13 Chapter Chemical Equilibrium
Up to now we have assumed that our chemical reactions would go to completion. That is, we would mix two reactants, and they would form product. The amount of product formed was calculated from a limiting reactant, and this reactant would disappear entirely form the reaction mixture. In reality, chemistry is not an all or none process, but an equilibrium process, that is, there is always some starting material left over - not all of it is converted into product. and, in fact some of the product is always trying to convert back into starting material! In the reactions we worked with up till now this equilibrium between starting material and product has been so much in favor of the products that we didnt have to worry about this refinement, but now we will study reaction where the balance between products and reactants is significant. Further the study of reactions and their equilibria teaches us much additional useful chemistry. 13.1 The equilibrium Condition A system is said to be in equilibrium when no changes are occurring in the concentrations of the products or reactants. This does not, however, mean that the reaction has stopped. On the contrary, reactant molecule are continually being transformed into product, while the product is continuously breaking down back into starting material. Equilibrium is a dynamic situation, a point where the rate of the forward reaction is equal to the rate of the backward reaction. Lets examine the reaction of water vapor and Carbon Monoxide gas H2O (g) + CO(g)6H2(g) + CO2(g) Notice that this reaction has a 1:1 Stoichiometry, so 1 molecule of CO is combines with 1 molecule of water to form 1 molecule each of hydrogen gas and carbon dioxide. Lets put an equal number of moles of water and carbon monoxide in a closed container at thigh temp and watch what happens Figure 13.2 (Note that starting with equal moles and 1:1 so line of [CO]=[H2O], [CO2]=[H2]) The concentration of reactant drops and the concentration of product builds up until at some point there is no net change. This is the equilibrium condition. Equilibrium incases that as long as the condition dont change, the concentrations of the reactant and the product wont change either. To study equilibria we have to be conscious that there are TWO reaction occurring at any time the FORWARD reaction with reactants going to products, and the REVERSE reaction with products falling back down into the initial reactants. Thus our reaction arrow actually is a double arrow, one for forward and another for reverse reactions.
2 Notice what is happening here Figure 13.4. Initially we only have reactant so the product is formed rapidly. As the reaction occurs the concentration of the reactants falls and the rate of the forward reaction (Show the forward arrow) begins to slow. Conversely, at the initial point there are no product present so the reverse reaction is negligible. However, as the reaction proceeds and product builds up, the rate of the reverse reaction can start to build up, and you se here the equilibrium is achieved where the rates of the forward and reverse reaction are equal. One of the characteristics of a system at chemical equilibrium is that there is no net change in the concentration of materials in the mix. This does not, however, always indicate that a mixture is at equilibrium. Sometimes the rate of the forward and reverse reaction can be so slow that change cannot be detected. For instance the equilibrium between Fe0 and Fe2O3 (iron oxide - rust) lies heavily on the rust side, thus all iron wants to turn to rust. However if you are careful with iron, you can keep it from rusting for 100's of years. 13.2 The equilibrium Constant & Equilibrium Expression Based on observation it was discovered in the mid 1800's that any reaction of the form jA + kB W lC + mD reactants W Products reached equilibrium, the concentration of the product and the reactants and their stoichiometric coefficients could be combined into an equation called the equilibrium expression:
where [] are the concentration of product and reactants, the exponents correspond to the stoichiometric coefficients, and the K is called the equilibrium constant. What does this expression mean? It means that you can reach equilibrium under virtually millions of different . But when you combine the concentration of the reactants and products using the above equation you will always get the same equilibrium constant, no matter what the individual concentrations of products and reactants is.
EXAMPLE writing equilibrium expression 4NH3 + 7O2 W 4NO2 + 6H2O K = [NO2]4[H2O]4/ [NH3]4[O2]7
3 P4 + 5O2 W P4O10 K= [P4O10]/ [P4] [O2]5 Calculating equilibrium Constants Lets look at the equilibrium that describes waht happens when acetic acid dissociates in water. CH3COOH W CH3COO- + H+ K = [CH3COO-][H+]/[CH3COOH] Under different conditions you can get different sets of concentrations that fulfill this equilibrium Condition 1 Condition 2 Condition 3 [CH3COOH] .053 .36 .8475 [CH3COO ] .947 .64 .1525 [H+] 1x10-6 1x10-5 1x10-4 [CH3COO-][H+]/[CH3COOH] .947(1x10-6) .64(1x10-5) .053 .36 = 1.79x10-5 1.78x10-5 .1525(1x10-4) .8475 1.80x10-5
So under these three very different sets of equilibrium conditions, we end up with the same answer to the equilibrium expression (give or take round off error)
Notice that if the equilibrium expression of the forward reaction jA + kB W lC + mD is:
What do you think the expression for the reverse reaction lC + mDW jA + kB is? Krev = [A]j[B]k /[C]l[D]m = 1/K
4 Along the same lines, what is we have a multiplier for the entire equation i.e. N2 + 3H2 W2NH3 ; K=[NH3]2/[N2][H2]3 = 3.8x104 L2/mol2 (previous example) Suppose we want the equilibria expression of the reaction N2 + 3/2 H2 WNH3 instead? [NH3]/[N2]1/2 [H2]3/2 = (N2 + 3 H2 W2NH3) = ( [NH3]2 /[N2] [H2]3) = K1/2 we see that we simply take the K for the expression and take it to the power of the coefficient
SUMMARIZING 1. Units of K are determined by powers of various concentration terms, the units for K will be different for every reaction 2. the equilibrium constant for a reverse reaction is the reciprocal of that of the forward reaction 3. when a balanced equation for a reaction is multiplied by a factor, n, the equilibrium expression for that reaction is raised to the power of N, and the equilibrium Constant for that reaction is also raised to the power of n. 13.3 Equilibrium Expression involving Pressure So far when we have used equilibria expression we have been dealing with [A] or the concentration of A. When we are dealing with gases we can vary the pressure as well as the concentration. Lets see what that will do to our equilibrium expression. Obtaining Kp from KC If you remember (Chapter 5) our ideal gas law is : PV=nRT and P = (n/V) RT N/V is number/divided by volume so is a concentration or molar concentration of the gas (mole/Liter) so Pressure/RT = concentration Lets look at a gas phase reaction: N2(g) + 3H2(g) W2NH3(g)
5 Note when we are dealing with gas reactions we can have K based on concentrations K = [NH3]2 / [N2][H2]3 as in our previous example. We can also take advantage of the above relationship between pressure and concentration to write K in terms of partial pressure Kp = PNH32 / PN2 PH23 In this class we will always assume a K is in terms of concentration except in the special case where we have a Kp to refer to a K in terms of partial pressure. How do we convert between one and the other? From the above we have that P/RT = concentration Thus all we have to do is to plug in P/RT for each concentration term. For example lets look at K = [NH3]2 / [N2][H2]3
Notice how K and Kp are related by a factor of 1/RT, but the power of the RT factor depends on the exact stoichiometry of the reaction. Here is a reaction where K= KP CH4 + 2O2 W CO2 + 2H2O K = [CO2][H2O]2/ [CH4] [O2]2 = PCO2(PH2O)2/PCH4(PO2)2 (1/RT)3 / (1/RT)3 = KP (1) For the general reaction jA + kB W lC + mD Kp = K(RT))n ; )n = (l+m) - (j+k)
6 Example 2NO(g) + Cl2(g) W 2NOCl(g) K = [NOCl]2/ [NO]2[Cl2] Kp = K(RT))n ; )n = (2) - (2+1) = -1 Kp = K(RT)-1 Kp = K/RT 13.4 Heterogeneous Equilibria The equilibria expression we have used so far works for gases or solids dissolved in liquids. These are called homogenous equilibria because all of the reactants and products are in the same phase. Equilibria that involve more than one phase are called heterogenous equilibria. Lets examine what happens here. Say we have the reaction Ca(IO3)2(s) WCa+2(aq) + 2IO3-(aq) (The reaction we are doing in lab!) What we have learned up to now, you would say K = [Ca+2][IO3-]2/[Ca(IO3)2] While this is what we would expect based on the law of mass action, what we find is the position of this equilibrium does NOT depend on the amount of pure solids(Ca(IO3)2) or the amounts of pure liquids (not this reaction) . What happens is that when you have a pure solid or pure liquid in a reaction, the concentration of the pure solid or liquid does not change in the reaction. Since it does not change, it is essentially constant, and this constant can be incorporated into the equilibrium constant This we have K = [Ca+2][IO3-]2/Constant; K=[Ca+2][IO3-]2
7 This kind of apparent constant will be seen over and over in reaction that occur in water CH3COOH + H2O WCH3COO- + H3O+ K=[CH3COO-][H3O+]/ [CH3COOH][H2O] But [H2O] is 55.5M for water (1000g/18g/mol) the little bit of water that is used in this reaction wont change that so: K =[CH3COO-][H3O+]/[CH3COOH]55.5 K = K55.5 =[CH3COO-][H3O+]/[CH3COOH] (Or substituting H+ for H3O+] K = [CH3COO-][H+]/[CH3COOH] 13.5 Application of the Equilibrium Constant We can use the equilibrium constant in many ways. I. We can use it to determine the tendency of a reaction to occur (But NOT the rate) The Extent of the reaction- does it favor products or reactants II. We can determine if a mixture is at equilibrium III. Given a set of starting concentrations we can determine what the final equilibrium concentrations should be.
Now lets look at each of these three in detail I. Determining the extent of a reaction What is the definition of the equilibrium constant? Concentration of product over reactants. If K is >1 which will you have more of product or reactant? Products If K <1 Will have more reactants. Thus size of K is an indication of how far to right (Products) or left(reactants) a given reaction will go. (Note however that kinetic, how fast it will go is something entirely different. Remind of activation energy. Can have reaction that are very favorable, but if activation energy too high will not occur. For instance Gas can sit in the tank of a car for years without burning up)
8 EXAMPLES The reaction of a weak acid (acetic) with water is CH3COOH + H2O W CH3COO- + H3O+ K = [CH3COO-][H3O+] / [CH3COOH] [H2O] K = 1.74x10-5= [CH3COO-][H3O+] / [CH3COOH] Is this reaction going to favor the reactants (right) or products(left) ? CH3COOH + H2O7 CH3COO- + H3O+ (K <1 so reaction favors reactants) The reaction of EDTA and a metal ion EDTA-4 + Mg+2W MgEDTA-2 K = 6.14x108 = [MgEDTA-2]/[EDTA-4][Mg+2] Again does this reaction favor reactants or products? (Products) II. Determining if a reaction is at equilibrium As we have emphasized over and over, just because a chemical system does not change, does not mean it is necessarily at equilibrium, how can you tell if equilb has been achieved?
One way is to plug your concentrations into the equilibrium expression and see if your values give you the Keq. If they do, fine you are at equilibrium. If they dont, then you arent at EQ. If you arent at equlib you can use the number you just calculate to see if the equilibrium constant favors your reaction moving to the right or to the left (toward Product or reactants) The number you get when you plug your initial concentrations in to the equilibrium expression is called the Reaction Quotient (Q). Very simply if Q=K, you are at equilibrium. What does it mean if Q>K Q>K ; K = product/reactants, so product in excess, will shift toward reactants Q<K K=product/reactants Product < reactants, will shift toward products
9 EXAMPLES Lets go back to our ammonia problem Say we have the following sets of concentrations [NH3] [N2] [H2] -3 -5 A. 1.0x10 1.0x10 2.0x10-3 B 2.0x10-4 1.5x10-5 3.54x10-1 -4 C 1.0x10 1.0x10-2 5.0 Reaction = N2 + 3H2 W 2NH3 K=[ NH3]2 / [N2][H2]3 Q=[ NH3]20 / [N2]0 [H2]30 =(1.0x10-3)2 / [1.0x10-5 (2.0x10-3)3] = 1.3 x107 L2/mol2 [NH3] [N2] [H2] Q 1.0x10-3 1.0x10-5 2.0x10-3 1.3x107 -4 -5 -1 2.0x10 1.5x10 3.54x10 6.01x10-2 1.0x10-4 1.0x10-2 2x10-3 5.0
A. B C
K was 6.0x10-2 So in case A Q>K so conc of product will DECREASE In B K=Q at equilib In C Q < K will move towards product (to the right)
Note - What will happen to our Expression if the concentration of a product is 0? O/X =0 Q=0 Q<K will shift toward products If a reactant conc is 0? K =X/0, K = 4, Q>K will shift toward Net reactants. of the above if either product or reactant is 0, reaction will shift to give that substance a non-zero conc.
III. Given a set of starting concentrations we can determine what the final equilibrium concentrations should be. This is where you have to start being careful of the algebra. Basically you have a set of initial conditions, you then decide which way the equilibrium is going to shift. You then have to add an unknown, x, to the appropriate parts of the equation that will describe how the equilibrium will shift and find out what that X value is.
10 EXAMPLE Lets start out with the reaction CO(g) + H2O(g) W CO2(g) + H2(g) K = [CO2][H2] / [CO][H2O] = 5.10 (no units) Note: why doesnt water drop out?? This is a gas phase reaction, not deaing with liquid water Say we start out with 2.0 mol/L of each reactant A. Will the reaction favor reactants or products B. What will the final concentration of the chemicals be? Answer A. Find Q = 2X2/ 2X2 =1; 1<K will move toward products Answer B. Since will move toward product lets call X, the change (increase) in concentration that CO2 will undergo as reaction goes to equilibrium. Thus [CO2]f = [CO2]i + X = 2.0 +X Because we have 1:1 stoichiometry, for each X moles of CO we make, we also have to make X moles of H2 thus [H2]f = [H2]i +X = 2.0 +X And for each mole of CO2 we make we DECREASE the conc. of CO ans H2O thus [CO]f = [CO]i -X = 2.0-X; [H2O]f = [H2O]i -2.0
Putting this al together we have K =5.10 = [CO2]f[H2]f / [CO]f[H2O]f 5.10 = (2.0+x)(2.0+x) / (2.0-x)(2.0-x) = (2.0+x)2/(2.0-x)2 = [(2+x)/(2-x)]2 Sqrt(5.10) =2.26 = (2+x)/(2-x) 2.26(2-x) =2+x 4.52-2.26x=2+x 4.52-2=X+2.26X 2.52=3.26X X=2.52/3.26 = .773 M so [CO2]f = [H2O]f = 2+.773 = 2.773 mol/L and [CO]f = [H2]f = 2-.773 = 1.227 mol/L
11 Problems like this can get harder or easier depending on how much information is given and how complicated the stoichiometry is. 13.6 Solving Equilibrium Problems Lets summarize the steps used in solving equilibrium problems 1. Write the balanced reaction 2. Write the equilibrium expression that corresponds to the balanced reaction 3. list initial concentrations 4. Calculate Q to determine what direction(if any) the reaction will shift 5. using a single variable(X), define change for each concentration needed to reach equilibrium using X. 6. Substitute modified concentration terms into the equilibrium expression 7. Solve the equilib. expression for the unknown. 8. Plug final values back in to see if they check out. Lets try this entire procedure on another problem . EXAMPLE We want to synthesize HF form H2 and F2 Say we start with 4 moles of H2 and 2 moles of F2 in a 3.0 L container 1. Balance Reaction H2 +F2W 2HF 2. Equilib expression K = [HF]2/ [H2][F2] = 1.15x102 (Must be looked up) 3. Initial Conc [H2] = 4mol / 3.0 L = 1.333M [F2] = 2 mol / 3.0 L = .667 M [HF] = 0 4. Calculate Q Actually is not needed. Since [HF] = 0 know equilib will shift to right to form HF 5. X will be amount of HF formed [HF] = 0+X = X [H2] = ? If [HF] = X, the mole conversion factor based on the equation is: X mole HF X (1 mole H2/2 moles HF) So moles of H2 left = initial -X/2 [H2] = 1.33 - 1/2X = 1.33-.5X
12 [F2] = ? Again, X mole HF X 1 mole F2/2 mole HF So [F2] = initial -X/2 [F2] = .667-1/2X = .667 -.5X 6. 1.15x102 = X2/[(1.33-.5X)(.667-.5X)] 7. Solve the above equation for X. Notice this problem gets hairy (as do most real problems) 1.15x102 = X2 / (.887 - .999X + .25X2) 1.15x102(.887 - 1.0X + .25X2) = X2 1.02x102 - 1.15x102 X + .2875x102 X2 = X2 102 - 115X +28.75X2 = X2 Remember how to solve this? The quadratic Equation? If you have aX2 + bX + c=0
So we have to get the above equation into this form 27.75 X2 -115X +102 =0 x= 115sqrt[1152-4(27.75)102] / 2(27.75) = 115 sqrt(13,225 - 11,320) / 55.5 = 115 sqrt(1905) / 59.5 =115 43.6 / 59.5 (115 + 43.6) /59.5 = 2.858 mol/L and (115 -43.6) 59.5 = 1.286 mol /L Which of these answers is correct? Lets keep going If [HF] = 2.858 M then [H2] = 1.33 - 1/2X = 1.33-(2.858/2) = 1.33-1.429 = Negative number This cant exist so this must be the wrong root.
13 Lets try the other If [HF] = 1.286 Then [H2] = 1.33 - 1.286/2 = 1.33 - .643 = .687 M and [F2] = .667-1.286/2 = .667-.643 = .024M 8. Check to see if these concentrations really work K = (1.286)2/ .687(.024) = 100 Not exactly correct, but round off errors are usually pretty significant in this kind of problem.
Treating systems with very large or small Ks As you can see the last problem got pretty hairy because we had a large K and so we ended up having to use a quadratic equation to get it solved. If you have a small K, you can sometimes makes some assumptions that simplify the problem and make it easier to solve. Lets try the reaction 2NOCl(g) W 2NO(g) + Cl2(g) K = [NO]2[Cl] / [NOCl]2 = 1.6x10-5 mol/L Where does this equilibrium lay? Does it favor products or reactants? (K<1, favors reactants, very little product formed) Lets say we start out with 1.5 mol of NOCl in a 1L flask. What are the concentrations of the various components after the reaction has come to equilibrium? [NOCl]i = 1.5 mol in 1L = 1.5 M [NO]i = 0 M [Cl2]i = 0M Initially no products present so Q = 0/(1.5)2 = 0. 0<K so reaction will move to form product. How far? Lets call the amount of NO formed X. Mole Cl2 = mole [NO] x 1 mole Cl2/2 mol NO = Mole NO/2 = X/2 Mole NOCl used = mole [NO] x 2 mole NOCl/2 mole NO =X
14 [NO]f = 0 +X [Cl2]=0+X/2 [NOCl] = 1.5-X Our equilibrium expression is then K = 1.6x10-5 = [NO]2[Cl] / [NOCl]2 = X2(X/2) / (1.5-X)2 = X2(.5X)/(1.5-X)2 = .5X3/(1.5-X)2 You can see that we will have a term with X3 ,X2, etc in the numerator so this is going to get very ugly. However, lets think about the fact that we have such a small K. This means that the reaction really doesnt favor the product very much, so you really dont expect much product to be formed. This also means that not much of the reactant will be used up. If we start out with 1.5M reactant, and very little of this reactant will be used up, we can make the assumption mathematically that 1.5-X .1.5 Plugging this in to the equation we have K = X2(X/2) /(1.5-X)2 .X2(X/2) /1.52 = X3/2(1.5)2 1.6x10-5 =X3/4.5 7.2x10-5 = X3 X=4.2x10-2 Our equilibrium concentration are then: [NO]f = 0 +X = 4.2x10-2 M [Cl2]=0+X/2 = 2.1x10-2 M [NOCl] = 1.5-X = 1.5-.042 = 1.46 which is only .04/1.5 x100 or 2-3% error Lets check these by plugging in to the equilibrium expression ? =[NO]2[Cl] / [NOCl]2 = (4.2x10-2)2 (2.1x10-2)/1.52 = 1.6x10-5 So our answer checks out Bottom line when the value of K is small you can often make some assumptions that simplify solving the problem.
15 13.7 Le Chteliers Principle When a system is at equilibrium it is balanced and forward reaction are exactly countered by back reactions. What happens, however, when we introduce some change to the equilibrium system? Le Chteliers principle says that is a system that is at equilibrium is changed, the position of the equilibrium will shift in a manner that will tend to reduce the change. This is a simplification, but it usually works well Lets see how in can be applied in different situations The Effect of a Change in Concentrations Lets continue with our NOCl system. We have just calculated that at equilibrium we have (2NOClW 2NO + Cl2) [NO] = 4.2x10-2 M [Cl2] = 2.1x10-2 M [NOCl] = 1.46 M What happens to the equilibrium if we pump NOCl into the system and increase its concentration to 3M does the equilibrium move toward the product or the reactants (is more product formed?) Well the way you have been handling this is to attack it with a Q calculation Q = [NO]2[Cl2] /[NOCl]2 = (4.2x10-2)2.1x10-2/ 32 (new value) Q = 4.1x10-6 ;K is 1.6x10-6; Q>K; Reaction moved toward product Now that we proved this with math, Lets think if we could have predicted this without the math Have reactant and products. Increase concentration of reactant. More reactant forward rate of reaction increases, start to make more product. After a while product builds up so reverse reaction catches up to forward reaction, but overall equilib has shifted toward product. Now Using Le Cs. Increase reactant, the relieve this change will form more product. The Le C principle also work in the other direction. Say we had some special way we could get rid of NOCl, say by making it react in a second different reaction, so the concentration goes to .5M instead of the 1.46M we have at equilibrium By Le C, we have lowered the con of reactants in the system, so the system will respond by shifting the equilb toward the reactants trying to relieve the change on the system. (You might confirm this with a Q calculation) In General we can state If a reactant or product is added to a system at equilib, the system will shift away form that component. If a product or reactant is removed form a system the system will
16 respond by shifting toward the missing component. The effect of pressure If we are working in the gas phase we can change the overall pressure in the system as well as the concentration of the individual components. We can do this by either adding an inert gas to the system that increases the overall pressure or by changing the volume of the container we are doing the reaction in. The effect of changes in pressure on gas phase reactions can also be understood using Le Cs Principle
Case 1 addition of inert gas When we change the pressure in a system by adding an inert gas we do not change either the K of the reaction or the concentration of the components or the partial pressure of the components. Thus nothing significant changes and you remain at equilibrium Case 2 Change in volume when the volumes changes, the concentrations change. Once concentration change we have to start all over on our Q calculations to predict what will happen. However if we are clever we dont have to do all that work. Lets focus on the volume of the system and look at how the volume of the system may be used in LeC Principle If we change the volume of the system, LeC s principle say that the system will respond to try to minimize that change Thus if the volume increases the equilb will shift in a manner that increases the volume. If the volume is decreased, the system will respond in a way to decrease the overall vol of the system OK that sounds good, but how do we tell how whether a shift in equilibrium favors an increased volume or a decreased volume? Remember PV=nRT? V=n RT/P if T and P are constant, then V is directly proportional to n i.e. volume is proportion to number of molecules (DUH) Thus all you have to do is to look at the total number of reactant gas molecules and product gas molecules to see if a reaction increases in volume or decreases in volume. In our example 2NOClW2NO + Cl2 we have 2 gas molecule as reactants and 3 as product, so the volume here increases with the reaction. If we increase the size of the container we will favor the product, if we decrease the volume we will favor the reactants
17 How about an even earlier example N2(g) + 3H2(g) W3NH3(g) How will this change with a change in volume? (4 molecules of reactants, 3 of product, product has smaller vol that reactant, decrease vol favor product, increase vol favor reactants The effect of Temp In studying the effect of temp, vol and ) conc on a reaction we have may change the conditions under which a reaction occurs but we don not change the actual K of the reaction. If we change the temperature at which a reaction occurs, then the K of the reaction changes and your first guess is that all bets are off. you have to star over because you need a new K value before you can do anything. However, to a first approximation you can make a good guess as to what will happen to K si you treat heat like a reactant or a product. Remember the concept of Enthalpy or heat of reaction? We found that many reaction release heat (- )H) while other reaction take heat form the environment (+)H). We can apply Le Cs princ to the heat in a reaction to predict which way K will move. If e add heat to a system the system will respond by trying to minimize the added heat. If we remove heat form the system, the system will respond by trying to generate heat to minimize the change Lets take the ammonia example N2(g) + 3H2(g) W3NH3(g) This reaction s exothermic and releases 92 kJ of heat, ()H =-92kJ) If we heat the system the system response is to remove the heat. it does this by decreasing the amount of product, so the heat is generated in the reaction in lowered thus K is decreased to favor reactants. Note book writes the E as a product N2(g) + 3H2(g) W3NH3(g)+92 kJ This way can see that increase in heat should push equilib to left, so K should go down, but this approach can be confusing. Lets try come other examples Heat + N2(g) + O2(g) W 2NO(g) )H = +181 (endothermic) 2SO2(g) + O2(g) W SO3(g) + Heat )H=-198 (exothermic) If the temp increases the K of reaction 1 will increase while the K of reaction 2 will decrease.