy • The entropy of liquid water is ___________ than the entropy of solid water (ice) at 0˚ C. S˚(H2O sol) < S˚(H2O liq)
Entropy Entropies of ionic solids depend on ___________________________. So (J/K•mol) MgO 26.9 NaF 51.5 Mg2+ & O2- Na+ & F- The larger coulombic attraction on MgO than NaF leads to a lower entropy.
Which substance has the higher entropy, why? • O2 (g) or 03 (g) • SnCl4 (l) or SnCl4 (g)
Arrange the substances in order of increasing entropy. Assume 1 mole of each at standard conditions. HCOOH(l) CO2(g) Al(s) CH3COOH(l)
Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. CO(g) + 3 H2(g) CH4(g) + H2O(g) 2 H2O(l) 2 H2(g) + O2(g) I2(g) + Cl2(g) 2 ICl(g)
S increases slightly with T S increases a large amount with phase changes Entropy Change
Entropy Change • Entropy usually increases when a pure liquid or solid ______________ in a solvent. • Entropy of a substance ____________ with temperature.
Entropy Change • The entropy change is the sum of the entropies of the products minus the sum of the entropies of reactants: DS0system = S S0(products) – S S0(reactants) You will find DSo values in the Appendix L of your book.
Calculate the standard entropy changes for the evaporation of 1.0 mol of liquid ethanol to ethanol vapor. C2H5OH(l) C2H5OH(g)
Calculate the standard entropy change for forming 2.0 mol of NH3(g) from N2(g) and H2(g) N2(g) + 3 H2(g) 2 NH3 (g)
Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.35 moles of NO(g) react at standard conditions. 2 NO(g) + O2(g) 2 NO2(g)
Calculate the standard entropy change for the oxidation of ethanol vapor (CH2H5OH (g)).
Entropy in the Universe DS0univ = DS0sys + DS0surr • 2nd Law of Thermodynamics: DSuniv is positive for a spontaneous process. • For a nonspontaneous process: DS0univ < 0 (negative) • If DSuniv = 0 the system is at equilibrium. • Calculate first the DS0sys, then DS0surr. DS0surr = qsurr/T = -DH0sys/T DH0sys = S H0(products) – S H0(reactants)
Show thatDS0univis positive (>0) for dissolving NaCl in water DS0univ = DS0sys + DS0surr 1) Determine DS0sys 2) Determine DS0surr NaCl(s) NaCl (aq)
Predicting whether a Process will be Spontaneous – Table 19.2 Based on the values of DH0sys and DS0sys there are 4 types: 1) DH0sys < 0 Exothermic & DS0sys > 0 Less order DS0univ > 0 ________________ under all conditions. 2) DH0sys < 0 Exothermic & DS0sys < 0 More order Depends on values, more favorable at __________ temperatures. 3) DH0sys > 0 Endothermic & DS0sys > 0 Less order Depends on values, more favorable at __________ temperatures. 4) DH0sys > 0 Endothermic & DS0sys < 0 More order DS0univ < 0 ___________________ under any conditions. Remember that –∆H˚sys is proportional to ∆S˚surr An exothermic process has ∆S˚surr > 0.
Classify the following as one of the four types of Table 19.2 DH0 (kJ) DS0 (J/K) CH4 (g) + 2 O2(g) 2 H2O (l) + CO2(g) -890 -242.8 2 FeO3(s) + 3 C (graphite) 4 Fe(s) + 3 CO2(g) +467 +560.7
Calculate the entropy change of the UNIVERSE when 1.890 moles of CO2(g) react under standard conditions at 298.15 K. Consider the reaction6 CO2(g) + 6 H2O(l) C6H12O6 + 6O2(g)for which DHo = 2801 kJ and DSo = -259.0 J/K at 298.15 K. • Is this reaction reactant or product favored under standard conditions?
Gibbs Free Energy DSuniv= DSsurr+ DSsys DSsurr= -DHsys/T DSuniv= -DHsys/T + DSsys Multiply equation by –T -T DSuniv = DHsys –TDSsys J. Willard Gibbs (1839-1903) DGsys = -T DSuniv DGsys = DHsys –TDSsys DGsys< 0, a reaction is ____________ DGsys= 0, a reaction is _____________ DGsys> 0, the reaction is ____________
Gibbs Free Energy and Spontaneity • J. Willard Gibbs (1839-1903) • Gibbs free energy, G, “free energy”, a thermodynamic function associated with the ________________. • G = H –TS • H- Enthalpy • T- Kelvin temperature • S- Entropy • Changes during a process: DG • Use to determine whether a reaction is __________. • DG is ___________related to the value of the _____________________________ , and hence to product favorability.
“Free” Energy • DG = w max • The free energy represents the maximum energy ____________________________. • Example: C(graphite) + 2 H2 (g) CH4 (g) • DH0rx = -74.9 kJ; DS0rx = -80.7 J/K • DG0rx = DH0 – TDS0 • = -74.9 kJ – (298)(-80.7)/1000 kJ • = -74.9 kJ + 24.05 kJ • DG0rx = - 50.85 kJ • Some of the energy liberated by the reaction is needed to “order” the system. The energy left is energy available energy to do_________, “free” energy. • DG < 0, the reaction is _______________.
Calculate DGo for the reaction below at 25.0 C. P4(s) + 6 H2O(l) → 4 H3PO4(l) DG0rx = DH0 – TDS0
Standard Molar Free Energy of Formation • The standard free energy of formation of a compound, DG0f, is the free energy change when forming __________of the compound from the __________________, with products and reactants in their __________________. • Then, DG0f of an element in its standard states is _________.
Gibbs Free Energy DG0rxn is the increase or decrease in free energy as the reactants in their standard states are converted completely to the products in their standard states. * Complete reaction is not always ________________. * Reactions reach an _____________. DG0system = S G0(products) – S G0(reactants)
Calculating DG0rxn from DG0f • DG0system = S G0(products) – S G0(reactants) • Calculate the standard free energy change for the oxidation of 1.0 mol of SO2 (g) to form SO3 (g). • DG0system = DGf0 (kJ/) SO2(g) -300.13 SO3(g) -371.04
Free Energy and Temperature • G = H – TS • G is a function of T, DG will change as T changes. • Entropy-favored and enthalpy-disfavored • Entropy-disfavored and enthalpy-favored
Consider the reaction below. What is DG0at 341.4 K and will this reaction be product-favored spontaneously at this T? CaCO3(s) CaO(s) + CO2(g) Thermodynamic values: DHf0 (kJ/mol) S0 (kJ/Kmol) CaCO3(s) -1206.9 +0.0929 CaO(s) -635.1 + 0.0398 CO2(g) -393.5 + 0.2136
Estimate the temperature required to decompose CaSO4(s) into CaO(s) and SO3(g). Thermodynamic values: DHf0 (kJ/mol) S0 (J/Kmol) CaSO4(s) -1434.52 +106.50 CaO(s) -635.09 + 38.20 SO3(g) -395.77 + 256.77 CaSO4(s) CaO(s) + SO3(g) DH0sys = S H0(products) – S H0(reactants) DH0sys = S S0(products) – S S0(reactants)
For the reaction: 2H2O(l) 2H2(g) + O2(g)DGo = 460.8 kJ and DHo = 571.6 kJ at 339 K and 1 atm. • This reaction is (reactant,product) _____________ favored under standard conditions at 339 K. • The entropy change for the reaction of 2.44 moles of H2O(l) at this temperature would be _________J/K. • DGorxn = DHorxn - DT Sorxn DSo = (DHo - DGo)/T
DG0, K, and Product Favorability • Large K – ____________ favored • Small K – ____________favored • At any point along the reaction, the reactants are not under standard conditions. • To calculate DG at these points: • DG = DG0 + RT ln Q • R – Universal gas constant • T - Temperature (kelvins) • Q - Reaction quotient