616 CHAPTER 13 SPONTANEITY OF CHEMICAL PROCESSES SECTION EXERCISES 13.3.1 13.3.2 13.3.3 Of the following pairs, which has the greater entropy? Explain wliy in \ (a) 1 g of dew or 1 g of frost; (b) 1 mol of gaseous hydrogen atoms oi ■ *• n gaseous hydrogen molecules; (c) "Perfect" diamond or flawed diamon I V4 carat; (d) 5 mL of liquid ethanol at 0 °C or 5 mL of liquid ethanol <" Explain the following differences in entropies in molecular terms (su standard conditions unless otherwise noted): (a) 1 mol of 02 has less entropy than 1 mol of 03. (b) 3 mol of 02 has more entropy than 2 mol of 03. (c) 1 mol of I2 has less entropy than 1 mol of 02. (d) 1 mol of HCl, , in concentrated solution (12 M) has less cnlro i ■ ■■ J of HCl m in dilute solution (0.100 M). Draw molecular pictures to illustrate your answers to part b of Sectio 13.3.1 and part b of Section Exercise 13.3.2. 13.3.4 Compute the standard entropy change for the following reaction: 12 NH3 (gl 4- 21 031g) ^ 8 HN03{g, + 4 NO (g) + 12 H20„, I n 13.4 SPONTANEITY AND FREE ENERGY The second law of thermodynamics states that the entropy of the ■ u ■ increase during a spontaneous process. Consequently, the sign of A> . n, positive for any chemical process to be spontaneous. To determine wl ill i ular chemical process is spontaneous, we must calculate the value ol .l1* accompanies the process. Unfortunately, this is not practical for most i n . usually possible to calculate the entropy change for a system, bul iht ." i' may undergo complicated changes of state for which AS cannot be determin I The surroundings include virtually all the universe, and keeping hack of cli i the universe is a tricky matter. It would be much more convenient if we] il way to determine the direction of spontaneous change using just the \\sten i surroundings. No such criterion can be applied to all processes. If we restrict the C"i Jil sufficiently, however, there is a state function whose change for the systen [■■ spontaneity. This new state function is called free energy (G) and is d< n Equation 13-6: Free energy — G — H — TS (1 where H is enthalpy, T is temperature, and S is entropy.* The American J. Willard Gibbs introduced free energy into chemical thermodynamics professor of maihematical physics at Yale University from 1871 until his death in 190 was the first to show how the laws of thermodynamics apply to chemical processes B Gibbs published in a litfle-known American Journal at a time when most scientific woi1 being done in Europe, his outstanding contributions to chemical thermodynamics wei recognized until 13 years after ihey were first published. In the 1890s, however, his 4 of elegant mathematical development of this subject were translated into French end and European scientists quickly recognized the greatness of his work. To commnmorc free energy is symbolized G and is sometimes called Gibbs free energy. 05 13.4 SPONTANEITY AND FREE ENERGY 617 ■ ifinition of free energy gives us an equation that relates the change in free .' i any system to other thermodynamic changes: AGsys = A//sys-Af7-SJsys -' '.,..t0 predict spontaneity, we must relate this equation to the total entropy (he universe. This cannot be done unless some restrictions are placed on ■ linns. First, the process must occur at constant temperature. This lets us - S;^toASsys: A(TS) = (TS),nsl -(TS),,ti3l = T(Sf-S.) = 7ASsy5 (constant T) AGsys = Atfsys - TASsys (13-7) the enthalpy change for the system can be related to the entropy change for ■■ jiulmgs by restricting the conditions to constant pressure. Recall that i U q when P is constant: AH™ = <7S. (constant P) ■r also that the heat flow for a system is always equal in magnitude but in sign to the heat flow of the surroundings: %s = -&UT and AHsys = -qmn -■-'■ arc, we have already restricted the process to constant temperature, so i How of the surroundings measures the entropy change of the surroundings. ^ = ASSĽrr and qmn = TASSU!r , n'lg these equalities gives an equation that relates A#sys and ASsurr: Atfsys = ~TASsun (constant P and T) i substitute this result into Equation 13-7: AG = -rA5surr - TAS = -T(AS^ + A5SVS) ''ec;mseA5surr + A5sys = A5universe: AG.„ - ~TAS„ (constant T and P) -1 powerful result because it states that the free energy of the system changes lhai mirrors the entropy change of the universe in any process that occurs at ľ and P. By defining a new function and imposing some restrictions, we und a way to use properties of a system to determine whether a process is (his. Because T is always positive, and A5universe is positive for any sponta- 11 iccss, 4GS>1J is negative for spontaneous processes under constant T and P ■■us. Although the restrictions of constant T and P are stringent, they are met important chemical processes. For example, the human body has a constant C. Any biochemical reaction that occurs in the body occurs under conditions ■ the i mmediate surroundings are at constant T and P. 618 CHAPTER 13 SPONTANEITY OF CHEMICAL PROCESSES Because free energy is a state function, its values can be tabulated foi i i chemical calculations. As with standard heats of formation, the standard molai in energy of formation (AG°f) for any substance is defined to be the change o f energy when 1 mol ofthat substance is formed from elements in their standaidii -Following the same reasoning we used for enthalpy changes, we obtain an cqi ■ c for calculating the free energy change for any chemical reaction: AG"™™™ = S coeff AG°f (products) - X coeffr AG°r(reactants) (L-íi The form of Equation 13-8 should be familiar because it is analogous to Equatic lJ 11 for reaction enthalpies and Equation 13-5 for reaction entropies. The standard free energy change for a reaction can also be calculated fioir ~'f and AS° for the reaction by making use of Equation 13-7 under standard conclit AG° - AH" - TAS° (i:n Either of these equations can be used to find standard free energy changes W ■!• equation we use depends on the available data. Sample Problem 13-6 illustrate; types of calculations. Pay close attention to units when using Equation 13-9. The values of AH° and AG" are usually given in kj or kj/mol, but entropies are usually expressed in J/K or J/mol K. Thus entropies must be multiplied by 10 3 kJ/J before adding the two terms. SAMPLE PROBLEM 13-6 FREE ENERGY OF REACTION Find the standard free energy change for the acryionitrile synthesis discussed in Sample Problem 13-5. METHOD: There are two ways to calculate AG°rsn. The first method uses standard free gies of formation and Equation 13-8. The second method uses Equation 13-9 and the w of AH° and A5° calculated earlier. We perform both calculations to show they give the s result. First, recall the balanced equation for the acryionitrile synthesis: 2 C3H6 (g) + 2 NH3 ig, + 3 02 {g) -> 2 C3H3N (l] + 6 H20 (1) (a) Using AG°f values from Appendix E: AG°rxn= [(6 mol)(-237.13 kJ/mol) + (2 mol)(208.6 kJ/mol)] - [(3 mol)(0 kJ/mol) + (2 mol)(-16.45 kJ/mol) + (2 mol)(74.62 kJ/mol)] AGarsn-~1122U (b) Using the results of Sample Problem 13-5: AG°rxn = A/T - TAS° AG°rxn = [-1318 U] - [(298K)(-659J/K)(l(r3kJ/J)3 =-1122kJ The large negative AG0IX„ found in this problem indicates that the production of acrj lorn is highly spontaneous under standard conditions. CHANGE IN FREE ENERGY UNDER NONSTANDARD CONDITIONS Standard conditions refer to unit concentrations and 298 K, but chemical ica( i occur at many different concentrations and temperatures. To use AG as a measn i-f spontaneity under nonstandard conditions, we must understand how free e .# depends on temperature and concentration. First, Equation 13-9 is valid ai ,1 temperature as long as the temperature is constant. However, to apply the equat n a temperature different from 298 Kf we must have the appropriate values loi L 'u and AS. SPONTANEITY AND FREE ENEROY 619 Chemical substances become more disordered as temperature increases. These langes in entropy as a function of temperature can be calculated, but the techniques [uire calculus. Fortunately, temperature affects the entropies of reactants and prod-in the same way. In other words, changes in the disorder of the reactants are i almost the same as changes in the disorder of the products. As a result, the ■ rature effect on the net entropy change of a reaction is usually small. In fact, in ;ases we can assume that ASmi is independent of temperature. I icall that Ai/rxn also does not change rapidly with temperature. As a result, we ,timate free energy changes at temperatures other than 298 K by assuming that I ud enthalpies and entropies at 298 K also apply at any other temperature: The temperature variation of the change in entropy during a reaction (AS„n) can often be neglected, but temperature variations in absolute entropies of individual substances (Sm0|0,) are never negligible. AGC xn.T — **" „„298 TAS* (13-10) n immediate consequence of Equation 13-10 is that any reaction with a large very sensitive to temperature. Sample Problem 13-7 shows a simple example. AMPLE PROBLEM 13-7 TEMPERATURE AND SPONTANEITY igen tetroxide can decompose into two molecules of nitrogen dioxide: N204(E)-^2N02(g) ow that this reaction is not spontaneous under standard conditions. id the temperature at which the reaction becomes spontaneous at standard pressures. '-"-IOD The key word here is spontaneous, which suggests that we need to work with nergies to solve this problem. Recall that the criterion for spontaneity is AGrsn <0. We ind this free energy change at standard temperature. Then, using Equation 13-10, we ilculate the temperature that makes AGtsn <0. ,e Equation 13-8 and values for AGöf from Appendix E to show that the decomposition on is not spontaneous under standard conditions: AG° = (2 mol)(51.31 kJ/mol) - (1 mol)(97.89 kJ/mol) = 4.73 kJ ositive value for AG0 indicates that this reaction is not spontaneous under standard f' (ions In fact, the calculation tells us that the reaction will be spontaneous in the oppo-■'irection. Under standard conditions, N02 reacts to form N204: AG0 = -4.73 kJ 2 NO, .-»N204W se Equation 13-10 to find the temperature at which N204 decomposition becomes spon-us when the partial pressures of both gases are 1 atm: AG^T=Aff^29&-TAS°^m mperature changes, AG must become zero before it becomes negative. We must find mperature at which AG = 0. Begin by calculating AW and AS0 using tabulated data: AH0 = (2 mol)(33.18 kJ/mol) - (1 mol)(9.16 kJ/mol) = 57.2 kJ AS0 = (2 mol)(240.1 J/mol K) - (1 mol)(304.3 J/mol K) = 175.9 J/K AS0 - 0.1759 kJ/K AH0 = 57.2 kJ lit? equal to zero and rearrange the equation to solve for temperature: AG = AH" - TAS° - 0 or TAS° = AH° AH0 57.2 kJ T = AS"1 0.1759 kJ/K -325 K 15 K, AG0 = 0. Therefore at all temperatures greater than 325 K, AG0 is negative, and ■imposition of N204 is spontaneous at 1 atm partial pressures. In Equation 13-10 we use the superscript "o" to denote standard concentrations (1 atm, 1 M) of all reagents, even though temperature is nonstandard (ľ * 298 K). When the temperature is not specified, "o" means 298 K and standard concentrations. Therefore AG° means "free energy change at 298 K, all reagents at unit concentration," whereas AG%0(( means "free energy change at 500 K, all reagents at unit concentration." 620 CHAPTER 13 SPONTANEITY Of CHEMICAL PROCESSES CHANGING CONCENTRATION The substances participating in a chemical reaction typically are at concentr ■ different from 1 M or pressures different from 1 atm. For example, a biochemii ■ ■! ■ wants to know what processes are spontaneous under physiological condition 1 find that the substances dissolved in biological fluids are rarely at 1 M conceni, ■ i How does change in free energy vary with changes in molarity and pressure? 1 -'l that enthalpy is virtually independent of concentration but that entropy obeys I ■! ■ tion 13-4. To see how entropy affects AG, consider the synthesis of ammonia carried " a pressurized reactor containing N2, H2, and NH3 at partial pressures (p) did .u from 1 atm: N2(g) + 3H2(g)->2NH3(g) Equation 13-4 gives the molar entropy of each gas as a function of its | i ' pressure: S = S°~Rlnc = S°~R Inp For example, 5(N2) = 5°(N2) - R ln{p(N2)} The entropy change for the reaction is the difference in entropy between pn ' i and reactants, obtained by multiplying each corrected entropy by the apprc i stoichiometric coefficient: ASrxn-2 5(NH3)-3 5(H2)-S(N2) Now we substitute each of the corrected entropies: ASan = 2[S°(NH3) - R InlpiNH,)}] - 3[S°(H2) - R ln{p(H2)}] - [S°(N2) - R ln{p(\ ■ Next, rearrange the equation so that all the logarithmic terms are together: ASrxn = 25°(NH3) - 3S°(ti2) - S°(N2)-2Rln{p(NHy)} +3R ln{p(H2)} + Rln{p , ■ The first three terms are the standard entropy change for the reaction, allowin ■ ■ ■ simplify: &Srxn = A5°rsn - 2Rln{p(NU})} +3R /n{p(H2)} + Rln{p(N2)} The properties of logarithms can be used to combine the In terms. First, x, In y\ giving: 2 R /h{/?(NH3)} = R ln{p2(NH$)} and 3 R ln{p(H2)} = R ln{p\H2)} With these changes, the equation becomes: A5„n = A^rxn - R [ln{p2(NH,)} - tn{p\H2)} - ln{p(NH,)}] A second logarithmic property, (In x-In y) = ln(x/y), lets us put all the logar ' ■ ■ terms into a single ratio: SPONTANEITY AND FREE ENERGY 621 AS = AS°rv„ -Rln P2(NH3) p(N2)p\B2) ; that the pressure ratio has product concentrations raised to their stoichio-c coefficients in the numerator and reactant concentrations raised to their chiometric coefficients in the denominator. The form of this equation applies to reactions, not just the synthesis of ammonia. Thus for the general reaction, aA + bB^cC + dD entropy change is: AS,. AS0..., -Rln [C]c[D]d [AY[B]b ratio in the logarithmic term is called the concentration quotient (Q): [Af[Bf = Q (13-11) entropy change for a reaction under nonstandard concentrations can be ressed in terms of the standard entropy change and Q: ASrxn = AS^ň-JU«ô s allows us to write an equation for the free energy change when concentrations nonstandard: , AGrxn = A/frxn - r(AS°ran -RlnQ) xethat Ar/° — TAS0 is just AG0, so this equation reduces to Equation 13-12: AGr,„ = AG°rxn + RTln Q (13-12) An immediate consequence of Equation 13-12 is that the direction of spontaneity reaction depends heavily on the concentrations of reactants and products. concentrations appear in the numerator of Q, so when the concentration of a 'i increases, In Q increases, as well. Increasing the In Q term makes AG less e, so a reaction becomes less spontaneous as product concentrations increase. sely, reactant concentrations appear in the denominator of <2, so increasing centration of a reactant also increases In Q, which in turn makes AG more e. Thus a reaction becomes more spontaneous as reactant concentrations . .n, however, it ■ )t be possible to change the value of Q sufficiently to make AGrxn <0. . iber that the term In Q changes much more slowly than Q itself. A 10-fold in Q, for instance, only changes In Q by a factor of 2.3. Nonetheless, reac- . at are almost spontaneous under standard conditions can be driven forward by appropriate changes in concentration. Sample Problem 13-8 shows how this A concentration quotient always contains the concentrations that are variable, that is, those of gases and solutes. The concentrations must be expressed relative to standard conditions, so gas concentrations are in aim, whereas solute concentrations are in mol/L. 622 CHAPTER 13 SPONTANEITY Of CHEMICAI. PROCESSES The mol unit in the value of AG* refers to "per mole of reaction" and comes from the concept of the reaction unit. The mol unit in AG" is included in the calculation to cancel the mol unit in R. For a review of the reaction unit, see Section 12.4. Recall that when y = Inx, x = er, where e is the basis for natural logarithms. SAMPLE PROBLEM 13-8 EFFECT OF CONCENTRATION ON SPONTANEITY The decomposition of dinitrogen tetroxide follows: N204(g)-^2N02(g) (a) Find the minimum partial pressure of N204 at which the reaction is spontaneous if P(N02) = 1 atm and T = 298 K. (b) Find the maximum partial pressure of N02 at which the reaction is spontaneous if /?(N204) = 1 atm and T = 298 K. METHOD: This is a two-part problem, so each part should be solved independently. Bot parts require us to relate change in free energy to concentrations, so we must use Equaiio 13-12: AG = AG0 + RTln Q = AG0 + RTln P2(NQ2) P(N204) (a) We are asked to find the partial pressure of N204 that will make the decomposition sp taneous when T = 298 K and p of N02 = 1 atm. The value of AG must be zero before it become negative. Therefore to find the threshold pressure of N204 that makes the decom sition spontaneous, we set AG = 0 and p(N02) = 1 atm, and then rearrange to solve for I partial pressure of N204: (1 atm)2 n 0 = AG° + RTln P(N204) In (1 atm)2 MN204) AG0 _ 4.73 X 103J/mol RT ~ (8.314J/molK)(298K) = -1.909 (1 atm)2 P(N204) _ „-I.909 _ = 0.1482 and p(N204) = 1/0.1482 - 6.75 atm This decomposition is spontaneous as long as the pressure of N204 is greater than 6.75 al As always, we have to be careful about units. Standard free energy was converted to joul to use the appropriate value of the gas constant (R) in J/mol K. Our final number is an antilogarithm, which has no dimensions, but the pressure must have the same units as the standard state for gases, which is atmospheres. (b) We are asked to find the maximum partial pressure of N02 below which the decoinpc tion is spontaneous when T = 298 K and p of N204 = 1 atm. The procedure is analogcw the one we just developed. You should be able to show that the desired pressure is 0.385 This problem shows that a reaction with a small positive AG0 can be made spontaneous I relatively small changes in concentrations. If neither temperature nor concentrations are at their standard values, free ciii calculations must be done in two steps. First, correct for temperature to obtain L using Equation 13-10. Second, use that result in Equation 13-12 to complete calculation of AG. INFLUENCING SPONTANEITY Suppose we want to design a particular chemical synthesis, but we find that the r tion has a positive value for AG°. The thermodynamic calculation indicates Ilia reaction is spontaneous in the wrong direction under standard conditions, but 13.4 SPONTANEITY AND FREE ENERGY not prevent it from occurring under all conditions. What can we do to make the tion go in the desired direction? Sample Problem 13-8 illustrates that changing the concentration quotient "ges the entropy change of the system. In particular, reducing the pressure of N02 ow 0 386 atm or increasing the pressure of N204 above 6.71 atm would cause ntaneous decomposition of N204, even though this reaction is not spontaneous 'er standard conditions. As with this gas-phase reaction, a reaction in liquid solu-may be induced to proceed spontaneously by increasing the concentrations of tants or by reducing the concentrations of products. Changing the temperature of the system is another way to influence the spon-ity of a reaction. The equation for AG has two parts, AH and TAS, which can t together or in opposition: AG\ = AH° - TAS° 'A positive AS0 promotes spontaneity because it makes AG0 more negative. This ects the fact that a positive A5° means the system becomes more disordered 'tig the reaction. A negative AH° promotes spontaneity, as well, because it also es AG0 more negative. This reflects the fact that the surroundings become more ordered when a reaction releases energy. Thus a reaction that has a positive AS0 da negative AH° is spontaneous at any T. The combustion of propane is an example of a reaction that is spontaneous at all ■peratures: C3H8(g| + 5 02(g) -4 3 C02(g) + 4 H20 (gi AH0 = -897 kJ AS" = +145 J/K products of this reaction are more disordered than the reactants, and the reaction eases energy. Consequently, AG° is negative at all T, so the reverse reaction cannot made spontaneous by altering T. By the same reasoning, a negative AS° and a positive AH0 oppose spontaneity, so ction that meets these criteria is nonspontaneous regardless of T. The system and surroundings would experience decreases in entropy if such a process were to •ur, and this would violate the second law of thermodynamics. A reaction that has the same sign for AS0 and AH° will be spontaneous at some peratures but nonspontaneous at others. At low temperature, AS0 is multiplied by mall value for T, so at sufficiently low temperature, AH° contributes more to AG0 an TAS°. At high temperature, AS° is multiplied by a large value for T, so at suffi-"-Jitty high temperature, AS° contributes more to AG" than AH°. Reactions with positive AH° and positive AS" are favored by entropy but disfa-red by enthalpy. Such reactions are spontaneous at high T, where the TAS° term 'minates AG0. The reactions are nonspontaneous at low T, where the AH° term minates AG°. These reactions are spontaneous at high temperature by virtue of the creased disorder in the system. The opposite situation holds for reactions that have negative values for AH° and . These reactions are spontaneous at low T by virtue of the increased disorder in surroundings. The favorable AH0 dominates AG° as long as T does not become , large. At high T, however, the unfavorable AS0 dominates AG0, and the reaction no longer spontaneous. The effects of temperature on spontaneity are summarized Table 13-3. 624 CHAPTER 13 SPONTANEITY OF CHEMICAL PROCESSES TABLE 13 -3 THE INFLUENCE OF TEMPERATURE ON SPONTANEITY AH° AS° AG°(HIGH T) AG° [LOW T ) SPONTANEOUS - + _ - All r + - + + NoT + + - + High r — __ + — Low r A/ŕ°, Standard enthalpy change; AS°, standard entropy change; AG", standard change in free enemy Calcium sulfate, the solid used to absorb water in desiccators, provides p example of this sensitivity to temperature. Anhydrous calcium sulfate absorbs w ■ r vapor from the atmosphere to give the hydrated salt. The reaction has a negative - ■ because the system becomes more ordered when gaseous water molecules move ■ ■ the solid state. The reaction also has a negative AH0 because of the coulombic toi of attraction between the ions of the salt and the polar water molecules. CaS04(i) + 2 H20 (gl -> CaS04-2H20 w Atf° = -104.9 kJ AS0 = -290.2 J/K At 300 K, the favorable AH° contributes more to AG° than the unfavorable ú' AG°300K - (-104.9 kJ) - (300 K)(-290.2 J/K)(10"-3 kJ/J) = -17.8 kJ Thus at room temperature, anhydrous calcium sulfate acts as a "chemical spon trapping water vapor spontaneously to form calcium sulfate dihydrate. The calcium sulfate in a desiccator is effective at removing water vapor onl long as some anhydrous salt remains. When all the anhydrous salt has 1 ■ converted to the dihydrate, the desiccator can no longer maintain a dry atmospr ■ Fortunately, the thermodynamics of this reaction makes it possible to regenei ate i drying agent. At 450 K, A5° contributes more to AG° than does AH°: AG°450K = (-104.9 kJ) - (450 K)(-290.2 J/K)(10~3 kJ/J) = +25.7 kJ At this T, the reverse reaction is spontaneous. Calcium sulfate dihydrate cat ■ converted to anhydrous calcium sulfate by placing it in a drying oven at 450 K 1 i ■ it can be cooled and returned to a desiccator, ready once more to act as a chen i sponge for water. SECTION EXERCISES 13.4.1 Estimate AG0 for the formation of gaseous water at T = 373 K. 13.4.2 Using Sample Problem 13-8, find the minimum partial pressure of N204 at vdtii decomposition of N204 occurs, if T is 400 K and p of N02 = 0.50 atm. 13.4.3 Does a temperature exist at which the water formation reaction becomes nonsni neous under standard pressure? If so, compute this 7". If not, explain why in molecular terms. SOME APPLICATIONS OF THERMODYNAMICS 625 13.5 SOME APPLICATIONS OF THERMODYNAMICS NITROGEN FIXATION distribution of nitrogen between the Earth's crust and the atmosphere is very en. In the crust, nitrogen is present at the level of 19 parts per million (ppm) ass. four orders of magnitude less than oxygen (4.55 X 10s ppm) and silicon X 105 ppm). In contrast, 80% of the atmosphere is molecular nitrogen. Para-;ally, nitrogen is absolutely essential for all life, but the sea of atmospheric gen is virtually inaccessible to higher life forms. Most biochemical systems lack bility to break the strong triple bond between the nitrogen atoms in N2. Molec-nittogen must be converted to some other form, usually ammonia (NH3) or ie (N03~), before most life forms can incorporate nitrogen atoms into their nemical molecules. This process, known as nitrogen fixation, is accomplished irious algae and bacteria, including a special group of bacteria that live in the of certain leguminous plants. 'he thermodynamics of nitrogen chemistry helps explain why nitrogen is so dant in our atmosphere and yet remains inaccessible to most life forms. Table shows that most of the abundant elements react with molecular oxygen under lard conditions. This is why many of the elements are encountered in the Earth's as their oxides. Nitrogen, however, is resistant to oxidation, as shown by the ive AG°f for N02. ecause of their resistance to chemical attack, nitrogen atoms are not "locked in any solid or liquid substances as are other elements, such as Si, Al, md H. On the Earth the most stable form of the element nitrogen is a ms diatomic molecule. Therefore the element nitrogen is concentrated in the i's gaseous atmosphere even though it is only a trace element in overall abun- :N=N: Bond energy = 940 kj/mol I veiy breath of air we take is 80% nitrogen, but our bodies must rely on the sen found in the proteins we eat to supply the elemental nitrogen required for 'nthesis. In the plant kingdom, the most important sources of nitrogen are NH3 he ammonium cation (NH4+). LE 13-4 SURFACE-ABUNDANT ELEMENTS AND THEIR OXIDES ELEMENT % BY MASS OXIDE AG°f (kJ/mol) 0 49.1 o2 0 Si 26.1 Si02 -856 Al 7.5 A1,03 -1376 Fe 4.7 Fe304 -1013 Ca 3.4 CaO -604 Na 2.6 Nap -377 K 2.4 K02 -239 Mg 1.9 MgO -570 H 0.88 H20 -237 Ti 0.58 Ti02 -885 CI 0.19 C120 +98 C 0.09 C02 -394 N <0.1 N02* +51 According to Tafale 13-4, chlorine (CI) Is also resistant to oxidation. Unlike nitrogen, however, chlorine reacts spontaneously with metals to generate such salts as NaCI and MgCI2. Thus among abundant elements on Earth, nitrogen is uniquely stable in its elemental form. ral other oxides of nitrogen exist. All have even more positive free energies of formation than N02.