# Entropy, Free Energy, and Equilibrium

## Presentation on theme: "Entropy, Free Energy, and Equilibrium"— Presentation transcript:

Entropy, Free Energy, and Equilibrium

Spontaneous Processes and Entropy
One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration). Entropy

A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous. Entropy

A waterfall runs downhill, but never up, spontaneously.
We observe spontaneous physical and chemical processes every day, including many of the following examples: A waterfall runs downhill, but never up, spontaneously. A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form. Water spontaneously freezes below 0 oC, and ice melts spontaneously above 0 oC (at 1 atm). Iron exposed to water and oxygen spontaneously forms rust, but rust does not spontaneously change back to iron. Entropy

These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction. Entropy

If we assume that spontaneous processes occur so as to decrease the energy of a system, we can explain why a ball rolls downhill and why springs in a clock unwind. Similarly, a large number of exothermic reactions are spontaneous. An example is the combustion of methane CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH = kJ Entropy

But consider a solid-to-liquid phase transition such as this spontaneous process that occurs above 0 oC H2O(s) H2O(l) DH = 6.01 kJ In this case, the assumption that spontaneous process always decrease a system’s energy fails. Entropy

This process is spontaneous, and yet it is also endothermic.
Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: NH4NO3(s) NH4+(aq) + NO3-(aq) DH = 25 kJ H2O This process is spontaneous, and yet it is also endothermic. Entropy

From the study of the examples mentioned and many more cases, we come to the following conclusion: Exothermicity favors the spontaneity of a reaction but does not guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is possible for an exothermic reaction to be nonspontaneous. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy. Entropy

Entropy In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, DH. The other is change in entropy, (DS). Entropy is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system, the smaller its entropy. Entropy

Ssolid < Sliquid << Sgas
For any substance, the particles in the solid state are more ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state. So for the same molar amount of a substance, we can write Ssolid < Sliquid << Sgas In other words, entropy describes the extent to which atoms, molecules, or ions are distributed in a disorderly fashion in a given region of space. Entropy

It is possible to determine the absolute entropy of a substance, something we cannot do for enthalpy. According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero (0 K). If the crystal is impure or if it has defects, then its entropy is greater than zero even at 0 K because it would not be perfectly ordered. Entropy

The important point about the third law is that it allows us to determine the absolute entropies of substances. The standard entropies (So) that are listed on your reference sheets are the absolute entropies of substances at 1 atm and 25 oC. These are the values that are generally used in calculations. Entropy

The units of entropy are J/K or J/K. mol for 1 mole of the substance
The units of entropy are J/K or J/K.mol for 1 mole of the substance. We use joules rather than kilojoules because entropy values are typically quite small. Entropy

Consider a certain process in which a system changes from some initial state to some final state. The entropy change for the process, DS, is DS = Sfinal - Sinitial If the change results in an increase in randomness or disorder, then the change in entropy is positive (+DS) Entropy

At absolute zero, a substance has a zero entropy value (assuming that it is a perfect crystalline solid). As it is heated, its entropy increases gradually because of greater molecular motion. At the melting point, there is a sizable increase in entropy as the more random liquid state is formed. Further heating increases the entropy of the liquid again due to enhanced molecular motion. At the boiling point there is a large increase in entropy as a result of the liquid to gas transition. Beyond that temperature, the entropy of the gas continues to rise with increasing temperature. Entropy increase of a substance as the temperature rises from absolute zero. Entropy

Processes that lead to an increase in the entropy of the system include
melting vaporization dissolving* heating Entropy

*Dissolving ionic compounds in water does not always result in an increase in entropy. For ionic compounds that contain Al+3 or Fe+3, hydration (the process of surrounding the ions by water molecules) can increase the order of the water molecules so much that the entropy change for the overall process can actually be negative (-DS). Entropy

The Second Law of Thermodynamics
The connection between entropy and the spontaneity of a reaction is expressed by the second law of thermodynamics: the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. DSuniverse = DSsystem + DSsurroundings > 0 DSuniverse = DSsystem + DSsurroundings = 0 Entropy

What if for some process we find that DSuniv is negative?
For a spontaneous process, the second law says that DSuniv must be greater than zero, but it does not place a restriction on either DSsys or DSsurr. Thus it is possible for either DSsys or DSsurr to be negative, as long as the sum of these two quantities is greater than zero. What if for some process we find that DSuniv is negative? The reaction is spontaneous in the opposite direction. Entropy

To calculate DSuniv, we need to know both DSsys and DSsurr.
Entropy

Entropy Changes in the System
Suppose that the system is represented by the following reaction: aA + bB cD + dD As in the case for enthalpy of a reaction, the standard entropy of reaction, DSorxn is given by DSorxn = [cSo(C) + dSo(D)] - [aSo(A) + bSo(B)] or, in general DSorxn = SnSo(products) - SnSo(reactants) Entropy

To calculate DSrxn (which is DSsys), look up the values on your reference sheets.
Calculate the standard entropy for the formation of ammonia from nitrogen gas and hydrogen gas at 25 oC. N2(g) + 3H2(g) NH3(g) DSorxn = SnSo(products) - SnSo(reactants) DSorxn = (2 mol)(193 J/K.mol) - [(1 mol)(192 J/K.mol) + (3 mol)(131 J/K.mol)] = -199 J/K Does this reaction result in an increase or decrease in order? Decrease because DS is negative Entropy

Typically, if a reaction produces more gas molecules than it consumes, DSo is positive, likewise, if the total number of gas molecules diminishes, DSo is negative. If there is no change in the total number of gas molecules, then DSo may be positive or negative, but will be relatively small numerically. Entropy

(b) NH4Cl(s) NH3(g) + HCl(g) (c) H2(g) + Br2(g) 2HBr(g) -DS
Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H2(g) + O2(g) H2O(l) (b) NH4Cl(s) NH3(g) + HCl(g) (c) H2(g) + Br2(g) HBr(g) -DS +DS ?DS it will be small Entropy

Entropy Changes in the Surroundings
When an exothermic process takes place in the system, the heat transferred to the surroundings enhances motion of the molecules in the surroundings. Consequently, there is an increase in disorder of the surroundings at the molecular level, and the entropy of the surroundings increases. Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the surroundings because molecular motion decreases. Entropy

For constant-pressure processes the heat change is equal to the enthalpy change of the system. Therefore, the change in entropy of the surroundings, DSsurr, is proportional to DHsys. DSsurr a -DHsys The minus sign is used because if the process is exothermic, DHsys is negative and DSsurr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, DHsys is positive and the negative sign ensures that the entropy of the surroundings decreases. Entropy

The change in entropy for a given amount of heat also depends on the Kelvin temperature. If the temperature of the surroundings is high, the molecules are already quite energetic. Therefore, the absorption of heat from an exothermic process in the system will have relatively little impact on molecular motion and the resulting increase in entropy will be small. However, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a more drastic increase in molecular motion and hence a larger increase in entropy. Entropy

From the inverse relationship between DSsurr and temperature (in Kelvins) we can rewrite the relationship between DH, T and DS as DSsurr = -DHsys T Entropy

Would you predict the synthesis of gaseous ammonia from nitrogen gas and hydrogen gas to be spontaneous at 25 oC? (Will it be +DSuniv ?) DSuniv = DSsys + DSsurr 1/2 N2(g) + 3/2 H2(g) NH3(g) DH = kJ or J DSsys = (1)(193.0) - [(1/2)(191.5) + (3/2)(131.0)] = J/K.mol DSsurr = -DHsys T DSsurr = -( J) = 155 J/K 298 K DSuniv = DSsys + DSsurr DSuniv = J/K.mol J/K.mol = J/K Because DSuniv is positive, we predict that the reaction is spontaneous at 25 oC. Entropy

It is important to keep in mind that just because a reaction is spontaneous does not mean that it will occur at an observable rate. Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur. Entropy

Gibbs Free Energy The second law of thermodynamics tells us that a spontaneous reaction increases the entropy in the universe (+DS), but in order to determine the sign of DSuniv, we would need to calculate both DSsys and DSsurr. In order to express the spontaneity of a reaction more directly, we can use another thermodynamic function called Gibbs free energy (G). Entropy

The change in free energy (DG) of a system for a constant-temperature process is
DG = DHsys - TDSsys If a particular reaction is accompanied by a release of usable energy (-DG), the reaction will occur spontaneously, if DG is equal to zero, the system is at equilibrium. Entropy

-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 kJ and 3.05 J/K, respectively. Would this reaction be spontaneous at 25 oC? DG = DH - TDS = J - (298 K)(3.05 J/K) = -394,000 J or -394 kJ -DG, therefore the reaction would be spontaneous Entropy

The standard free-energy of reaction (DGorxn) is the free-energy change for a reaction when it occurs under standard-state conditions, when reactants in their standard states are converted to products in their standard states. Entropy

DGorxn = SnGf(products) - SnGf(reactants)
To calculate DGorxn we start with the equation aA + bB cC + dD the standard free-energy change for this reaction is given by the equation DGorxn = [cDGf(C) + dDGf(D)] - [aDGf(A) + bDGf(B)] or, in general, DGorxn = SnGf(products) - SnGf(reactants) Entropy

Calculate the standard free-energy changes for the combustion of 1 mol of methane at 25 oC.
DGorxn = [1DGf(CO2) + 2DGf(H2O)] - [1DGf(CH4) + 2DGf(O2)] DGorxn = [(1 mol)( kJ/mol) + (2 mol)( kJ/mol)] - [(1 mol)(-50.8 kJ/mol) + (2 mol)(0 kJ/mol)] DGorxn = kJ Entropy

DG < 0 The reaction is spontaneous in the forward direction
Summarizing the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of DG DG < 0 The reaction is spontaneous in the forward direction DG > 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction. DG = 0 The system is at equilibrium. There is no net change. Entropy

Temperature and Chemical Reactions
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. Entropy

Estimate the temperature at which decomposition becomes spontaneous.
Calcium oxide (CaO) also called quicklime, is an extremely valuable inorganic substance. It is prepared by decomposing limestone (CaCO3) in a kiln at a high temperature according to the following reaction CaCO3(s) CaO(s) + CO2(g) Estimate the temperature at which decomposition becomes spontaneous. DG = DH - TDS DH = [(1 mol)( kJ/mol) + (1 mol)( kJ/mol)] - [(1 mol)( kJ/mol)] DH = kJ DS = [(1 mol)(39.8 J/K.mol) + (1 mol)(213.6 J/K.mol)] - [(1 mol)(92.9 J/K.mol)] DS = J/K.mol Entropy

DG = DH - TDS DG = 177.8kJ - (298 K)(.1605 kJ/K) DG = kJ Since DG is a large positive quantity, we conclude that the reaction is not spontaneous at 25 oC. In order to make DG negative, we first have to find the temperature at which DG is zero, the point at which the system is at equilibrium. At 835 oC the system is at equilibrium. At temperatures higher than 835 oC, DG becomes negative, indicating that the decomposition is spontaneous. Reactions with +DH and +DS are spontaneous at high temps! 0 = DH - TDS -DH = -TDS DH = TDS DH = T DS T = J = 1108 K or 835 oC 160.5 J/K Entropy

Two things need to be kept in mind regarding the previous calculation
Two things need to be kept in mind regarding the previous calculation. First, we used DH and DS values at 25 oC to calculate changes that occur at a much higher temperature, so the value of DG will just be close to the actual value. Second, some decomposition will occur at below 835 oC (just like some water will evaporate at temps below 100 oC. Entropy

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

DG = DH – TDS 0 = DH – TDS DH = TDS T = DH DS
= 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

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. Entropy

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, J/mol K, T is the absolute temperature, and Q is the reactant quotient. Entropy

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 Entropy

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. Entropy

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 = kJ 3H2 + N2  2NH3 (PNH3)2 (PN2)(PH2)3 (.50)2 (1.0)(3.0)3 = 9.3 x 10-3 Q = = Entropy

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

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 Entropy

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 Entropy

The End Entropy