Gibbs Free Energy Study Guide Name: Date: Period:

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1 Gibbs Free Energy Study Guide Name: Date: Period: The basic goal of chemistry is to predict whether or not a reaction will occur when reactants are brought together. Ways to predict spontaneous reactions or processes: a. if ΔH < 0 reactions tend to go spontaneously. (exothermic rxn.) ; \ b. if K >1 reactions tend to go (they are product favored ) c. If ΔG is negative "Nature tends to move spontaneously from a state of lower probability to one of higher probability." Entropy: describes the randomness involved in energy transactions. A system that is more random has a higher S. So, if ΔS is +, products are more random than reactants. Entropy, ΔS is a State function You need to know what a state function is in order to know the significance of this revelation. I. Spontaneous Reactions A. Examples: H2O(s) H2O(l) at 25 C CH4(g) + 2O2(g) CO2(g) + 2H2O(l) B. Factors Affecting spontaneity: Energy factor: at 25 C, 1 ATM, exothermic reactions are ordinarily spontaneous. Randomness Factor: Other things being equal, systems tend to move from a more ordered to a more random structure. temp(k) II. Entropy Changes ΔS = ΔS products ΔS reactants A. Measure of Change in ordering Solid liquid; ΔS + Liquid Gas; ΔS + ΔS is usually + for a reaction in which the # of moles or gas increases. 2SO3(g) 2SO2 (g) + O2(g); Δng =? so ΔS=?

2 N2(g) + 3H2(g) 2NH3 (g); Δng =? so ΔS =? B. Calculations of ΔS o ( o means 1 atm, 1M, 25 o C) ΔS o = S o products - S o reactants Example: Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(g) ΔS o = 2S o Fe(s) + 3S o H2O(g) - S o Fe2O3(s) - 3S o H2(g) = 2(27.3 J/K) + 3(188.7 J/K) J/K - 3(130.6) = J/K ΔS o is nearly temp. independent C. Reactions for which ΔS is + tend to be spontaneous, at least at high temp. H2O(s) H2O(l) ΔS= +/- ΔH= +/- H2O(l) H2O(g) ΔS= +/- ΔH= +/- At what conditions do these reactions become spontaneous? III. Second law of thermodynamics: In a spontaneous process, the entropy of the universe increases. ΔSuniv > 0 1. ΔStotal = ΔSsystem + ΔSsurr > 0 For spontaneous Process B. Calculation of ΔSsurr at constant T, P : ΔSsurr = qsurr/t (note: q is heat, which is related directly to enthalpy, H) ΔSsurr = -ΔH/T (constant T, P) Show by calculation that the vaporization of water at 25 0 C and 1 atm is non- spontaneous H2O(l) H2O(g). ΔSsystem = S o H2O(g) - S o H2O(l) = kJ/K ΔSsurr = -[ΔHf H2O(g) - ΔHf H2O(l)] / T = -44.0/298 ΔStotal = = kj/k Where would ΔStotal = 0? IV. Free Energy Changes G o = H o - T S o (Gibbs/Helmholtz Equation) Note that G, like S, is dependent on P. Unlike H o and S o, G o is strongly temperature dependent because of T in the equation. If G O is -, reaction spontaneous at standard conditions If G O is +, reaction non-spontaneous at standard conditions If G O is 0, reaction at equilibrium at standard conditions Free Energy: "Real" energy change not only from breaking and forming bonds (Enthalpy), but taking organization changes (Entropy) in account as well. The striving to achieve a state of minimum free energy may be interpreted as the driving force for a chemical reaction.

3 Comments Regarding "Standard Free Energy Changes" The free energy of a substance (as well as its enthalpy and entropy) depends upon the state of the substance, i.e., whether it is a solid, liquid, or gas. In addition, the free energy depends upon the temperature, the pressure, and in the case of solutions, on the concentration. Consequently, the change in free energy that occurs during a chemical reaction depends upon the state (s, l, g), the conditions (T,P), and the concentrations of the reactants and products. For convenience, standard conditions have been chosen for reference. These conditions are: T= 25 C P= 1 atm Concentration = 1 M Standard Free Energies of Formation, Gf O Definition: The free energy change when one mole of a substance in its free state is formed from its elements in their free states. This definition is obviously analogous to our earlier definition on enthalpy of formation, and the calculation of the free energy change which occurs during a reaction is accomplished in a similar manner. Note: The superscript o, above is a thermodynamic symbol that generally indicates free or standard state. A QUALITATIVE LOOK at what is Implied by the Equation ΔG = ΔΗ - TΔS Since there are two possible signs for both ΔH and ΔS (positive or negative), there are four possible combinations of these signs. If both ΔH is negative (good) and ΔS is positive (good), then ΔG must be negative, and a process for which this is true will be spontaneous at all temperatures. If ΔH is positive (bad) and ΔS is negative (bad), then ΔG must be positive, and a process for which this is true will not be spontaneous at any temperature. If both are + or -, then ΔG can be either negative or positive depending upon which has which sign and whether ΔH is greater or less than TΔS. These ideas are summarized in the following chart. Gibbs-Helmholtz equation: ΔG O = ΔH O TΔS O If ΔH O is +, ΔS O is -, then ΔG O is + at all T, nonspontaneous at 1 atm. * ΔH O is -, ΔS O is +, then ΔG O is - at all T, spontaneous at 1 atm. ** ΔH O is +, ΔS O is +, then ΔG O is + at low T, becomes negative at high T ΔH O is -, ΔS O is -, then ΔG O is - at low T, becomes positive at high T. (*)Reverse reaction is spontaneous (**) Reverse reaction is nonspontaneous Practice Problem #1: Consider: H2O(l) H2O(g) at P = 1atm If ΔH = kJ, and ΔS = J/K mol. What is the boiling point of H2O(l)? First, write the EQUATION! ΔG = ΔH-TΔS At equilibrium, ΔG = 0, so 0 = ΔH TΔS

4 therefore ΔH = TΔS Thus, T = ΔH/ΔS = 40630/108.8 = 373 K Practice Problem #2: Consider the reaction: 3/2 O2(g) O3(g) Calculate ΔG o for this reaction, given: T = 25 o C and ΔHf o [O3(g)] = kj/mol S o [O2(g)] = J/K mol and S o [O3(g)] = J/K mol Equation: ΔG f o = ΔH o TΔS o ΔS o = /2(205.0) = J/K ΔG = +142,200 - (298)(-69.8) = +163,000 J = +163 kj Practice Problem#3: Calculate the free energy change that occurs during the combustion of 6.48 grams of propane, C3H8, to form carbon dioxide and liquid water C, 1atm Given: ΔGf o [C3H8(g)] = kj/mol ΔGf o [CO2(g)] = kj/mol ΔGf o [H2O(l)] = kj/mol Balanced equation is: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔG o = Σ ΔGf o (Products) - Σ ΔGf o (Reactants) ΔG o = [3 ΔGf o (CO2) + 4 ΔGf o (H2O)] - [ΔGf 0 (C3O8) + 5ΔGf 0 (O2)] ΔG o = [3 ( ) + 4 ( )] [(-23.47) + 0] ΔG o = kj/mol C3H8 Note: normally you wouldn t use ΔG o s: they are very temperature dependent. Instead, use ΔH o s and ΔS o s. ΔG= ΔH -TΔS Calculate ΔG o at 25 o C and at 500 o C. Don t use appendix (for ΔG o s).

5 ΔS o = 2S o Fe +3S o H20(g) - (S o Fe2O3 + 3S o H2) At: Thus, the reaction is spontaneous at 500 o C, not at 25 o C. At what temperature does the reduction of Fe2O3 by H2 become spontaneous at 1 atm? Relationship Between ΔG and K Since both the free energy change for a process and the equilibrium constant for a process describe the tendency for the process to occur, it would seem that these two quantities should be related to each other. If K > 1, ΔG o < 0 K < 1, ΔG o > 0 K = 1, ΔG o = 0 Reaction spontaneous at stand. cond. Reaction nonspontaneous at stand cond. Reaction at equilibrium at stand cond. ************************************************* Calculation of ΔG o for the following reaction: H2O(l) H + (aq) + OH - (aq) ΔG o = -( )(298) ln (1.0 x ) = kj Nonspontaneous at standard conditions (1 M H +, 1 M OH - )

6 What is the second law of thermodynamics? You can t break even! The entropy of the universe increases! What is the first law of thermodynamics? Practice Problem If the free energy of formation of HI at 490 o C is kj/mol. Calculate K for H2(g) + I2(g) 2HI(g) ΔG o = -RTln K ln K = -ΔG o RT ln K = -(2)(-12.1 x 10 3 )/(8.314)(763) ln K = K = e = 45.4 Absolute Entropies And The 3 rd law of Thermodnamics The 3 rd Law simply states that: The entropy of perfect crystals of all pure elements and compounds is zero at the absolute zero of temperature. One significant result of this law is that it allows scientists to calculate (and publish for our use) tables of absolute entropies of substances. NOTE: The calculation of absolute entropies gets rather involved. Essentially, the calculations use the relationship ΔS = ncpln T2/T1 where Cp is Heat Capacity but are actually much more complicated, because Cp itself is a function of temperature, and often a very complicated one.