Thermodynamics: Spontaneity of Chemical Reactions Entropy Free Energy Direction of Chemical Reactions

Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions Spontaneous process Calculating Entropy Change of a Reaction Entropy, Free Energy, and Work Free Energy, Equilibrium, and Reaction Direction The Second Law of Thermodynamics: Predicting Spontaneous Change

A ball is placed on a slippery slope? Spontaneous Change A spontaneous change occurs without a continuous input of energy from outside the system. What will happen if A ball is placed on a slippery slope? Natural gas is mixed with air and lit with spark? Ice is placed at room temperature? Alkaline battery is connected to a electronic device? An inflated balloon is connected to a evacuated container? activation energy still needs to be overcome

Non-spontaneous Change A nonspontaneous change occurs only if the surroundings continuously supply energy to the system. As if the previous processes is filmed and played reversed A ball rolled UP on a slippery slope Carbon dioxide and water react to give natural gas and oxygen gas at room temperature Depleted alkaline battery resumes fully charged Water freezes at 10°C If a change is spontaneous in one direction, it will be _____spontaneous in the reverse direction.

System vs. Surroundings in Thermodynamics System: Collection of matters undergoing specific chemical/physical changes. Example: Ice cubes melt in the room temperature air; mixture of natural gas and oxygen gas Surroundings: Since System + Surroundings = Universe, Surroundings = Universe – System Example: The air surrounding the melting ice cube.

The First Law of Thermodynamics Energy is conserved. It is neither created nor destroyed, but is transferred in the form of heat and/or work. DE = q + w The total energy of the universe is constant: DEsys = -DEsurr or DEsys + DEsurr = DEuniv = 0 The law of conservation of energy applies to all changes, and does not allow us to predict the direction of a spontaneous change.

Many spontaneous Changes are Exothermic Examples: ball rolling down the slope combustion reactions freezing and condensation at low temperatures oxidation of iron in the air acid-base neutralization reactions

Many spontaneous Changes are Endothermic Some spontaneous processes are endothermic: melting and vaporization at higher temperatures dissolving of most soluble salts. Sr(OH)2(s) + 2NH4Cl(s) → SrCl2(aq) + 2NH3(g) + H2O(l) The sign of DH does not by itself predict the direction of a spontaneous change.

A spontaneous endothermic chemical reaction Ba(OH)2 ·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l) DH°rxn = +62.3 kJ water This reaction occurs spontaneously when the solids are mixed. The reaction mixture absorbs heat from the surroundings so quickly that the beaker freezes to a wet block.

Spontaneity beyond Enthalpy: Freedom of Particle Motion All spontaneous endothermic processes result in an increase in the freedom of motion of the particles in the system. solid → liquid → gas crystalline solid + liquid → ions in solution less freedom of particle motion more freedom of particle motion localized energy of motion dispersed energy of motion A change in the freedom of motion of particles in a system affects the direction of a spontaneous process.

Macrostates Definition: A particular set of values of energy, number of particles and volume of an isolated thermodynamic system. Example: Four chlorine gas molecules in a two-chamber container at 298 K and 1.00 atom. A grain of salt crystal submerged in pure water

Microstates Given a system with a certain macrostate, the particles can have many ways to exist: Like the snapshots of the system taken from a ultrafast camera Example: Four chlorine gas molecules in a system of two compartments. A total of sixteen possibilities. Each possibility represents one microstate. If the volume increases, more compartments are available, more microstates!

Microstates and Dispersal of Energy At any instant, the total energy of the system is dispersed throughout one microstate. At a given set of conditions, each microstate has the same total energy as any other. Each microstate is therefore equally likely. The more possible microstates, the more ways in which a system can disperse its energy, the more freedom of particle motion.

Entropy = Freedom of Particle Motion The number of microstates (W) in a system is related to the entropy (S) of the system. S = k lnW Where k = 8.314 J / Avogadro’s number = 1.38 × 10-23 J/K A system with more microstates has higher entropy All spontaneous endothermic processes exhibit an ___________ in entropy Entropy is a state function: independent of the path taken between the final and initial states.

Measurement of Entropy Change A reversible process is one that occurs in such tiny increments that the system remains at equilibrium, and the direction of the change can be reversed by an infinitesimal reversal of conditions. The tiny increase in entropy (DS) equals the heat absorbed over the temperature of the system while at equilibrium. DS = qrev T

Factors Affecting Entropy Entropy depends on temperature. For any substance, S° increases as temperature increases. Entropy depends on the physical state of a substance. S° increases as the phase changes from solid to liquid to gas. The formation of a solution affects entropy. Entropy is related to atomic size and molecular complexity. Remember to compare substances in the same physical state.

The increase in entropy during phase changes from solid to liquid to gas

Entropy change accompanying the dissolution of a salt: NaCl(s)  NaCl(aq) S > 0 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. pure solid MIX pure liquid solution The entropy of a salt solution is usually greater than that of the solid and of water, but it is affected by the organization of the water molecules around each ion.

Entropy change accompanying the dissolution of gas: O2(g)  O2(aq) S < 0

Entropy and Atomic Size S° is higher for larger atoms or molecules of the same type. Li Na K Rb Cs Atomic radius (pm) 152 186 227 248 265 Molar mass (g/mol) 6.941 22.99 39.10 85.47 132.9 S°(s) 29.1 51.4 64.7 69.5 85.2 HF HCl HBr HI Molar mass (g/mol) 20.01 36.46 80.91 127.9 S°(s) 173.7 186.8 198.6 206.3

Spontaneous without enthalpy change: Expansion of a gas When the stopcock is opened, the gas spontaneously expands to fill both flasks. Increasing the volume increases the number of microstates – and the entropy – increases.

PROBLEM: Predicting Relative Entropy Values Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: (a) 1 mol of CO2(s) or 1 mol of CO2(g) (c) 3 mol of O2(g) or 2 mol of O3(g) (d) 1 mol of KBr(s) or 1 mol of KBr(aq) (e) seawater at 2°C or at 23°C (f) 1 mol of CF4(g) or 1 mol of CCl4(g) B; A; B; B; B

The Second Law of Thermodynamics Considering both parties (the system vs. the surroundings), all real processes occur spontaneously in the direction that increases the entropy of the universe (Suniv). For a process to be spontaneous, any decrease in the entropy of one party must be offset by a larger increase in the entropy of the other party. DSuniv = DSsys + DSsurr > 0

Comparing Energy and Entropy The total energy of the universe remains constant. DEsys + DEsurr =DEuniv = 0 DH is often used to approximateDE. For enthalpy there is no zero point; we can only measure changes in enthalpy. The total entropy of the universe increases. DSuniv =DSsys +DSsurr > 0 For entropy there is a zero point, and we can determine absolute entropy values.

Perfect crystal? A “perfect” crystal has flawless alignment of all its particles. At absolute zero, the particles have minimum energy, so there is only one microstate. Perfect crystal of solid NO Misaligned NO molecules

The Third Law of Thermodynamics A perfect crystal has zero entropy at absolute zero. Ssys = 0 at 0 K A “perfect” crystal has flawless alignment of all its particles. At absolute zero, the particles have minimum energy, so there is only one microstate. S = k lnW = k ln 1 = 0 To find the entropy of a substance at a given temperature, we cool it as close to 0 K as possible. Then it is heated in small increments, measure q and T, and calculate DS for each increment. The sum of these DS values gives the absolute entropy at the temperature of interest.

Standard Molar Entropies S° is the standard molar entropy of a substance, measured for a substance in its standard state in units of J/mol·K. The conventions for defining a standard state include: 1 atm for gases 1 M for solutions, and the pure substance in its most stable form for solids and liquids.

Entropy Changes in the System The standard entropy of reaction, DS°rxn, is the entropy change that occurs when all reactants and products are in their standard states. DS°rxn = SmS°products - SnS°reactants where m and n are the amounts (mol) of products and reactants, given by the coefficients in the balanced equation. DS°rxn is also the entropy change of the system DS°sys

Predicting the Sign of DS°rxn We can often predict the sign of DS°rxn for processes that involve a change in the number of moles of gas (Dng) DS°rxn > 0 if the amount of gas increases; e.g., H2(g) + I2(s) → 2HI(g). DS°rxn < 0 if the amount of gas decreases; e.g., DS° > 0 ? if a new structure forms that has more freedom of motion. N2(g) + 3H2(g) 2NH3(g)

Calculating the Standard Entropy of Reaction, DS°rxn Predict the sign of DS°rxn and calculate its value for the combustion of 1 mol of propane at 25°C. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Given: S° of CO2= 213.7J/K·mol S° of H2O = 69.9J/K·mol S° of C3H8 = 269.9 J/K·mol S° of O2= 205.0 J/K·mol negative -374 J/K

Entropy Changes in the Surroundings The surroundings function as a heat source or heat sink. In an exothermic process, the surroundings absorbs the heat released by the system, and Ssurr increases. qsys < 0; qsurr > 0 and DSsurr > 0 In an endothermic process, the surroundings provides the heat absorbed by the system, and Ssurr decreases. qsys > 0; qsurr < 0 and DSsurr < 0

Determination of DSsurr For any reaction, qsys = -qsurr. The heat transferred is specific for the reaction and is the same regardless of the temperature of the surroundings. The heat released by the system is absorbed by the surrounding, and vice versa. At constant pressure: qsurr = -DHsys DSsurr = - DHsys T for a process at constant P. DSsurr = qsurr T

Problem: Determining Reaction Spontaneity At 298 K, the formation of ammonia has a negative DS°sys; N2(g) + 3H2(g) → 2NH3(g) DS°sys = -197 J/K, DH°rxn = -91.8 kJ Calculate DS°univ, and state whether the reaction occurs spontaneously at this temperature. DSsurr = - DHsys T = 308 J/K DS°univ = DS°sys + DS°surr = 111 J/K

DS and Equilibrium State For any process approaching equilibrium, DS°univ > 0. At equilibrium, there is no further net change, and DS°sys is balanced by DS°surr. DS°univ = DS°sys + DS°surr = 0, so DS°sys = - DS°surr When a system reaches equilibrium, neither the forward nor the reverse reaction is spontaneous, so there is no net reaction in either direction.

Components of DS°univ for spontaneous reactions exothermic For an exothermic reaction in which DSsys > 0, the size of DSsurr is not important since DSsurr > 0 for all exothermic reactions. The reaction will always be spontaneous. exothermic For an exothermic reaction in which DSsys < 0, DSsurr must be larger than DSsys for the reaction to be spontaneous.

Components of DS°univ for spontaneous reactions (contd.) endothermic For an endothermic reaction in which DSsys > 0, DSsurr must be smaller than DSsys for the reaction to be spontaneous. DSsurr is always < 0 for endothermic reactions.

Gibbs Free Energy The Gibbs free energy (G) combines the enthalpy and entropy of a system; G = H – TS. The free energy change (DG) is a measure of the spontaneity of a process and of the useful energy available from it. DGsys = DHsys - T DSsys DG < 0 for a spontaneous process DG > 0 for a nonspontaneous process DG = 0 for a process at equilibrium

Calculating DG° DG°sys = DH°sys - T DS°sys At temperature T, DG°rxn can be calculated using the Gibbs equation: At 298 K, DG°rxn can also be calculated using values for the standard free energy of formation of the components. DG°rxn = SmDG°products - SnDG°reactants

Calculating DG°rxn from Enthalpy and Entropy Values Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. 4KClO3(s) 3KClO4(s) + KCl(s) -391.2 kJ/mol -433.5 kJ/mol -435.9 kJ/mol 143.0 J/mol K 151.0 J/mol·K 82.7 J/mol·K Δ DHf° S°

Calculating DG°rxn from DG°f Values PROBLEM: Use DG°f values to calculate DG°rxn for the reaction 4KClO3(s) 3KClO4(s) + KCl(s) Δ DG°f -296.3 kJ/mol -304.2 kJ/mol -408.3 kJ/mol

DG° and Useful Work DG is the maximal useful work done by a system during a spontaneous process at constant T and P. In reality, some free energy is lost to the surroundings as heat. EXAMPLE: Discharge of rechargeable battery produces heat due to the resistance. DG is the minimal work required to make a nonspontaneous process occur at constant T and P. EXAMPLE: Charging of rechargeable battery requires energy input. A reaction at equilibrium (DGsys = 0) can no longer do any work.

Reaction Spontaneity: Enthalpy vs. Entropy DGsys = DHsys - T DSsys Spontaneous reactions (DGsys < 0) are favored when The reaction is exothermic (DHsys ___ 0) Entropy of the system increases (DSsys ___ 0) A “bipartisanship” in the system (“enthalpy party” vs. “entropy party”) to determine the direction of change Spontaneous reactions are possible only if Both parties approve, or The “Yes” party outweighs the “Nay” party. High Temperature empowers ___________ Party.

Effect of Temperature on Reaction Spontaneity DGsys = DHsys - T DSsys Reaction is spontaneous at all temperatures If DH < 0 and DS > 0 DG < 0 for all T Reaction becomes spontaneous at high T If DH > 0 and DS > 0 DG  as T . Reaction is nonspontaneous at any temperatures If DH > 0 and DS < 0 DG > 0 for all T Reaction becomes spontaneous at low T If DH < 0 and DS < 0 DG  as T decreases.

Reaction Spontaneity and the Signs of DH, DS, and DG -TDS DG Description – + Spontaneous at all T Nonspontaneous at all T + or – Spontaneous at higher T; nonspontaneous at lower T Spontaneous at lower T; nonspontaneous at higher T

Determining the Effect of Temperature on ΔG A key step in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g): 2SO2(g) + O2(g) → 2SO3(g) At 298 K, DG = -141.6 kJ; DH = -198.4 kJ; and DS = -187.9 J/K (a) Is this reaction is spontaneous at 25°C? Predict how DG will change with increasing T. (b) Assuming DH and DS are constant with increasing T, is the reaction spontaneous at 900.° C?

Practice : Assume both the enthalpy change and entropy change of the following reactions remains constant at all temperatures, find the temperature at which each reaction becomes spontaneous (DG < 0) : DH = 60 kJ, DS = 200 J/K; DH = -60 kJ, DS = -200 J/K DH = 60 kJ, DS = -200 J/K DH = -60 kJ, DS = 200 J/K T > 300 K T < 300 K T < -300 K, so not spontaneous at any temperatures. T > -300 K, so spontaneous at all temperatures

The coupling of a nonspontaneous reaction to the hydrolysis of ATP A spontaneous reaction can be coupled to a nonspontaneous reaction so that the spontaneous process provides the free energy required to drive the nonspontaneous process. The coupled processes must be physically connected.

DG, Equilibrium, and Reaction Direction A reaction proceeds spontaneously to the right if Q < K; < 1 so ln < 0 and DG < 0 Q K A reaction proceeds spontaneously to the left if Q > K; > 1 so ln > 0 and DG > 0 Q K A reaction is at equilibrium if Q = K; = 1 so ln = 0 and DG = 0 Q K

DG, Q, and K Q DG = RT ln = RT lnQ – RT lnK K If Q and K are very different, DG has a very large value (positive or negative). The reaction releases or absorbs a large amount of free energy. If Q and K are nearly the same, DG has a very small value (positive or negative). The reaction releases or absorbs very little free energy.

ΔG and the Equlibrium Constant For standard state conditions, Q = 1 and DG° = -RT lnKP A small change in DG° causes a large change in K, due to their logarithmic relationship. As DG° becomes more positive, K becomes smaller. As DG° becomes more negative, K becomes larger. DG = DG° + RT lnQP To calculate DG for any conditions:

{ { { The Relationship Between DG° and K at 298 K DG° (kJ) K Significance 200 9x10-36 Essentially no forward reaction; reverse reaction goes to completion 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 Forward and reverse reactions proceed to same extent -1 1.5 -10 5x101 -50 6x108 -100 3x1017 Forward reaction goes to completion; essentially no reverse reaction -200 1x1035 { FORWARD REACTION REVERSE REACTION { {

Calculating Equilibrium Constant KP from DG° The oxidation of SO2(g) to SO3 is used in the manufacture of sulfuric acid: 2SO2(g) + O2(g) → 2SO3(g) Calculate K at 298 K and at 973 K. Given DG°298 = -141.6 kJ/mol, DG°973 = -12.12 kJ/mol K at 298 K = e57.2 = 7x1024 K at 973 K = e1.50 = 4.5

Calculating DG at Nonstandard Conditions The oxidation of SO2(g) with oxygen gas: 2SO2(g) + O2(g) → 2SO3(g) K at 298 K = 7x1024 K at 973 K = 4.5 (b) Two containers are filled with 0.500 atm of SO2, 0.0100 atm of O2, and 0.100 atm of SO3; one is kept at 25°C and the other at 700.°C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate DG for the system in part (b) at each temperature. Q < K at both temperatures, the reaction will proceed toward the products (to the right) At 298 K, -138.2 kJ/mol At 973 K, -0.9 kJ/mol

Free energy and the extent of reaction. Each reaction proceeds spontaneously (green curved arrows) from reactants or products to the equilibrium mixture, at which point DG = 0. After that, the reaction is nonspontaneous in either direction (red curved arrows). Free energy reaches a minimum at equilibrium.

Why Equilibrium Constant depends on Temperature? DG° = -RT lnK = DH° - TDS° Assuming DH° and DS° remains constant over temperature, then K depends on temperature. Experimental determination of equilibrium constant change versus temperature change gives thermodynamic information (DH° and DS° )

The sign of DG switches at The effect of temperature on reaction spontaneity. DH ΔS T = The sign of DG switches at