The first law of thermodynamics (referred to in Chapter 7) can be stated as energy can neither be created nor destroyed, which can also be expressed as the total amount of energy that exists is a constant. Modern science has confirmed the law by recognizing that one type of energy can be converted into another (electricity running a light bulb, which produces both light and heat) and has found that the total energy of the system has not changed. However, the state of the system can change.
E and H are properties of a system that, along with others, define the state of the system. Those properties are state functions. If any of these properties change, the system is said to have undergone a change in state. Any change in E must be equal to the amount of heat absorbed by the system plus the amount of work performed on the system. A formal statement of the first law is:
In equation (16-1), q is the heat absorbed from the surroundings and w is the work done on the system. A few examples of work are (1) if a chemical reaction occurs within a system, work may be done upon it if gases are consumed and its volume is decreased, or (2) the system may perform work if gases are produced, or (3) work may be done if the system delivers an electric current to an external circuit.
Pertinent terminology related to these studies includes system, which refers to what specifically is under investigation, while the surroundings is the location under investigation. Practically, the surroundings is the portion of the environment that is affected by the system. The encompassing term for these factors is the universe (system + surroundings). Although universe usually refers to everything outside the system, practically, the system and immediate surroundings are more often the subject of the considerations. Utilizing these terms, the second law of thermodynamics states that the entropy (disorder) of the universe increases during spontaneous reaction. For example, a reaction taking place in a test tube (the system) giving off energy to the air around the test tube (the surroundings) tends to heat up that air. We could say that the test tube and reaction, along with the air that is measurably affected by the reaction, are the universe.
The existence of an energy balance is not sufficient to answer all questions about a chemical reaction. Does a given reaction take place at all? If so, to what extent does it proceed? Questions relating to the processes and extent of chemical reactions require the introduction of some new thermodynamic functions which, like E and H, are properties of the state of the system. These new functions are entropy, S, and Gibbs free energy, G. In order to answer these and other questions, a mathematical statement of the second law of thermodynamics is required:
In words: when a system undergoes a change, the increase in entropy of the system is equal to or greater than the heat absorbed in the process divided by the temperature. On the other hand, the equality, which provides a definition of entropy increment, applies to any reversible process, whereas the inequality refers to a spontaneous (or irreversible) process, defined as one which proceeds without intervention from the outside. Example 1 illustrates the reversible and irreversible reactions.
EXAMPLE 1 Consider a mixture of liquid and solid benzene at its normal freezing point, 5.45°C. If the temperature is raised by the slightest amount, for instance 0.01°C, the solid portion will gradually melt. But if, instead, the temperature were lowered by that amount, the liquid would gradually crystallize. The freezing process (as well as the melting process) at 5.45°C is reversible.
It is possible to cool liquid benzene carefully to a temperature below its normal freezing point, say to 2.00°C, without crystallization occurring. The liquid is then said to be supercooled. If a tiny crystal of solid benzene is added, the entire mass will crystallize spontaneously and irreversibly. Raising the temperature by 0.01 °C (or even 1.00°C) will not stop the crystallization. One would have to raise and maintain the temperature above 5.45°C to restore the liquid state. The crystallization of liquid benzene at 2.00°C is an example of an irreversible process.
The above statement of the second law (16-2) implies that there is a difference between those reactions that occur spontaneously (for which ΔS > Q/T) and those that cannot (ΔS < Q/T).
The spontaneity of a reaction can be predicted with a mathematical model (an equation/formula fitting known data) to explain the state function, G, the free energy. A usable model is
Equation (16-2) leads by complex argument to the following free-energy principle:
In other words, equation (16-4) states that the change in free energy at constant temperature and pressure can be negative or zero—negative in the case of a spontaneous irreversible process, or zero in the case of a reversible process. This rule is restricted to processes in which the only form of work is the increase in volume pressing against the surroundings (or the reverse). Processes for which ΔGT, p is positive can occur only with the application of work (or energy) from some external source (such as the process of electrical decomposition by the application of an electrical potential or coupling with another chemical reaction for which ΔGTp is negative). Another statement of the free-energy principle is that the maximum amount of work which a system can perform at constant temperature and constant pressure in forms other than expansion or contraction is equal to the decrease in the free energy of the system.
For all practical purposes, we can state that all spontaneous processes in nature result in an increase in the entropy of the universe. We can also generalize with the statement that any system (even the universe) will tend to run down over time (tend in increase in entropy until total chaos—disorder—is reached).
Entropy is often defined as an increase in disorder. A way of understanding entropy is to think in terms of the increase in entropy as an increase in the chaotic state (disorganized, untidy, or hectic state) of the system. Then, the greater the number of arrangements in a system, the greater the entropy.
The value of the entropy, S, is positive above absolute zero with the numerical value becoming higher with an increase in temperature. Using (16-2), changes in entropy, ΔS, can be calculated from gross thermal measurements of reversible processes. However, changes in entropy may also be related to the molecular properties of matter. Some general statements and examples relating to entropy are:
1. Liquids have more entropy than their corresponding crystalline forms. Every atom or molecule in a crystal is in a position in the lattice. In a liquid, the positions of the atoms or molecules are not as fixed in space and in those positions. Many liquids are composed of loosely attracted atoms or molecules in a crystalline structure, but that structure is not fixed throughout the mass of the liquids, as it is in solids.
2. Gases have more entropy than their corresponding liquids (or solids in the case of sublimation). Although the molecules in a liquid are free to occupy a variety of positions, they are held in close contact with their nearest neighbors. In a gas, the number of possible positions for the molecule is much greater because there is much more free space available per molecule.
3. Gases at low pressure have higher entropy than at high pressure. The argument for this factor is similar to 2 above—the amount of space per molecule is greater at low pressure than at high pressure.
4. A large molecule has a greater entropy than any of its submolecular fragments existing in the same phase of matter. The internal vibrations and rotations of atoms within molecules give many possibilities for distribution of intramolecular motions when considering the larger molecule as opposed to those available within each of the fragments.
5. The entropy of a substance always increases when its temperature is raised. The temperature is a measure of the average energy per molecule and, therefore, the total energy. The higher the temperature, the greater the total energy. The greater the total energy, the greater the number of ways of apportioning this energy among the fixed number of molecules. So, we can expect a different number of collections of molecular (energy) states if the temperature is higher, rather than lower.
6. If a chemical reaction is accompanied by a change in the number of gas molecules, ΔS is positive in the reaction direction that results in an increase in the number of gas molecules.
7. When one substance dissolves in another, ΔS is positive. The number of possible configurations of randomly arranged, unlike particles (solvent and solute) is greater than the number of configurations of separately packaged, like particles.
Equation (16-2) allows the calculations of changes in the entropy of a substance, specifically by measuring the heat capacities at different temperatures and the enthalpies of phase changes. If the absolute value of the entropy were known at any one temperature, the measurements of changes in entropy in going from that temperature to another temperature would allow the determination of the absolute value of the entropy at the other temperature. The third law of thermodynamics provides the basis for establishing absolute entropies. The law states that the entropy of any perfect crystal is zero (0) at the temperature of absolute zero (0K or -273.15°C). This is understandable in terms of the molecular interpretation of entropy. In a perfect crystal, every atom is fixed in position, and, at absolute zero, every form of internal energy (such as atomic vibrations) has its lowest possible value.
The use of standard entropies and free energies of formation to determine the value of ΔG for a reaction requires a little understanding. Consider the following:
1. Although enthalpies of substances are relatively independent of pressure (for gases) and of concentration (for dissolved species), their entropies (and free energies, too) depend on these variables. The values in tables for So and Go usually refer to the idealized state of 1 bar or 1 atm pressure for gases, 1M for concentrations in solution, and to the pure substances for liquids and solids.
2. Table 16-1 compiles some data for So, the molar entropy, and Δ
, the free energy of formation from the elements. All values in Table 16-1 are presented at 25°C and at standard states. Notice that the units of entropy and free energy are stated per mole, mol-1. This means that the moles used to balance a chemical reaction are included by the multiplication of the coefficient (mol in balanced equation) and the value from the table so that unit mol cancels. This is also the way in which we handled calculations involving ΔHvalues.
Table 16-1 Standard entropies and free energies of formation (25°C and 1 atm)
3. The third law allows the determination of the entropy of a substance without reference to its specific elements. However, if elements are important to the procedure, ΔG in the ground state is zero. As an example: the standard state of Br2 is a liquid at 25°C (the temperature of Table 16-1’s entries) and its ΔG is 0; but bromine can exist as a liquid at 25°C, which is not the ground state. The ΔG for Br2(l) is 3.14kJ/mol.
4. Unless otherwise stated, standard state values are provided at 1 atm. Although 1 bar can also be used, there is a slight difference in the values. We choose to use atm, rather than bar.
In general, for the reaction
the concentration dependence of the free-energy change may be expressed as
In this equation, the square brackets [substance] are used to state that the substance is expressed in concentration units of mol/L. Further, the mathematical relationship relies on the natural log, abbreviated In, not the log base 10, log. ΔG is the free-energy change at the given concentrations. ΔGo (standard free energy change) is the free-energy change for the hypothetical reaction in which all reactants and products are in their standard states. A point of interest is, if the concentration of every species is 1, the logarithmic term in (16-5) becomes 0. This makes ΔG = ΔGo, which is as it should be, because unit concentration implies the standard state.
In (16-5), if ΔG is to be expressed in joules, the universal gas constant must be taken as
Because the units included in Table 16-1 are in joules and kilojoules, the value of R must be stated in terms of joules, not calories. However, if there is a necessity for expressions in calories, the conversion from joules to calories is 4.184 J = 1 cal. (Note: 1 kcal = 1 Cal = 4.184 kJ.)
On the right side of equation (16-5), the term ΔGo lacks the unit mol-1, which must be removed from the second term as well, having been absorbed into the logarithmic term (the fraction) by mathematical manipulation.
The generalization of (16.5) to any number of reactants and products is
where Q is the reaction quotient, which is the term following the “In” (natural logarithm) in (16-5), but, of course, must include all reactants and products in the chemical reaction, except pure solids and liquids.
In theory, any chemical reaction could proceed at the same time in the reverse direction to some extent. In practice, this is not usually the case. Often, the driving force of a reaction favors one direction so greatly that the extent of the reverse reaction is so small that it is impossible to measure. The driving force of a chemical reaction is the change in free energy accompanying the reaction and it is an exact measure of the tendency of the reaction to go to completion. The possibilities are:
1. When the magnitude of ΔGo is very large and the sign negative, the reaction may go practically to completion in the forward direction. (Note that the positive sign on ΔGo indicates that the reaction will tend to go in the reverse direction from that written.)
2. If ΔG° is only slightly negative, the reaction may proceed to a small extent until it reaches a point where ΔG (calculated from ΔG° values) for any further reaction would be zero. The reaction could be reversed with a slight change in concentrations.
In the last case, the reaction is said to be thermodynamically reversible. Since many organic and metallurgical reactions are of this reversible type, it is necessary to learn how conditions should be altered to obtain economically advantageous yields, speed up desirable reactions, and cut down on undesirable reactions.
A chemical system which has reached the reversible thermodynamic state shows no further net reaction, since ΔG° = 0. This does not mean that nothing is occurring. Actually, the chemical reactions in both directions continue, but at the same rate. This means that each move to the right has an equal move to the left, which is the definition of a dynamic equilibrium. (Dynamic means that something is happening, as opposed to a static situation, which is unchanging.) Dynamic reactions are characterized by the forward and reverse reactions occurring at the same rate, but in opposite directions. The result is no net change in concentrations, but there is still a lot going on chemically.
For a reversible reaction at equilibrium,
or
Equation (16-7) is a remarkable statement. It implies that Qeq, the value of the reaction quotient under equilibrium conditions, depends only on thermodynamic quantities that are constant in the reaction (the temperature, and the standard free-energy change for the reaction at that temperature), and is independent of the actual starting concentrations of reactants or products. For this reason, Qeq is usually denoted the equilibrium constant, K, and (16-7) is rewritten as
EXAMPLE 2 Consider the reversible reaction below in which the participants are gases:
A reaction mixture could be made up by starting with H2 and I2 alone, HI and I2, HI alone, or a mixture of all three substances. Regardless of how we start, a net reaction would occur in one direction or the other until the system comes to an eventual state of no further net change (an equilibrium). That equilibrium could be described by specifying the concentration of the three substances. Because of the variety of ways of making up the initial mixture (differing relative amounts of substances), there are an infinite number of equilibrium states. Each of these equilibrium states can be described by a set of concentrations of the three substances. However, a ratio describes the relationships of the concentrations K:
That is, the particular function of the three concentrations defined by Q (reaction quotient) is always the same at equilibrium. This is true even though any individual concentration may vary by as much as a magnitude of 10. This unifying principle allows the calculation of conditions at equilibrium under virtually any set of conditions.
Experimental measurements show that molecules in highly compressed gases or highly concentrated solutions, especially if electrically charged, abnormally affect each other. In such cases the true activity or effective concentration may be greater or less than the measured concentration. Therefore, when the molecules involved in equilibrium are relatively close together, the concentration should be multiplied by an activity coefficient, which is determined experimentally. At moderate pressure and solutions, the activity coefficient for nonionic compounds is close to unity, indicating little in the way of molecular interactions. In any event, the activity coefficient correction will not be made in the problems in this book.
The equilibrium constant, K, is a pure number (no units) whose magnitude depends not only on the temperature, but also, generally, on the standard-state concentration to which all concentrations are referred. In this chapter, the standard state for dissolved substances will be taken as a 1M concentration, unless a statement is made to the contrary. The magnitude of K is independent of the choice of standard concentration in the special case where the sum of the concentration exponents in the numerator equals the sum of the exponents in the denominator.
The concentration of a gas is proportional to its partial pressure (n/V = P/RT where [gas] = n/V). Consider the following equilibrium in which all the participants are gases:
The equilibrium constant for this reaction can be written as
Kp can replace K in equation (16-8). If ΔG° is obtained from a table based on the standard state of 1 atm, Kp will normally be correct only for P in atmospheres. If the values are for a standard state of 1 bar, then Kp will normally apply only to P in bars. When the equation shows no change in the total number of moles of gas as the reaction proceeds (e.g., N2 + O2 
2NO), Kp will be the same, regardless of the pressure units used and will be identical with the K expressed in molar concentrations.
The values for pure solids and liquids are usually omitted from the K expression because their concentrations vary only slightly. Additionally, the concentrations of pure solids and liquids are extremely large in comparison to the concentrations of gases and dissolved materials (normally, as measured in mol/L). This includes situations when the solvent is one of the products or reactants, as in the following, the hydrolysis of urea in aqueous solution:
1. Notice that water has been left out of the K expression; this is because it is the solvent. The concentration of water cannot be much less than that of pure water, 55.6 mol/L. This concentration is much larger than the solute, especially in moderately dilute to very dilute solutions (nearly all the solutions that are used in laboratories).
2. Also, notice that the molar concentration of ammonia, NH3, is squared. The reason for this is that the K expression is the product of the products divided by the product of the reactants and ammonia appears twice (coefficient of 2). Notice that there are three products and one usable reactant (no water).
The statement of Le Chatelier’s principle is generally if a system at equilibrium is stressed, the reaction will shift to relieve that stress. Chemical reactions can be exposed to stresses including a change in temperature, a change in pressure, a change in concentration of one or more of the participants, and others.
The effect of an increase in temperature of a system at equilibrium is a shift in the direction which absorbs heat.
EXAMPLE 3 In the equation for the synthesis of methanol, where all substances are gases,
the forward reaction liberates heat (–ΔH, exothermic), while the reverse reaction absorbs heat (+ΔH, endothermic, numerically to the same extent). If the temperature of the system is raised, the reverse reaction (right to left) occurs in an attempt to absorb the heat being supplied. Eventually, a new equilibrium will be established, which would have a higher concentration of reactants (CO and H2) and less product than at the initial temperature. Of course, if the production of methanol is important commercially, a lower temperature would favor the forward reaction, producing more methanol.
There are calculations that will support the above argument. The equation required is
or, if all are in standard states,
Combination of (16-8) and (16.10) gives
If ΔH° and ΔS° are fairly independent of temperature, as they are for most reactions, (16-11) shows that In K is a decreasing function of T for ΔH° < 0 (exothermic reaction). Then, K itself is a decreasing function of T. A decrease in K means a shift of the equilibrium to favor the formation of the substances whose concentrations appear in the denominator of the K expression (the reactants).
In the event that ΔH° > 0 (endothermic), K increases with T, and the equilibrium shifts to favor the formation of the products.
When the pressure of a system at equilibrium increases, the reaction occurs in the direction that lowers the pressure by reducing the volume of gas.
EXAMPLE 4 In the synthesis of methanol (all substances gases), the forward reaction is accompanied by a decrease in volume. The total number of moles gas is less on the right than on the left of this reaction:
Then,
A shift to the right will relieve the increase in pressure. It is important to recognize that the value for K does not change with the increase in pressure; only the pressures change.
Another way to increase the yield of methanol is shown in Fig. 16-1. This a way to increase the pressure by reducing the volume of the system. The system will respond to reduce the pressure by shifting to the right, increasing the yield of methanol.
Fig. 16-1
A pressure change will not affect the relative amounts of the substances at equilibrium in any gaseous system where the number of molecules reacted equals the number produced. For example, there is no change in concentration in the reaction, H2 + CO2 CO + H2O, because there is no advantage offered by a lower number of molecules (moles of gas) on a side. In other words, neither the forward nor the reverse reaction changes the net number of gas molecules.
Any effect of pressure on equilibrium systems involving mixed physical states (gas and solid or gas and liquid) is due to a change in the concentration of the molecules of gas. Generally, pure solids and pure liquids are not compressible and, therefore, there should not be a change in concentration of them with a change in pressure. It is possible to have changes in pressure that change the volume of solids or liquids, but the pressures involved are at or above thousands of atmospheres (not commonly encountered).
For reactions that take place in solution, increasing the amount of solvent (dilution) will displace the equilibrium in the direction of forming the larger number of dissolved particles. This is pretty much the same concept as discussed above in relation to the behavior of gases.
EXAMPLE 5 Consider the production of a dimer (a molecule composed of two building blocks) of acetic acid in benzene solution.
Let us imagine this reaction at equilibrium. If the solution is suddenly diluted to twice its original volume, and if the reaction has not yet occurred, the concentrations are 
what they were before the dilution. In the equilibrium constant expression, the numerator would be its original value and the denominator would be 
its former value ( squared). The ratio of the numerator to the denominator would become 2 times its original value ( divided by ). But this ratio must return to its original value of K (K does not change under these conditions). It can do this if the numerator becomes smaller and the denominator larger. In other words, some of the dimer, (HC2H3O2)2, must decompose to form acetic acid, 2HC2H3O2.
Changes in the amount of solvent will not affect the equilibrium in any system where the number of dissolved particles of reactants is the same as the dissolved particles of products. For example, the esterification of methyl alcohol with formic acid in an inert solvent, as below, does not respond to concentration changes. Note that, if the solvent is not water, then H2O must be in the K expression.
Increasing the concentration of any component of a system at equilibrium will cause a shift resulting in using up some of the added substance. For examples, suppose we were to add some hydrogen to the reaction below in which all participants are gases.
The reaction is no longer at equilibrium (more iodine than originally); the response to this stress is to shift to the right (using up some of the iodine). Naturally, some hydrogen is used up and some HI is produced during the shift. Since the value of K does not change, the final concentrations are calculated from it by substituting into the K expression.
A note of caution: Many of problems where concentrations change cannot be solved by a simple algebraic technique because there are exponents involved. For those solutions that involve a square, the quadratic formula may be necessary [Problem 16.12(c)]. Memorization of the formula is a good idea because many professors expect you to know it during a test (assuming you cannot use a formula sheet).
Catalysts accelerate both forward and reverse reaction rates equally. However, catalysts can be used to shorten the amount of time it takes to reach equilibrium when the original concentrations do not match equilibrium concentration.
16.1. Without consulting entropy tables, predict the sign of ΔS for each of the following:
(a) O2(g) → 2O(g)
(b) N2(g) + 3H2(g) → 2NH3(g)
(c) C(s) + H2O(g) → CO(g) + H2(g)
(d) Br2(l) → Br2(g)
(e) N2(g, 10 atm) → N2(g, 1 atm)
(f) Desalination of seawater.
(g) Devitrification of glass.
(h) Hard-boiling an egg.
(i) C(s, graphite) → C(s, diamond)
(a) Positive. There is an increase in the number of gas molecules.
(b) Negative. There is a decrease in the number of gas molecules.
(c) Positive. There is an increase in the number of gas molecules.
(d) Positive. S is always greater for a gas than for its corresponding liquid.
(e) Positive. Entropy increases on expansion.
(f) Negative. Desalination is the opposite of solution; a solute must be removed from a solution.
(g) Negative. Devitrification is the onset of crystallization in a supercooled liquid.
(h) Positive. The process of “boiling” an egg is not boiling, but the denaturation of the egg protein (the egg does not boil). A protein is a large molecule which exists in a particular configuration in the so-called native state. But it may occupy a large number of almost random configurations in the denatured state resulting from rotations around the bonds. The change in configuration during the cooking process requires energy and results from rotations around the bonds.
(i) Negative. Diamond, being a harder solid, would be expected to have more restricted atomic motions within the crystal. Therefore, diamond is more dense and has less entropy than graphite.
16.2. Calculate ΔS for the following phase transitions: (a) melting of ice at 0°C, and (b) the vaporization of water at 100°C. Use data from Chapter 7.
(a) 
Since the melting of ice at 0°C is a reversible process, (16-2) may be used with the equals sign. (Recall that q = ΔH at constant pressure.)
(b) 
Since the vaporization at 100°C is reversible, (16-2) may be used with the equals sign.
16.3. After comparing data in Table 7-1 and the answer to Problem 16.1(i), how do you account for the fact that ΔH and ΔS for the phase transition from diamond to graphite are not related by the same equation that applied in Problem 16.2?
From Table 7-1, the formation of diamond from graphite (the standard state of carbon) is accompanied by a positive ΔH of 1.88 kJ/mol at 25°C. From Problem 16.1(i), ΔS for the same process is negative. Since 25°C is not the transition temperature, the process is not a reversible one. In fact, it is not even a spontaneous irreversible process, and (16-2) does not apply with the inequality sign. On the contrary, the opposite process, the conversion of diamond to graphite at 1 atm, is thermodynamically spontaneous. The ΔS for this process would obey (16-2) with the inequality sign. This means that “diamonds are NOT forever!” The term “spontaneous” does not cover the speed of reaction; in this case, the reaction is so slow at normal life temperatures that it is not considered to be happening. So, the diamonds will be around for a long, long time.
16.4. Calculate Δ
for C2H5OH(g).
For the special process in which a substance in its standard state is formed from its elements in their standard states, (16-12) gives
Δ can be calculated from (16-12) with the use of data from Table 16-1. Write the balanced equation for 1 mol C2H5OH(g) produced from the elements. Write the S° values. Each of the S° values must be multiplied by the number of moles, n, necessary to balance the equation.
Keeping in mind that we lose the substances on the left and gain those on the right,
Now, from (16-12),
Note that Gibbs free energy is listed as a single entry for each substance in Table 16-1, but entropy must be calculated by taking the difference of the tabulated absolute entropies of the substance and its elements from which it is made.
16.5. (a) What is the ΔG° at 25°C for this reaction?
(b) What is ΔG at 25°C under conditions where the partial pressures of H2, CO2, H2O, and CO are 10, 20, 0.02, and 0.01 atm, respectively?
(a) The first step is to record the free energy values under each substance in the balanced equation.
Then, the calculation of ΔG° is done in the same manner as that of ΔH° (Problem 7.12).
(b) The calculation for ΔG° for the reaction including the partial pressures depends on
Note that the reaction, although not possible under standard conditions, becomes possible (G < 0) under this set of experimental conditions. Note the conversion from natural to common logarithms is by the factor of 2.3026.
16.6. Calculate the absolute entropy of CH3OH(g) at 25°C.
Although there is no S° entry in Table 16-1 for CH3OH, the Δ value for this substance listed in Table 16-1, the Δ value listed in Table 7-1, and the S° values for the constituent elements may be combined to yield the desired value. From (16-12),
From the equation for the formation of 1 mol of CH3OH under standard conditions,
we can write
Solving for S°(CH3OH), we received 239.6 J/K · mol
16.7. Estimate the boiling point of PCl3.
The boiling point is the temperature at which G° for the following reaction is zero.
The reaction is not spontaneous at 25°C, where, according to Table 16-1,
If we assume that ΔH° and ΔS° are both independent of temperature between 25°C and the boiling point, then the temperature dependence of ΔG° is given by the factor T in (16-10). Further, if ΔH° and ΔS° are known from the data at 25°C, T can be calculated to satisfy the condition of ΔG° equals zero.
Now,
and, from Table 7-1,
Then, after converting so that the values are in the same units (we chose J, not kJ),
or an estimated temperature of 73°C. The observed boiling point, 75°C, is very close to the estimated.
16.8. For the gaseous reaction
explain the effect on the equilibrium of (a) increased temperature; (b) increased pressure; (c) higher concentration of Cl2;(d) higher concentration of PCl5; and (e) presence of a catalyst.
(a) The effect of the increase in temperature is to shift the equilibrium in the direction that absorbs heat. Table 7-1 can be used to determine that the forward reaction as written is endothermic,
So, increasing the temperature will cause the dissociation of PCl5.
(b) When the pressure of a system at equilibrium is increased, the equilibrium point is displaced in the direction of the smaller volume. One volume each of PCl3 and Cl2 (2 gas volumes) forms one volume of PCl5. The pressure increase will shift the reaction to the left.
(c) Increasing the concentration will shift the equilibrium in the direction that will reduce that concentration in the effort to re-establish the equilibrium. The shift will be to the right.
(d) Using the same line of logic in (c), the shift will be to the left.
(e) Since catalysts speed up both the forward and reverse reactions to the same extent and do not favor either direction, there is no difference in the equilibrium.
16.9. What conditions would you suggest for the manufacture of ammonia by the Haber process?
From the sign of ΔH, the forward reaction is exothermic (gives off heat) and we know that the forward reaction is favored. However, we can shift the equilibrium to the right by dropping the temperature, except that reactions tend to progress more slowly at low temperatures. A catalyst would help increase the rate of reaction, but by itself the catalyst would not increase the output of ammonia.
We can shift the reaction to the right by increasing the concentration of N2 or H2, or both reactants. A shift to the right will also occur if NH3 is removed. We can also cause a shift to the right by running the reaction at high pressure (left, 4 volumes; right, 2 volumes). If the reaction were to be run at high pressure and high temperature (to increase the rate of reaction), the shift to the right would be expected.
16.10. All reactants and products are gases.
(a) Calculate ΔG° and Kp for this reaction at 25°C:
(b) Calculate ΔG° and Kp for the reverse reaction:
(c) Calculate the ΔG° and Kp for the forward reaction (a) written with different coefficients:
(d) Repeat calculation (a) for a standard state pressure of 1 bar (a little above 1 atm). For NO2(g), Δ is 51.32 kJ/mol, and for N2O4(g) it is 97.89 kJ/mol (both are at 1 bar).
Since ΔG° = –RT In Kp and by rearrangement,
(b)
Since ΔG° = –RT ln Kp and by rearrangement,
Parts (a) and (b) illustrate the general requirements that ΔG of a reverse reaction is the negative of ΔG for the forward reaction and K for a reverse reaction is the reciprocal of K (i.e., 1 divided by K) for the forward reaction.
(c)
Since ΔG° = -RT ln Kp and by rearrangement,
Parts (a) and (b) illustrate the general result that ΔG for a reaction with halved coefficients is half the ΔG value for the standard coefficients. Further, K for the reaction with halved coefficients is the one-half power (i.e., square root) of the K for the standard reaction. Note that the value of the following ratio at equilibrium must be independent of the way we write the balanced equation:
Any equilibrium constant for the reaction must involve the two partial pressures in exactly this way. For part (a), K is equal to this ratio. For part (c), K is equal to the square root of this ratio.
(d)
Since 1 atm and 1 bar are close (1 atm = 1.013 bar), the differences are small, but not insignificant. Kp (bar) can also be calculated from Kp (atm) by simply converting units.
16.11. A quantity of PCl5 was heated in a 12-L vessel at 250°C and reached the equilibrium below:
At equilibrium, the vessel contained 0.21 mol PCl5, 0.32 mol PCl3, and 0.32 mol Cl2. (a) Calculate the equilibrium constant, Kp, for the dissociation of PCl5 at 250°C when pressures are referred to the standard state of 1 atm. (b) What is ΔG° for the reaction? (c) Estimate ΔG° from the data in Tables 7-1 and 16-1, assuming constancy of ΔH° and ΔS°. (d) Calculate Kp from the original data using SI units and a standard state of 1 bar.
(a)
(b)
(c)
At 298.2 K,
At 523 K,
The estimate is close to the experimentally determined value based on the equilibrium measurement. This is despite the large temperature range over which ΔH° and ΔS° were assumed constant.
16.12. When 1 mol of pure ethyl alcohol is mixed with 1 mol of acetic acid at room temperature, the equilibrium mixture contains 
mol each of ester and water. (a) What is the equilibrium constant? (b) What is ΔG° for the reaction? (c) How many moles of ester are formed at equilibrium when 3 mol of alcohol are mixed with 2 mol of acid? All substances are liquids at the reaction temperature.
(a) A table is a convenient way of doing the bookkeeping for equilibrium problems. Use the balanced equation because the amounts included in the lines must reflect the coefficients in the balanced equation. The first line (1) shows the amounts of starting materials. The second line (2) records changes, including the signs (– for materials lost and + for those appearing). The third line (3) is the simple addition of the other two lines. Line (3) is used to calculate the value of the equilibrium constant for the reaction.
Let v = liters of mixture. (Since we have not been given information from which we can calculate volume, we can take 1 mol/L as the standard-state concentration.) The calculation of K becomes
Note that the concentration of water in this experiment is not so large compared with the other reaction components that it remains constant under the reaction conditions. Water’s concentration becomes extremely large when water is the solvent, but that is not the case with this reaction as the reactants are pure materials. Water must be included in the calculations for K.
(b) 
Because K is independent of the choice of standard-state concentration (the product of the exponents in the numerator equals the product of the exponents in the denominator), so is ΔG°.
(c) Let x = number of moles of alcohol reacting. Note that the amount of alcohol lost during the reaction is the same as that of the acid reacted. That amount is the same as the ester gained and the water gained. These gains and losses are dictated by the balanced equation, 1:1:1:1 ratio in this case.
The above becomes x2 = 4(3 – 4x + x2) or 3x2 – 16x + 12 = 0. This equation requires the quadratic formula to solve for x.
Of the two values provided by the quadratic formula, only one has physical meaning and can usually be chosen easily. Keep in mind that the value of x is substituted back into the third line (3) in the table. Since 4.4 cannot be subtracted from 3 (look at the alcohol column) and give us an amount that makes sense, 0.9 is the value we must use. So, the answer to the original question is: 0.9 mol ester produced.
Note that, in more complex problems, especially if the equation to be solved exceeds the abilities of the quadratic formula (cubed values and higher), it may be best to find the equilibrium concentrations by a method of successive approximations. You would select a consistent set of concentration values near where it is guessed the answer will be. Then, the calculations for K are performed and repeated until a sufficiently precise result is reached.
It is always a good idea to check your work. The check for (c) is to substitute back into the K expression.
Note that the number of moles of ester formed is greater than the number of moles of ester in (a), 0.9 compared with 0.67 (from ). This result was to be expected because the increased concentration of alcohol, one of the reactants, will shift the reaction to the right. More alcohol can be added to increase the yield of ester. Regardless of the amount of alcohol added, only 1 mole of ester can be produced before running out of acid (1 mole provided). In reality, there is a good chance that the reaction will not proceed as indicated by the equation (common with organic reactions), resulting in less ester than predicted. If the alcohol and acid are cheap compared to the ester, the yield of ester could be increased by adding either alcohol or acid (the cheaper of the two).
16.13. In a 10-L evacuated chamber, 0.5 mol H2 and 0.5 mol I2 are reacted at 448°C.
At the given temperature, and for a standard state of 1 mol/L, K = 50. (a) What is the total initial pressure in the chamber and the pressure at equilibrium? (b) How many moles of I2 remain unreacted at equilibrium? (c) What is the partial pressure of each component in the equilibrium mixture?
(a) Before the reaction progresses, the total number of gas molecules is 0.5 + 0.5 = 1. At equilibrium, there has been no change in the total number of moles. The total initial pressure and that at equilibrium are the same and can be calculated from the ideal gas law.
(b) Let x be the number of moles iodine reacting.
Note that the reaction ratio of H2:I2:HI is 1:1:2 and that must be reflected in the K expression. Regardless of how complete or incomplete the reaction may be, 2 mol HI are always formed for each mol H2 and/or I2 that react.
Since the number of moles of reactant gas equals the number of moles of product gas, moles may be used in place of concentrations (as in Problem 16-12), and Kp = K.
The above calculation is substituted into line (3), 0.5 —0.39, providing the information that 0.11 moles of I2 are in excess at equilibrium.
A check of the solution is
(c)
or
16.14. Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having the formulas 
, 
, 
, and so on. The equilibrium constant for the formation of is 12, for is 130, and both are formed from S and S2–. What is the equilibrium constant for the formation of from and S?
To avoid confusion, let us refer to the equilibrium constants for the various reactions by subscripts. Also, we note that only the ion concentrations appear in the equilibrium constant equations because solids (sulfur, in this case) are omitted from the calculations.
The desired constant, K3, expresses the equilibrium ratio of and concentrations in a solution in equilibrium with solid sulfur. Such a solution must also contain sulfide ion, S2–, resulting from the dissociation of (the reverse of the first reaction). Since all four species (S, S2–, , and ) are present, all the equilibria represented above must be satisfied. The three equilibrium ratios are not all independent because:
The result, K2 = K1 K3, is a general one for any case where one chemical equation (the second in this case) can be written as the sum of two other equations (the first and third).
16.15. At 27°C and 1 atm, N2O4 is 20% dissociated into NO2. (a) Find Kp. (b) Calculate the percent dissociation at 27°C and a total pressure of 0.10 atm. (c) What is the extent of dissociation in a 69-g sample of N2O4 confined in a 20-L vessel at 27°C?
(a) When 1 mol N2O4 dissociates completely, 2mol NO2 are formed. Since this problem does not specify a particular size vessel or a particular weight of sample, we are free to choose 1 mol (92 g) as the starting amount of N2O4. For the given total pressure of 1 atm, the table set up with the reaction is
We can now calculate Kp using the partial pressures.
(b) Let a = fraction of N2O4 dissociated at equilibrium, 0.1 atm total pressure.
or 0.4a2 = 0.167(1 – a2). Solving, a = 0.54, which translates to 54% dissociated at 27°C and 1 atm.
Note that a larger fraction of the N2O4 is dissociated at 0.1 atm than at 1 atm. This is in agreement with Le Chatelier’s principle as decreasing the pressure should favor the side with the greater volume (2NO, rather than N2O4).
(c) If the sample were all N2O4, it would contain 69 g/92 g/mol = 0.75 mol. Let a be the fraction dissociation. The table we set up is as follows:
Because the total pressure is unknown, it is simplest to calculate the partial pressure directly from Dalton’s law (Chapter 5).
Then,
or
The quadratic formula is required to solve for a.
The negative is discarded (negative matter?… not likely!). The extent of dissociation is 19%.
16.16. (a) Under what conditions will CuSO4 · 5H2O be efflorescent at 25°C? (b) How good a drying agent is CuSO4 · 3H2O at the same temperature? The reaction is:
Kp at 25°C is 1.086 × 10-4. The vapor pressure of water at 25°C is 23.8 torr.
(a) An efflorescent salt is one that loses water to the atmosphere. This will occur if the water vapor pressure in equilibrium with the salt is greater than the water vapor pressure in the atmosphere. The mechanism by which CuSO4 · 5H2O could be efflorescent is that the salt would lose 2 molecules of water and simultaneously form 1 formula unit of CuSO4 · 3H2O for each unit of the original salt that dissociates. Then, the above equilibrium equation would apply since all three components would be present.
Since CuSO4 · 5H2O and CuSO4 · 3H2O are both solids,
Taking the square root of both sides,
where P(H2O) is the partial pressure of water vapor (relative to the standard state pressure of 1 atm) in equilibrium with the two solids. A conversion to torr is necessary for the comparison:
Since the P(H2O) of 7.92 is less than the vapor pressure of water at the same temperature (23.8 torr), CuSO4 · 5H2O will not always effloresce. It will effloresce only on a dry day when the partial pressure of water in the air is less than 7.92 torr. This will occur when the relative humidity is less than
(b) CuSO4 · 3H2O could act as a drying agent by reacting with 2 molecules of water to form CuSO4 · 5H2O. Because of the equilibrium partial pressure of water discussed above, CuSO4 · 3H2O cannot absorb water from the air below that partial pressure, 7.92 torr. Many other drying agents can reduce the partial pressure of water below 7.92 torr (Problem 16.44).
To find the conditions under which CuSO4 · 3H3O would be efflorescent, we would have to know the equilibrium constant for another reaction. This reaction shows the dehydration of CuSO4 · 3H2O,
Tables 7-1 and 16-1 are to be used for the solution of the problems below. The data in the tables are based on experimental measurements and are subject to reevaluation. That means that the tables in different books may have some entries that do not agree with those in this book. However, the problems in this book were written from Tables 7-1 and 16-1 and the answers were calculated from them.
16.17. Calculate S° at 25°C for PCl5.
Ans. 364 J/K · mol
16.18. Calculate ΔH° for Cl2O(g) at 25°C.
Ans. 80.2kJ/mol
16.19. Predict the phase-transition temperature for the conversion of gray to white tin at one atmosphere using the thermodynamic data in Table 16-1.
Ans. 9°C (the observed value is 13°C)
16.20. Consider the production of water gas: C(s, graphite) + H2O(g) CO(g) + H2(g). (a) What is ΔG° for this reaction at 25°C? (b) Estimate the temperature at which ΔG° = 0.
Ans. (a) 91.44 kJ; (b) 982 K. The extrapolation for this estimate is extended over such a large temperature range that an appreciable error might be expected. The experimental value is 947 K, which is not all that far from the estimate.
16.21. The Δ for the formation of HI(g) from its gaseous elements is –10.10 kJ/mol at 500 K. When the partial pressure of HI is 10 atm and that of I2 is 0.001 atm, what must the partial pressure of hydrogen be to reduce the magnitude of ΔG to 0 at the same temperature?
Ans. 7.8 × 102atm
16.22. Under what conditions could the decomposition of Ag2O(s) into Ag(s) and O2(g) proceed spontaneously at 25°C?
Ans. The partial pressure of oxygen must be kept below 0.090 torr.
16.23. The S° for silver is 42.72 J/K · mol (a) Estimate the minimum temperature at which the decomposition of Ag2O(s) will proceed spontaneously (refer to Problem 16.22) when the oxygen pressure is 1.00 atm. (b) Make the same estimate when the partial pressure of oxygen is 0.21 atm.
Ans. (a) 466 K; (b) 425 K
16.24. The effect of changing the standard state from 1 atm to 1 bar is to raise the molar entropies, S°, of all gaseous substances by 0.109 J/k · mol at 298.2 K. Convert Δ of (a) CH3OH(g); (b)CH3OH(g); (c)H2O(l); and (d) Sn(s, gray) to the standard state of 1 bar. Assume that within the precision of the tables in this text, Δ of all substances is unchanged by this small change in standard state.
Ans. (a) –166.28kJ/mol; (b) –161.95kJ/mol; (c) –237.14kJ/mol; (d) 0.12kJ/mol
16.25. Given: N2 + O2 2NO state the effect on the reaction equilibrium of (a) increased temperature; (b) decreased pressure; (c) higher concentration of O2;(d) lower concentration of N2;(e) higher concentration of NO; (f) presence of a catalyst.
Ans. Favors (a) forward reaction; (b) neither reaction; (c) forward reaction; (d) reverse reaction; (e) reverse reaction; (f) neither reaction.
16.26. Predict the effect upon the following reaction equilibria of (a) increased temperature; (b) increased pressure:
1. CO(g) + H2(g) CO2(g) + H2(g)
2. 2SO2(g) + O2(g) 2SO3(g)
3. N2O4(g) 2NO2(g)
4. H2O(g) H2(g) + O2(g)
5. 2O3(g) 3O2(g)
6. CO(g) + 2H2(g) CH3OH(g)
7. CaCO3(s) CaO(s) + CO2(g)
8. C(s) + H2O(g) H2(g) + CO(g)
9. 4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g)
10. C(s, diamond) C(s, graphite)*
*This equilibrium can only exist under very special conditions. Density of diamond = 3.5; density of graphite = 2.3g/cm3.
Ans. F = favors the forward reaction, B = favors backward (reverse) reaction.
1.(a)B, (b) neither
2.(a)B, (b)F
3.(a)F, (b)B
4.(a)F, (b)B
5.(a)B, (b)B
6.(a)B, (b)F
7.(a)F, (b)B
8.(a)F, (b)B
9.(a)B, (b)F
10.(a)B, (b)B
16.27. Assuming that ΔH° and ΔS° remain constant, derive from equation (16-11) an expression relating K1 at temperature T1 to K2 at temperature T2.
Ans. ln(K2/K1) = (ΔH°/R)(T2 – T1)/T2 T1
16.28. For the neutralization reaction:
The ΔH° is –55.8 kJ. For the ionization of water, which is the reverse of the above, the equilibrium constant at 25°C is: Kw = [H+][OH–] = 1.0 × 10–14. This is a very important constant (Memorize it!), which appears throughout Chapter 17. Using the result of Problem 16.27, calculate Kw at (a) 37°C, normal human body temperature, and at (b) 50°C.
Ans. (a) 2.4 × 10–14;(b) 5.7 × 10–14
16.29. When α-D-glucose is dissolved in water, it undergoes a partial conversion to β-D-glucose, a sugar of the same molecular mass but slightly different physical properties. At 25°C, this conversion, called mutarotation, stops when 63.6% of the glucose is in the β-form. Assuming that equilibrium has been attained, calculate K and ΔG° for the reaction, α-D-glucose β-D-glucose, at 25°C.
Ans. 1.75, –1.38 kJ
16.30. The equilibrium constant is 0.09 for the reaction, H3BO3 + glycerin (H3BO3-glycerin). How much glycerin would be added per liter of 0.10 molar H3BO3 solution so that 60% of the H3BO3 is converted to the boric acid-glycerin complex?
Ans. 1.7 mol
16.31. The equilibrium,
p-xyloquinone + methylene white p-xylohydroquinone + methylene blue
may be studied conveniently by observing the difference in color between methylene blue and methylene white. One mmol of methylene blue was added to a liter of solution that was 0.24M p-xylohydroquinone and 0.0120M p-xyloquinone. It was found that 4.0% of the added methylene blue was reduced to methylene white. What is the equilibrium constant for the reaction? Note: The equation is balanced with 1 mole of each of the four substances.
Ans. 4.8 × 102
16.32. When SO3 is heated to a high temperature, it decomposes by
A sample of pure SO3 was sealed in a large cylinder fitted with a piston and heated to a high temperature, T. The equilibrium ratio SO2:SO3 was 0.152 and the total pressure was 2.73 atm. If the piston is pushed in to cut the volume in half, what will be the new pressure? (Note: Because of the degree of the equations involved, a solution is most readily obtained by successive approximations.)
Ans. 5.40 atm
16.33. Consider the equilibrium described in the previous problem. Instead of pushing in the piston, a quantity of SO3(g) equal to that already present is injected. This addition momentarily doubles the concentration of SO3. (a) What is the new SO3 pressure when equilibrium is restored with no temperature change? (b) What is its ratio of the equilibrium pressure to that pressure momentarily present at the time of injection (before reaction)? (c) Calculate a similar ratio for SO2.
Ans. (a) 4.26 atm; (b) 0.96; (c) 1.54
16.34. A saturated solution of iodine in water contains 0.33 g I2 per liter. More than this can dissolve in a KI solution because of the following equilibrium:
A 0.100 M KI solution (0.100 M I–) actually dissolves 12.5g iodine per liter, most of which is converted to 
(a) Assuming that the concentration of I2 in all saturated solutions is the same, calculate the equilibrium constant for the above reaction with 1 molar solutions as standard states. (b) What is the effect of adding water to a saturated solution of I2 in the KI solution?
Ans. (a) 7.1 × 102; (b) The reverse reaction is favored.
16.35. Consider the reaction: H2(g) + I2(g) 2HI(g). When 46 g of I2 and 1.00 g of H2 are heated to 470°C, the equilibrium mixture contains 1.90 g of I2. (a) How many moles of each gas are present at equilibrium? (b) Calculate the equilibrium constant.
Ans. (a) 0.0075 mol I2, 0.32mol H2, 0.35 mol HI; (b) K = 50
16.36. Exactly 1 mol each of H2 and I2 are heated in a 30-L previously evacuated chamber to 470°C. Using the value of K from Problem 16.35, determine (a) how many moles of each gas are present at equilibrium; (b) the total pressure in the chamber; (c) the partial pressures of I2 and of HI at equilibrium. (d) If one additional mole of H2 is introduced into the equilibrium system, how many moles of the original iodine will remain unreacted?
Ans. (a) 0.22 mol; (b) 4.1 atm; (c) P(H2) = P(I2) = 0.45 atm, P(HI) = 3.2 atm; (d) 0.065 mol
16.37. Consider the reaction: PCl5(g) PCl3(g) + Cl2(g). Calculate the number of moles Cl2 present at equilibrium when 1 mol PCl5 is heated at 250°C in a 10-L vessel. At 250°C, K = 0.041 for this dissociation based on the 1 mol/L standard state.
Ans. 0.47 mol
16.38. PCl5 is introduced into an evacuated chamber and comes to equilibrium (see Problem 16.37), at 250°C and 2.00 atm. The equilibrium gas contains 40.7% Cl2 by volume.
(a1) What are the partial pressures of the gaseous components at equilibrium?
(a2) From these data, calculate Kp at 250°C on the basis of the 1 atm standard state for the reaction written in Problem 16.37.
If the volume of the gas mixture is increased so that it is at 0.200 atm at 250°C, calculate:
(b1) The percent of PCl5 that would be dissociated at equilibrium.
(b2) The percent by volume of Cl2 at equilibrium.
(b3) The partial pressure of Cl2 at equilibrium.
Ans. (a1) P(Cl2) = P(PCl3) = 0.814 atm, P(PCl5) = 0.372 atm; (a2) 1.78 (b1) 94.8%; (b2) 48.7%; (b3) 0.0974 atm
16.39. At 46°C, Kp for N2O4(g) 2NO2(g) is 0.67 at 1 bar. (a) Calculate the percent dissociation of N2O4 at 46°C and a total pressure of 0.507 bar. (b) What are the partial pressures of N2O4 and NO2 at equilibrium?
Ans. (a) 50%; (b) P(N2O4) = 0.17 bar, P(NO2) = 0.34 bar
16.40. Consider the reaction: 2NOBr(g) 2NO(g) + Br2(g). If nitrosyl bromide (NOBr) is 34% dissociated at 25°C and a total pressure of 0.25 bar, calculate Kp for the dissociation at this temperature based on 1 bar standard state.
Ans. 1.0 × 10-2
16.41. At 986°C, the equilibrium constant for the reaction below is 0.63:
A mixture of 1 mol of water vapor and 3 mol of CO is allowed to come to equilibrium at a total pressure of 2 atm. (a) How many moles H2 are present at equilibrium? (b) What are the partial pressures of the gases at equilibrium?
Ans. (a) 0.68 mol; (b) P(CO) = 1.16 atm, P(H2O) = 0.16 atm, P(CO2) = P(H2) = 0.34 atm
16.42. Consider the reaction: SnO2(s) + 2H2(g) 2H2O(g, steam) + Sn(l).
(a) Calculate Kp at 900 K; the equilibrium steam-hydrogen mixture was 45% H2 by volume.
(b) Calculate Kp at 1100 K; the equilibrium steam-hydrogen mixture was 24% H2 by volume.
(c) Would you recommend higher or lower temperatures for more efficient reduction of tin?
Ans. (a) 1.5; (b) 10; (c) higher
16.43. The preparation of quicklime from limestone is CaCO3(s) CaO(s) + CO2(g). Experiments carried out between 850°C and 950°C led to a set of Kp values fitting an empirically determined (resulting from experimentation) equation
where T is the absolute temperature. If the reaction is carried out in quiet air, what temperature would be predicted from this equation for the complete decomposition of the limestone? In quiet air, assume it is necessary to build up the CO2 pressure to 1 atm to ensure continual removal of the product.
Ans. 894°C
16.44. The moisture content of a gas is often expressed in terms of the dew point, the temperature to which the gas must be cooled before it becomes saturated with water vapor. At this temperature, water (liquid or solid depending on the ambient temperature) will be deposited on a solid surface.
The efficiency of CaCl2 as a drying agent was measured by a dew-point experiment. Air at 0°C was allowed to pass slowly over large trays containing CaCl2. The air was them passed through a glass vessel through which a copper rod was sealed. The rod was cooled by immersing the part of the rod outside the vessel in a dry ice bath. The temperature of the rod inside the glass vessel was measured by a thermocouple. As the rod was cooled slowly, the temperature at which the first crystals of frost were deposited was observed to be –43°C. The vapor pressure of ice at this temperature is 0.07 torr. Assuming that the CaCl2 owes its desiccating (ability to remove water from a gas mixture) properties to the formation of CaCl2 · 2H2O, calculate Kp at 0°C for the reaction,
Ans. 8 × 10–9
16.45. At higher temperatures, the following equilibria result from reactions in a mixture of carbon, oxygen, and their compounds:
If it were possible to measure K1 and K2 independently, how could K3 and K4 be calculated?
Ans. K3 = K2/K1 and K4 = K1/K3 = K2/K2
16.46. Consider this reaction including urea, CO(NH2)2, CO2(g) + 2NH3(g) CO(NH2)2(s) + H2O(g). The Δ for urea is –197.2kJ/mol. (Δ for NH3 is –16.7; refer to Table 16-1 for others.) (a) Calculate the ΔG for the reaction. (b) Is the direction of the reaction spontaneous as written?
Ans. (a) ΔG = +2.03 kJ; (b) No, AG is negative for a spontaneous reaction.