Spontaneity of Reaction Chapter 16

Key Concepts of Chapter 16 • Identifying Spontaneous Processes. • Identifying reversible and irreversible processes. • Entropy and its relation to randomness. • Second Law of Thermodynamics. • Predicting Entropy Changes of a Process. • Third Law of Thermodynamics. • Relate temperature change to entropy change. • Calculating change in standard entropy.

• Free energy in terms of enthalpy and entropy. • Relating free energy change to spontaneity. • Calculating standard free energy change. • Relationship between free energy and work. • Calculating free energy Δ under nonstandard conditions.

The First law of Thermodynamics From Chapter 8 Chemical thermodynamics is the study of energy relationships in chemistry. The First law of Thermodynamics - energy cannot be created or destroyed only converted from one form to another.

Enthalpy heat transfer between the system and its surroundings under constant pressure. Enthalpy is a guide to whether a reaction is likely to proceed. It is not the only factor that determines whether a reaction proceeds.

Spontaneous Processes Occur without outside intervention Have a definite direction. The reverse process is not spontaneous. Temperature has an impact on spontaneity. Ex: Ice melting or forming Ex: Hot metal cooling at room temp.

KI (aq) + Pb(NO3)2 (aq)  PbI2(s) + KNO3 (aq) When mixed  Precipitate forms spontaneously. *It does not reverse itself and become two clear solutions.

Reversible & Irreversible System changes state and can be restored by reversing original process. Ex: Water (s) Water (l) Irreversible: System changes state and must take a different path to restore to original state. Ex: CH4 + O2  CO2 + H2O State functions: temperature , internal energy, and enthalpy These define a state and do not depend on how the state was reached. was reached. Q( heat transferred between systems and surroundings) and w (work done by or on the system) do depend on the path taken from one state to another. Reversible Irreversible-

*Scrambled eggs don’t unscramble* Whenever a system is in equilibrium, the reaction can go reversibly to reactants or products (water  water vapor at 100 º C). In a Spontaneous process, the path between reactants and products is irreversible. (Reverse of spontaneous process is not spontaneous). *Scrambled eggs don’t unscramble* Discuss the idea that at a constant temperature of 100 degrees Celsius liquid water molecules evaporate as gaseous water molecules are recaptured by the liquid into liquid form. (They are in equilibrium).

The Second Law of Thermodynamics - The entropy of the universe always increases in a spontaneous process and remains unchanged in an equilibrium process.

“But ma, it’s not my fault… the universe wants my room like this!”

Entropy (S) A measure of randomness or disorder S = entropy in J/K·mole Increasing disorder or increasing randomness is increasing entropy. Three types of movement can lead to an increase in randomness.

Entropy is a state function Change in entropy of a system S= Sfinal- Sinitial Depends only on initial and final states, and not the pathway. -S indicates a more ordered state (think: < disorder or - disorder) Positive (+) S = less ordered state (think: > disorder or + disorder)

Entropy, S - a measure of disorder Ssolid  Sliquid  Sgas

Increasing Entropy

Increasing Entropy

Increasing Entropy

Net increase in the entropy of the universe If entropy always increases, how can we account for the fact that water spontaneously freezes when placed in the freezer? • Movement of compressor + • Evaporation and condensation of refrigerant • Warming of air around container Net increase in the entropy of the universe

On the AP exam, you will likely be asked to: predict whether a process leads to an increase in entropy or a decrease in entropy. Determine if ΔS is + or – Determine substances or reactions that have the highest entropy.

Processes that lead to an Increase in Entropy When a solid melts. When a solid dissolves in solution. When a solid or liquid becomes a gas. When the temperature of a substance increases. When a gaseous reaction produces more molecules. If no net change in # of gas molecules, can be + or -, but small.

Freezing liquid bromine Predict whether the entropy change is greater than or less than zero for each of the following processes: Freezing liquid bromine Evaporating a beaker of ethanol at room temperature Dissolving sucrose in water Cooling N2 from 80ºC to 20ºC S<0 S>0 System becomes more ordered b) MORE DISORDER C.) More disordered d.) Cooling decreases molecular motion therefore S<0 S>0 S<0

1.) Ag+(aq)+ Cl-(aq)AgCl(s) 2.) NH4Cl(s) NH3(g)+ HCl(g) Predict whether the entropy change of the system in each of the following reactions is positive or negative: 1)S – 2)S+ 3)S? 1.) Ag+(aq)+ Cl-(aq)AgCl(s) 2.) NH4Cl(s) NH3(g)+ HCl(g) 3.) H2(g) + Br2(g)2HBr(g) AgCl is solid and there are less particles moving from left to right Solid is converted to 2 gasses. Gas molecules are the same on each side, not known whether it is positive or negative; it is a small number.

According to the 2nd law of thermodynamics; the entropy of the universe always increases. ? What if the entire senior class assembles in the auditorium? Aren’t we decreasing disorder, and therefore decreasing entropy? If so, how can the second law of thermodynamics be true?

The magnitude of the entropy increase of the surroundings will If we consider the senior class as the system, the ΔS of the system would indeed decrease. ΔS of the system is – In order for the students to gather, they would: Metabolize food (entropy increase of surroundings) Generate heat (entropy increase of surroundings) The magnitude of the entropy increase of the surroundings will always be greater than the entropy decrease of the system.

+ means entropy increases Theoretical values Suniverse = Ssystem + Ssurroundings Suniverse = (-10) + (+20) Suniverse = +10 + means entropy increases

The same can be considered in a chemical process. When a piece of metal rusts: 4Fe(s) + O2(g)  2Fe2O3(s) The entropy of the solid slowly decreases. The decrease in entropy of the system is evident just by looking at the fact that the product is more ordered. Although this is a slow process, it is exothermic, and heat is released into the surroundings causing an overall increase in entropy of the universe!

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0C and ice melts above 0 0C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous

spontaneous nonspontaneous

Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions at 25 °C CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = -890.4 kJ H+ (aq) + OH- (aq) H2O (l) DH0 = -56.2 kJ H2O (s) H2O (l) DH0 = 6.01 kJ NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ H2O

TWO Trends in Nature Order  Disorder   High energy  Low energy 

The Driving Forces Energy Factor (ΔH) Randomness Factor (ΔS) or more lately the “dispersal” factor.

Ssolid < Sliquid << Sgas Entropy (S) is a measure of the randomness or disorder of a system…the tendency to spread energy out – disperse energy. disorder S order S DS = Sf - Si If the change from initial to final results in an increase in randomness Sf > Si DS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state Ssolid < Sliquid << Sgas H2O (s) H2O (l) DS > 0

First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0

The Second Law of Thermodynamics - The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Suniverse = Ssystem + Ssurroundings Spontaneous = irreversible, equilibrium means that it can return to exact same state without a net change in the surroundings or the universe (ex: water at 100 degrees celcius) Equilibrium = reversible Suniverse > 0 for spontaneous rxn Suniverse = 0 at equilibrium

In terms of temperature, how would you describe an object that has an entropy value of 0? 0 K Perfect solid crystal with no motion Only Theoretical It is not possible to reach absolute 0! Entropy of universe is always increasing!

3rd Law of Thermodynamics the entropy of a perfect crystalline substance is zero at absolute zero *Based on 0 entropy as a reference point, and calculations involving calculus beyond the scope of this course, data has been tabulated for Standard Molar Entropies ΔSº Pure substances, 1 atm pressure, 298 K

Standard Molar Entropies Standard molar entropies of elements are not 0 (unlike ΔHºf). (0 entropy is only theoretical; not really possible) S.M.E of gases > S.M.E of liquids and solids. (gases move faster than liquids) 3) S.M.E. increase with increasing molar mass. (more potential vibrational freedom with more mass) 4) S.M.E. increase as the number of atoms in a formula increase. (same as above)

Since we are considering ΔS° Calculating the Entropy Change Sorxn = n So(products) - m So(reactants) Units for S S=J/mol•K Since we are considering ΔS° J/K are often used because moles are assumed and cancel in the calculations when considering standard states. n and m are the coefficients form a balanced equation

Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g) Calculate the standard entropy change (Sº) for the following reaction at 298K Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g) Substance Sº(J/mol-K) at 298K Al 28.32 Al2O3 51.00 H2O(g) 188.8 H2(g) 130.58

Sorxn = n So(products) - m So(reactants) Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g) So = [2Sº(Al) + 3Sº(H2O)] - [Sº(Al2O3) + 3Sº(H2)] 180.3 J/K = 180.4 J/K

Predict the sign of ΔSº of the following reaction. 2SO2(g) + O2(g) 2SO3(g) Entropy decreases, - Lets’ Calculate

Calculate the standard entropy change (Sº) for the following reaction at 298K 2SO2(g) + O2(g) 2SO3(g) Substance Sº(J/mol-K) at 298K SO2(g) 248.1 SO3(g) 256.7 O2(g) 205.0 ΔSº = -187.8 J K-1

Predicting spontaneous reactions Spontaneous reactions result in an increase in entropy in the universe. Rx’s that have a large and negative  tend to occur spontaneously. Spontaneity depends on enthalpy, entropy, and temperature.

Gibbs Free Energy (G) Provides a way to predict the spontaneity of a reaction using a combination of enthalpy and entropy of a reaction.

G is (-), forward rxn is spontaneous. G is 0, rxn is at equilibrium. If Both T and P are constant, the relationship between G and spontaneity is: G is (-), forward rxn is spontaneous. G is 0, rxn is at equilibrium. G is (+) forward rxn is not spontaneous (requires work) reverse rxn is spontaneous.

Coefficients from equation (sum of standard free energies of formation of products) minus (sum of standard free energies of formation of reactants) the sum of Coefficients from equation

The values of ΔGºf of elements in their most stable form is 0, just as with enthalpy of formations.

Substance Gº(KJ/mol) at 298K CH4 -50.8 CO2 -394.4 H2O(l) -237.2 Calculate the ΔG°rxn for the combustion of methane at 298K and determine if the reaction is spontaneous. Substance Gº(KJ/mol) at 298K CH4 -50.8 CO2 -394.4 H2O(l) -237.2 -818.0 KJ Spontaneous

Calculate the (Gº) for the thermite reaction (aluminum with iron(III) oxide). Substance Gº(KJ/mol) at 298K Al2O3 (s) -1576.5 Fe2O3 (s) -740.98 -835.5 KJ spontaneous

Gibbs Free Energy: Equation 2 This equation allows us to determine if a process provides energy to do work. A spontaneous reaction in the forward direction provides energy for work. If not spontaneous, ΔG equals the amount of energy needed to initiate the reaction. Allows us to calculate the value of ΔG as temperature changes. Gibbs free energy (G°) is a state function defined as: ΔG° = ΔH° – TΔS° (Given on AP Exam) T is the absolute temperature ΔG° = ΔH° – TΔS°  ΔG = ΔH – TΔS (when nonstandard) If given a temperature change and asked to determine spontaneity or value of ΔG, this is the equation you would use. Value of ΔG tells us if a reaction is spontaneous.

G = H – T(S) If we know the conditions of ΔH and ΔS, we can predict the sign of G. We will see that: Two conditions always produce the same result, and two conditions depend on temperature.

Predicting Sign of ΔG in Relation to Enthalpy and Entropy - + Always negative (spontaneous) + - Always positive (nonspontaneous) - - Neg. (spontaneous) at low temp Pos. (nonspontaneous) at high temp. + + Pos. (nonspontaneous) at low temp Neg. (spontaneous) at high temp.

- + + - - - + + ΔH ΔS ΔG Different sign, not temperature dependent. - + Different sign, not temperature dependent. + - - - Freezing Same sign, temperature dependent. + + Freezing: This process is exothermic, so enthalpy is -. Freezing causes a decrease in entropy, so entropy is -. Freezing is only spontaneous at low temperatures. Opposite is true for melting. Melting

G = H – T(S) Some reactions are spontaneous because they give off energy in the form of heat (ΔH < 0). Others are spontaneous because they lead to an increase in the disorder of the system (ΔS > 0). Calculations of ΔH and ΔS can be used to probe the driving force behind a particular reaction.

Ag+(aq) + Cl-(aq)  AgCl(s) Example – The entropy change of the system is negative for the precipitation reaction: Ag+(aq) + Cl-(aq)  AgCl(s) Ho = -65 kJ Since S decreases rather than increases in this reaction, why is this reaction spontaneous?

Yes, entropy increases, which goes against the second law Yes, entropy increases, which goes against the second law. However, in this case, the entropy decrease is minimal compared to the magnitude of change in enthalpy. Therefore, the release of heat drives the reaction to stability, which is why it is spontaneous. Theoretical Values G = H – TS G = -65 – 298(-.030) = -73.94

Problem For a certain reaction, ΔHº = -13.65 KJ and ΔSº = -75.8 J K-1. What is ΔGº at 298 K? B) Will increasing or decreasing the temperature make the reaction become spontaneous? If so, at what temperature will it become spontaneous?

Given: ΔHº = -13.65 KJ ΔSº = -75.8 J /K T = 298 K A) What is ΔGº at 298 K? At 298 K the free energy is ΔG° = ΔH° – TΔS° ΔG° = -13.65 KJ – 298(-.0758 KJ K-1) = +8.94 KJ (Reaction is not spontaneous at 298 K)

Reaction is spontaneous below 180 K, not spontaneous above 180 K B) Will increasing or decreasing the temperature make the reaction become spontaneous? If so, at what temperature will it become spontaneous? Because enthalpy and entropy have the same signs, spontaneity is indeed temperature dependent. *Since going from not spontaneous to spontaneous crosses the point of equilibrium, and ΔG° = 0 at equilibrium, we can make ΔG° = to 0 to find the temperature at which equilibrium is crossed. 0 = ΔH° – TΔS° 0 = -13.65 KJ – T(-.0758 KJ K-1) T = 13.65 KJ 0.0758 KJ K-1 T = 180 K Reaction is spontaneous below 180 K, not spontaneous above 180 K

Entropy Changes in the System (DSsys) The standard entropy of reaction (DS0rxn) is the entropy change for a reaction carried out at 1 atm and 250C. aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) DS0 rxn nS0(products) = S mS0(reactants) - What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

Entropy Changes in the System (DSsys) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) The total number of gas molecules goes down, DS is negative.

Entropy Changes in the Surroundings (DSsurr) Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0

Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

Gibb’s Free Energy --The energy of a system that is available to do work. ΔGorxn = ΔHorxn – TΔSorxn . Hess’ Law also applies: ΔGrxn = ΣΔGfo(products) – ΣDGfo (reactants) ΔGfo has a similar definition as ΔHfo If ΔGorxn is positive, the rxn is nonspontaneous at that temperature. If ΔGorxn is negative, the rxn is spontaneous at that temperature. If ΔGrxn is zero, the rxn is at equillibrium.

For a constant-temperature process: Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant-temperature process: Gibbs free energy (G) DG = DHrxn -TDSrxn DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

DGo of any element in its stable form is zero. The standard free-energy of reaction (DGorxn ) is the free-energy change for a reaction when it occurs under standard-state conditions. aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - Standard free energy of formation (DGfo) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DGo of any element in its stable form is zero. f

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) What is the standard free-energy change for the following reaction at 25 0C? 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0 (CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12(–394.4) + 6(–237.2) ] – [ 2(124.5) ] = -6405 kJ Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous

ΔGfo ΔHfo TΔSfo -@ high T + + ? - - -@ low T ? - +@ all T + ? - +

DG = DH - TDS

Temperature and Spontaneity of Chemical Reactions CaCO3 (s) CaO (s) + CO2 (g) DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ DG0 = 0 at 835 0C

Gibbs Free Energy and Phase Transitions T = ΔH/ ΔS DG0 = 0 = DH0 – TDS0 H2O (l) H2O (g) DS = T DH = 40.79 kJ 373 K = 109 J/K

DGo = - RT lnK DG = DGo + RT lnQ Gibbs Free Energy and Chemical Equilibrium Non-standard Conditions: DG = DGo + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium DG = 0 Q = K 0 = DGo + RT lnK DGo = - RT lnK

DGo = - RT lnK

Chemistry In Action: The Thermodynamics of a Rubber Band TDS = DH - DG High Entropy Low Entropy

Phase Changes The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.

H2O (s) H2O (l) The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium Melting Freezing

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance.

H2O (s) H2O (g) Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid. Sublimation Deposition DHsub = DHfus + DHvap ( Hess’s Law)