4. Influence of Temperature and Pressure on Chemical Changes

Transcription

1 4. Influence of Temerature and Pressure on Chemical Changes Toic: The influence of temerature and ressure on the chemical otential and drive and therefore the behaviour of substances. 4.1 Introduction Until now, the tabular values we have used were the so-called standard values based uon room temerature and standard ressure (298,15 K and 100 kpa). For dissolved substances, the standard concentration is 1 kmol/m 3. U to this oint, our statements about the ossibility of a chemical change have been valid for these conditions only. However, temerature and ressure often have a decisive influence uon the chemical otential and thereby, uon the course of chemical rocesses. Water freezes in the cold and evaorates in the heat. Cooking fat melts in a frying an and udding gels while cooling, ice melts under the blades of ice-skates and butane gas (the fuel of a cigarette lighter) becomes liquid when comressed. The chemical otential μ is not a material constant, but deends uon temerature, ressure, etc. 4.2 Temerature deendence of chemical otential and drive To begin, let us consider as a tyical examle the change with temerature in the chemical otential of table salt μ(nacl). For comarison, the grahic also shows the temerature deendence of the chemical drive for decomosition of table salt into the elements A(NaCl Na Cl 2 ). Examle: μ(nacl) and A(NaCl Na Cl 2 ) It is striking that the chemical otential falls more and more steely with increasing temerature. Excet for a very few excetions of dissolved substances (f.e. Ca 2+ in aqueous solution), all substances exhibit this behaviour. The tendency of a substance to change generally decreases when it is ut into a warmer environment. The chemical drive A(T) which is calculated from the temerature deendent otentials, exhibits a noticeably more linear gradient than the μ(t) curves. The decline in otential aears, at first glance, to contradict the observation that reactions rogress more readily and more quickly at higher temeratures than at lower ones. But it 1

2 should be noted that a higher velocity does not necessarily means a stronger chemical drive. This can be caused by a smaller or even vanishing inhibition as is actually the case in chemical reactions. The strong decrease of inhibition resulting from an increase of warming masks the mostly weak change to the drive A. Moreover, it should be remembered that A is determined by the difference of the chemical otentials of the starting substances and the roducts, and not by the absolute levels of otentials. Since the otentials of the initial substances as well as of the roducts decrease as a result of an increase of temerature, the otential difference which is alone resonsible for the reaction drive does not necessarily decrease. It can remain constant or even increase, as in our examle. In order to describe the decrease of otential, we will be content with a simle aroach at first. For examle, if one wishes to show how the length l of a rod changes with temerature, this can be done with the hel of a temerature coefficient which tells us by how much the length increases when its temerature is changed by 1 K. The increase in length for a temerature increase from an initial value of T 0 to a value of T can be described by a linear equation as long as ΔT = T T 0 is not too large: l = l 0 + ε ΔT. The initial value is reresented by l 0 and ε reresents the temerature coefficient. To indicate the change of chemical otential as a result of warming, we roceed exactly in the same manner: μ = μ 0 + α ΔT. Here, μ 0 characterises the initial value of the chemical otential. This reresents a value at arbitrarily chosen values of temerature T 0, ressure 0 and concentration c 0 (in contrast to the standard value μ ). But standard values often serve as the initial values of a calculation, so that in secial cases, μ 0 = μ. The temerature coefficient α reresents the sloe of the function μ(t) at the oint (T 0 ; μ 0 ), and is therefore almost always negative, as we have seen. For the temerature deendence of the drive A of a chemical change we obtain analogously: B + B +... D + D +... A = A 0 + α ΔT. The temerature coefficient α of the drive can be calculated by the same easy to remember rocedure as the drive itself: α = α(b ) + α(b ) + α(d ) α(d ).... (Remember that A = μ(b ) + μ(b ) +... μ(d ) μ(d )....). If we take standard room conditions as starting oint, the error is about 1 kg for lowmolecular substances for ΔT values of about ±100 K. This aroach remains useful for rough estimates u to ΔT 1000 K and above, although µ(t) falls sharly with rising temerature. This remarkable and (for alications) imortant circumstance is based uon the fact that it is 2

3 initial final not the otentials which are decisive in chemical rocesses, but the drives. When taking the difference A = μ μ, the rogressive contributions of the functions μ(t) largely cancel. If higher recision is desired, the aroach can be easily imroved by adding more terms to the equation: μ = μ 0 + α ΔT + α (ΔT ) 2 + α (ΔT) Of course, there are other ossible aroaches; recirocals for instance, or logarithmic terms. However, we do not wish to go into mathematical refinements of this tye here because it is astounding how far one can actually go with the linear aroach. It is our goal here to show this. The following table shows the chemical otential μ as well as its temerature coefficient α for some substances at standard conditions (298,15 K, 100 kpa, dissolved substances at 1 kmol/m 3 ): Substance Formula µ kg α G/K Iron Fe s Fe l Fe g Grahite C Grahite Diamond C Diamond Iodine I 2 s I 2 l I 2 g I 2 w Water H 2 O s H 2 O l H 2 O g Ammonia NH 3 l NH 3 g NH 3 w Calcium(II) Ca 2+ w Along with the already mentioned basic rule which states that the temerature coefficient α is (almost) always negative, another rule (which almost all substances follow) becomes aarent when the α-values are comared for hase transformations. The temerature coefficient α of the chemical otential of a substance B becomes increasingly negative when the hase changes from solid to liquid and finally to the gaseous state. The jum corresonding to the second transition (reresented by the sign >>) is considerably greater than the one corresonding to the first one. For a substance in an aqueous solution, α is mostly similar to that of the liquid state. The values scatter more strongly, though, so that we cannot easily fit α(b w) into the other α-values: 3

4 0 > α(b s) > α(b l) >> α(b g) α(b w) For clarification, we will single out the values for iodine at standard conditions given in G K 1 from the table above: 260,7 < < 150,4 < 116,1 < ,2 (As we will see in Chater 8, the temerature coefficient α corresonds to the negative molar entroy S m, i.e., α = S m. Anticiating this can hel with remembering the rules more easily: First, in Chater 2, we demonstrated that the molar entroy is always ositive; the negative sign of the temerature coefficient easily results from this. Second, the fact that the molar entroy of a liquid is greater than that of a solid, and the molar entroy of a gas is much greater than that of a liquid, leads to the sequence above (see Section 2.9)). The chemical otential of gases therefore decreases esecially fast with increase in temerature. Their tendency to transform decreases most strongly so that, by comarison to other states, the gaseous state becomes more and more stable. This only means that, as a result of temerature increase, all other states must eventually transform into the gaseous state. At high temeratures, gases ossess the weakest tendency to transform and therefore reresent the most stable form of matter. We will use water to take a closer look at this behaviour. Under standard conditions, the chemical otential of ice, water, and water vaour has the following values: H 2 O s H 2 O l H 2 O g μ / kg One sees here that under these conditions, ice melts, and water vaour condenses because water in its liquid state has the weakest tendency to transform. However, this changes if the temerature is raised or lowered sufficiently. For easy calculation, we will consider a temerature change of ±100 K. The following results are obtained using the linear aroach: H 2 O s H 2 O l H 2 O g α / G K μ(398 K) / kg μ(198 K) / kg We see that at 398 K (125 C), the chemical otential of water vaour has the smallest value and that water vaour must result from the other forms, while at 198 K ( 75 ), ice must develo. This result is reresented grahically in the figure on the right. Taking the ste to calculate the hase transition temeratures now aears obvious: If a substance like lead is solid at normal temerature, this is because its chemical otential has its lowest value in the solid state. The otential of liquid lead must exceed that of solid lead, otherwise, at room temerature, it would be liquid like mercury. We will now visualize this in a diagram (the lowest chemical otential in each case 4

5 is highlighted): μ(pb s) as otential of an element at room temerature (and standard ressure) is equal to zero since this value has been arbitrarily chosen as the zero oint of the μ- scale. Under these conditions, μ(pb l) must lie above this. The chemical otentials decrease with warming. This haens more quickly in the liquid state than in the solid. For this reason, the curves must intersect at some oint, say at the temerature T sl. This T sl is the melting oint of lead because below T sl, the most stable state of lead is the solid state, above T sl, however, it is the liquid state. In order to indicate the hase transformation in question, the symbols for the corresonding aggregation states are inserted. We can calculate the temerature T sl. In order to do this we have to consider the melting rocess Pb s Pb l. T sl is the temerature at which the chemical otentials of solid and liquid hase are equal, μ s = μ l. At this temerature, the two hases are in equilibrium. The temerature deendence of μ is exressed by the linear formula: By transforming this we obtain μ s,0 + α s (T sl T 0 ) = μ l,0 + α l (T sl T 0 ). and resectively (α l α s )(T sl T 0 ) = μ s,0 μ l,0 μ μ A T = T = T α. s,0 l,0 0 sl 0 0 αs αl The derivation is somewhat shortened when the following equivalent of the first equation A = 0 is used as a starting oint for the existence of a state of equilibrium. If the temerature deendence of the driving force is taken into account, we have and therefore A + α ( T T ) = 0 0 sl 0 T 0 sl = T0 A α. Of course, strictly seaking, our result is not accurate because our formula for temerature deendence is only an aroximation. The smaller ΔT (:= T sl T 0 ) is, the more exact the calculated value will be. The melting oint of lead is actually 601 K. Based on the tabulated 5

6 standard values, our calculation yields G T sl = 298 K = 620 K. 1 ( 64,8) ( 71,7) G K The result is surrisingly good for this rough aroximation. We will now comlete our diagram by adding the chemical otential of lead vaour. At room temerature, the chemical otential of vaour lies much higher than that of the liquid hase. However, with rising temerature, μ(pb g) falls rather steely, as is usual in all gases. At some temerature T lg the otential of lead vaour intersects with that of liquid lead. When this temerature is exceeded, the melted lead transforms into vaour because now vaour is the most stable state. T lg is nothing other than the boiling temerature of lead melt. The boiling temerature can be calculated in the same manner as the melting temerature, only now the otentials and their temerature coefficients for liquid and gaseous states will be used. There are substances for which the chemical otential of the vaour is relatively low comared to that of the melt. The otential of the vaour can then intersect that of the solid below the melting oint. This means that there is no temerature (for a given ressure) at which the liquid hase exhibits the lowest chemical otential and is therefore stable. Such substances do not melt when warmed but transform immediately into the vaour state. This henomenon is called sublimation. An excellent examle of such a substance is frozen carbon dioxide which has the characteristic of vaorising without melting. Because of this it is also called dry ice. Sublimation temeratures T sg can be calculated based on the same rocedure as above. Other transformations can be dealt with in the the same way. A good object for demonstration is the already mentioned mercury iodide: HgI 2 yellow HgI 2 red μ / kg α / G K

7 Exeriment: Heating HgI 2 to above 398 K When heated, the temerature coefficient of the yellow form decreases more quickly than that of the red one because α(hgi 2 yellow) < α(hgi 2 red) < 0, so that above a certain temerature, μ(hgi 2 yellow) falls below μ(hgi 2 red), making the yellow form the more stable modification. The temerature of transformation (398 K or 125 C) can be calculated just like the melting oint of lead and can be easily verified by exeriment. Chemists are mostly interested in real chemical reactions. Because the temerature changes in gases have the strongest effect on their otentials, they are what shaes the behaviour of conversions. Processes which roduce more gas than is used u (so-called gas forming reactions) benefit from the strongly negative temerature coefficients α of gases when the temerature rises. In contrast, the chemical drive of a gas binding reaction is weakened by the rise in temerature. Consider the examle of thermal decomosition of silver oxide: 2 Ag 2 O s 4 Ag s + O 2 g μ : 2 ( 11.3) kg A = 22.6 kg α: 2 ( 121) 4 ( 43) 205 G K 1 α = +135 G K 1 The decomosition does not take lace at room temerature due to the negative drive. However, since a gas should be formed, we exect that this rocess begins at a high enough temerature. The minimum temerature T D for the decomosition of Ag 2 O is obtained from the condition that the combined chemical otentials of the initial and final substances must be equal and the chemical drive A changes its sign: In analogy to the equations above, we obtain A = A + α ( T T ) = 0. 0 D 0 T 0 D = T0 A α. Inserting the A and α values which are calculated according to α = 2 α 4 α α results in T D 465 K. Ag2O Ag O2 Exeriment: Annealing of Ag 2 O When the blackish brown silver oxide is heated by a burner, the oxygen that forms can be demonstrated with a glowing slint. White shiny silver metal remains in the test tube. The same rocedure can be used, for examle, to calculate how strongly a comound containing crystal water must be heated in a drying oven in order to dehydrate it. Industrially imortant rocesses such as smelting of iron ore in a blast furnace can also be made accessible to a descrition. If the technical details are left out, a blast furnace can be considered a chemical reactor where iron ore, coal, and oxygen are introduced 7

8 and furnace gas and ig iron exit. If this rocess uses the minimum amount of coal (in the reaction equation simlistically reresented by carbon) Fe 2 O 3 s + 3 C grahite 2 Fe s + 3 CO s μ : ( 137.2) kg A = 329,4 kg α: 87 3 ( 6) 2 ( 27) 3 ( 198) G K 1 α = +543 G K 1 it cannot take lace at room temerature due to its negative chemical drive. However, a gas is formed, so we exect that the reaction should be ossible at higher temeratures. If one wishes to find out if the 700 K in the uer art of the shaft of the furnace is hot enough, the drive must be aroximated according A = A + α ( T T ) 0 0 for this temerature. With a value of 111 kg, the drive is noticeably less negative, i.e., the otential difference between the reactants and roducts has become smaller, but the reaction still cannot take lace. Again, the minimum temerature needed can be aroximated by an equivalent to the equation above: T 0 R = T0 A α. We therefore obtain a value for T R of 900 K. Extra coal is needed for the furnace to reach this temerature. Of course, all of these calculations deend uon access to the necessary data. 4.3 Pressure deendence of chemical otential and drive As reviously stated, the value of the chemical otential of a substance deends not only uon temerature, but uon ressure as well. Moreover, the otential generally increases when the ressure increases. In a small range of ressures, all the curves can be aroximated as linear, comarable to how we described the influence of temerature: μ = μ 0 + β Δ. μ 0 is the starting value of the chemical otential for the initial ressure 0. The ressure coefficient β is almost always ositive. 8

9 Analogously, the ressure deendence of drive A of a reaction results in B + B +... D + D +... A = A 0 + β Δ, where β = β(b ) + β(b ) + β(d ) β (D ) The linear aroach is useful for solid, liquid, but also dissolved substances and for the drives of the corresonding conversions u to Δ 10 5 kpa (= 1000 bar). For obtaining general aroximations, it is useful even u to 10 6 kpa (= bar). In the case of gases and the drives of reactions in which gases articiate, Δ/ < 10 % is considered accetable because the sloe β of the corresonding curve changes relatively strongly with ressure. For greater ranges of ressure Δ, the mass action relation must be alied. We will be introduced to this in Section 4.5. The following table shows the β values for the substances of the table above. Substance Formula µ kg β μg Pa 1 Iron Fe s Fe g Grahite C Grahite Diamond C Diamond Iodine I 2 s I 2 l I 2 g Water H 2 O s H 2 O l H 2 O g Ammonia NH 3 l NH 3 g NH 3 w A rule similar to the one for temerature coefficients is valid for ressure coefficients. It is very useful for qualitative considerations: 0 < β(b s) < β(b l) <<<< β(b g). β(b w) For clarification, we will single out again the values for iodine at standard conditions given this time in μg Pa 1 : 0 < 51,5 < 60,3 <<<< 24, (In this case, as well, there is a relation to a molar quantity, namely the molar volume V m. Therefore, we have β = V m (comare Chater 8). Because all molar volumes are basically ositive, the ressure coefficient always has a ositive sign. The molar volume of a gas is far greater (by a factor of 1000) than that of the condensed hases (liquid and solid). On the other 9

10 hand, the molar volume of a liquid hase is usually greater than that of the solid hase so that the sequence above results). Like any rule, this one has excetions. For instance, the β for some ions in an aqueous solution is negative and sometimes as in the case of water the β in the solid state is greater than in the liquid state. This is exactly the oosite from what the rule would lead us to exect. Raising the ressure generally causes the chemical otential to increase although, as already stated, the increase varies for the different states of aggregation. In the solid state, it is smallest and in the gaseous state, greatest. As a rule, the higher the ressure is, the more stable the solid state is comared to the others and the greater the tendency of the substance to transform to the crystalline state. Conversely, a ressure reduction results in a reference of the gaseous state. Let us once more consider the behaviour of water from this new viewoint. The following table summarizes the necessary chemical otentials and ressure coefficients: H 2 O s H 2 O l H 2 O g μ / kg β / 10 6 μg Pa One sees that lukewarm water can boil at low ressure, although, at room conditions, μ(h 2 O g) > μ(h 2 O l), liquid water is the stable hase. If the ressure is lowered enough by uming the air above the water out of a closed container, μ(h 2 O g) will fall below μ(h 2 O l), at some oint because β is esecially great for the gaseous state. The reduction of ressure becomes noticeable by a strong decrease of chemical otential and the water begins to transform into water vaour by boiling. But low ressure can also be created by using ice-water to cool a closed flask containing only hot water and steam. In the rocess, a art of the vaour condenses, leading to a decrease in ressure. Exeriment: Boiling of warm water at low ressure Exeriment: Boiling by cooling We shall take a closer look at a further examle of the transformation of a substance under ressure. Diamond is a high ressure modification of carbon which should never aear at normal ressure. The most stable modification of carbon, the one with the lowest chemical otential, is grahite which we know from encils. A characteristic of grahite is that its 10

11 chemical otential increases more strongly with ressure than the otential of diamond so that, at one oint, μ(c grahite) should exceed μ(c diamond) making it ossible for diamond to form. At normal ressure and room temerature, μ(c grahite) equals zero because this value has been arbitrarily set as the zero oint of the μ scale. The μ() curve is steeer for grahite than for diamond. Therefore, the two curves must intersect at a ressure αβ, which we will call the transformation ressure. The index αβ indicates that the transformation of a modification α (here grahite) into another modification β (here diamond) is considered. Below αβ, grahite is more stable, above it, diamond is more stable. The ressure αβ can be calculated because αβ is the ressure for which μ α = μ β. The ressure deendence of μ is exressed by a linear relation: resulting in μ α,0 + β α ( αβ 0 ) = μ β,0 + β β( αβ 0 ), and corresondingly ( β β β α)( αβ 0 ) = μ α,0 μ β,0 μ μ A = =. α,0 β,0 0 αβ 0 0 βα ββ ß The exression shows a great formal similarity to the one for determining a transformation temerature whether it alies to a hase change, a decomosition, or something else. Inserting the tabulated values results in αβ kpa (= bar). Strictly seaking, this result cannot be accurate because the linear relations only reresent aroximations. However, as a general tool for orientation, it is quite useful. 4.4 Simultaneous temerature and ressure deendence There is nothing stoing us from exanding our ideas to reactions in which temerature and ressure change simultaneously. In this case the chemical otential can be exressed as follows: μ = μ0 + α ΔT + β Δ. Corresondingly, the chemical drive takes the form A = A0 + α ΔT + β Δ. But also the deendence of transformation temeratures from ressure can be determined by these equations. Here is a familiar examle reresentative of many others. Ice melts under high ressure (if it is not too cold). By nature, the otential of ice equals that of ice-water 11

12 (μ(h 2 O s) = μ(h 2 O l)) at 273 K (0 C) and standard ressure. However, because of β(h 2 O s) > β(h 2 O l), the value of μ(h 2 O s) exceeds that of μ(h 2 O l) as the ressure increases, and the ice begins to melt. This is how a wire loo with a weight hanging from it slowly melts its way through a block of ice. Exeriment: Ice melting under ressure As mentioned, water is among the few excetions where β in the solid state is greater than in the liquid state. This secial characteristic of ice is resonsible for the ability of a glacier to flow downward a few meters er day in a mountain valley like slow moving dough. Where the ice is under esecially high ressure, it melts and becomes liable so that it gradually moves around obstacles. But a block of ice does not totally melt when comressed because it cools down during melting. The chemical otentials increase because of the negative temerature coefficients α. Because of 0 > α(h 2 O s) > α(h 2 O l), the effect is stronger in water than in ice. The otential difference due to excess ressure is comensated and the rocess of melting stos. Again, there is equilibrium between the solid and the liquid hase, but this time at a lower melting oint. Only when the ressure is further increased does the ice continue to melt until additional cooling balances the otentials again. To illustrate this, let us have a look at the figure on the left. If the ressure is increased, the chemical otential of the solid hase as well as that of the liquid hase increase; but this increase is much more ronounced for the solid state than for the liquid one (because of β(b s) > β(b l) > 0). Thus the intersection oint of the curves (T sl ) shifts to the left, i.e. the freezing oint is lowered by the amount ΔT sl. It is easy to calculate the lowering of temerature in comressed ice which is nothing else than the freezing-oint deression of water under ressure. The condition for equilibrium μ s = μ l takes the following form: μ + α ΔT + β Δ = μ + α ΔT + β Δ. s,0 s s l,0 l l If the freezing oint of water at standard ressure (T 0 = 273 K) is chosen as the initial value, then μ s,0 and μ l,0 are equal and dro out of the exression. The following relation remains: β β β T. s l Δ = Δ = Δ αs αl α For Δ = 10 4 kpa (100 bar), the lowering of the freezing oint due to ressure results in ΔT = 0,67 K. 12

13 However, for most substances the melting temerature increases under ressure (because of β(b l) > β(b s) > 0) (see the figure on the right). Corresondingly, the shifts in otentials cause higher ressure to raise the boiling oint and a lower ressure to lower the boiling oint (because of β(b g) >> β(b l) > 0). This is also valid for water as we have seen in exeriments above. Again, the change ΔT can be aroximated with the formula derived above. The value of β for evaoration is roughly 10 4 greater than for melting, whereas the α values do not vary so drastically. Therefore, even small changes of ressure are enough to noticeably shift the boiling oint. To achieve a comarable change of the melting oint, much higher ressures are necessary. A ressure increase of about 10 kpa (0.1 bar) already results in a shift of the boiling oint of water of about +2.0 K, while for a comarable change of the melting oint (ΔT = 2.0 K), a ressure increase of more than kpa (300 kbar) is necessary. We will close this section with a look at our home lanet. The temerature increase towards the middle of the Earth (> 5000 K) causes the iron core to melt. The ressure which grows to kpa, turns it into a solid again at the very center (standard melting and boiling oints of iron are about 1809 K and 3340 K, resectively). This behaviour clearly shows the great influence of temerature and ressure uon the chemical otential. 4.5 Behaviour of gases under ressure As already stated, the chemical otential of gases is esecially sensitive to changes of ressure. For this reason, the ressure coefficient is greater by several owers of ten than those of solid or liquid substances. At the same time, β itself is strongly deendent uon ressure. For these reasons, the linear aroximation is only alicable to a narrow range of ressures (Δ/ < 10 %). This is far too limiting for most alications so a formula must be sought that sans a much wider range of ressures. A look at the tabulated values shows that β has not only a large value but the same value for all gases at standard conditions. Aarently, the ressure coefficient β of gases is a universal quantity. For given T and, it is the same for all gases in all milieus. Moreover, it is directly roortional to the absolute temerature T and indirectly roortional to the ressure of the gas in question. This remarkable fact can be exressed as follows: 13

14 RT β = where R = G K 1. R is a fundamental constant and is the same for all substances. It is called the universal gas constant because it was first discovered in a law valid for gases (Section 9.2). The relation above is based uon the henomenon called mass action in chemistry. We will go more deely into this in the next chater. (Mentioned in assing: β corresonds here to the molar volume of a so-called ideal gas as we will see in section 9.1). Inserting β into the relation above yields the following equation: RT μ = μ0 + Δ. Those roficient in mathematics immediately see that there is a logarithmic relation between μ and : μ= μ0 + RT ln 0. The ressure coefficient β of gases is nothing other than the derivative of the function μ() with resect to. If we take the derivative with resect to of the function above, we retrieve the first equation. {For those interested in mathematics: The equation above can be transformed to result in RT Δμ = Δ. For small (infinitesimal) changes, the relation is RT dμ = d. If we wish to calculate the change of the chemical otential from the initial value μ 0 to the final value μ for a change of ressure from 0 to, we must integrate both sides. The following elementary indefinite integral will serve well for this: Inserting the limits results in: 1 dx = lnx + C. x μ 1 d μ= RT d, μ0 0 and resectively, μ μ0 = RT ln.} 0 14

15 In contrast to the linear aroximation, this still relatively simle logarithmic formula sans the much wider range of ressures from between zero and 10 4 kpa (100 bar). The range of validity will be discussed in more detail in section 5.5. Let us take a closer look at these relations using the examle of butane (the fuel in a gas lighter). The μ() curve of gaseous butane shows the exected logarithmic relationshi. Futhermore, we can see from the figure that, when comressed at room temerature, butane turns into liquid relatively easily. The so-called boiling ressure lg, i.e., the intersecting oint of the otentials for the liquid and the gaseous hase, lies only a little above 200 kpa. This intersecting oint characterises the state of butane in a lighter at room temerature. However, further imortant information can be derived from the figure: The μ() curve for a liquid is an almost horizontal line (its sloe is very small). For this reason, the chemical otential of condensed hases (liquids and solids) can be considered nearly indeendent of ressure in most cases. Furthermore, the chemical otential of a gas continues to decrease with falling ressure. The μ value aroaches negative infinity if the ressure aroaches zero. This leads to the following remarkable conclusions. We can infer, for examle, that calcium carbonate CaCO 3 cannot be stable if the CO 2 ressure in the environment falls to zero. In this case, the chemical otential of CO 2 would have the value. The reaction CaCO 3 s CaO s + CO 2 g μ : 1128,8 603,3 394,4 kg A = 131,1 kg α: G K 1 α = +159 G K 1, which cannot take lace at standard conditions, would have a ositive drive. The sum of otentials on the left would be higher than on the right. However, decomosition roduces CO 2, so that the CO 2 ressure must rise in a closed system. The rocess continues until the CO 2 ressure has reached a value for which the chemical otentials on the left and right sides balance. This CO 2 ressure is called the decomosition ressure of calcium carbonate. The decomosition ressure can be easily calculated. If the chemical otentials satisfy μ = μ + μ, CaCO CaO 3 CO 2 we have equilibrium. We ignore the ressure deendence of solid substances because, in comarison to gases, it is smaller by three orders of magnitude. We only take the deendence for CO 2 into account: μ = + + CaCO,0 μcao,0 μco,0 RT ln

16 This results in μ = CaCO,0 μ CaO,0 μ CO,0 RT ln A 0 and A0 ex RT = ex ln 0 or, resectively, 0 = 0 ex A. RT In order to calculate the decomosition ressure for a temerature different from standard temerature, the μ values in the exonents only need to be converted to the new temerature. The linear formula for temerature deendence is generally good enough for this: A + α ( T T ). RT 0 0 = 0 ex With the hel of corresonding data, the (T) curve can be determined. This curve gives the decomosition ressure as a function of temerature: ( T / K 298) = 100 kpa ex T / K. 16