Thermodynamics 2: Gibbs Free Energy and Equilibrium

Reading: Moore chapter 18, sections 18.6-18.11Questions for Review and Thought: 62, 69, 71, 73, 78, 83, 99, 102.

Key Concepts and skills:• definitions of Gibbs Free Energy (ΔG), Equilibrium constant (Keq), Reaction

Quotient (Q), Standard State Gibbs Free Energy (ΔG°)• Review the methods of calculation of ΔG°, the prediction of reaction spontaneity

from the sign of ΔG, Prediction of the effect of temperature on reaction direction,the relation between ΔG and the equilibrium constant (K) and reaction quotient(Q).

• Understand the following concepts: 1.) the relation between ΔG and work; 2.)how “coupling” reactions can make practical use of Gibbs energy “work”, and 3.)the distinction between thermodynamic stability and kinetic stability.

Lecture Topics:

Review:Prediction of reaction spontaneity: ΔSuniverse>0 reaction spontaneous in forward directionΔSuniverse=0 reaction at equilibrium (a reversible process)ΔSuniverse<0 reaction spontaneous in the reverse direction

• Useful to define reaction spontaneity by considering properties of the system only:New parameter: ΔGsystem= -TΔSuniverseThus, ΔGsystem= ΔHsystem-TΔSsystemUnder standard state conditions, ΔG°system= ΔH°system-TΔS°system (1 atm pressure or 1Mconcentrations)

DefinitionsThe Gibbs Free Energy (G) of a substance measures a molecule’s inherentlikelihood or potential to undergo a change or reaction.In general, at constant T and P,when ΔGsystem< 0, the process/reaction is spontaneous in the forward directionwhen ΔGsystem= 0, the process/reaction is in equilibrium/ reversiblewhen ΔGsystem> 0, the process/reaction is spontaneous in the reverse direction

• The Gibbs energy of a substance depends on its temperature, pressure, phase, andconcentration.

• Gibbs energy changes for chemical reactions are often compared under a well-defined set of standard-state conditions (ΔG°rxtn): 1 atm gases and solutes at 1Mconcentation.

•ΔG°rxtn values are computed using tabulated ΔG°f values for all reactants andproducts: ΔG°rctn= Σ[(moles of product)* ΔGf°product] - Σ[(moles of reactant)* ΔGf° reactant]

• as for ΔHf°, ΔGf° = 0 for an element in its standard state.• Chemical equations can be added together, with their ΔGf° values combined.

Effect of Temperature on ΔG

• Values of ΔGrxn° calculated from Appendix J are accurate only at 298 K. Values ofΔGrxn° can be estimated for reactions at other temperatures and at P=1atm using

ΔG°= ΔH°-TΔS°and tables of standard entropies and enthalpies of formation. The estimates will beclose to the true values if it is assumed ΔH° and ΔS° are not strongly dependent on T.• The value of ΔG° depends strongly on T. There exists a special temperature T*

defined by: T* = ΔH°/ ΔS° at which ΔG°=0•If both ΔH° and ΔS° are positive, the reaction will be spontaneous at temperatureshigher than T• If both ΔH° and ΔS° are negative, the reaction will be spontaneous at temperatureslower than T.

Effects of Concentration on ΔG(for non-standard conditions)For the reaction: a A + b B ––> cC + dD where Q= [A]a[B]b/[C]c[D]d

From the empirical study of the effects of concentration on G, it has been found thatΔG = ΔG° + RT ln Q

For standard conditions (concentrations of all reactants and products 1mol/L or 1 atm)Q=1 and ΔG = ΔG°. When ΔG=0, the reaction is at equilibrium (there is no tendencyfor the reaction to occur in either the forward or revese direction), thus

ΔG°= -RT ln K where K is the thermodynamic equilibriumconstant for the reaction at a specific temperature T:

Q (at equilibrium) =K=[A]eqa[B]eq

b/[C]eqc[D]eq

d = e-ΔG°/RT

Combining the above equations gives the simple result:ΔG = RT lnQ/K When Q>K, ΔG>0, forward reaction is non-spontaneous

When Q<K, ΔG<0, forward reaction is spontaneousWhen Q=K, ΔG=0, the reaction is at equilibrium.

(Dependence of K on temperature:Since -RT ln K= ΔG°=ΔH°-TΔS°; lnK = -ΔH°/RT+ΔS°/RA plot of K vs. 1/T gives straight line with slope = -ΔH°/R and intercept ΔS°/RFurthermore: lnK2/K1= -ΔH°/R [1/T2 – 1/T1] )

ΔG and Work; Coupling reactions

For ΔG°>0, ΔG° represents the minimum amount of work that must be done toforce a reactant-favored process to produce products.For ΔG°<0, ΔG° represents the maximum amount of useful work that can be doneby a product-favoring system on its surroundings.Example: C8H18(l)+ 25/2 O2(g)––> 8 CO2(g) + 9 H2O (l) ΔG° = -5295 kJPart of the Free energy given off in this combustion powers the motion of theautomobile

The “useful work” can also be used to drive a reactant-favored process in thereverse “product-favored” direction.

Example: Coupling processes for the reduction of iron ore:

2Al(s) + 3/2O2 (g) ––> Al2O3(s) ΔG° = -1582 kJFe2O3(s)––> 2 Fe(s) + 3/2O2(g) ΔG° = 742 kJ

Net: 2Al (s) + Fe2O3(s) ––> 2 Fe(s) + Al2O3(s) ΔG° = -840 kJ (a favored reaction)

Thermodynamic/Kinetic stability Ea1 Ea2

Consider the following process : P1 <–– SM ––> P2 Ea1<Ea2Where for SM––>P1 ΔG= -50kJ/molAnd SM ––> P2 ΔG= -150kJ/mol

A Chemical process at low temperature is under kinetic control if it leadspreferentially to the less thermodynamically stable product P1

The same chemical process at high temperature is under thermodynamiccontrol if it leads preferentially to the more thermodynamically stable product P2.

Problems Set #4

1. Decide if each of the following statements are true or false. If false, correct thestatement to make it a valid assertion:

(a) Reactions with a positive ΔHrxn and a positive ΔSrxn can never bespontaneous

(b) At absolute zero (0 K) an endothermic reaction is never spontaneous(c) A reaction with a negative Gibbs energy change is predicted to be

spontaneous with a rapid transformation of reactants to products(d) The entropy of a substance increases from the liquid to the vapor state(e) When ΔG° is positive for a reaction, the reaction cannot take place.

2. The oxidation of SO2 (g) is too slow at 298K to be useful in the manufacture ofsulfuric acid, so the process is conducted at an elevated temperature (973K) toproduce product at a faster rate: 2 SO2(g) + O2(g) <-> 2SO3 (g)(a) Given one mole of each of the starting materials and sufficient time, at what

temperature would you obtain the greater amount of SO3(g), 298K or 973K?Use thermodynamic data from Appendix J in support of your answer.

(b) What is the equilibrium constant for the reaction at 298K and at 973K,assuming a temperature-independent enthalpy and entropy of reaction?

(c) In experiments to determine the effect of temperature on reaction spontaneity,sealed containers are filled with 0.50 atm SO2, 0.010 atm O2 , and 0.20 atmSO3 and kept at 298K and 973K, respectively. By calculating ΔGsystem at eachtemperature, determine in which direction the reaction will proceed to reachequilibrium at each temperature

3. At 25°C and 1 atm, the following reactions are NOT spontaneous in the forwarddirection. Indicate whether an INCREASE, a DECREASE, or NOT POSSIBLECHANGE IN TEMPERATURE will make the forward reaction spontaneous atconstant T and P.

(a) Br2(liq) Br2 (g)(b) 2 NO2(g) N2O4(g)(c) CdS(s) Cd(s) + S(s)

4. One promising method of making coal a cleaner fuel involves the conversion ofcoal to methane. One such scheme follows:

(1) C (s) + H2O (g) CO(g) + H2(g)(2) CO (g) + H2O (g) CO2(g) + H2(g)(3) CO (g) + 3 H2(g) CH4 (g) + H2O (g)

(a) For each of the three reactions, calculate ΔH°, ΔS°, and ΔG° at 298K and1atm.

(b) Over what temperature ranges will each of these reactions be spontaneous?(c) Write the equation for the overall reaction, with the net equation consuming

the intermediate CO (you need to carry out step 1 twice for this to occur)(d) What is the Enthalpy and Gibbs Free Energy change for the overall (net)

reaction? Is this reaction spontaneous at 298K? If not, over what temperaturerange will it be spontaneous. Do you think this will be a practical process?

5. For the melting of one mole of ice at 10°C and 1 atm, what is the sign of eachquantity below? Assume that the density of ice is less than that of liquidwater at 10°C. Provide a brief explanation of your responses.

>0 =0 <0 Can’t tellΔEΔVΔTΔPqw

ΔSsysΔSsurrΔHsysΔSuniverseΔGsys

Remember: qp=ΔHsys, and wirrev=-PextΔVNote the enthalpy of fusion (melting) of water is 6 kJ/mol

Bonus:6. For the dimerization reaction 2NO2(g) N2O4(g), use the data below tographically determine ΔH°rxtn and ΔS°rxtn (assume that both ΔH°rxtn and ΔS°rxtn aretemperature independent):

T (K) 282.2 293.2 298.2 306.2 325.2 333.2 343.2

Keq 33.2 13.1 8.79 4.77 1.26 0.751 0.408