Equilibrium Reaction Systems

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Winfred Parks
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1 Equilibrium Reaction Systems Equilibrium defines a perfectly reversible process. We have found in an earlier chapter, that we have a function that, when negative, will give us a criteria for determining whether or not a reaction will occur. This is, of course, the Gibbs Free energy. Thus, let us begin with this function. Components of a system In order to examine what is happening within a system of a single component within phases or within a multi-component system (mixture, such as a solution), we need to look at ar quantities. Let's examine a single component with our new Gibbs Free Energy function. At a fixed temperature, we know that ΔG = VΔP We also know that the volume of a substance is dependent on the pressure applied. If we accept, as a first approximation, the ideal gas law, we have.. P 2 P 2 nr T P 2 ΔG = V dp = dp = nr Tln = G P P P 2 G P 1 Now, let us define a standard state as 1 bar. We will then reference G to this state. Letting G 1 be equal to G o and P 1 be equal to Po which will be set to 1 bar, we can rewrite the above expression as: G = G o nr T ln P P G o = standard Gibbs Free Energy o It is important to note that this is true for any single component in a system of components. As we will see in a later chapter, the thermodynamic properties, per e, of any given substance is important. Without evidence of it's importance, I introduce, here the concept of a ar Gibbs Free energy, known as the "chemical potential" and given the symbol, μ. For any single component in a system, we then write.. G G o = RT n n ln P r μ = μ o RT ln P o If μ is a single component in a multi-component system, we write: where P i is the partial pressure of the component in the system μ = μ i oi RT ln P i 1

2 The above expression is interesting and important. For easy comparison, consider for a moment, Newtons Third Law of Dynamics. This law, simplified, says that any action results in an equal and opposite reaction. The chemical potential is an equivalent relation in chemistry. Any internal change in the chemical potential of a component will be accompanied by an equal and opposite change an the chemical potential of another component. This fact will be of great use later. Now, let us put the chemical potential to use as, because it is teh Gibbs Free Energy, it is supposedly THE criteria for spontaneity... Let us consider the Gibbs energy change for a reaction system of interest. These are obviously important as we want to know the reaction likelihood and equilibrium condition. ΔG Now, for the reaction: aa + bb -> cc + dd, we have = ν μ i i i ν i μ( T) i o = RT i lnp i = i ν μ( T) i i o RT ln c d P C PD P A a PB b but i ν μ( T) i i o = ΔG( T) o So now.. ΔG rxn = ΔGT ( ) o RT ln c d P C PD P A a PB b By convention. c d P C PD P A a PB b = Q rxn = Reaction Quotient If reaction is at equilibrium, then G = 0 and c d ΔG( T) o P C PD = RTln = RTln K a b P A PB eq or K eq = e ΔG( T) o RT 2

3 Note... K p = P C P A c a P D P B d b c d c d n C RT n D RT M C M D V C P o V D P o = RT Δn M o M o = = K a b a b c RT Δn n A RT n B RT M A M B V A P o V B M o M o Important!!! It is critical to note all the assumptions and standards used. We assumed... A standard state of 1 bar pressure. All materials are gases All gases are ideal How on EARTH are we to apply these assumptions to, say, an acid equilibrium in solution?! We will address this after we examine some properties of ideal gaseous equilibria. Calculation of the Equilibrium Constant at 298 K Calculate the Equilibrium Constant for the reaction SO 2 + 1/2 O > SO 3. Strategy: Calculate G rxn using formation tables, then calculate Keq from tables, G f (SO 2 ) = / and G f (SO 3 ) = /. Now G rxn is 1000joule ΔG rxn ΔG e e rxn e R joule T 298K ek ΔG rxn RT K eq e K eq

4 Temperature Dependence of the Equilibrium Constant It is obvious that the equilibirum constant depends on temperature. If we want to examine a reaction system at other temperatures, we must know the requirements. Without undue mathematics, we can get a reasonable temperature relationship for the equilibrium constant. Consider K eq at two different temperatures. From above, we have.. K eq T 1 e substituting, we have that ΔG rxn RT 1 = but recall that ΔG = ΔH TΔS K eq T 1 e ΔHT 1 ΔS RT 1 = = e ΔH rxn ΔS rxn RT 1 R A similar expression can be said for a second temperature.. K eq T 2 e ΔH rxn T 2 ΔS rxn RT 2 = = e ΔH rxn ΔS rxn RT 2 R Dividing these two expressions and taking the natural logarithm of both sides gives... ln KT 2 KT 1 = ΔH rxn R 1 1 T 2 T 1 assuming H and S are independent of temperature!!! As usual, this expression will work if the temperature ranges are small. if we allow the enthalpy to be temperature dependent and the heat capacities to be constant, we get... ln KT 2 KT 1 ΔCpT 1 = ΔH rxn_298 R 1 1 T 2 T 1 ΔCp T 2 R ln T 1 4

5 Consider the equilibrium reaction: N 2 O 4 (g) = 2 NO 2 (g). Calculate the equilibrium constant at 595 K given the Kp at 298 and constant Temperature Cp values. J J Kp ΔCp 2.88 ΔH K T 1 298K T 2 595K Kp 595_literature R J K Method I - Assume complete temperature independence. Kp T ln Kp( T) Kp 298 ΔH 298 = R 1 1 T 2 T 1 Kp 298 exp Kp I T 2 ΔH 298 R 1 1 T 2 T Kp I T 2 Method II - H = f(t) ln Kp T Kp 298 ΔCp T 1 = ΔH 298 R 1 1 T 2 T 1 ΔCp T 2 R ln T 1 Kp II T 2 Kp 298 exp ΔCpT 1 ΔH 298 R 1 1 T 2 T 1 ΔCp T 2 R ln T 1 Kp II T

6 Method III This method is used for illustrative example only. Do not try this at home..leave it to the professionals. In it, the heat capacities are assumed to be temperature dependent. cal cal a NO b K NO c NO K 2 cal cal a N2O b K N2O c N2O K 2 cal K 3 cal K 3 Cp NO2 ( T) a NO2 b NO2 T c NO2 T 2 Cp N2O4 ( T) a N2O4 b N2O4 T c N2O4 T 2 Δa 2a NO2 a N2O4 Δb 2b NO2 b N2O4 Δc 2c NO2 c N2O4 Kp III T 2 Kp 298 exp T 2 T 1 T ΔH 298 Δa ΔbT ΔcT 2 dt T 1 RT 2 dt Kp III T T 300K350K 1000K Equilibrium Constants vs. Temperature for the 3 Methods Kp I Kp II Kp III Kp Temperature (K) 6

7 Pressure Dependence of Equilibrium Although the equilibrium constant is independent of pressure, the equilibrium is not! These examples illustrate that fact. The following is called the "extent of reaction" approach. In this method, "x" is the "extent" of the reaction, or a number between 0 and 1 that tells us how far the reaction has proceeded. If x = 1, then the reaction has gone to completion for example Consider, first the reaction: PCl 5 ( g) = PCl 3 ( g) Cl 2 ( g) K p PCl 2 P PCl 3 for which K P = PPCl 5 Now, consider that x fraction of reactant decompose. If this is the case, then the ICE diagram approach from General Chemistry gives: in terms of TOTAL system pressure will be: PCl 5 ( g) = PCl 3 ( g) Cl 2 ( g) 1 x x x Notice that if x = 0, nothing has happened and if x = 1, the reaction is completed. It is our desire to get the partial pressures of all species at equilibrium. To do this, we will use Dalton's Law of Partial pressures where the partial pressure of any component in a mixture of gases is the e fraction multiplied by the total pressure or P i = X i P tot To get the e fractions, we need the total amount of material in the mixture. We do this from the reaction stoichiometry using "x" to get x Total = ( 1 x) x x = 1 x Since x is prop[ortional to the es, we can get the e fraction as; XPCl 3 x = XCl 1 x 2 x = and xpcl 1 x 5 = 1 x 1 x 7

8 Now, applying Dalton's Law, the partial pressures will be: PCl 5 ( g) = PCl 3 ( g) Cl 2 ( g) ( 1 x) P T 1 x xp T 1 x xp T 1 x and the equilibrium expression can be now written as: K p = xp T ( 1 x) ( 1x) P T ( 1x) 2 x 2 P T x 2 P T = or K 1 x 2 P p = o 1 x 2 Now, consider the condition of compressing the reaction vessel such that the total pressure of the system increases. The extent of the reaction, x, may be found as a function of total pressure.. Solving for x(p) and plotting PCl 5 and PCl 3 fractions vs. total pressure, P demonstrates the total pressure effect on the equilibrium From our K p expression, we do a little algebra.. x 2 P T = K p 1 x 2 = K p K p x 2 or x 2 P T K p = K p and x 2 K p = P T K p Finally K p x = Lets Plot! P T K p 8

9 From above, we see that 1bar PPCl 3 xp T = and PPCl 1 x 3 = ( 1 x) P T 1 x P T 1bar 2bar 20bar xp T P T K p K p PCl 5 P T 1 xp T P T 1 x P T PCl 3 P T xp T P T 1 xp T Partial Pressures PCl5 PCl3 Partial Pressures vs. Total Pressure Pressure Total Notice the agreement with LaChatelier's Principle!! 9

10 Gibbs Connection. Let us not forget where this all came from..that of finding the minimum Gibbs Free Energy. If one calculates and plots the Gibbs Free Energy as a function of the progress of the reaction, we see an interesting result. The Gibbs Free Energy drops, the reaches a minimium and then finally climbs. The equilibrium point should be at the minimum! This calculation was done at a pressure of 10.0 bar. When computed properly, it is found that the bottom of the curve is at a reaction coordinate of x = If I place that value into the K p expression we developed above at 10.0 bar... x bar x K p K 1 x 2 P p o We get the correct K p!! So we see that the Gibbs Free Energy does rule the game! 10

11 Constant Volume Conditions - Illustration Only An initial number of es, n init, of N 2 O 4 is placed in a 1.0 L vessel held at constant volume, V. Calculate and plot the equilibrium pressure of NO 2 as a function of Volume for varying initial es of 1, 10 and 100 es. Using ICE diagrams as introduced in Gen Chem... N 2 O 4 = 2 NO 2. n init - x 2x Now, since nr T P = we have V K p = P NO2 P N2O4 2 = V ( 2x ) RT 2 n init x RT V = 4x 2 n init x RT VP o Solving for x yields... 4x 2 n init x K p V = = A where A = RT K p V RT 4x 2 = An init Ax or 4x 2 Ax An init = 0 This is a quadratic equation which yields.. 11

12 x = A A 2 44 ( ) An init 8 Evaluating.. V 1L R Lbar T 325K K K p V A A n RT init 1 x( V) A A 2 44 ( ) An init 8 P NO2 ( V) 2xV ( ) RT V P N2O4 ( V) n init xv ( ) RT V P T ( V) P NO2 ( V) P N2O4 ( V) V 1.0L 1.01L 5L Partial Pressures (bar) Partial Pressures vs Volume N2O4 NO Volume (L) 12

13 X NO2 ( V) P NO2 ( V) P T ( V) X N2O4 ( V) P N2O4 ( V) P T ( V) 0.8 Mole Fractions vs. Volume 0.7 Mole Fraction N2O4 NO2 Volume (L) As mentioned above, everything we've done here involve ideal gases using a standard pressure of 1 bar. What if the gas is non-ideal, or what if the system is a solution? We will deal with solutions in the next section. For now, let us introduce the approach used to deal with all non-ideal systems.. 13

14 Non-Ideal Systems To this point, we have largely concerned ourselves systems that are expected to behave ideally. The result is that we have ignored the effect of concentrations and non-ideal effects on macroscopic properties. More often it is mixtures with which we are concerned. As might be expected, the interaction amongst the components can be treated ideally, in which case the system may be treated component by component and the results obtained for a component mutiplied the number of es and added to the results of the other components. This is what we have done so far. However, for more exacting work and for solutions that are far from ideal, we need to be able to find information about the system such as the boiling point, vapor compositions above mixtures, osmotic pressures, equilibria, etc. The best tools we have yet to work with systems is the chemical potential. However, the chemical potential is defined in terms of ideal gases! This poses a problem. Were we to rederive expressions in terms of real equations-of-state, (even if we HAD one for complex systems) the resulting expressions would be too cumbersome to use and would be inconsistent in structure from system to system. We thus recognize that we LIKE the form of the equilibrium constant and wish to keep it. The problem is that non-ideal systems do not produce this form. The answer to this problem lies with recognizing how the form of the equilibrium constant came about. That form came from the expression: μ = μ o RT ln P If this form is not retained, then teh form of the equilibrium constant is lost and our work becomes greatly magnified! Note for example: We used the ideal gas law as.. V = RT P What if we had used a non-ideal gas law? Let us pick a simple one such as: V = RT P αp where α is a parameter specific to a chosen gas. 14

15 Were we to use this to re-derive the equilibrium conditions, the result would be: ΔG( T) o = RTln c d P C PD P A a PB b cα C P C dα D P D aα A P A bα B P B Note how the simplicity of the equilibrium constant expression is totally lost even with this simple equation of state correction! Clearly this is too cumbersome of an approach and depends on our choice of the equation of state. We must take a better approach to this. To do this, let us return to the expression that led us to our simple expression: μ = μ o RT ln P What will will do is to make every attempt to retain this form. Clearly, we can not rely on the ideal pressures for non-ideal systems. Instead, we define a new unit called the "activity", a and use it in the place of the pressure. μ = μ o RT ln( a) If we retain this form, then we also retain the form of the equilibrium constant. The only change is that the quantities expressed are no longer pressures, but activities. For example... Example: Write down the equilibrium constant expression for the following reaction: H 2 (g) + Cl 2 (g) = 2 HCl(g) 2 P HCl Ideal answer... K eq = P H2 P Cl2 New NON-ideal answer K eq = 2 a HCl a H2 a Cl2 15

16 Note: The equilibrium constant has not changed here. We still calculate it from ΔG rxn RT K eq = e as before. But what is the activity??? It is important to recognize that: The activity of a substance, solution, etc., is a thermodynamic quantity that connects the system to the Free Energy and thus to spontaneity. But what is this quantity and, most important, how does it relate to the measured pressures and concentrations that we see in a chemical system? The only problem that remains is that we must connect this thermodynamic quantity of activity to an experimentally measured quantity. One we have done this, then we can go into the lab, make a measurement and connect it to the thermodynamics. From our new Gibbs equation, we can clearly see that we have replaced the activity, a for P/. Thus, if the system were an ideal gas, the we know that: But what if the system is non-ideal?? To contend with this, we introduce a correction, of sorts, called the activity coefficient, show here as" a = P a = γ P From this definition, we see that if we have an ideal gas, then the activity coefficient should equal "1" and the activity will be equal to P/ as before. Thus, we have then defined a reference point for gases as the point at which the gas becomes ideal, which is as P goes to 0. a γ B = where P P lim 0 a P = 1 16

17 So what is this value and how do we calculate it? This is not an easy answer and involves a great deal of work. We will see how this value is determined and ultimately how it is used. Let's begin with a simple example. Example: Consider the following equilibrium at 1500 K: CaCO 3 (s) = CaO(s) + CO 2 (g) ΔG f ΔH f J J C p K K J K Calculate the non-ideal equilibrium pressure of CO 2 at this temperature. We begin by calculating the equilibrium constant, first at K, then at 1500 K using the methods previously introduced. T o ΔG Rxn ΔG Rxn K ΔG Rxn K eq_o exp K RT eq_o o Value at K Now, calculating at 1500 K... ΔH rxn ΔH rxn J J C p_caco C K p_cao C K p_co J K 17

18 ΔC p C p_co2 C p_cao C p_caco3 ΔC p 1.34 J K T K K eq K eq_o exp ΔC p T o R ΔH rxn 1 1 T 2 T o ΔC p T 2 R ln T o K eq Now, solving the system... The equilibrum constant expression for this system is simple as all but one item are solids. We thus write. K eq = a CO2 Now, inserting the correct expression for the activity, we have... P CO2 K eq = γ P 1bar o where we have used the reference standard = 1 bar and P CO2 will be in bars. From the K calculated at the desired temperature, we can solve for the pressure of CO 2 as.. P CO2 = K eq γ BUT... The only problem is, of course, we don't know a value for γ! The value can be calculated or measured and I simply present the result below. 18

19 Activity Coefficient (unitless) Activity Coefficient for Carbon Dioxide vs. Pressure Pressure (bar) Here, we see a problem!!..the activity coefficient is different depending on the pressure of carbon dioxide! So, what value do we pick?? This is a standard problem and is solved in the following manner Solve for the pressure of CO 2 assuming ideal conditions. Using the ideal pressure, find a value for the activity coefficient and using that value, solve the non-ideal expression to find a new pressure for CO 2. Repeat step 2 until the answer converges. Let us proceed.. K eq Step 1: P CO2 P 1 CO2 4bar Step 2: From the plot, we see that at 115 bar, the activity coefficient has an approximate value of Resolving K eq Step 3: P CO2 P 0.45 CO2 9bar We have a new answer. We will now take this pressure and, examining the plot again, find a new value for the activity coefficient, which at that pressure is approximately

20 K eq Repeating.. P CO2 P 0.31 CO2 12bar K eq And again P CO2 P 0.29 CO2 13bar At 396 bar, the coefficient is about the same value, so we have arrived at the correct answer of about 396 bar which is quite different from ideal!! This process is carried out the same way for all systems. We only need to know how to find the activity coefficient. This is done largely through experiment. The advantage of this definition for the activity is that it can be applied to ANY system, gaseous or not! We will now move on to examine non-ideal solution systems. 20

aa + bb ---> cc + dd

17 Chemical Equilibria Consider the following reaction: aa + bb ---> cc + dd As written is suggests that reactants A + B will be used up in forming products C + D. However, what we learned in the section