Chapter 19: Chemical Thermodynamics Joe DiLosa Hannah Robinson Justin Whetzel Emily Penn Casey White

19.1 – Spontaneous Processes Vocabulary Spontaneous Process – a process that proceeds on its own without any outside assistance Reversible Process – A process that can go back and forth between states along exactly the same path Irreversible Process – A process that is not reversible Isothermal Process – A process that occurs at a constant temperature

19.1 – Spontaneous Processes Occur in a definite direction Ex: When a brick is dropped and it falls down (gravity is always doing work on the brick) Opposite direction (reverse) would be nonspontaneous (takes work to go against gravity) If the surroundings must do work on the system to return it to its original state, the process is irreversible Any spontaneous process is irreversible

19.1 – Spontaneous Processes http://wps.prenhall.com/wps/media/objects/3312/3392504/blb1901.html

19.1 – Spontaneous Processes A spontaneous reaction does not mean the reaction occurs very fast Iron rusting is spontaneous Experimental conditions are important in determining if a reaction is spontaneous (temperature and pressure) To determine spontaneity, distinguish between the reversible and irreversible paths between states

19.1 – Spontaneous Processes Which of the following processes are spontaneous and which are nonspontaneous: (a) the melting of ice cubes at -5°C and 1 atm pressure, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form N2 molecules at 25°C (d) alignment of iron filings in a magnetic field, (e) formation of CH4 and O2 molecules from CO2 and H2O at room temperature and 1 atm of pressure?

19.1 – Spontaneous Processes (a) is nonspontaneous because the ice will not change state (b) is spontaneous because to get the sugar back, the temperature of the system must be altered. (c) is spontaneous because work must be done to form N2 and it is an irreversible process (d) is spontaneous because the system must do work against the magnetic field to go back to its original state (e) is nonspontaneous because a change in temperature and work must be done to break the products up and form the original reactants

19.1 – Spontaneous Processes Practice Problems - #7, 8, and 9

19.2 – Entropy and Second Law of Thermodynamics Vocabulary Entropy – A thermodynamic function associated with the number of different equivalent energy states or spatial arrangements in which a system may be found Second Law of Thermodynamics – in general, any irreversible process results in an overall increase in entropy whereas a reversible process results in no overall change in entropy Infinitesimal difference - an extremely small difference

19.2 – Entropy and Second Law of Thermodynamics Entropy = q/T = S A reversible change produces the maximum amount of work that can be achieved by the system on the surroundings (wrev = wmax) Reversible processes are those that reverse direction whenever an infinitesimal change is made in some property of the system

19.2 – Entropy and Second Law of Thermodynamics Example of Reversible Process Flow of heat between a system and its surroundings Example of Irreversible Process Gas that has been limited to part of a container expands to fill the entire container when the barrier is removed

19.2 – Entropy and Second Law of Thermodynamics ∆S = Sfinal – Sinitial ∆S = qrev/T (where T = constant, isothermal process) qrev = ∆Hfusion when calculating ∆Sfusion for a phase change

19.2 – Entropy and Second Law of Thermodynamics ∆Ssystem = qrev/T ∆Ssurroundings= -qrev/T ∆Stotal = ∆Suniv = ∆Ssystem + ∆Ssurroundings For a reversible process ∆Stotal = 0 For an irreversible process ∆Stotal > 0

19.2 – Entropy and Second Law of Thermodynamics The normal boiling point of methanol (CH3OH) is 64.7˚C and its molar enthalpy of vaporization is ∆Hvap=71.8 kJ/mol. When CH3OH boils at its normal boiling point does its entropy increase or decrease? Calculate the value of ∆S when 1.00 mol CH3OH (l) is vaporized at 64.7 ˚C Increases, the process is endothermic ∆S = qrev/T = ∆Hvap/T 64.7 ˚C + 273 = 337.7 K (1 mol)(71.8 kJ/mol)/(337.7 K)= 0.213 kJ/K = 213 J/K entropy increases

19.2 – Entropy and Second Law of Thermodynamics Practice Problems- # 20 and 21 http://ffden-2.phys.uaf.edu/212_fall2003.web.dir/Anca_Bertus/page_7.htm

19.3 – The Molecular Interpretation of Entropy Vocabulary Translational Motion – movement in which an entire molecule moves in a definite direction Vibrational Motion – the atoms in the molecule move periodically toward and away from one another Rotational Motion – the molecules are spinning like a top Microstate – the state of a system at a particular instant, one of many possible states of the system Third Law of Thermodynamics – the entropy of a pure crystalline substance at absolute zero is zero

19.3 – The Molecular Interpretation of Entropy Vibration Vibration Vibration Rotation Img src = wps.prenhall.com

19.3 – The Molecular Interpretation of Entropy S = k ln W (W = number of microstates) k = Boltzmann’s constant = 1.38 x 10-23 J/K Entropy is a measure of how many microstates are associated with a particular macroscopic state ∆S= k ln Wfinal - k ln Winitial = k ln Wfinal/Wintial Entropy increase with the number of microstates of the system

19.3 – The Molecular Interpretation of Entropy The # of microstates available to a system increases with Increase in volume Increase in temperature Increase in the number of molecules Because any of these changes increases the possible positions and energies of the molecules of the system

19.3 – The Molecular Interpretation of Entropy Generally expect the entropy of a system to increase for processes in which Gases are formed from either solids or liquids Liquids or solutions are formed from solids The # of gas molecules increases during a chemical reaction

19.3 – The Molecular Interpretation of Entropy At absolute zero there is no thermal motion so there is only one microstate As the temperature increases the molecules gain energy in the form of vibrational motion When the molecules increase their motion they have a greater number of microstates

19.3 – The Molecular Interpretation of Entropy http://wpscms.pearsoncmg.com/au_hss_brown_chemistry_1/57/14649/3750314.cw/content/index.html

19.3 – The Molecular Interpretation of Entropy Practice Problems- # 28, 30, 38, 40

19.4 – Entropy Changes in Chemical Reactions Vocabulary Standard Molar Entropy – The molar entropy value of substances in their standard states

19.4 – Entropy Changes in Chemical Reactions The standard molar entropies of elements are not 0 at 298K The standard molar entropies of gases are greater than those of liquids and solids standard molar entropies generally increase with increasing molar mass Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance

19.4 – Entropy Changes in Chemical Reactions Pictures taken from p820 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten

19.4 – Entropy Changes in Chemical Reactions ∆S˚= ΣnS˚(products) - ΣmS˚(reactants) ∆Ssurr= -qsys/T For a reaction at constant pressure qsys= ∆H

19.4 – Entropy Changes in Chemical Reactions Calculate ∆S˚ values for the following reaction: C2H4 (g) + H2 (g)  C2H6 (g) ∆S˚= ΣnS˚(products) - ΣmS˚(reactants) = 229.5 J/molK – (219.4 J/molK + 130.58 J/molK) = -120.5 J/molK

19.4 – Entropy Changes in Chemical Reactions Practice Problems - # 48

19.5 - Gibbs Free Energy Vocabulary Gibbs free energy- a thermodynamic state function that combines entropy and enthalpy Standard free energies of formation- the change in free energy associated with the formation of a substance from its elements under standard conditions

19.5 - Gibbs Free Energy Free energy (G) is defined by: G = H-TS Where T is absolute temperature, S is entropy, and H is enthalpy

19.5 - Gibbs Free Energy When temperature is constant the equation for the change in free energy for the system is ΔG=ΔH-TΔS If T and P are constant then the sign of ΔG and spontaneity of a reaction are related. If ΔG < 0, reaction proceeds forward If ΔG = 0, reaction is at equilibrium If ΔG > 0, the forward reaction is not spontaneous because work must be supplied from the surroundings but the reverse reaction is be spontaneous.

19.5 - Gibbs Free Energy In any spontaneous process at constant temperature and pressure the free energy always decreases

19.5 - Gibbs Free Energy Conditions of Standard free energies of formation (implied) 1 atm pressure for gases pure solids pure liquids 1M solutions 25˚ C (normally) The free energy of elements is 0 in their standard states.

19.5 - Gibbs Free Energy ΔGf˚ = standard free energies of formation Calculated the same way as ΔH ΔG˚ = ΣnΔGf˚ (products) - ΣmΔGf˚ (reactants) ΔG = -wmax

19.5 - Gibbs Free Energy http://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Gibbs_free_energy_graph

19.5 – Gibbs Free Energy Practice Problems - #49, 51, 53, 54, 56, 58, 63, 66

19.5 – Gibbs Free Energy For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K? ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J

19.6 – Free Energy and Temperature Vocabulary No new vocabulary

19.6 – Free Energy and Temperature When ΔH and –TΔS are negative, ΔG is negative and the process is spontaneous When ΔH and –TΔS are positive, ΔG is positive and the process is non-spontaneous When ΔH and –TΔS are opposite signs, ΔG will depend on the magnitudes http://www.chem1.com/acad/webtext/thermeq/TE4.html

19.6 – Free Energy and Temperature Picture taken from p828 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten

19.6 – Free Energy and Temperature For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. is the reaction exothermic or endothermic? Does the reaction lead to an increase or decrease of disorder in the system? Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K?

19.6 – Free Energy and Temperature A) The reaction is exothermic because ΔH is negative B) It leads to a decrease in the disorder of the system (increase in the order of the system) because ΔS is negative C) ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J D) Yes, if all reactants are in their standards states because ΔG is negative

19.6 – Free Energy and Temperature Practice problems (same as 19.5) - #49, 51, 53, 54, 56, 58, 63, 66

19.7 – Free Energy and Equilibrium Constant Vocabulary Nonstandard conditions – conditions other than the standard conditions

19.7 – Free Energy and Equilibrium Constant ΔG = ΔGo + RT ln Q Where R = 8.314 J/mol-K, T = absolute temperature, Q = reaction quotient ΔGo = ΔG under standard conditions Because Q = 1 and lnQ = 0

19.7 – Free Energy and Equilibrium Constant ΔGo = -RT ln K At equilibrium, ΔG = 0 K = e –(delta G standard)/RT Smaller (more negative ΔG0), the larger the K value

19.7 – Free Energy and Equilibrium Constant Direction of Reaction Keq ΔG Toward forming more products >1 negative Toward forming more reactants <1 positive Products and reactants equal 1 0 http://www.wiley.com/legacy/college/boyer/0470003790/reviews/thermo/thermo_keq.htm

19.7 – Free Energy and Equilibrium Constant #71 – Explain Quantitatively how ΔG changes for each of the following reactions as the partial pressure of O2 is increased: A) 2 CO(g) + O2(g) → 2 CO2 (g)

19.7 – Free Energy and Equilibrium Constant 2 CO(g) + O2(g) → 2 CO2 (g) -137.2 kJ/mol 0 kJ/mol -394.4 kJ/mol ΔG = products – reactants ΔG = -394.4 kJ/mol + 137.2 kJ/mol ΔG = -257.2 kJ/mol

19.7 – Free Energy and Equilibrium Constant Practice Problems: #71,76,79

Additional Chapter Problems #82, 87, 88, 89, 90, 92