Major Concept Area: Equilibrium. In this chapter the dynamic equilibrium established in chemical reactions is explored, and its similarity to phase equilibrium is emphasized. The chapter culminates in a relationship between two signposts of spontaneity: the standard free energy of reaction and the equilibrium constant.
Specific Concepts in this Chapter:In Chapter 10, we discussed phase equilibrium as a balance of opposing rates. This state of balance gives the appearance of stasis at the macroscopic level, despite the continuous and dynamic activity at the microscopic level. Phase equilibrium is dynamic equilibrium. We summarize here the characteristics of dynamic equilibrium:
In this chapter, we extend the dynamic equilibrium concept to chemical reactions. We will see that, like a pure substance in a closed container, a chemical reaction in a closed system reaches a state of dynamic equilibrium, a state in which conversion of reactants to products is balanced by conversion of products to reactants. At the macroscopic scale, there is no change in the amount of a particular reactant or product with time. In contrast, at the molecular level there is a continuous reshuffling of atoms.
12-1 Chemical Equilibrium--Some Thought Experiments. We introduce chemical equilibrium with a simple reaction, the dimerization of nitrogen dioxide to dinitrogen tetroxide, in equation 12-1-1:
As we have done several times before, we will study the reaction by carrying out several thought experiments designed to reveal the behavior of the reaction under various conditions. Throughout the following discussion, bear in mind that the experiments could be (and in fact have been) carried out in the laboratory (although in a somewhat different manner than described). Figure 12-1 shows an initially evacuated container of volume 1.0 L, with an attached manometer. The manometer allows the total gas pressure in the container to be measured. The temperature of the container is 298 K. The container is fitted with a piston that can be freely moved or locked in place. The following experiments are sequentially performed.
Expt 1. With the piston locked in place, 0.0460 g (0.0100 moles) of NO2 gas (red-brown in color, and toxic!) is injected into the container through a septum cap. The manometer quickly registers a pressure of 0.248 atm (188 mm of Hg). Then, over several minutes, the red-brown color fades and the pressure of the system falls, at first rapidly, then more and more slowly, until it levels off at a value of 0.175 atm. This is the sum of the partial pressures of NO2, 0.102 atm (measured from the intensity of the red-brown color) and N2O4, 0.073 atm (calculated by difference). Figure 12-2a shows the manner in which the total pressure and the partial pressures of NO2 and N2O4 vary with time after injection of NO2.
Expt 2. The piston is kept locked in place and the container is pumped out. Then 0.0460 g (0.0050 moles) of N2O4 (colorless) is injected. The manometer quickly registers a pressure of 0.124 atm. Over several minutes, a red-brown color develops and the pressure of the system increases, at first rapidly, then more slowly, until it levels off at a value of 0.175 atm. The partial pressures of NO2 and N2O4 are again 0.102 and 0.073 atm, respectively, once the situation of constant pressure is reached. Figure 12-2b shows how the pressures vary with time after injection of N2O4. Note that the final pressures are the same as in Experiment 1.
Expt 3. With the piston still in place, the container is pumped out. Simultaneously, 0.0230 g (0.0050 moles) NO2 and 0.0230 g (0.0025 moles) N2O4 are injected to the container. The manometer quickly registers a pressure of 0.186 atm, but over several minutes, the red-brown color slightly fades and the total pressure drops to and levels off at 0.175 atm. The partial pressures of the two gases are the same at the end as in the first 2 experiments.
Expt 4. After evacuating the container, 0.0460 g of NO2 and 0.0460 g of N2O4 are injected simultaneously. The manometer quickly registers 0.372 atm. Over several minutes, the red-brown color fades and the pressure of the system decreases, at first rapidly, then more and more slowly, until it levels off at 0.325 atm. The final partial pressures of NO2 and N2O4 are 0.155 and 0.170 atm. Figure 12-2c shows how the pressures vary with time after injection of the gases.
Expt 5. The piston is now quickly pushed in until the volume of the container is cut in half. Immediately after this is done, the manometer registers 0.650 atm (twice the final pressure in the previous experiment). However, over an interval of time the pressure and the red-brown color decrease until they eventually level off again. The final total pressure is 0.611 atm. The pressures of NO2 and N2O4 are 0.232 and 0.379 atm, respectively.
We now draw what conclusions we can from the experimental results. First, whether we start with NO2, N2O4, or some arbitrary mixture, a dynamic process occurs, the result of which is a situation in which both gases are present. Second, the dynamic process leads to a situation in which there is no further change in macroscopic variables (i.e., the amounts of the 2 gases). Third, experiments 1-3 show us that the final amounts of NO2 and N2O4 are the same as long as we start with the same total moles of nitrogen and oxygen atoms (in all 3 experiments, we begin with 0.0100 moles of nitrogen atoms and 0.0200 moles of oxygen atoms). It appears that there is some regularity in the result of the dynamic process. Fourth, experiment 5 shows that we affect the reaction system by putting a stress on it, but that it adjusts to the stress and returns to equilibrium. We conclude that the system exhibits the characteristics of dynamic equilibrium.
For liquid-vapor phase equilibrium, we saw that the equilibrium state at a fixed T is characterized by a constant pressure (concentration) of vapor. No matter what we did to the system, other than change T, the pressure returned to this value. Is there a similar quantity for the reaction 12-1-1, a quantity that characterizes the equilibrium state as long as T is constant, and which in some way depends on the pressures (amounts) of the gases? We cannot use the sum or the product of the gas pressures, because these are obviously different at equilibrium in experiments 1 and 4. The values in experiment 4 are both larger than those in experiment 1, which suggests that some ratio of pressures might be constant. Table 12-1 shows the initial and final pressures, and the ratios PN2O4/PNO2 and PN2O4/(PNO2)2 for Experiments 1-5:
| NO2: | N2O4: | |||||||
| Expt | Pinitial | Pfinal | DeltaP | Pinitial | Pfinal | DeltaP | PN2O4/PNO2 | PN2O4/(PNO2)2 |
|---|---|---|---|---|---|---|---|---|
| 1 | .248 | .102 | -.146 | 0 | .073 | .073 | 0.716 | 7.0 |
| 2 | 0 | .102 | .102 | .124 | .073 | -.051 | 0.716 | 7.0 |
| 3 | .124 | .102 | -.022 | .062 | .073 | .009 | 0.716 | 7.0 |
| 4 | .248 | .155 | -.093 | .124 | .170 | .046 | 1.10 | 7.1 |
| 5 | .310 | .232 | -.078 | .340 | .379 | .039 | 1.63 | 7.0 |
Experiments 4 and 5 demonstrate that a simple ratio does not work. However, the modified ratio, in which the pressure of NO2 is squared, does remain constant for all experiments. This ratio is interesting, because the powers to which the pressures are raised are the stoichiometric coefficients of the gases in equation 12-1-1. The number, 7.0 atm-1, is called the equilibrium constant for reaction 12-1-1 at 298 K, and is given the symbol Keq. The ratio, PN2O4/(PNO2)2, is called the equilibrium constant expression for the reaction:
12-2 Chemical Equilibrium--A General Concept. The results of experiments on many different reactions over many years allow us to generalize as follows. For the generic reaction 12-2-1, in which the participating species are gases, the partial pressures at equilibrium obey the simple expression in 12-2-2, independent of the initial pressures of the gases.
In words, the equation says that if we divide the product of the pressures of the products, each raised to the power of its stoichiometric coefficient in the equation for the reaction, by a similar product of reactant pressures, the result is a constant, Keq, that depends only on temperature. Every chemical reaction, simple or complex, obeys an equilibrium constant expression of the form of 12-2-2. This is a very simple result that makes possible calculations involving the direction and extent of reaction. If the value of Keq for a reaction is large, the pressures of products will be large relative to those of reactants at equilibrium -- the equilibrium lies to the right. If Keq is small, pressures of products will be small relative to those of reactants at equilibrium -- the equilibrium lies to the left.
Our conclusions are not restricted to gas phase reactions. An expression analogous to 12-2-2 applies to reactions in solution, except that molarities must be used in place of pressures. In fact, when we recognize that gas pressure is proportional to n/V (molarity) through the ideal gas law, we realize the equivalence of expressions in terms of pressure and molar concentration. Equation 12-2-3 therefore applies to all reactions, in the gas phase, in solution, or in some combination:
The brackets signify molarity. One caution is in order for gas phase reactions. The numerical value and the units of the equilibrium constant depend on whether pressures or molarities are used. If Keq is specified in units of bar, pressures are used in calculations; if in moles/L, concentrations are used. If Keq has no units, either pressure or molarity may be used.
Rules for Writing Equilibrium Constant Expressions. It is important to become facile in writing and using equilibrium constant expressions for reactions. Several guidelines will help in this regard.
Example 12-1. Write the equilibrium constant expression for the reaction
Solution. All species are gases, so we use their partial pressures:
Example 12-2. Write the equilibrium constant expression for the reaction
Solution. Only a pure solid and dissolved species are involved. We leave the pure solid out and use molarity for the ions:
Example 12-3. Write the equilibrium constant expression for the "reaction"
Solution. Although this is not a chemical reaction, since no new substances are produced, we can treat it by the same rules. In writing the equilibrium constant expression, we leave out the pure liquid and use the partial pressure of the water vapor:
Thus the equilibrium vapor pressure of a pure liquid (or solid) is a special case of an equilibrium constant.
Example 12-4. The reaction of hydrogen and oxygen to form water may be written in many equivalent ways, two of which are
How are K1 and K2 related?
Solution. Reaction (b) is clearly twice reaction (a). We see that K2 is K12. Generally,
Example 12-5. The reaction of nitrogen and hydrogen to form ammonia may be written in either direction:
How are the two Keq's related?
Solution. Reaction (b) is the negative of reaction (a): (b) = -(a). At the same time, K2 = 1/K1. This is a special case of equation 12-2-4.
Example 12-6. The third reaction below can be obtained by adding the first two reactions.
How is K3 related to K1 and K2?
Solution. Inspection of the equilibrium constant expressions shows that K3 is the product of K1 and K2. Generalizing, when two reactions are added to give a third, their Keq's must be multiplied to obtain Keq for the third:
Example 12-7. What are the units of the Keq expression for the reaction
Solution. For pressures in atm, the units of the numerator of Keq are atm2, while those of the denominator are atm3. The overall units are thus atm-1.
Example 12-8. What are the units of Keq for the reaction
Solution. Units of both the numerator and denominator are bar2, so Keq is dimensionless. In general, Keq is dimensionless when the total moles of gas (or dissolved species) is the same in reactants and products.
Unless Keq is dimensionless, we must use concentration units consistent with the specified units of Keq.
12-3 The Reaction Quotient, Q. We again consider reaction 12-2-1:
Assuming that all species are gases, this has the equilibrium constant expression 12-2-2:
The "e" subscripts denote equilibrium partial pressures. Assume that we initially mix A, B, D, and F with arbitrary pressures. Then we define the reaction quotient, Q, by equation 12-3-1:
where the pressures are the arbitrary initial values. We now raise two questions: Is the reaction at equilibrium with the pressures as mixed? If not, which way will reaction proceed to reach equilibrium? If Q is numerically equal to Keq, the reaction is already at equilibrium. But if Q and K have different values, the reaction is not at equilibrium with the arbitrary pressures and will proceed either to the right or left to get to equilibrium. Suppose that with the initial pressures Q > Keq. This means that the product pressures in the numerator of the Q expression are too large relative to the reactant pressures in the denominator, and reaction must proceed to remedy this. It will proceed right-to-left to decrease product pressures (numerator) and increase reactant pressures (denominator):
Reaction continues from right to left until reactant and product pressures satisfy equation 12-2-2, at which point equilibrium will have been reached. If initially Q < K, pressures in the numerator are too small relative to those in the denominator, and reaction must proceed left-to-right to increase Q:
When Q becomes equal to Keq, equilibrium will have been reached and no further change in concentration occurs. In summary,
The manner in which Q changes with time for the case Q > Keq is illustrated in Figure 12-3. Note that the rate of change of Q (the slope of the curve) is largest when the difference between Q and Keq is greatest.
Example 12-9. For the reaction below, Keq = 55.3 at 700 K. Suppose that initially 3.0 atm of H2(g), 0.10 atm of I2(g), and 10.0 atm of HI(g) are mixed at 700 K. Will reaction occur? If so, in which direction?
Solution. We must calculate Q from the specified initial pressures and compare it with Keq:
Since Q > Keq, reaction must proceed right-to-left (i.e., some HI must be consumed and some H2 and I2 formed) to reach equilibrium.
Example 12-10. Consider the NO2/N2O4 reaction:
Keq = 7.00 atm-1 at 298 K. Initially 1.5 atm NO2 and 3.8 atm N2O4 are mixed in a closed container. Will reaction occur, and if so, which way?
Solution. Calculate Q and compare with Keq:
Since Q < Keq, reaction proceeds left-to-right to equilibrium. Q varies with time as shown in Figure 12-4.
12-4 Equilibrium Calculations -- Extent of Reaction. In the previous section we used Q and Keq to predict the direction in which a reaction will proceed from specified starting conditions. We now wish to calculate the extent to which it will proceed. We do this by example.
Example 12-11. Keq for the dimerization of nitrogen dioxide is 7.0 atm-1 at 298 K. Suppose that sufficient NO2 is placed in a flask to give an initial pressure of 1.5 atm. What will be the NO2 and N2O4 pressures in the flask when the reaction has reached equilibrium?
Solution. In solving this problem we develop a general approach that is applicable to any chemical equilibrium situation. Since we will study chemical equilibrium for a variety of reaction types, it is important for you to learn now how to apply the approach.
| 2NO2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 1.5 | 0 |
The word "initial" is written to the left of the initial pressures as a reminder of the significance of the numbers that follow. Since there is initially no N2O4 present, we write 0 under its formula. It is important to note that there is no stoichiometric restriction on the initial pressures. The initial pressures need not and usually do not bear any relationship to the stoichiometric coefficients for the reaction.
| 2NO2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 1.5 | 0 | |
| D P: | -x | x/2 |
| 2NO2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 1.5 | 0 | |
| D P: | -x | x/2 | |
| equilibrium: | 1.5-x | x/2 |
At equilibrium we have less NO2 and more N2O4 than we started with, consistent with expectations based on the reaction quotient, Q.
Expansion and rearrangement gives a quadratic equation in x:
This is solved with the quadratic formula (Appendix D) to give
There can be only one correct answer to our problem because physically there can be only one position of equilibrium resulting from the given initial state. Examination of the two roots shows that 1.87 cannot be correct; if it is subsituted for x in the expression for the equilibrium pressure of NO2, 1.5 - x, a negative value, -0.37 atm, is obtained. This is physically meaningless. It is impossible to use up more NO2 than is initially present. Thus x = 1.21 atm, and the equilibrium pressures of the two participants are
These results are readily checked by substituting them into the Keq expression:
Within the limits of round-off error, this is the same as the given value of Keq. We conclude that the answer is correct.
A graphical interpretation of this problem, showing the manner in which the initial pressures evolve to the equilibrium values, is given in Figure 12-5.
Example 12-12. Suppose that the volume of the reaction container for Example 12-11 is instantly decreased by a factor of 2. In which direction will reaction shift? What will the equilibrium pressures of NO2 and N2O4 be?
Solution. To answer the first question we must compare Q with Keq. To answer the second requires a calculation similar to that in the previous example. The strategy is
These are the initial pressures for the problem at hand. We calculate Q:
The reaction must therefore shift right to reattain equilibrium.
Now we set up a table similar to that in the previous example:
| 2NO2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 0.58 | 1.2 | |
| D P: | -2x | x | |
| equilibrium: | 0.58-2x | 1.2+x |
In this case we have defined x differently than in Example 12-11. Previously we defined x as the pressure of NO2 that disappears during reaction. Here it is the pressure of N2O4 formed during reaction. In both examples, the pressure of NO2 that disappears is twice the pressure of N2O4 formed, as stoichiometry demands. It doesn't matter how we define x, as long as the change line obeys the reaction stoichiometry.
You should proceed as in the previous example to set up the quadratic equation and solve it to obtain x = 0.55, 0.076 atm. The root, 0.55, is physically unreasonable (why?) so x = 0.086 atm. Substituting into the expressions for the equilibrium pressures gives
You should check that these values satisfy the Keq expression. The effect of the volume decrease and the subsequent return of the system to equilibrium are shown in Figure 12-6.
Example 12-13. Consider the reaction N2 + 3H2 ---> 2NH3. Suppose that the initial pressures of N2, H2, and NH3 are 1.0, 1.5, and 3.2 atm, respectively, and that PNH3 at equil is 0.12 atm. What are PN2 and PH2 at equil? What is Keq for the reaction?
Solution. This problem can be approached using the system introduced earlier: write the initial pressures first; then enter the one final (equilibrium) pressure known; calculate the change in pressure of NH3; use the reaction stoichiometry to calculate pressure changes for N2 and H2; and finally, calculate the equilibrium pressures of N2 and H2. Once all equilibrium pressures are known, the equilibrium constant may be calculated.
| N2 | +3 H2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 1.0 | 1.5 | 3.2 | |
| Change: | ||||
| equilibrium: | 0.12 |
The change in pressure of ammonia during reaction is easily calculated to be 3.2 - 0.12 = 3.08 atm. Because the pressure of NH3 decreases, this change is negative. The changes in PH2 and PN2 must be 1.5 and 0.5 times this value, respectively, and are positive. Thus PN2 = 0.5 * 3.08 = 1.54 atm; and PH2 = 1.5 * 3.08 = 4.62 atm. The changes are now inserted into the change line and the equilibrium pressures calculated:
| N2 | +3 H2 ---> | N2O4 | Keq = PN2O4/PNO22 = 7.0 atm-1 | |
| initial: | 1.0 | 1.5 | 3.2 | |
| Change: | 1.54 | 4.62 | -3.08 | |
| equilibrium: | 2.54 | 6.12 | 0.12 |
Finally, Keq is calculated:
We make one final and very important point about the entries in the table under the reaction in the example just completed: only the entries in the DP line of the table are governed by the reaction stoichiometry. Neither the initial pressures (which are in general completely arbitrary) nor the equilibrium pressures are related to the coefficients of the chemical equation. This will in general be true. Although the stoichiometry governs the relative changes in concentration (pressure) that occur as the system passes from one state to another, it is not necessarily related to the relative amounts of participants present in a particular state (e.g., initial and equilibrium states in the above examples).
12-5 Le Chatelier's Principle and Chemical Equilibrium. This principle applies as well to chemical equilibrium as to phase equilibrium. In fact, the principle is a qualitative manifestation of the quantitative characteristics of the equilibrium constant. Three stresses are particularly relevant to chemical equilibrium.
Assume the reaction to be initially at equilibrium. Suppose that we add some NH3 to the system. This imposes a stress on the system that will be relieved, according to Le Chatelier, by a shift that relieves the stress -- i.e., uses up NH3. The reaction shifts left until enough NH3 is used and N2 and H2 produced to again make Q = Keq. When the new equilibrium is reached, the pressures of all three gases will be larger than they were before the stress. Note that addition of NH3 to the system does not change the value of Keq. The same value applies before and after the stress. Figure 12-7 shows the effect of NH3 addition on the equilibrium. Please note in the figure that the changes in pressure of the three gases following the perturbation are in accord with the reaction stoichiometry: DP for NH3 is negative and is twice as large in magnitude as DP for N2; and DP for H2 is three times the value for N2.
Now instantly cut the system volume in half. This instantly doubles the total pressure of the system. Le Chatelier predicts that, if possible, reaction will shift to offset the effect of the stress; to decrease the total system pressure. This may be accomplished by a shift toward the right, which will decrease the total number of gas molecules in the container (4 molecules of reactant are converted to only 2 of product). Moles of gas decreases, and so does total pressure.
In general, decreasing the volume (increasing pressure) of a reaction system involving one or more gases results in a shift toward the side with the least moles of gas. Apply this idea to a decrease in system pressure for the reaction
A change in system volume (pressure) has no effect on the magnitude of Keq for the system, as long as temperature is kept constant.
12-6 A Molecular View of Chemical Equilibrium. Consider now the general reaction in equation 12-6-1.
Why do the stoichiometric coefficients appear as exponents in the expression for Keq? Chemical equilibrium is like phase equilibrium; it results from a balance of opposing rates. In this case, the rates involved are those of the reaction in the forward and reverse directions:
We assume now, and explore more fully later (Chapter 15), that molecules must collide in order to react. This is certainly sensible, because it is not possible for atom rearrangements to occur unless the reacting atom aggregates are in close proximity. The rate of a chemical reaction is thus expected to be proportional to the number of collisions between molecules per unit time:
For the forward and reverse processes in 12-6-2, we write
The rate constants, kf and kr', are proportionality constants in the same way that kE and kC are. If there are NA molecules of A and NB of B in a container, a particular A molecule may collide with any of the NB molecules of B. The total number of ways in which A and B may collide is thus NA x NB. Then
NC molecules of C can collide in a total of NC(NC-1)/2 ways. This arises because each molecule can collide with NC-1 others. However, the simple product of NC and NC-1 double counts the collisions (i.e., it includes collisions of molecule x with y and of y with x; however, there can be only one collision between two particular molecules). We divide by 2 to eliminate double counting. Finally, because NC is very large, NC - 1 = NC. It follows that
At chemical equilibrium, then
Rearranging gives
Since pressure is proportional to number of molecules, this equation is readily translated to pressure terms, giving the equilibrium constant expression in equation 12-6-1. Notice that, as was true for phase equilibrium, the equilibrium constant is a ratio of rate constants for forward and reverse processes. Chemical equilibrium, like phase equilibrium, is dynamic rather than static. Although it appears at the macroscopic level that nothing is happening since concentrations remain constant, both forward and reverse reactions continue at equal rates at the molecular level.
12-7 Free Energy and Equilibrium. To this point we have encountered two quantities that indicate the extent to which a chemical reaction will proceed: its spontaneity. These are Keq and the standard free energy change of reaction, DGRo. We now explore the relationship between these quantities.
The sign of DGRo indicates the spontaneity or non-spontaneity of a process in which reactants in their standard states are converted to products in their standard states. The magnitude of DGRo tells us how far the standard state is from the equilibrium state. A process for which DGRo is negative is spontaneous under standard conditions, and will proceed left to right to reach equilibrium. At equilibrium, the reactants will be present at less than the standard state pressure (1 atm) or concentration (1 M), because they have been used up to some extent in achieving equilibrium. The products will be present at greater than the standard state pressure or concentration, because they have been formed in achieving equilibrium. In the equilibrium state, products will be present in relatively large amounts, reactants will be present in relatively small amounts, consistent with Keq > 1. Thus DGRo < 0 corresponds to Keq > 1. Further, for a process with DGRo = -50 kJ, the standard state is much farther away from equilibrium than is the standard state for a process with DGRo = -5 kJ. The first reaction will have to proceed further to reach the equilibrium state than will the second reaction, and will thus have a larger Keq. We can make very similar statements about processes for which DGRo > 0, except in reverse. The larger the magnitude of DGRo, the further the standard state is from the equilibrium position, and the further reaction will have to proceed to reach equilibrium from the standard state. For DGRo > 0, reaction must proceed from right to left toward equilibrium, decreasing the pressures or concentrations of products and increasing those of reactants. Thus Keq < 1. The larger the magnitude of DGRo, the smaller Keq. Finally, when DGRo = 0, the process is equally spontaneous in both forward and reverse directions; the standard state is the equilibrium state.
Example 12-14. Given are several reactions and their standard free energies of reaction. Which reaction is furthest from equilibrium in the standard state? In which direction will each reaction have to proceed to reach the equilibrium state?. Which reaction has the largest value of Keq?
| Process | DGRo |
|---|---|
| N2(g) + 3H2(g) 2NH3(g) | -33 kJ |
| 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) | 2880 kJ |
| CO2(g) + 2H2O(l) CH3OH(l) + 3/2 O2(g) | 702.4 kJ |
Solution. The magnitude of DGRo is largest for the second reaction; it is furthest from equilibrium under standard conditions. The first reaction will proceed left to right toward equilibrium, because it is spontaneous under standard conditions; the other two will proceed right to left. The most spontaneous reaction has the largest Keq. This is the first reaction.
It would be useful to have available a way to estimate DG for a reaction under arbitrary (not necessarily standard) conditions of pressure or concentration. This would allow us to predict the direction of equilibrium from any desired starting concentrations or pressures. In fact, an expression relating the free energy change for a reaction under arbitrary conditions to the standard free energy change is developed in Appendix F. Interested readers are referred there for the full details. Here, we present only the results. The desired relationship is 12-7-1.
Here Q is the reaction quotient introduced earlier in the chapter. Equation 12-7-1 says that the free energy change under arbitrary concentration conditions can be obtained by correcting the standard free energy change, DGRo, by a term (Q) whose value depends upon what the non-standard conditions are. This seems quite reasonable. When Q = 1 (i.e., all concentrations are standard at 1 M), the equation tells us that DGR = DGRo, as it should. When Q < 1 (reactants are present in greater relative amounts than products), the equation tells us that DGR < DGRo; in other words, that reaction is more spontaneous than under standard conditions. This too is reasonable; reaction should have a greater tendency to proceed left to right if reactant concentrations are increased. When Q > 1 (reactants are present in lesser relative amounts than products), DGR > DGRo; reaction is less spontaneous than under standard conditions. Reaction has a lesser tendency to proceed left to right if product concentrations are increased.
Suppose now that we choose some arbitrary initial concentrations of reactants (they could be, but need not be, 1 atm or 1 M) and products, and allow reaction to proceed from there to equilibrium. At equilibrium, DGR must be zero because the forward and reverse spontaneities are the same. Further, once equilibrium is attained, the pressures (or concentrations) of all reactants and products must be equilibrium values, so Q = Keq. Equation 12-7-1 becomes
This is one of the most important results in all of thermodynamics. It gives a direct relationship between our two criteria of spontaneity. Substitution of DHo - TDSo for DGo gives the temperature dependence of the equilibrium constant.
Equation 12-7-3 is called the van't Hoff relationship. Note that it is identical in form to the Clausius-Clapeyron equation for the temperature dependence of the vapor pressure. This reinforces our earlier recognition that Pvap is simply an equilibrium constant, and gives substance to the statement that all physical and chemical processes exhibit temperature dependence of the same form.
Example 12-15. Consider the familiar NO2/N2O4 reaction, with the given thermodynamic information:
| N2O4 | 2NO2 | (12-7-4) | |
|---|---|---|---|
| DHfo | 9.16 | 33.18 kJ/mole | |
| So | 304.2 | 240.0 J/K-mole |
Solution.
Finally, we examine the physical sense of the calculated enthalpy and entropy changes for reaction 12-1-1. The reaction is shown in Figure 12-8, with participant structures made explicit. The figure makes it clear that N2O4 is the result of the joining together of two 1-bond NO2 fragments, as discussed in Chapter 3. The forward reaction involves the breaking of the N-N bond, which involves an increase of PE. This is consistent with DHRo > 0. When the bond is broken, 1 molecule becomes 2. Disorder increases, consistent with DSRo > 0. A plot of DGRo versus T should be linear with positive intercept and negative slope. This is consistent with the conclusion above that reaction becomes more spontaneous as T increases.
The relationship between DG, DGo, and Keq can be seen most clearly by plotting DG versus ln Q, according to equation 12-7-1. Figure 12-9 shows such a plot for a process with negative DGo. Several features of the plot are noteworthy:
We close this section with an interesting feature of equation 12-7-2, DGo = -RT ln Keq. When this equation is plotted as DGo/T versus ln Keq, the result is a straight line with slope -R. The interesting feature is that the data for all chemical reactions and physical processes fall on the same line! Thus physical and chemical processes are in some sense all the same.
12-8 Spreadsheets in Equilibrium Calculations. In this chapter we have introduced and applied a systematic approach to the calculation of equilibrium pressures in gas phase reactions. However, we have been careful to choose examples that generate no higher than a second-order polynomial -- a quadratic equation -- which is readily solved. There are many chemical reactions that generate polynomials of order 3 and higher when our general method is applied. Such polynomials cannot be solved analytically. That is, there is no analog of the quadratic formula that gives the roots of a polynomial of order > 3. Consider the very important reaction
for which the equilibrium constant expression is 12-8-2:
Application of our approach to this reaction generates an equation in x4 (from the denominator of Keq) which cannot be solved analytically.
However, the availability of personal computers and modern Spreadsheet programs (such as Lotus 1-2-3, Excel, Works, QuattroPro, etc.) enables such problems to be easily solved. We consider a specific example:
| N2(g) + | 3H2(g) ---> | 2NH3(g) | Keq = 15.0 atm-1 | |
| initial | 10.5 | 0.38 | 0.24 |
For the initial conditions, Q = 0.10, so reaction will proceed right to equilibrium. The change and equilibrium lines then follow:
| Change | -x | -3x | 2x |
| equil | 10.5-x | 0.38-3x | 0.24+2x |
Now, instead of substituting into the Keq expression as we normally do (which would generate a polynomial of order 4), we set limits on the value of x; i.e., we figure out the minimum and maximum values that x can have. The minimum is clearly zero, which would correspond to no reaction. The maximum value is calculated by assuming that reaction proceeds to the right until the limiting reagent is used up. The limiting reagent is clearly H2, so the maximum possible x value is 0.38/3 = 0.127 atm. The numerical range in which x must lie is thus 0 < x < 0.127. We now explore this range with a trial-and-error procedure that goes as follows:
Table 12- 2 shows the variation of Q with x as x is incremented in steps of 0.01.
| x | Q |
|---|---|
| 0 | 0.1 |
| 0.01 | |
| . | |
| . | |
| 0.09 | 12.7 (Q < K) |
| 0.10 | 36.3 (Q > K) |
| . | |
| . | |
| 0.127 | infinite |
Incrementation of x in steps of 0.01 allows us to greatly narrow the range of possible x values to 0.09 < x < 0.10. We now "zoom" in on this range by incrementing x by 0.001:
| 0.09 | 12.7 |
| 0.091 | 14.0(Q < K) |
| 0.092 | 15.4(Q > K) |
| . | |
| . | |
| . | |
| 0.10 | 36.3 |
We have now limited x to the range between 0.091 and 0.092. If desired, we could continue this zooming process to obtain x to any desired degree of precision. However, since the initial hydrogen pressure is given to only 2 significant figures, the value x = 0.092 atm is adequately precise (0.092 is chosen rather than 0.091 because 15.4 is closer to Keq than is 14.0).
The trial and error calculation illustrated above can be carried out manually. However, it is much faster and more reliable to use a spreadsheet. Further, once the spreadsheet is set up for a particular reaction, it can be used for any specified initial conditions. Figure 12-10 shows a spreadsheet set-up for the ammonia reaction. The cell entries are as follows:
| First few lines | explanatory text |
| B12-B14 | initial pressures |
| B15 | initial Q: B14^2/(B12)(B13^3) |
| E15 | minimum x value (0) |
| G15 | maximum x value, calculated as PoH2/3 |
| A17-19 | reminders of relationship between x and pressures |
| B21 | increment of x (Del x) |
| Row 25 ff | x values, calculated concentrations, and Q/K comparison |
| A26 | new x calculated by adding del x ($B$21) to old value of x in A25. $ tells spreadsheet not to change the row and column. |
| B26 | new value of PN2 for x in A26 |
| C26, D26 | PH2, PNH3 |
| E26 | Value of Q/K |
To get subsequent entries in columns A-E, copy row 26. Once the value of x is narrowed down, define a new "del x" for zooming. Copy the table headings and proceed as in the first table.
The spreadsheet method can be used for any equilibrium problem, involving any power of x. It is therefore very general and powerful.
12-9 An Everyday Example of Dynamic Equilibrium. We have discussed dynamic equilibrium at the molecular level in connection with phase equilibrium in Chapter 10, and again in this chapter in the context of chemical reactions. Dynamic equilibrium is conceptually difficult because there are not many examples of it on a macroscopic scale. In this section, we consider a macroscopic example of dynamic equilibrium that, although artificial in some respects, nonetheless serves to clarify the concept. In the process, we introduce the finite difference method of calculation, which will surface again in Chapter 15.
We consider two cities, A and B. City A has initially a large population of 1 million people, but has substantial crime and unemployment rates. Consequently, people tend to want to leave A to find a better life. In contrast, city B has an initially small population of 20000 people; is quiet and relatively crime free; and features a low unemployment rate. Thus there is a much smaller tendency for people to leave B. Further, B is just the sort of place that dissatisfied A citizens seek. We make the following assumptions about the cities and their populations: 1)Cities A and B form a closed system; that is, people who leave A must go to B and vice versa. There is no leakage of population out of the system. 2) People leave A at a rate that is proportional to the population of A: Rate (A->B) = RA = kAPA, where PA is the population of city A and kA is the proportionality constant relating rate to population. People leaving A enter B. 3) People leave B in the same manner that they leave A: RB = kBPB. People leaving B enter A. 4) kA > kB. This means simply that there is a greater inherent tendency for people to leave A than to leave B because of the undesirable aspects of A and the desirable aspects of B.
Method of Finite Differences. We now want to calculate the manner in which the populations of A and B change during each of a succession of equal time intervals, starting with the initial populations at time t = 0. For each finite interval of time, we will calculate the changes in the populations of A and B during that time, using the rates of population change that hold at the beginning of the interval. By considering a sufficiently large number of such time intervals, we will show that the populations of A and B eventually stop changing, even though people are still moving back and forth between them. This is exactly what we mean by dynamic equilibrium. Our specific sequence of operations is as follows.
In words, the first of these equations says that the new population of city A is calculated by correcting the old population by the outflux of people from A to B, and the influx of people from B. The population lost via RA is simply the product of the time interval and the rate of movement of people from A to B at t0:
List PA, PB, RA, and RB at time t1.
We will apply this procedure to the first couple of cycles to illustrate the method. We choose the following initial populations and proportionality constants:
Figure 12-11 shows plots of the populations of cities A and B, PA and PB, as a function of time. The plots are reminiscent of plots of vapor pressure versus time in Chapter 10. It is clear from the figure that the populations of the cities level off and remain constant after about 30 months. The final ratio of populations, PB/PA, stabilizes at 2.5, which is exactly the ratio of the proportionality constants, kA and kB. Representing the flow of population between the two cities in equation form as
the "equilibrium constant" for the process, K = PB/PA, is just kA/kB. The most important aspect of the results in Table 12-3 is that at population equilibrium, the values of RA and RB are equal, and are NOT zero. Even though the populations of A and B remain constant, there is a constant exchange of people between them; the equilibrium is dynamic, involving a balance of opposing processes.
Supplement.
ApplicationsOver a time interval between t0 and t1, the reaction is at equilibrium at some temperature, with constant partial pressures of reactants and products as shown.
At time t1, the volume of the reaction system is instantly halved. Show on the graphs the immediate effect of the volume decrease on the partial pressures of species, and the manner in which these pressures change as equilibrium is reestablished. Your graphs should be as quantitatively correct as possible.
12-3. You are studying the equilibrium P4(g) ---> 2P2(g). In a previously evacuated vessel, you place a quantity of diatomic phosphorus gas. What is the value of the reaction quotient Q? Which direction will reaction proceed to achieve equilibrium?| A + | 3B ---> | 2D | Keq = PD2/PAPB3 = 9.6 | |
| initial | 0.10 | 0.10 | 0.50 atm |
| 2A + | 2B ---> | 3D | Keq = PD3/PA2PB2 = 0.84 |
| initial | 0.15 | 0.05 | 0.22 atm |
Use finite difference methods in a spreadsheet to show whether or not equilibrium is established. In this system, equilibrium can be taken to mean that the heights of water in the two tanks remain the same in time.