Chemical Equilibrium The Concept of Equilibrium Sometimes you can visually observe a certain chemical reaction. A reaction may produce a gas or a color change and you can follow the progress of the reaction by noting the volume of gas, or the intensity of the color. Often a reaction may appear to stop even though some reactants remains in the sample The condition where the concentrations of all reactants and products no longer change with time is called chemical equilibrium What is actually going on? At chemical equilibrium, the forward rate of the reaction, that produces product(s), and the backwards rate of reaction, that produces reactant(s) are equal One condition for reaching equilibrium is that there is no process by which the reactant(s) or product(s) are removed from the system Familiar examples of equilibrium You have already seen several examples of physical processes in equilibrium: Establishment of a vapor pressure results from equilibrium of molecular movement between the liquid and gas phase. The rate of molecules leaving the liquid phase is equal to the rate of molecules entering the gas phase. In a saturated aqueous solution containing an ionic solid a crystal of the ionic solid may neither dissolve or grow, but stay the same size. The rate of ions leaving the crystal is equal to the rate of ions colliding with the crystal and entering the crystal lattice. Chemical equilibrium is similar, but deals with rates of chemical reactions The Concept of Equilibrium At equilibrium the rate at which products are formed from reactants equals the rate at which products break down to form reactants Consider the unimolecular elementary reaction process: A B Since this reaction as written has been defined as an elementary process, we now know something about the rate of the (forward) reaction: Reaction rate = k [A] What about the reverse reaction? The reverse reaction is also a unimolecular elementary process and, therefore, the reaction rate is equal to some rate constant times the concentration of B. But it is a potentially different rate constant o Recall that the rate is proportional to the activation energy, E a, and the activation energy for the reverse reaction will be equal to (E a + ΔE rxn ) o The forward rate constant (k f ) will therefore be different from the reverse reaction rate constant (k r ) Forward reaction rate = k f [A] 1
Reverse reaction rate = k r [B] Consider how this reaction would look at the molecular level if we started with pure A: At the beginning there is no B, therefore, the reverse reaction rate is 0 (since [B] = 0). At the beginning there is a lot of A, so the forward reaction rate is high (since [A] = large) As time goes by (as it always does) the concentration of A decreases (as it is used up). Therefore the forward reaction rate slows down. Also, the concentration of B starts to build up. Therefore, the reverse reaction rate starts to increase. At some point, the forward and reverse reaction rates will balance out and equal each other. Note that this does not necessarily mean the concentrations of A and B equal each other. For the rates to be equal, the product of (k f * [A]) must equal( k r *[B]). Therefore, the concentrations of A and B will be equal at equilibrium only if k f = k r. Forward Rate = Reverse Rate k f [A] = k r [B] This equation can be rearranged to relate the concentration of A to B at equilibrium: What this means is that at equilibrium the ratio of the concentration of B to A will always have the same value Once equilibrium is established, the concentrations of A and B do not change This does not mean that all reactions have stopped. Rather it means that the rates of the forward and backward reactions are equal to each other, and therefore, there is no net change in the concentration of reactant and product This is known as a dynamic equilibrium. The forward and reverse reactions are represented by a double arrow in the chemical equation: At dynamic equilibrium: The concentrations of A and B are constant The forward and reverse reaction rates are equal 2
The Equilibrium Constant The Haber Process Human agriculture requires a whole bunch of ammonia-based fertilizer. Natural deposits of nitrate compounds, and bird and bat guano have been rich sources of nitrogen from which to produce ammonia based fertilizer (and also, nitrogen based explosives). Although the atmosphere is about 70% nitrogen, it was not until the early 1900's that a chemical method was developed to allow the chemical production of ammonia from nitrogen gas. This method was developed by Fritz Haber in Germany in 1912, a method known as the Haber Process: In the Haber Process, N 2 and H 2 are placed together in a high-pressure tank (at several hundred atmospheres pressure), and at a temperature of several hundred C (and in the presence of a catalyst also). Under these conditions the two gases react to produce ammonia: N 2 (g) + 3H 2 (g) 2NH 3 (g) Extreme conditions are required because we have to break the N-N bond in N 2 and this bond is a (strong) triple bond When the N 2 (g) and H 2 (g) are combined in the Haber process, the reaction proceeds and ammonia, NH 3 (g) is produced The reaction seems to stop at a certain point, and some N 2 (g) and H 2 (g) remain in the sample (along with the ammonia that is produced) Another way of saying the reaction appears to stop, is to say that at some point in time the concentrations of H 2, N 2 and NH 3 reach a steady state (i.e. they don't appear to change) 3
Curiously enough, the same equilibrium concentrations of H 2, N 2 and NH 3 were observed even when the reaction was started with the vat containing only pure NH 3 (i.e. pure product!) The same equilibrium condition for the concentrations of reactants and products was reached from either direction (i.e. starting from either pure reactants, or pure product) For a simple AB unimolecular reaction Forward Rate = Reverse Rate k f [A] = k r [B] This equation can be rearranged to relate the concentration of A to B at equilibrium: A similar equation relates the concentrations of H 2, N 2 and NH 3 at equilibrium in the Haber reaction The Law of Mass Action In 1864 Guldberg and Waage postulated the Law of Mass Action which expresses the relationship between the concentrations of reactants and products at equilibrium in any reaction 4
Given the following general equilibrium equation According to the Law of Mass Action, the equilibrium condition is expressed by the equation: Where the brackets indicate molar concentration of the reactants and products at dynamic equilibrium. o This expression is called the equilibrium expression for the reaction o K c is called the equilibrium constant. Its value is what we get when we put in the observed concentrations of the reactants and products, at equilibrium, into the equilibrium expression. o The numerator (the stuff on top) of the equilibrium expression is the product of all concentrations of products raised to their coefficients in the balanced equation. The denominator (the stuff on the bottom) of the equilibrium expression is the product of all concentrations of reactants raised to their coefficients in the balanced equation. Back to Fritz Haber and his famous process to make ammonia from N 2 and H 2 : From the Law of Mass Action, the equilibrium expression would be based upon the balanced equation: N 2 (g) + 3H 2 (g) -> 2NH 3 (g) Note that once we have a balanced equation, although we may not know the reaction mechanism (i.e. underlying elementary steps or reaction intermediates), we can determine the equilibrium expression. Thus, the equilibrium expression depends only upon the balanced equation and is independent of the reaction mechanism Yet another note: The convention is to write equilibrium constants as dimensionless values Expressing Equilibrium Constants in Terms of Pressure, K p Kc indicates that the equilibrium constant is in terms of concentration (in molar units) When reactants and products in a reaction are all gases we can use partial pressures in the equilibrium equation The Magnitude of Equilibrium Constants Given the following general equation for a reaction, and the associated equilibrium expression 5
What can we conclude about an equilibrium constant that is LARGE? The value of the numerator (the stuff on top) must be larger than the value of the denominator (stuff on bottom) This will happen if the equilibrium concentrations of the products are larger than the reactants Thus, for a reaction with a large equilibrium constant, the equilibrium "lies to the right", meaning that the equilibrium mixture comprises mostly product What can we conclude about an equilibrium constant that is SMALL? The value of the numerator (the stuff on top) must be smaller than the value of the denominator (stuff on the bottom) This will happen if the equilibrium concentrations of the reactants are larger than the products Thus, for a reaction with a small equilibrium constant, the equilibrium "lies to the left", meaning that the equilibrium mixture comprises mostly reactants To summarize the interpretation for the magnitude of the equilibrium constant: K >> 1 Products favored K << 1 Reactants favored The Direction of the Chemical Equation and K By definition, equilibrium implies that we have both a "forward" and a "backward" reaction in a balanced chemical equation. "Forward" and "backward" are thus relative terms Consider the following reaction: Experimentally, the value of K c for this expression equals 0.212 (@100 C) We could, however, just as equally valid, consider the reaction to be the following: The equilibrium expression in this case would be: The value of Kc in this case would be (1/0.212) = 4.72 The equilibrium expression for a reaction written in one direction is the reciprocal of the one for the reaction written in the reverse direction 6
Therefore, an equilibrium constant, in the absence of information about the direction of the reaction, is ambiguous (i.e. meaningless) Heterogeneous Equilibria Homogeneous reactions have all reactants and products existing in the same phase. A homogeneous reaction at equilibrium is termed a homogeneous equilibrium Heterogeneous reactions will have at least one reactant or product in a different phase from the other components A heterogeneous reaction at equilibrium is termed a heterogeneous equilibrium Calcium carbonate, CaCO 3 (a solid) can decompose to produce calcium oxide, CaO (another solid) and carbon dioxide, CO 2 (a gas) The system involves a gas and two solids in equilibrium: How do we calculate the concentration (i.e. moles/volume) of a solid? The concentration of any pure liquid or solid is equal to its density (mass per unit volume) divided by its molar mass (mass per mole) o o o The density of a liquid or solid changes little with temperature, and is essentially a constant. The molar mass is a characteristic of the compound in question (i.e. is a constant). Therefore, the value of (density/molar mass), i.e. the concentration of a solid or liquid, is a constant, regardless of the amount of the solid or liquid present. With this information, the equilibrium constant for the decomposition of calcium carbonate reduces to: 7
Thus, we ignore the concentrations of solids and liquids in a heterogeneous equilibrium expression (even though these compounds must be there for the equilibrium to be established) The concentrations of gases (and solutes in solution) are included in the equilibrium expression because their concentrations can change What does the heterogeneous equilibrium expression for the decomposition of calcium carbonate (given above) tell us? At a given temperature, an equilibrium mixture of calcium oxide, carbon dioxide and calcium carbonate will always result in the same concentration of CO 2 (g). Since CO 2 is a gas, this means that the pressure of CO 2 at equilibrium will be the same value (at a given temperature) Since CaO and CaCO 3 are solids, the pressure of CO 2 at equilibrium is independent of the amounts of these compounds (but if one or both are missing, we won't have equilibrium, and the pressure of CO 2 won't be the expected equilibrium value) Calculating Equilibrium Constants The Haber process revisited: Haber and his coworkers were concerned with figuring out what the value of the equilibrium constant, K c, was at different temperatures. If a temperature for the reaction was chosen such that the value of K c was small, then it would mean that the reverse reaction is favored (equilibrium lies to the left) and very little ammonia would be produced when the reaction reaches equilibrium. This is not good, because the goal is to make ammonia. Haber did experiments where he started with various mixtures of N 2, H 2 and NH 3 and allowed them to come to equilibrium at various temperatures He would measure the equilibrium concentrations of the various gases and determine K c at a variety of temperatures 8
In one experiment a mixture of H 2, N 2 and NH 3 was allowed to reach equilibrium conditions at 472 C. The concentration of gases at equilibrium was analyzed and found to contain 0.1207M H 2, 0.0402M N 2 and 0.00272M NH 3. What value did Haber come up with for the equilibrium constant, K c? K c = (0.00272) 2 /(0.0402)(0.1207) 3 K c = 0.105 (equilibrium lies to the left, not much ammonia will be present at equilibrium conditions) Often we do not know (or are not easily able to determine) the concentration of all products and reactants at equilibrium However, if we know the concentration of reactants and products at the start of the experiment, and the concentration of at least one product or reactant at equilibrium, and we know the stoichiometry of the balanced chemical equation, we can determine the values of the other reactants and products The steps in this method are as follows: 1. List the known initial and equilibrium concentrations of all reactants and products involved in the equilibrium 2. For those reactants or products, for which both the initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium 3. Use the stoichiometry of the reaction to calculate the predicted changes in concentration for all the other reactants and products in the equilibrium 4. From the initial concentrations, and the changes in concentration, calculate the equilibrium concentrations (and K c ) Relating K c and K p For a gas we can express the equilibrium constant in terms of concentration (molarity) or in units of pressure. How are these related? Molarity relates the number of moles per unit volume (n/v) The ideal gas law, PV = nrt, includes terms for both number of moles and volume PV = nrt P = (n/v)rt P = MRT For a gaseous substance, A, the partial pressure of A is equal to the molarity times the gas constant R and temperature T in Kelvin P A = [A](RT) 9
Note: P = MRT can be manipulated to solve for R, the gas constant: R =P/MT. This provides an interpretation for the meaning of the gas constant - it is a way to relate the molarity of a gas sample to its pressure and temperature Applications of Equilibrium Constants This section of our study of equilibrium constants deals with various types of calculations, including: 1. How to figure out in which direction a reaction will go (i.e. towards making product, or more reactant) 2. Calculating equilibrium concentrations. This may involve knowing equilibrium values for some of the reactants and products and determining the concentration of an unknown. 3. Alternatively, we may be provided with the starting concentrations of reactants and products and may be asked to find the equilibrium concentrations The Magnitude of K The magnitude of the equilibrium constant, K, indicates the extent to which a reaction will proceed: If K is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products) If K is a small number, it means that the equilibrium concentration of the reactants is large. In this case, the reaction as written will proceed to the left (resulting in an increase in the concentration of reactants) Knowing the value of the equilibrium constant, K, will allow us to determine: The direction a reaction will proceed to achieve equilibrium The ratios of the concentrations of reactants and products when equilibrium is reached Predicting the Direction of a Reaction The value of K c for the Haber reaction at 472 C is 0.105. If we place the following amounts of H 2 (g), N 2 (g) and NH 3 (g) in a 3.0L container at 472 C will the N 2 (g) and H 2 (g) react to form more NH 3 (g)? N 2 (g) + 2H 2 (g) 2NH 3 (g) H 2 (g) = 0.5 mol N 2 (g) = 8.3 mol NH 3 (g) = 1.8 mol First of all, we need to convert the amounts of the components into concentrations (mol/l or M). Thus, we will divide the mol value by the volume of the container (gases expand to fill their container) Concentration of H 2 (g) = 0.5mol/3.0L = 0.167M Concentration of N 2 (g) = 8.3mol/3.0L = 2.77M Concentration of NH 3 (g) = 1.8mol/3.0L = 0.600M We need to state the equilibrium expression for this reaction, based upon the balanced equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) 10
Inserting the initial values of reactants and products into the equilibrium expression we get the following: = (0.600) 2 /((2.77)*(0.167) 3 ) = 27.9 How does this number compare to the value of the equilibrium constant at this temperature? The "initial value of the equilibrium constant" is 27.9, at equilibrium we know that this must decrease to 0.105 For the value of K c to decrease to the experimentally expected value at equilibrium, the concentration of NH 3 (g) must decrease, and/or the concentrations of N 2 (g) and H 2 (g) must increase o Therefore, the reaction would have to proceed to the left (increasing reactants and decreasing product concentrations) When we substituted the initial values for the concentrations of the reactants and products into the equilibrium expression the number we came up with is called the Reaction Quotient (Q) (i.e. "the initial value of the equilibrium constant") If Q = K c, then the system is already at equilibrium If Q > K c, then essentially we have too much product and the reaction will proceed to the left (to reduce the concentration of product and increase the concentration of product) If Q < K c, then essentially we have too little product and the reaction will proceed to the right (to produce more product and decrease the concentration of reactant) Calculation of the Equilibrium Concentration of a Reactant or Product Many types of equilibrium problems deal with determining how much of a product (or reactant) we will have once a reaction reaches equilibrium. The following example involves K p and partial pressures for our friend, the Haber reaction: At 500 K p = 1.45 x 10-5 for the Haber reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) In a sample of N 2 (g), H 2 (g) and NH 3 (g) at equilibrium in a Haber reaction the partial pressure of the H 2 (g) is 1.32 atm, and the partial pressure of the N 2 (g) is 0.648 atm. What is the partial pressure of the NH 3 (g) in the equilibrium mixture? In this problem we have a homogeneous equilibrium of gases and the equilibrium constant, K p, is given in terms of partial pressures of the component gases. For this reaction, the definition of K p is given as The equilibrium partial pressures of H 2 (g) and N 2 (g) are given, as is the value for K p 1.45 x 10-5 = (P NH3 ) 2 / ((0.648)*(1.32) 3 ) We can solve for the partial pressure of NH 3 (g) P NH3 = (2.16 x 10-5 ) 1/2 11
P NH3 = 4.64 x 10-3 atm If there is any doubt that we have done the calculations correctly, we can substitute this value (along with the equilibrium partial pressures for H 2 (g) and N 2 (g)) into the equilibrium expression and make sure that we get the correct value for K p Le Châtelier's Principle In order to try to figure out how to optimize the production of ammonia from hydrogen and nitrogen, Haber studied the equilibrium concentrations of ammonia in his famous process: N 2 (g) + 3H 2 (g) 2NH 3 (g) He noted the equilibrium concentration of ammonia at different temperatures (while keeping pressure constant) He also noted the equilibrium concentration of ammonia at different pressures (while keeping the temperature constant) Haber observed that the equilibrium concentration of ammonia: o decreased with increasing temperature, o and increased with increasing pressure The underlying basis behind both of these phenomena was described by Henri-Louis Le Châtelier; Le Châtelier's principle: If a system is in equilibrium, and this equilibrium is perturbed by a change in temperature, pressure or the concentration of a reactant or product, then the system will shift its equilibrium so as to counteract the effect of this perturbation A Change in Reactant or Product Concentration At equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. Le Châtelier's principle states that if the concentration of one of the components of the reaction (either product or reactant) is changed, the system will respond in such a way as to counteract the effect If a substance (either reactant or product) is removed from a system, the equilibrium will shift so as to produce more of that component (and once again achieve equilibrium) If a substance (either reactant or product) is added to a system, the equilibrium will shift so as to consume more of that component (and once again achieve equilibrium) 12
As an example, consider the Haber reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) What would happen if we started with a Haber reaction at equilibrium, and then suddenly added some H 2 (g) to the reaction mix? The following is a graphical representation of how the concentrations of the individual components, and the overall system, would react in response to adding H 2 (g): What has happened to the equilibrium of the system in response to the added H 2 (g)? Prior to the addition of H 2 (g) the system is in equilibrium. This can be seen because the concentrations of the different components do not change with time (the forward and reverse rates must be equal) When the H 2 (g) is added, we see that the system responds by consuming N 2 (g) and producing NH 3 (g). Forward rate = k 1 [N 2 ] [H 2 ] 3 The production of NH 3 (g) requires both N 2 (g) and H 2 (g) as reactants. Therefore, the production of NH 3 (g) not only consumes N 2 (g) but also H 2 (g) The addition of H 2 (g) causes the balance of the system to shift in favor of the forward reaction (i.e. the production of NH 3 (g) from N 2 (g) and H 2 (g). Thus, some of the added H 2 (g) is consumed, and thus, the system responds to counteract the perturbation caused by the added H 2 (g) N 2 + 3H 2 2NH 3 (reaction is driven "to the right" by the effects of added H 2 ) After some time, the system reaches a new state of equilibrium. It will not be identical to the original state, however. Although the system has responded to resist the effects of the added H 2 (g), the new equilibrium state contains a slightly higher concentration of NH 3 (g), and slightly lower concentration of N 2 (g) (as well as a slightly higher concentration of H 2 (g). The overall ratio of [NH 3 ] 2 /([N 2 ]*[H 2 ] 3 ) is the same as before at equilibrium (i.e. the value of K c does not change) What would happen if we repeated the experiment, but added NH 3 (g) instead of H 2 (g)? 13
The system would respond by decomposing some of the added NH 3 (g) and a new equilibrium condition would be established (with slightly higher concentrations of H 2 (g) and N 2 (g) - as well as slightly higher equilibrium concentrations of NH 3 (g). Reverse rate = k -1 [NH 3 ] 2 At the new equilibrium the concentrations of reactants and product is slightly different from before, but K c has the same value Effects of Volume and Pressure changes A chemical system in equilibrium can respond to the effects of pressure also. According to Le Châtelier's Rule, if the pressure is increased on a system, it will respond by trying to reduce the pressure. How does it do this? We are primarily concerned with homogeneous gaseous reactions The stoichiometry of the reaction may lead to a greater number of molecules on one side of the equation. For example, in the Haber reaction, N 2 (g) + 3H 2 (g) 2NH 3 (g) there are twice as many moles of reactants as products If the Haber reaction were in equilibrium, and the pressure was increased, the reaction would respond to oppose the increase in pressure. It could accomplish this by shifting the equilibrium to the right (producing NH 3 (g)) o This would reduce the overall number of moles in the reaction, and therefore, lower the pressure An interesting point about pressure effects is that they do not cause a change in the value of the equilibrium constant, K (as long as T is held constant). Their affects are upon concentration of reactants and products For the Haber process at 472 C, the value of K is 0.105. An example of one such system in equilibrium at this temp might include [H 2 ]=0.121M, [N 2 ]=0.0402M and [NH 3 ]=0.00272M K for the Haber process is defined as [NH 3 ] 2 /([N 2 ]*[H 2 ] 3 ) If the volume of the sample is suddenly decreased by half, from the ideal gas equation the concentrations of the components in the sample have doubled (and the pressure has doubled): PV = nrt P = (n/v)rt The concentration of the components in this reaction at this new pressure would therefore be [H 2 ]=0.242M, [N 2 ]=0.0804M and [NH 3 ]=0.00544M The reaction quotient, Q, at this new condition would therefore be: (0.00544) 2 /(0.0804*0.242 3 ) = 0.0262. This is less than K=0.105, thus, in response, the reaction equilibrium shifts to the right, consuming reactants and producing NH 3 (g) product. This shift to the right reduces the number of molecules, therefore, the pressure is reduced. Effect of Temperature Changes The intrinsic value of K does not change when we increase concentrations or pressures of components in a reaction. However, almost every equilibrium constant (K) changes in response to changes in temperature. Once again, we can apply Le Châtelier's rule in order to predict the effects of temperature changes upon chemical reactions Most chemical reactions have some heat change associated with the reaction. (Note, the energy of a reaction can be used to either do work - i.e. accelerate an object against some force - or to change temperature. We will consider reaction conditions under which no work is done, and 14
therefore all energy changes associated with reactions will be manifested by temperature changes) Exothermic reactions are associated with heat release when the reaction proceeds in the forward direction Endothermic reactions are associated with heat release when the reaction proceeds in the reverse direction (i.e. heat is absorbed in the forward direction) These two types of reactions and their associated heat changes can be written as: Exothermic: Reactants Products + Heat Endothermic: Reactants + Heat Products If temperature is increased, the equilibrium will shift so as to minimize the effect of the added heat The reaction will shift in the appropriate direction such that the added heat is absorbed When heat is added to exothermic reactions at equilibrium, products will be consumed to produce reactants (shift to the LEFT) When heat is added to endothermic reactions at equilibrium, reactants will be consumed to produce products (shift to the RIGHT) Based on this behavior, what is the effect of T upon K? Assume K = 1.0 for an exothermic reaction at equilibrium. Added heat causes the reaction to shift to the left. Reactants Products + Heat Thus, 1.0 must represent a reaction quotient, Q, that is too large in comparison to the new value of K. Thus, the effect of increasing temperature on an exothermic reaction is to lower the value of K. Conversely, the effect of increasing temperature on an endothermic reaction is to increase the value of K The Effect of Catalysts A catalyst lowers the activation energy barrier, E a Although the activation energy barrier is a different magnitude for the forward and reverse reactions, the change a) is the same for both the forward and reverse reactions Therefore, a catalyst changes the rate at which equilibrium is achieved, but does not change the composition of the equilibrium mixture (i.e. does not alter the equilibrium constant, K) 2000 Dr. Michael Blaber 15