Chapter 17 Spontaneity, Entropy, and Free Energy Thermodynamics The study of energy and its transformations 1 st Law of Thermodynamics The total energy of the Universe is constant Energy can therefore not be created or destroyed but can only be converted from one form to another Example: Conversion of potential energy (stored in chemical bonds) into kinetic energy (as measured by the molecular motions of the products) during an exothermic reaction When applied to chemical reactions, the 1 st Law of Thermodynamics can tell us: 1. The magnitude of the energy change 2. The direction of energy flow 3. The nature of the energy change However, the 1 st Law tells us nothing about whether a reaction will actually occur! Spontaneous Reactions Reactions which occur on their own without outside intervention 1
Does a spontaneous reaction always occur rapidly? No! Spontaneous doesn t necessarily mean fast! Remember from Chapter 12 that thermodynamics tells us nothing about the rate of a reaction! That is the realm of kinetics! Combustion of Wood wood + O 2 (g) CO 2 (g) + H 2 O(g) + ash + energy Why do some changes occur spontaneously while others do not? However, the reverse process never occurs: CO 2 (g) + H 2 O(g) + ash + energy wood + O 2 (g) A gas trapped in one end of a vessel When the valve is opened the gas spreads evenly throughout the entire container 2
However, the reverse process never occurs! Does a process have to be exothermic to be spontaneous? No! Many endothermic processes are also spontaneous! Example: Entropy, S A measure of molecular randomness or disorder Describes the number of possible arrangements of energy and/or position for a system in a particular state The more randomness there is, the more possible arrangements there are and hence these will have the highest probability of existing Since a system will favor the states which have the highest probability of existing, these will always be the ones with the highest randomness or entropy! Result: H > 0 A system will always spontaneously move in the direction of increasing entropy! Positional Probability/Entropy Explains why gas spreads evenly throughout the entire container since it is the most probable arrangement! Process is spontaneous since positional entropy increases! microstates possible arrangements of microstates 3
Also explains why the reverse process does not occur since it is highly improbable arrangement! Process is not spontaneous since positional entropy decreases! Other Examples of Positional Entropy Changes of State Formation of Solutions S solid < S solution 4
2 nd Law of Thermodynamics The entropy of the Universe is always increasing Any spontaneous process always results in an increase in entropy of the Universe According to the 1 st Law energy is conserved while according to the 2 nd Law entropy is not The 2 nd Law Applied to Chemistry How S sys and S surr determine S univ In a chemical system, the change in entropy of the Universe, S univ is the sum of the change in entropy of the system, S sys and the change in entropy of the surroundings, S surr : S univ = S sys + S surr The sign of S univ determines whether a process is spontaneous or not: S univ > 0 S univ < 0 S univ = 0 spontaneous not spontaneous at equilibrium A process with S univ < 0 is not spontaneous in the forward direction but is spontaneous in the reverse direction Therefore, to determine whether a process is spontaneous we need to consider the changes in both the system and the surroundings Effect of Temperature on Spontaneity Factors that affect S sys and S surr Changes in S sys are largely determined by changes in positional probably associated with changes in state and/or numbers of particles Changes in S surr are largely determined by the flow of energy into the system from the surroundings or out of the system into the surroundings In exothermic reactions there is a flow of heat into the surroundings which increases the random motions of the particles and so the entropy increases ( S surr > 0) In endothermic reactions there is a flow of heat from the surroundings which decreases the random motions of the particles and so the entropy decreases ( S surr < 0) H 2 O(l) H 2 O(g) H 0 = +40.65 kjmol -1 S sys > 0 since going from a liquid to a gas S surr < 0 since process is endothermic The two terms are in opposition! Which dominates must depend on temperature since we know this process is spontaneous above 100 C (water boils) but not spontaneous below this temperature Since S surr is determined by heat flow, it is this term that is sensitive to temperature Its sign depends on the direction of energy flow while its size depends on temperature since a constant amount of energy will have at greater effect on slower moving molecules at a low temperature compared to faster moving molecules at a high temperature 5
Relationship between S surr and Temperature For a reaction that occurs at a constant temperature and pressure: S Surr H T where H is the enthalpy change and T is the temperature in Kelvin The negative sign is needed since when H < 0 (exothermic), the entropy change must be positive ( S surr > 0) Notice that S surr is inversely proportional to T as expected! Vaporization of Water Revisited What are the signs of S sys, S surr and S univ? H 2 O(l) H 2 O(g) H 0 = +40.65 kjmol -1 S sys > 0 S surr < 0 but gets smaller as the temperature increases Below 100 C S surr is more negative than S sys is positive so S univ is negative (not spontaneous) At 100 C and above S surr is less negative than S sys is positive so S univ is positive (spontaneous) H > 0 Josiah Willard Gibbs (1839-1903) Gibbs Free Energy, G Another thermodynamic function useful for predicting spontaneity of processes at constant temperature and pressure (true for most chemical reactions) For a process that occurs at constant temperature, the change in free energy, G is given by: G = H - T S where H is the enthalpy change in the system, T the temperature in Kelvin and S, the entropy change in the system Unless explicitly specified we can normally assume that S refers to the system ( S = S sys ) 6
Significance of Free Energy Therefore a process occurring at constant temperature and pressure will only be spontaneous if G < 0 Standard Thermodynamic Quantities When all reactants and products are in their standard states at a particular temperature, a symbol is placed after the corresponding thermodynamic quantity: Effect of Temperature on Spontaneity H 2 O(s) H 2 O(l) H 0 = +6.03 x 10 3 Jmol -1, S 0 = S sys = +22.1 JK -1 mol -1 G = H -T S When the temperature is below the melting point, S surr is more negative than S sys so S univ < 0 and G > 0 As temperature increases, S surr becomes less negative At the melting point, S surr =- S sys so S univ = 0 and G = 0 Above the melting point, S surr is less negative than S sys so S univ > 0 and G < 0 Processes change from being non-spontaneous to spontaneous of vice versa when G = 0: G = H -T S = 0 so: H = T S This can be used to determine the temperature at which a process changes its spontaneity: T = H / S 7
Summary G = H - T S Chemical Reactions and S sys ( S ) Changes in S sys are largely determined by changes in positional entropy and hence they depend on the physical states of the reactants and products: S sys (solid) < S sys (liquid) < S sys (gas) 8
Reactions Involving Gases When a reaction involves gaseous species, the change in positional entropy (and hence S sys ) is determined by the relative numbers of gaseous reactants and products This is because a greater number of molecules mean a greater number of possible configurations while a smaller number of molecules mean a smaller number of possible configurations S sys ( S ) for a reaction will be positive if: # molecules gaseous products > # molecules gaseous reactants S sys ( S ) for a reaction will be negative if: # molecules gaseous products < # molecules gaseous reactants N 2 (g) + 3H 2 (g) 2NH 3 (g) What is the sign of S for the complete combustion of cyclooctane (C 8 H 8 )? 2 product molecules < 4 reactant molecules S sys < 0 Cool Matter - temp speed Properties of Absolute Zero 1. The coldest possible temperature 2. Matter can never be cooled to absolute zero At absolute zero = -273 ºC or -460 ºF atomic and molecular motion essentially ceases! 9
3 rd Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero The internal structure of a perfect crystal is perfectly regular so that at absolute zero there is only one possible arrangement of particles Standard Entropy Values, S 0 These correspond to the increase in entropy that occurs when a substance is heated from 0 K to 298 K at a pressure of 1 atm Since S = 0 for a perfect crystal at 0 K, the entropy value for a substance at a given temperature can be calculated if we know how entropy depends on temperature Effect of Molecular Structure on S 0 H 2 O(g) has a higher S 0 value (189 JK -1 mol -1 ) than H 2 (g) (115 JK -1 mol -1 ) since it is a non-linear triatomic molecule and has more rotational and vibrations states (degrees of freedom) than the simpler diatomic H 2 molecule For molecules with the same number of atoms, S 0 generally increases with molecular mass Overall, the more complex the molecule, the higher the value of S 0 10
Calculating Entropy Changes for Chemical Reactions For each of the following pairs, which has the greater value of S at 25 C? a. H 2 (g) or He(g) b. H 2 O(l) or C 6 H 6 (l) c. CO 2 (g) or SO 2 (g) Like enthalpy, entropy is a state function (pathway-independent) so the entropy change for a chemical reaction can be calculated by taking the difference between the standard entropy values of the products and the reactants: S 0 reaction = Σn p S 0 products - Σn r S 0 reactants where n p and n r are the number of moles of a particular reactant or product respectively Standard Free Energy Change, G 0 Free energy change when reactants in their standard states are converted into products in their standard states Standard conditions of temperature and pressure are 298 K and 1 atm Cannot be measured directly, but can be calculated from other quantities Since G 0 values correspond to standard conditions, they allows us to compare the relative tendencies of different reactions to occur since the more negative G 0 is, the more likely the reaction is to occur However, kinetics is needed to determine how fast a spontaneous reaction will actually proceed Calculating G 0 There are three ways to calculate G 0 values: 1. From H 0 values (calculated from standard enthalpies of formation, H f0 ) and S 0 values (from standard entropy values, S 0 ) using the equation G 0 = H 0 - T S 0 2. Since free energy, like enthalpy is a state function (pathwayindependent) we can calculate G 0 of an unknown reaction from reactions with known G 0 values using techniques similar to Hess s Law for enthalpy 3. From the standard free energies of formation, G f 0 (the change in free energy that occurs when 1 mole of a substance forms from its elements in their standard states under standard conditions) of the reactants and products 11
Dependence of Free Energy on Pressure In order to understand how the free energy of a gas depends on pressure we need to know how the thermodynamic functions that are used to calculate free energy (enthalpy and entropy) depend on pressure (since G = H - TS) For an ideal gas, enthalpy does not depend on pressure Entropy does depend on pressure since positional entropy increases with decreasing pressure (and increasing volume) Calculations show that the relationship is as follows: G = G 0 + RTln(P) where: G is the free energy of the gas at P atm G 0 is the free energy of the gas at 1 atm R is the universal gas constant (8.3145 JK -1 mol -1 ) T is the temperature in Kelvin Dependence of G 0 for a Reaction on Pressure For the generic reaction: aa(g) + bb(g) cc(g) + dd(g) the relationship between G 0 and pressure is as follows: G = G 0 + RTln(Q) where: G is the free energy of the reaction when reactants and products are at pressures P A, P B, P C and P D respectively G 0 is the free energy of the reaction with all reactants and products at 1 atm R is the universal gas constant (8.3145 JK -1 mol -1 ) T is the temperature in Kelvin Q is the reaction quotient from the Law of Mass Action: Q = (P C c )(P D d ) / (P A a )(P B b ) 12
Free Energy and Equilibrium Although the G value for a reaction tells us whether the reactants or products are favored under a given set of conditions, it does not necessarily mean the system will proceed to pure products ( G < 0) or remain as pure reactants ( G > 0) A reaction at constant temperature and pressure will always proceed in the direction that lowers its free energy. The lowest possible free energy possible is always attained when a reaction reached equilibrium. This is why reactions proceed until they reach that state! Relationship between G 0 and the Equilibrium Constant, K When substances undergo a chemical reaction, the reaction always proceeds to the minimum possible free energy corresponding to the equilibrium position This will always be attained when: G products = G reactants or G = G products G reactants = 0 Since: G = G 0 + RTln(Q) and G = 0 and Q = K at equilibrium then: Qualitative Relationship between G 0 and K 1. G 0 = 0: In this case the free energies of the reactants and products are equal and the system is already in equilibrium with K = 1 so no shift will occur 2. G 0 < 0: In this case the free energies of the products are lower than the reactants so the equilibrium position will shift to the right with K > 1 3. G 0 > 0: In this case the free energies of the products are lower than the reactants so the equilibrium position will shift to the left with K < 1 G = 0 = G 0 + RTln(K) so G 0 = -RTln(K) 13
The Temperature Dependence of K Le Châtelier s Principle allows us to predict qualitatively how the value of K changes with temperature Thermodynamics, on the other hand, allows us to specify the quantitative dependence of K on temperature: This is a simple linear equation of the form y = mx + b so if you measure K at different temperatures a plot of ln(k) vs. 1/T will yield a straight line with a slope of H 0 /R and an intercept of S 0 /R Also, assuming that H 0 and S 0 are independent of temperature (which is generally true if the temperature range is small), this equation will allow you to calculate K at different temperatures if H 0 and S 0 are known Free Energy and Work Chemical processes are important because they can be used to do useful work e.g. the combustion of gasoline in a car engine The change in free energy not only tells us whether a process is spontaneous but also how much work can be done with it The maximum possible useful work from a process (at constant temperature and pressure) is equal to the change in free energy: w max = G The is why its called free energy since it s the energy free to do useful work! What is the maximum amount of work that can be extraction from the reaction of hydrogen and oxygen in a fuel cell if the reactants and products are in their standard states? How would the actual amount of work extracted compare to this value? However, a process is never 100% efficient so the actual amount of useful work is always less than the theoretical maximum In any real process, some work is always transformed into heat in the surroundings, ensuring that the entropy of the universe increases 14