Lecture 18. Chemical Reactions (Ch. 5)

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1 Lecture 8. Chemical Reactions (Ch. 5) In chemical reactions, the products of reaction are intermixed with the reacting substances (reactants). hus, the process is governed by two factors: (a) the energy change (V,const) or enthalpy change (,const), and (b) the entropy change: G U + V S For a reaction to be energetically favorable, the Gibbs energy for products should be lower than the Gibbs energy for reactants.

2 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. reactants Δ r Δ Δ s products he other wor (electrical, chemical, etc.) performed on a system at const and const in a reversible process is equal to the change in the Gibbs free energy of the system: δ W dv + δ W other ( δ W other ) dg, the enthalpy released in the reaction at,const reaction coordinate C4 + O CO + O OCO OO - C - -O- OO -O- During this reaction, some bonds should be broen and other bonds (with a more negative potential energy) should be formed. his process is characterized by a potential barrier thus, the oltzmann factor! For the direct reaction, the barrier is Δ r, for the reverse one - Δ s. he reactions are characterized by directionality (which free energy is lower, reactants or products), energy release, and rate.

3 Directionality of Chemical Reactions ΔG Δ ΔS he directionality of a chemical reaction at fixed, is governed by the Gibbs free energy minimum principle. wo factors are at play: the entropy and the enthalpy. Since the change in G is equal to the maximum useful wor which can be accomplished by the reaction, then ΔG< indicates that the reaction will proceed spontaneously. Δ ΔS ΔG Δ-ΔS reaction will go spontaneously reaction needs input of energy to go he sign of ΔG depends on the relative values of Δ, ΔS and If reaction is spontaneous, the reaction is entropy-driven: the increase in disorder is sufficient to drive the reaction despite the fact that it absorbs heat from its surroundings. C + 5O9 6CO + 5O +.5O - clearly, S increases. Also, the energy of the products of the reaction is lower than the reactant (the energy is released in the explosion): the reaction is shifted strongly toward the products. + - though S increases, the equilibrium is shifted to the left (Δ >ΔS) at K + O O - though S decreases, the equilibrium is shifted to the right because the decrease of Δ overpower the increase of -ΔS

4 Several channels of the reaction between CO and at K reaction Δ J/mol - ΔS J/mol ΔG J/mol CO+ C+ O CO+ C O CO+ C COO Examples: Several channels of the reaction between CO and at 6K reaction Δ J/mol - ΔS J/mol ΔG J/mol CO+ C+ O CO+ C O CO+ C COO at this, C O and C COO will spontaneously dissociate hese estimates tell us nothing about the reaction rate! (the process of transformation of diamond into graphite also corresponds to negative ΔG -.9 J/mol, but our experience tells us that this process is extremely slow).

5 Exchange with the Environment he difference between the internal energies (V, const) or enthalpies (, const) of reactants and products represent the heat of reaction: ( δ Q) Δ, < exothermic, > - endothermic If an exothermic reaction proceeds spontaneously, it means that (a) it is entropydriven, and (b) the system gets some energy necessary for the reaction from its environment (heat bath). he 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, ΔS ΔS + ΔS ΔS" Universe " Universe ΔS in the system system env ΔS in the environment environment system If the entropy of a system is increased in the process of chemical reaction, the entropy of the environment must decrease. he associated with this process heat transfer from the environment to the system in a reversible process: δ q ΔSenv < hus, an entropy increase pumps some energy out of the environment into the system.

6 roblem Molar values of Δ and ΔS for the reaction of dissolving of 4 Cl in water at standard conditions ( bar, 98 K) are 4.7 J/mol and.67 J/(K mol), respectively. () Does the reaction proceed spontaneously under these conditions? () ow does the entropy of the environment and the Universe change in this reversible process? ΔG Δ ΔS J/mol < ΔG <, thus the reaction proceeds spontaneously (despite its endothermic character). igh temperatures favor the spontaneity of endothermic processes. he reversible energy (heat) for this process: revers qsyst ΔSsyst J / mol ΔS q q env sys 49.8 J / mol Δ Senv.67 J /( K 98 K Universe ΔS syst + ΔS.67 J /( K mol).67 J mol) /( K mol) env.

7 Oxidation of Methane Consider the reaction of oxidation of methane: For this reaction Δ -64 J/mol, ΔS -6 J/mol K. (a) (b) (c) ΔG Δ ΔS < C O CO Find the temperature range where this reaction proceeds spontaneously. Calculate the energy transferred to the environment as heat at standard conditions ( 98 K, bar) assuming that this process is reversible. Calculate the change in the entropy of environment, ΔS env, 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, ΔG must be negative: 64 J / mol < 6 J / (b) he energy transferred to the environment as heat: (c) q revers syst q ( mol K ) ( 6J / K mol) 48.8 J / mol Δ ΔG ΔS 98K qsyst 48.8J mol qenv 48.8 J / mol Δ Senv 6J / K mol 98K env / For reversible processes: ΔS ΔS + ΔS total system env K

8 Mammals get the energy necessary for their functioning as a result of slow oxidation of glucose : Glucose Oxidation C6 O6 + 6O 6CO + 6O At standard conditions, for this reaction Δ -88 J/mol, ΔS 8.4 J/mol K. ΔG Δ ΔS J/mol < hus, at 98K, this reaction will proceed spontaneously. If this process proceeds as reversible at,const, the maximum other wor (chemical, electrical, etc.) done by the system is: W max ΔG 86 J/mol his wor exceeds the energy released by the system (Δ -88 J/mol). Clearly, some energy should come from the environment. For reversible processes: ( 8.4) J / mol 54.4 J mol qenv ΔSenv 98 / hus, the environment transfers to the system 54.4 J/mol as the thermal energy (heat). he system transforms into wor both the energy released in the system (Δ) and the heat received from the environment (q env ). For all reactions that are characterized by enthalpy decrease and entropy increase, W max exceeds Δ. Interestingly that the ature selected this process as a source of wor: it not only releases a great deal of energy, but also pumps the energy out of the environment!

reactants Δ r Δ Δ s products reaction coordinate he Rates of Chemical Reactions he rates of both direct and reverse rections are governed by their oltzmann factors (the activation over the potential

9 reactants Δ r Δ Δ s products reaction coordinate he Rates of Chemical Reactions he rates of both direct and reverse rections are governed by their oltzmann factors (the activation over the potential barrier). hus, each rate is an exponential function of. Catalysts reduce the height of an activation barrier. he rate is proportional to the probability of collisions between the molecules (concentration of reactants). r Δ Chemical equilibrium dynamical equilibrium, [ conc.reactants] exp the state in which a reaction proceeds at the same s rate as its inverse reaction. Δ [ ] conc.products exp rate association dissociation equilibrium t

10 Chemical Equilibrium rate association dissociation equilibrium t his plot shows that despite the fact that a particular reaction could be energetically favorable, it hardly ever go to comletion. At any non-zero, there is a finite concentration of reactants. G A reactants ΔG Δ ΔS no mixing ideal mixing x G products of reaction chemical equilibrium (strongly shifted to the right, but still there is a finite concentration of reactants) his can be understood using the concept of the minimization of the Gibbs free energy in equilibrium. he 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 % of reactants into products. owever, because the products are intermixed with the reactants, breaing just a few products apart into the reactant molecules would increase significantly the entropy (remember, there are infinite slopes of G(x) at x,), and that shifts the equilibrium towards x <.

11 Let s consider a general chemical equation: Chemical Equilibrium (cont.) a R R + a R R + ar R + reactants... a + a a... products In equilibrium, at fixed and, he Gibbs free energy is at minimum: Combining with the expression for d i : a he chemical potentials i are functions of,, and all i. ence this condition implies that in equilibrium, there is a definite connection between the mean numbers of molecules of each ind. In principle, the statistical physics allows one to calculate the chemical potentials i and thus to deduce explicitly the connection between the numbers i. m i i i m i a i i he sign of coefficients a i is different for the reactants and for the products: + Cl Cl + Cl he numbers of different inds of molecules, i, cannot change independently of each other the equation of chemical reaction must be satisfied: d i must be proportional to the numbers of molecules appearing in the balanced chemical equation: ere λ is a constant of proportionality, d i > (d i <) for molecules formed (disappeared) in the reaction. Cl ( ) Cl Cl + Cl Cl d Cl d stoichiometric coefficients : d m dg d i i i λ i a i : d : : Cl - the general condition for chemical equilibrium

12 Chemical Equilibrium between Ideal Gases Example: transformation of the nitrogen in air into a form that can be used by plants + ammonia 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 now (,) ln ln ln In equilibrium: ( ) ( ) ( ) + +, ln ln,, V G represents the chemical potential of a gas in its standard state, when its partial pressure is (usually bar). ( ) ( ) ( ) ln ln ln ln + ( ) ( ) ( ) ( ) ln G R A Δ x A - the tabulated change in G for this reaction at bar he chemical potential of an ideal gas:

13 R ln ( ) ( ) ( ) ΔG he Law of Mass Action ( ) ( ) ΔG ( ) exp R ( ) ( ) ( ) K In general, for a reaction a m i a a ΔG... exp R the equilibrium constant K a ( ) a a a i i ln... R K (, ) Δ G - all pressures are normalized by the standard pressure he product of the concentration of the reaction partners with all concentrations always taen 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 oltzmann factor: ΔG K exp R Generalization of this law for the concentrations of the reaction partners in equilibrium (not necessarily in the gas phase) is nown as the law of mass action (Guldberg-Waage, 864): exp ΔS ΔU exp R R a a a... exp ΔG R K

he calculation of the equilibrium constant K is only the first step in evaluating the reaction (e.g., its usefulness for applications).

14 Ammonia Synthesis + ( ) ( ) ( ) At 98K and bar, ΔG -.9 J for production of two moles of ammonia ΔG K exp R.9 exp ( 8. J/K)( 98K ) ΔG exp R J 5.9 hus, the equilibrium is strongly shifted to the right, favoring the production of ammonia from nitrogen and hydrogen. he calculation of the equilibrium constant K is only the first step in evaluating the reaction (e.g., its usefulness for applications). owever, the value of K tells us nothing about the rate of the reaction. For this particular reaction, at the temperatures below 7 C, the rate is negligible (remember, the rapture of - and - bonds is an activation process). o increase the rate, either a high temperature or a good catalist is required. aber rocess, developed into an industrial process by C. osch - a major chemical breathrough at the beginning of the th century (99): 5 C, 5 bar, plus a catalist (!!!). At this temperature, K (the drop of K can be calculated using van t off s equation and Δ -46 J, see r. 5.86). o shift the reaction to the right (higher concentration of the product), a very high pressure is needed. 5

15 Example of Application of the Law of Mass Action Let s loo at a simple reaction + CO O + CO otice that we have the same # of moles on both sides of the reaction equation. We start with n and n CO moles of the reacting gases and define as the yield y the number of moles of O that the reaction will produce at equilibrium: n + O y nco y n n y n CO n CO y n n n n CO he mass action law requires: equilibrium concentrations of and CO y ( n y)( n y) his is a quadratic equation with respect to y, the solution is straightforward but messy. What ind of starting concentrations will give us maximum yield? o find out, we have to solve the equation dy/dn. he result: CO n n n / K n / n CO - maximum yield is achieved if you mix just the right amounts of the starting stuff. his result is always true, even for more complicated reactions.

16 he emperature Dependence of K (van t off Eq.) It s important to now how the equilibrium concentrations are affected by temperature (r. 5.85). We also need this result for solving roblems 5.86 and ΔG K exp R ΔG R ( ln K ) R ΔG For either the reactants or the products, ΔS ( ) ( Δ ΔS ) ln K R Let s find the partial derivative of lnk with respect to at const ΔG Δ R ΔG G S Δ R ( ln K ) ΔS van t off s equation Δ is the enthalpy change of the reaction. If Δ is positive (if the reaction requires the absorption of heat), then higher 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). f f Δ d( K ) R i ln d ln K( ) ln K( ) i f i Δ R i f

17 G A Chemical Equilibrium in Dilute Solutions no mixing ideal mixing Water dissociation: G - O + + O ΔG 79.9 J Under ordinary conditions, the equilbrium is strongly shifted to the left, but still there is a finite concentration of ions + and O - dissolved in water. In equilibrium: + + O O x + +O - O the shift is exaggerated: Δx eq ~ -7 Assuming the solution is very dilute: solvent solute solvent solute (, ) (, ) (, ) + ln msolute + + ln m O O O ln m where are the chemical potentials for the substance in its standard state: pure liquid for the solvent, molal for the solutes. his differs from the reactions in the gas phase, where the standard state corresponds to the partial pressure bar. ( m + m - ) ( ) G R ln A Δ O O O m + m O - solvent solvent ΔG exp R 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 because the number of dissociated molecules is tiny. m 7 + m - O he 7 is called the p of pure water. -

18 Chemical Equilibrium between Gas and Its Dilute Solution he 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. O ( gas) O ( aqueous) ΔG 6.4 J he Gibbs free energy change ΔG for this reaction is for one mole of O dissolved in G of water at bar and 98 K. gas In equilibrium: ( ) ( ) For O gas: ( gas ), O ( aqueous), O gas, + ln m (, ) (, ) + ln For O dissolved in water: solute solute( ) solute ( ) ΔG / + ln solute + ln msolute R ln A solute m solute m / solute ΔG exp R For O in water: ΔG exp R enry s law (the amount of dissolved gas is proportional to the partial pressure of this gas) exp 6.4 ( 8. J/K)( 98 K) J. O. bar, m solute. - x..6-4 equivalent of 6.6 cm of O gas at normal conditions ( mol at bar ~5 liters). gas

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