Chemical Thermodynamics (Chapter 19)

The First law of Thermodynamics From Chapter 6 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.

Main Ideas The three laws of thermodynamics Qualitative assessment of spontaneous processes Calculating entropy changes for a reaction Calculating Gibbs free energy changes for a reaction Calculating free energy and the equilibrium constant

Formulas ΔG = ΔH – TΔS ΔS°rxn = ΣS°(products) - ΣS°(reactants) ΔG°rxn = ΣG°(products) - ΣG°(reactants) ΔG = ΔG° + RTlnQ ΔG° = -RTlnK

Thermodynamics Driving influences for any chemical and physical change are: Change in enthalpy ΔH Heat transfer between system and surroundings Change in entropy, ΔS Randomness or disorder of a system

Thermo vs. Kinetics Thermodynamics – Kinetics Considers initial and final states of reactants and products Lets us predict whether a process will occur Kinetics Focuses on the pathway between reactants and products Involves mechanisms and rates

Predictions? Is it important to be able to predict if a reaction will occur if two chemicals are combined? How might you make that prediction?

Spontaneous (of natural phenomena) arising from internal forces or causes; independent of external agencies; self-acting.

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 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

Spontaneity Knowing the change in enthalpy (∆H) is not enough to predict spontaneity. Enthalpy (S) obviously does not account for all the energy flow in the process. We must also know about the change in entropy (∆S).

Entropy (S) is a measure of the randomness or disorder of a system. 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

Entropy W = 1 W = 4 W = 6 W = number of microstates S = k ln W DS = Sf - Si DS = k ln Wf Wi Wf > Wi then DS > 0 Wf < Wi then DS < 0

Processes that lead to an increase in entropy (DS > 0)

Practice Choose the substance with the higher entropy per mole at a given temperature. Solid CO2 and gaseous CO2 N2 gas at 1 atm and N2 gas at 0.01 atm

How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 600C to 800C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0)

Entropy State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, enthalpy, pressure, volume, temperature , entropy Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

General trends in physical and chemical systems are towards --- a lower enthalpy and higher entropy

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.

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. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 10

(X)° ° - standard conditions ΔH usually does not equal ΔH°

Appendix Standard values for H, S, and G can be looked up in your book Starting on page A-8

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 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

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

Increasing Entropy

Increasing Entropy

Increasing 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.

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

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 Changes in the System (DSsys) The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn 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, ∆S is negative.

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

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

Absolute Entropy The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

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

Enthalpy (H) Review Endothermic system absorbs heat from surroundings ΔH is positive Exothermic releases heat into the environment ΔH is negative ΔH°rxn = Σ ΔH°f, products - Σ ΔH°f, reactants

Spontaneity By combining information regarding the enthalpy and the entropy of a system we can develop an overall picture of the energy transfer involved. We can use this information to predict the spontaneity of a reaction.

Gibbs Free-Energy Free-energy is the energy available to do work. G

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) ∆G = ∆Hsys -T▪∆Ssys 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.

ΔG A process, at constant T & P, is spontaneous in the direction in which free energy decreases (∆G ‹ 0).

Practice At what temperature is the following process spontaneous at 1 atm? Br2(l)  Br2(g) What is the normal boiling point of liquid bromine?

DG0 of any element in its stable form is zero. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn 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 (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f DG0 of any element in its stable form is zero. f

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 = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0C? DG0 = -6405 kJ < 0 spontaneous

DG = DH - TDS

For a constant temperature and constant pressure process: Gibbs Free Energy Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0 For a constant temperature and constant pressure process: Gibbs free energy (G) DG = DHsys -TDSsys 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.

-DG, therefore the reaction would be spontaneous For the reaction C(s) + O2(g) CO2(g) the values of DH and DS are known to be -393.5 kJ and 3.05 J/K, respectively. Would this reaction be spontaneous at 25 oC? DG = DH - TDS = -393500 J - (298 K)(3.05 J/K) = -394,000 J or -394 kJ -DG, therefore the reaction would be spontaneous Entropy 2009-2010

Spontaneity and Rate A spontaneous reaction does not always occur at a fast rate. They are two different things Thermodynamics can predict whether or not a reaction will occur. It cannot predict the rate.

The Whole Picture Thermodynamics and kinetics are needed to fully understand a reaction.

Practice Using the following data (at 25°C) C(diamond) + O2(g)  CO2(g) ΔG° = -397 kJ C(graphite) + O2(g)  CO2(g) ΔG° = -394 kJ Calculate the ΔG° for the reaction Cdiamond  Cgraphite

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

Phase Changes Phase changes will occur when a system is at equilibrium (DG = 0). At approximately what temperature are the liquid and gaseous bromine at equilibrium? (Or in other words, estimate the normal boiling point of liquid Br2.) Br2(l)  Br2(g) DH = 31.0 kJ/mol DS = 93.0 J/K.mol Entropy 2009-2010

DG = DH – TDS 0 = DH – TDS DH = TDS T = DH DS = 3.10 x 104 J/mol = 333 K 93.0 J/K . mol So this means that at any temp ABOVE 333 K, the above process will be spontaneous, therefore, 333 K is Bromine’s boiling point. Entropy 2009-2010

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

Gibbs Free Energy and Chemical Equilibrium DG = DG0 + 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 = DG0 + RT lnK ∆G0 = - RT lnK

Tip ΔG° is usually in kJ and R is in J/mol●K. Be sure to convert

∆G0 = - RT lnK

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 (-), rx is spon. forward G is 0, rx is at equilibrium If Both T and P are constant, the relationship between G and spontaneity is: G is (-), rx is spon. forward G is 0, rx is at equilibrium G is (+) forward rx is not spontaneous(requires work) reverse rx 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 there 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. CH4 (g) + 2 O2 (g)   →   CO2 (g) + 2 H2O (l) 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). Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s) 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° T is the absolute temperature ΔG° = ΔH° – TΔS° = G = ΔH – TΔS (when nonstandard) If given a temperature change and asked to determine spontaneity or value of G, 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 2 – 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-1 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)

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

∆G, ∆G˚, and Keq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 ---> N2O4 ∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

∆G, ∆G˚, and Keq Reactant-favored reaction. N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ Here ∆Gorxn is greater than ∆Grxn , so the state with both reactants and products present is more stable than complete conversion.

Thermodynamics and Keq Keq is related to reaction favorability. When ∆Gorxn < 0, reaction moves energetically “downhill” ∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

19.6 Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C: 2H2O(l) 2H2(g) + O2(g)

19.6 Strategy The equilibrium constant for the reaction is related to the standard free-energy change; that is, ΔG° = -RT ln K. Therefore, we first need to calculate ΔG°. Then we can calculate KP. What temperature unit should be used?

19.6 Solution ΔG°rxn = [2ΔG°f(H2) + ΔG°f(O2)] - [2ΔG°f(H2O)] = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-237.2 kJ/mol)] = 474.4 kJ/mol Using Equation

19.6 Comment This extremely small equilibrium constant is consistent with the fact that water does not spontaneously decompose into hydrogen and oxygen gases at 25°C. Thus, a large positive ΔG° favors reactants over products at equilibrium.

AgCl(s) Ag+(aq) + Cl-(aq) 19.7 Using the solubility product of silver chloride at 25°C (1.6 x 10-10), calculate ΔG° for the process AgCl(s) Ag+(aq) + Cl-(aq)

19.7 Strategy The equilibrium constant for the reaction is related to standard free-energy change; that is, ΔG° = -RT ln K. Because this is a heterogeneous equilibrium, the solubility product (Ksp) is the equilibrium constant. We calculate the standard free-energy change from the Ksp value of AgCl. What temperature unit should be used?

AgCl(s) Ag+(aq) + Cl-(aq) 19.7 Solution The solubility equilibrium for AgCl is AgCl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.6 x 10-10 we obtain ΔG° = -(8.314 J/K·mol) (298 K) ln (1.6 x 10-10) = 5.6 x 104 J/mol = 56 kJ/mol Check The large, positive ΔG° indicates that AgCl is slightly soluble and that the equilibrium lies mostly to the left.

19.8 The equilibrium constant (KP) for the reaction N2O4(g) 2NO2(g) is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium.

19.8 Strategy From the information given we see that neither the reactant nor the product is at its standard state of 1 atm. To determine the direction of the net reaction, we need to calculate the free-energy change under nonstandard-state conditions (ΔG) and the given ΔG° value. Note that the partial pressures are expressed as dimensionless quantities in the reaction quotient QP because they are divided by the standard-state value of 1 atm.

19.8 Solution Because ΔG < 0, the net reaction proceeds from left to right to reach equilibrium.

19.8 Check Note that although ΔG° > 0, the reaction can be made to favor product formation initially by having a small concentration (pressure) of the product compared to that of the reactant. Confirm the prediction by showing that QP < KP.

Free Energy and the Equilibrium Constant The temperature at which a reaction occurs (becomes spontaneous) is important to the practical chemist. We can make a reasonable estimate of that temperature using the data on your reference sheets.

We can use the value of DGo to calculate the value of DG under nonstandard conditions. DG = DGo + RT lnQ In this equation, R is the ideal gas constant, 8.314 J/mol K, T is the absolute temperature, and Q is the reactant quotient.

Under standard conditions, all the reactants and products are equal to 1. Thus, under standard conditions, Q = 1 and therefore, lnQ = 0, and DG = DGo, as it should under standard conditions. When the concentrations of reactants and products are nonstandard, we must calculate the value of Q to determine DG

Calculate DG for the formation of ammonia at 298 for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.

Calculate DG at 298 for a reaction mixture that consists of 1 Calculate DG at 298 for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3, DGo = -33.3 kJ 3H2 + N2  2NH3 (PNH3)2 (PN2)(PH2)3 (.50)2 (1.0)(3.0)3 = 9.3 x 10-3 Q = =

DG = DGo + RT lnQ DG = -33.3 kJ/mol + (.008314 kJ/mol K)(298 K)ln(9.3 x 10-3) DG = -33.3 kJ/mol + (-11.6 kJ/mol) = -44.9 kJ/mol

at equilibrium, DG = 0, and Q = k We can use the previous equation to derive the relationship between DGo and the equilibrium constant, k, for the reaction. DG = DGo + RT lnQ at equilibrium, DG = 0, and Q = k 0 = DGo + RT ln k DGo = - RT ln k

Given the value of DGo from the previous calculation, solve for the equilibrium constant for the formation of ammonia at 298 K. DGo = - RT ln k -33,300 J/mol = - (8.314 J/mol K)(298 K) ln k 13.4 = ln k Which makes sense, a –DG should equal a k > 1! Taking the inverse ln of both sides… 6.60x 105 = k

Hess’s Law Germain Henri Hess

Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ Step #3: Multiply reaction #2 by 2

Hess’s Law Example Problem Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O     Reaction Ho   C + 2H2  CH4 -74.80 kJ C + O2  CO2 -393.50 kJ H2 + ½ O2  H2O -285.83 kJ CH4  C + 2H2 +74.80 kJ C + O2  CO2 -393.50 kJ 2H2 + O2  2 H2O -571.66 kJ CH4 + 2O2  CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4: CH4 + 2O2  CO2 + 2H2O Hrxn =  Hf(products) -   Hf(reactants)     Substance Hf   CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Imagine this reaction as occurring in 3 steps which add up to the overall equation C3H8 (g)  3 C(graphite) + 4 H2 (g) H1 = + 103.85 kJ 3 C(graphite) + 3 O2 (g)  3 CO2 (g) H2 = -1181kJ 4 H2 (g) + 2 O2 (g)  4 H2O (l) H3 = -1143 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Then H = H1 + H2 + H3 = 103.85 – 1181 -1143 = -2220 kJ

Using Hess’s Law IF you are given an overall equation and a set of equations to arrange so they add up to the overall equation, you need to reverse them or multiply them to get the overall equation. EXAMPLE: Given this information: 2 CO(g)  2 C(s) + O2(g) H1 = 221 kJ CO(g) + 2 H2(g)  CH3OH(g) H2 = 91 kJ Find H for 2 C(s) + O2(g) + 4 H2(g)  2 CH3OH(g)

Hess’s Law To get the final equation, reverse the first equation and change sign of H1 : 2 C(s) + O2(g)  2 CO(g) -DH1 = 221 kJ Double the second equation and double H2 2[CO(g) + 2 H2(g)  CH3OH(g)] 2 DH2 = 2(91 kJ) 2 C(s) + 4 H2(g) + O2(g)  2 CH3OH(g) DHrxn = -DH1 + 2 DH2 DHrxn = 221 kJ + 2(91 kJ) = 403 kJ