Lecture 18. Chemical Reactions (Ch. 5)

1. Lecture 18. Chemical Reactions (Ch. 5) In chemical reactions, the products of reaction are intermixed with the reacting substances (reactants). Thus, the process is governed by two factors: (a) the energy change (V,T=const) or enthalpy change (T,P=const), and (b) the entropy change: For a reaction to be energetically favorable, the Gibbs energy for products should be lower than the Gibbs energy for reactants.

2. O=C=O O=O H-O-H O=O H-O-H H H - C - H H Chemical Reactions - processes of molecular transformations that involve at least one of the following changes: the number of atoms in a molecule, the type of atoms, their mutual positioning in a molecule (isomers), or their charge. The “other” work (electrical, chemical, etc.) performed on a system at T = const and P = const in a reversible process is equal to the change in the Gibbs free energy of the system: reac- tants  products reaction “coordinate” the enthalpy released in the reaction at P,T=const During this reaction, some bonds should be broken and other bonds (with a more negative potential energy) should be formed. This process is characterized by a potential barrier – thus, the Boltzmann factor! For the “direct” reaction, the barrier is , for the reverse one - . The reactions are characterized by “directionality” (which free energy is lower, reactants or products), energy release, and rate.

3. “Directionality” of Chemical Reactions The directionality of a chemical reaction at fixed P,T is governed by the Gibbs free energy minimum principle. Two factors are at play: the entropy and the enthalpy. Since the change in G is equal to the maximum “useful” work which can be accomplished by the reaction, then G<0 indicates that the reactionwill proceed spontaneously. - clearly, S increases. Also, the energy of the products of the reaction is lower than the reactant (the energy is released in the TNT explosion): the reaction is “shifted” strongly toward the products. - though S increases, the equilibrium is shifted to the left (H >TS) at 300K - though S decreases, the equilibrium is shifted to the right because the decrease of H “overpower” the increase of -TS

4. Several channels of the reaction between CO and H2 at 300K Examples: Several channels of the reaction between CO and H2 at 600K at this T, CH3OH and CH3COOH will spontaneously dissociate These estimates tell us nothing about the reactionrate! (the process of transformation of diamond into graphite also corresponds to negative G = -2.9 kJ/mol, but our experience tells us that this process is extremely slow).

5. Exchange with the Environment The difference between the internal energies (V,T –const) or enthalpies (P,T – const) of reactants and products represent the “heat” of reaction: < 0 – exothermic, > 0 - endothermic If an exothermic reaction proceeds spontaneously, it means that (a) it is entropy-driven, and (b) the system gets some energy necessary for the reaction from its environment (heat bath). The latter process happens whenever the entropy of the ptoducts is greater than that for the reactants. “Universe” = a nonisolated system + its environment For reversible processes, Sin the system Sin the environment environment If the entropy of a system is increased in the process of chemical reaction, the entropy of the environment must decrease. The associated with this process heat transfer from the environment to the system in a reversible process: system Thus, an entropy increase “pumps” some energy out of the environment into the system.

6. Problem Molar values of H and S for the reaction of dissolving of NH4Cl in water at standard conditions (P=1 bar, T=298 K) are 34.7 kJ/mol and 0.167 kJ/(K·mol), respectively. (1) Does the reaction proceed spontaneously under these conditions? (2) How does the entropy of the environment and the Universe change in this reversible process? G <0, thus the reaction proceeds spontaneously (despite its endothermic character). High temperatures favor the spontaneity of endothermic processes. The “reversible” energy (heat) for this process:

7. Oxidation of Methane Consider the reaction of oxidation of methane: For this reaction H = -164 kJ/mol, S = -162 J/molK. • Find the temperature range where this reaction proceeds spontaneously. • Calculate the energy transferred to the environment as heat at standard conditions (T = 298 K, P = 1 bar) assuming that this process is reversible. • Calculate the change in the entropy of environment, Senv, at standard conditions assuming that this process is reversible. What is the total entropy change for the “Universe” (the system + environment) if the process is reversible? (a) For this reaction to proceed spontaneously, G must be negative: (b) The energy transferred to the environment as heat: (c) For reversible processes:

8. Glucose Oxidation Mammals get the energy necessary for their functioning as a result of slow oxidation of glucose : At standard conditions, for this reaction H = -2808 kJ/mol, S = 182.4 J/molK. Thus, at T=298K, this reaction will proceed spontaneously. If this process proceeds as reversible at P,T=const, the maximum “other” work (chemical, electrical, etc.) done by the system is: This work exceeds the energy released by the system (H = -2808 kJ/mol). Clearly, some energy should come from the environment. For reversible processes: Thus, the environment transfers to the system 54.4 kJ/mol as the thermal energy (heat). The system transforms into work both the energy released in the system (H) and the heat received from the environment (qenv). For all reactions that are characterized by enthalpy decrease and entropy increase,Wmax exceeds H. Interestingly that the Nature selected this process as a source of work: it not only releases a great deal of energy, but also pumps the energy out of the environment!

9. reac- tants The Rates of Chemical Reactions The rates of both “direct” and “reverse” rections are governed by their Boltzmann factors (the activation over the potential barrier). Thus, each rate is an exponential function of T. Catalysts – reduce the height of an activation barrier. products reaction “coordinate” The rate is proportional to the probability of collisions between the molecules (concentration of reactants). Chemical equilibrium = dynamical equilibrium, the state in which a reaction proceeds at the same rate as its inverse reaction. association equilibrium rate dissociation t

10. Chemical Equilibrium This plot shows that despite the fact that a particular reaction could be energetically favorable, it hardly ever go to comletion. At any non-zero T, there is a finite concentration of reactants. association equilibrium rate dissociation t This can be understood using the concept of the minimization of the Gibbs free energy in equilibrium. The reason for the “incompleteness” of reactions is intermixing of reactants and products. Without mixing, the scenario would be straightforward: the final equilibrium state would be reached after transforming 100% of reactants into products. However, because the products are intermixed with the reactants, breaking just a few products apart into the reactant molecules would increase significantly the entropy (remember, there are infinite slopes of G(x) at x = 0,1), and that shifts the equilibrium towards x < 1. GA no mixing GB ideal mixing x  reactants products of reaction chemical equilibrium (strongly shifted to the right, but still there is a finite concentration of reactants)

11. Let’s consider a general chemical equation: Chemical Equilibrium (cont.) stoichiometric coefficients reactants products The sign of coefficients aiis different forthe reactants and for the products: The numbers of different kinds of molecules, Ni, cannot change independently of each other – the equation of chemical reaction must be satisfied: dNi must be proportional to the numbers of molecules appearing in the balanced chemical equation: Here  is a constant of proportionality, dNi>0 (dNi<0) for molecules formed (disappeared) in the reaction. In equilibrium, at fixed T and P, The Gibbs free energy is at minimum: - the general condition for chemical equilibrium Combining with the expression for dNi : The chemical potentials i are functions of T, P, and all Ni. Hence this condition implies that in equilibrium, there is a definite connection between the mean numbers of molecules of each kind. In principle, the statistical physics allows one to calculate the chemical potentials iand thus to deduce explicitly the connection between the numbers Ni.

12. Chemical Equilibrium between Ideal Gases Let’s consider the reaction that occurs in the gas phase, and assume that each reactant/product can be treated as an ideal gas. For this case, we know  = (T,P). Example: transformation of the nitrogen in air into a form that can be used by plants ammonia The chemical potential of an ideal gas: 0 represents the chemical potential of a gas in its “standard state”, when its partial pressure is P = P 0 (usuallyP 0 = 1 bar). In equilibrium: x NA • the tabulated change in G • for this reaction at P0 = 1 bar

13. The Law of Mass Action the equilibrium constant K In general, for a reaction - all pressures are normalized by the standard pressure P0 The product of the concentration of the reaction partners with all concentrations always taken to the power of their stoichiometric factors, equals a constant K which has a numerical value that depends on the temperature and pressure. In particular, - the exponential temperature dependence of the equilibrium constant K is due to the Boltzmann factor: Generalization of this law for the concentrations of the reaction partners in equilibrium (not necessarily in the gas phase) is known as the law of mass action (Guldberg-Waage, 1864):

14. Ammonia Synthesis At T = 298K and P = 1 bar, G= -32.9 kJ for production of two moles of ammonia Thus, the equilibrium is strongly shifted to the right, favoring the production of ammonia from nitrogen and hydrogen. The calculation of the equilibrium constant K is only the first step in evaluating the reaction (e.g., its usefulness for applications). However, the value of K tells us nothing about the rate of the reaction. For this particular reaction, at the temperatures below 7000C, the rate is negligible (remember, the rapture of N-N and H-H bonds is an activation process). To increase the rate, either a high temperature or a good catalist is required. Haber Process, developed into an industrial process by C. Bosch - a major chemical breakthrough at the beginning of the 20th century (1909): T = 5000C, P = 250 bar, plus a catalist (!!!). At this temperature, K = 6.9·10-5 (the drop of K can be calculated using van’t Hoff’s equation and H0 =-46 kJ, see Pr. 5.86). To shift the reaction “to the right” (higher concentration of the product), a very high pressure is needed.

15. Example of Application of the Law of Mass Action Let’s look at a simple reaction Notice that we have the same # of moles on both sides of the reaction equation. We start withn0H2  and n0CO2  moles of the reacting gases and define as the yieldy the number of moles of H2O that the reaction will produce at equilibrium: equilibrium concentrations of H2 and CO2 The mass action law requires: This is a quadratic equation with respect to y, the solution is straightforward but messy. What kind of starting concentrations will give us maximum yield? To find out, we have to solve the equation dy/dn0H2 = 0. The result: - maximum yield is achieved if you mix just the right amounts of the starting stuff. This result is always true, even for more complicated reactions.

16. The Temperature Dependence of K (van’t Hoff Eq.) It’s important to know how the equilibrium concentrations are affected by temperature (Pr. 5.85). We also need this result for solving Problems 5.86 and 5.89. Let’s find the partial derivative of lnK with respect to T at P=const For either the reactants or the products, van’t Hoff’s equation H0 is the enthalpy change of the reaction. If H0 is positive (if the reaction requires the absorption of heat), then higher T “shifts” the reaction to the right (favor higher concentrations of the products). For the exothermic reactions, the shift will be to the left (higher concentration of the reactants).

17. Chemical Equilibrium in Dilute Solutions Water dissociation: GA Under ordinary conditions, the equilbrium is strongly shifted to the left, but still there is a finiteconcentration of ions H+ and OH- dissolved in water. In equilibrium: no mixing GB ideal mixing Assuming the solution is very dilute: 0 x  1 H2O H++OH- the shift is exaggerated: xeq ~ 1·10-7 where 0 are the chemical potentials for the substance in its “standard” state: pure liquid for the solvent, 1 molal for the solutes. This differs from the reactions in the gas phase, where the “standard” state corresponds to the partial pressure 1 bar. In the final equation, partial pressures are replaced with the molalities. Most importantly, the water concentration vanishes from the left side – its “standard” concentration remains =1 because the number of dissociated molecules is tiny. The 7 is called the pH of pure water.

18. Chemical Equilibrium between Gas and Its Dilute Solution The same technique can be applied to the equilibrium between molecules in the gas phase and the same molecules dissolved in a solvent. Example:oxygen dissolved in water. The Gibbs free energy change G0for this “reaction” is for one mole of O2 dissolved in 1 kG of water at P = 1bar and T = 298 K. In equilibrium: For O2 gas: For O2 dissolved in water: Henry’s law (the amount of dissolved gas is proportional to the partial pressure of this gas) For O2 in water: PO2 =0.2 bar, msolute = 1.3·10-3 x 0.2 = 2.6 ·10-4 equivalent of 6.6 cm3 of O2 gas at normal conditions (1 mol at P=1 bar ~25 liters).