Remember next exam is 1 week from Friday. This week will be last quiz before exam.

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Jeremy Hunt
6 years ago
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1 Lecture Chapter 3 Extent of reaction and equilibrium Remember next exam is week from Friday. his week will be last quiz before exam. Outline: Extent of reaction Reaction equilibrium rxn G at non-standard state Relating G and ξ and K and p K and ξ as fctn of p Review Partition function values for typical molecule (HCl) huge for translational ( 6 ) a handful for rotational () just over for vibrational (.7) and exactly for electronic () I m not going to write it all down, but remember that we had that nice table that summarized the heat capacities of molecules. Where is the temperature dependence? all of them technically, but usually just vibration. nd Chapt ) Density of States: ) Fluctuations are critical: 3) Negative temp: 4) Graphical connection between energy functions. Check out p.6 carefully. Now on to Chapt 3 We have done a lot recently figuring out how to relate microscopic information (energy levels and such) to macroscopic quantities (H,, S, U ) through the partition function. oday we are going to start really applying this to find out some useful chemical information. We are getting into some real freshman chemistry equilibrium constants. Of course, we ll do it in just a bit more detail than you saw in CHEM. Let s define the simplest possible reaction: a z If we started with n and n moles of and then let s define the amounts of and at any time n n aξ n n + zξ

2 ξ (xi or squiggle) is the extent of reaction. It measures the progress of the reaction and has units of moles. Subtraction is because is a reactant. For instance: H + ½O H O t start: n H n O n HO Later: n H ξ n O ½ξ n HO + ξ Of course, we are really interested in the change in the amount of the substances: dn j dξ (negative if reactant) We can substitute this expression into what is called the Gibbs-Duhem equation: Gibbs-Duhem is for constant and p (recall dg -Sd + Vdp + µ dn + µ dn ) dg µ dn So, rewriting Gibbs-Duhem in terms of ξ gives: dg jµ dξ (where products have positive and reactants negative coefficients) his can be rewritten as: jµ ( ξ ) G where r G is the reaction Gibbs energy. his is defined for a balanced chemical equation and the reactants and products at their standard states. (add products, subtract reactants) r Let s look at a simple case: Isomerization Reaction: -cis retinal all trans retinal Everyone know what this does? vision µ trans ( ξ ) µ ( ξ ) cis Both of these chemical potentials change as the rxn proceeds, so this differential depends on the extent of reaction.

3 < µ cis > µ trans G µ cis µ trans > µ cis < µ trans % cis () ξ % trans () When µ > µ, or < the reaction -> is spontaneous (ξ must increase to minimize G) When µ < µ, or > the reaction -> is spontaneous (ξ must decrease to minimize G) he minimum in occurs when its value is zero, or µ µ. (really aµ zµ. his is when the reaction is at equilibrium! he reaction stops. ll right then, we can see that equilibrium is determined by the chemical potential. nother way of saying this is that the chemical potential is the molar G for each product/reactant. (Note that chemical potential depends on concentration,, etc.) Note that the standard molar reaction Gibbs energy is different than the reaction Gibbs energy because for the standard energy, the composition of reactants and products are fixed at their standards states ( M or atm), rather than an arbitrary composition: r G zµ - aµ Calculating the composition of reactions at equilibrium (what is the yield?) Remember that we are talking about equilibrium constants, so we want to try and get back to this. If and are assumed to be perfect gases, we can calculate r G just as we have done before (constant p and ): r G zµ - aµ 3

4 We now need to make use of another expression that shows the pressure (concentration) -dependence of the chemical potential. µ p µ + R ln p (comes from Chapt : q µ k B ln ) N p p z µ + R ln a µ + R ln p p p p zµ aµ + Rz ln Ra ln p p p + R ln p z a where r G zµ - aµ <PUSE> We will find it convenient (for more complicated reactions) to denote the argument of the logarithm as the reaction quotient, Q (errible notation of course. Q is not the partition function.) G G r r ( + R ln Q) t equilibrium r G and Q is equal to the equilibrium constant K (by definition): r G - R ln K his is one of the most useful equations in chemical thermodynamics because it provides a direct relationship between all that thermodynamic data and information on how actual chemical reactions proceed. Obviously, when K < (the reactant is favored), r G > and the reaction is not spontaneous and when K > (the product is favored), r G < and the reaction is spontaneous in the direction. (If we have equal concentrations of products and reactants.) Exergonic and endergonic reactions r G < : reaction is spontaneous and exergonic (work-producing); could drive other processes. e.g. P DP or glucose combustion r G > : reaction is not spontaneous; it is endergonic (work-consuming) e.g. DP P or electron transfer 4

5 Discussion of the general form of G vs. ξ If r G for a reaction is negative, the products are more thermodynamically stable than the reactants. Everyone agree? Do most reactions go all the way to products, or stop somewhere in equilibrium? Most stop short of % yield. Why? he standard molar reaction Gibbs energy ( r G ) does not account for the additional entropy introduced when substances are mixed together. In other words, r G does not include contributions from mix G. mix G? nr(x ln X + X B ln X B ) (hink back to chapt 6 homework) (X is mole fraction, ignore enthalpy) So, the minimum in mix G occurs when? our B reaction is 5 % complete. fter that point, our devious friend entropy fights the continued conversion of reactants to products. G react G Rxn G G ot G prod G Mix ξ Depending on the relative magnitudes of r G and mix G, the extent of reaction can vary dramatically. Now that we have a relationship between equilibrium constants and thermodynamic data, we can calculate the extent of reactions (ξ) for real systems. Use this format: N O 4 (g) <-> NO (g) Initial amounts Equilibrium amounts - ξ ξ otal amount + ξ Equilibrium mole fractions ( - ξ)/( + ξ) ξ/( + ξ) (Note that in this case ξ is now unitless and must be between and because of how we have set up our initial conditions in relation to the stoichiometry.) 5

6 OK so break into groups tell me what K p is for this rxn in terms of pressure of NO and N O 4, and then solve for K p in terms of ξ and P tot. What should the equilibrium constant look like in terms of pressures? products ( PNO ) K p reactants ( P ) N O4 Using Dalton s law and neglecting non-ideal effects: K p products reactants ( P ( P NO ) N O4 ) ( X P ) {[ ξ /( + ξ )] P} X NO N O4 ot P ot [( ξ ) /( + ξ )] P 4ξ P ξ K clearly is constant. So if p changes, ξ must change to compensate (Le Chatelier)! Solving for ξ ξ [ + (4 / K)P] his is an interesting result. he equilibrium constant does not depend on total pressure, but the extent of the reaction does. We will discuss this further later For the above reaction mixture at bar total pressure and 98 K and using the equilibrium expression: r G - R ln K to calculate K ( r G 4.73 k mol - ): K.48 ξ.9 So at room temperature and one atmosphere pressure, only about % to the product gets used up. Perhaps the surprise is that any NO gets consumed at all since r G is positive. 6

rate of reaction forward conc. reverse time P time Chemical Equilibrium Introduction Dynamic Equilibrium Dynamic Equilibrium + RT ln f p

rate of reaction forward conc. reverse time P time Chemical Equilibrium Introduction Dynamic Equilibrium Dynamic Equilibrium + RT ln f p Chemical Equilibrium Chapter 9 of Atkins: Sections 9.1-9.2 Spontaneous Chemical Reactions The Gibbs Energy Minimum The reaction Gibbs energy Exergonic and endergonic reactions The Description of Equilibrium