4. CHEMICAL REACTION EQUILIBRIA

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Jade McCormick
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1 4. CHEMICAL REACION EQUILIBRIA 4.1. Introduction In materials processing or materials in service, chemical reactions take place. For example, in the production of steel, liquid iron containing impurities are subjected a reaction with gaseous oxygen. When a material is exposed to an atmosphere, it may be capable of reacting with other constituents of the environment Chemical Reaction Equilibrium Consider a general equilibrium, aa + bb = cc + dd Reactants Products Under constant and P G = 0 (Criteria for equilibrium) G = Gproducts - Greactants = cg C + dg D - ag A - bg B For condensed phases: o Gi = G i + R ln ai For gases, o Gi = G i + R ln Pi herefore, activity is replaced by partial pressure for gases; hen G = cg C o + cr ln a C + dg D o + dr ln a D - ag A o - ar ln a A - bg B o - br ln a B G = G + R ln a a c C a A a a d D b B

2 Activity can be replaced by partial pressure for gases. For pure condensed phases activity can be taken as unity. his equation can be written as: G = G + R ln Q Q is called reaction quotient. G = f() tabulated Since, G = H - S at any temperature G = H o Cp d - ( S o Cp = a + b + c-2 Cp = a + b + c-2 where a = aprod - areact Cp d ) the same way b and c are differences between b and c constants of the products and the reactants. a b c G = H o ( a + b + c -2) d - ( S o G = H o b ( a 2 2 c / ( S o c ( a ln b 2 Replacement of the upper and the lower limits yields b c G = I o + I 1 - a ln where, I o = H o b ( a c/ 298) 2 I 1 = a - S o c a ln 298 b d ) 2 ) 298

3 4.3. Oxidation of Metals One important equilibrium; between condensed phases (metals and oxides) and a gaseous phase (oxygen containing) is oxidation of metals. Consider the oxidation of copper 4 Cu(s) + O2(g) = 2 Cu2O(s) H o 298 = J S o 298 = J/K Cp = x x105-2 J/K Using above values Io = I1 = G o, purely from thermochemical data is G o = ln - 9.2x x105-1 J (a) If pure Cu reacting with oxygen to form pure Cu 2 O; G o = R ln P O 2 (eqm') Experimental variation of G o with can be calculated from the measured oxygen partial pressure, P O 2 (eqm') that is in equilibrium with Cu and Cu 2O. When experimental G o vs. is fit to G o = A + B log + C for oxidation of copper G o = log (J) (b) he comparison of G o calculated from (a) and (b) is shown in able. he difference in the temperature range 400 to 1200 K is between 293 to 794 J.

4 2-term fit G o = A + B G o = (J) (c) he comparison of G o calculated from (c) with (b) is given in able. he difference between (b) and (c) in the 400 to 1200 K temperature range is between 286 to 788 J. able (K) G o (a) G o (b) G o (c) (a-b) (b-c) (a-c) able shows that overall difference between G o calculated from (a) and (c) in the above temperature range is at most 1.6 kj. his difference is within the experimental error for most of the G o measurements. For oxidation and sulfidation reactions, in general, the variation of G o with temperature could be approximated to linear forms Ellingham Diagrams Ellingham plotted the experimentally determined G o - for the oxidation and sulfidation of a series of metals. In spite of the terms ln, 2 and -1, the general forms of the relationships approximated to straight lines over temperature ranges in which no change in physical state occurred. Linear line representation of G o on these diagrams can be expressed by G o = A + B

5 Since, G o = H - S A, which is the intercept at 0 K, is the temperature independent H B, the slope, - S All these imply that Cp 0, for oxidation reactions. Another important characteristic of Ellingham Diagram is that; all the lines represent reactions involving one mole of oxygen. 2x/y M + O 2 = 2/y M x O y herefore, the ordinate ( G ) of all the oxidation reactions become R ln P ( eqm ) O2 Almost all the G lines have positive slopes. Slope = - S ; S < 0 For oxidation M(s) + O2(g) = MO2(s) o o o S = S S S MO2 M O2 Generally in the temperature range where M and MO2 are solid S O o2 is dominant. herefore, S -S O o2 Since all the standard entropies are positive, slopes are also positive. In addition, in the temperature range where metal and oxide are solid, the slopes are approximately equal. So, in this temperature range, almost all the lines are parallel to each other. Examples: Using standard entropies at 298 K S O = J/mol.K o2 Ni(s) : 29.8; NiO(s): Ni(s) + O2(g) = 2 NiO(s) S = J/K Sn(s): 51.49; SnO2(s) : Sn(s) + O2(g) = SnO2(s) S = J/K

6 he exceptions to above statements are: C(s) + O2(g) = CO2(g) S 0 and 2 C(s) + O2(g) = 2 CO(g) S > 0

7 Effect of Phase ransformation Lines on Ellingham diagrams often have sharp breaks in them. hese sharp breaks are caused by phase transformations. he straight line representation of G o vs. relationships are valid if there is no physical change taking place for any one of the components taking part in the reaction equilibrium. Straight line indicates that G o = A + B A = H emperature independent B = - S emperature independent which implies that Cp = 0 for oxidation reactions. Consider the transformation of M(s) to M(l) at m(m) H m(m) 0 S Since, G o - S = H - S m(m)

8 M(l) 0 M(s) m(m) G M(l) M(s) m(m) he net effect of phase transformation of a reactant from a low temperature to a high temperature phase is increase in slope. Now, consider the transformation of product MO 2 (s) to MO 2 (l) at m(mo 2 ) H m(mo) 0 S m(mo) Since, G o = H - S

9 - S MO(s) MO(l) 0 G m(mo) MO(l) MO(s) m(mo) he net effect of phase transformation of a product from a low temperature to a high temperature phase is decrease in slope. Effect of evaporation on G vs. plots is similar to melting, but more pronounced. his is because Hv is about 10 Hm. Of course, slope changes for allotropic transformations will be less than that for melting. G M(l) M(g) M(s) m(m) b(m)

10 Above figure illustrates the effects of melting and evaporation of a reactant metal on Ellingham diagram. It can be easily seen from the applicationof subtraction, that M(s) + O 2 (g) = MO 2 (s) H 1 M(l) + O 2 (g) = MO 2 (s) H 2 M(s) = M(l) Hm = H 2 - H 1 When there are more than one phase transformations;

11 Stability of Oxides he Ellingham line for the M/MO2 equilibrium is shown in figure. G o ( MO 2,O 2 ) M MO 2 ( M, O 2 ) All along the line G = 0 and oxygen partial pressure is P(O2)(eqm'). When a metal is exposed to an oxidizing atmosphere, let the partial pressure of oxygen in this environment as PO2(actual). hen

12 o G G R ln P O2 1 ( actual) hen, G R ln PO ( eqm ) R ln PO ( actual) 2 2 At any point above the line P ( actual)> P eqm O2 O2 ( ) hen, G < 0. his implies that, MO2 formation is spontaneous. herefore, MO2 is stable. At any point below the line P ( actual)< P eqm O2 O2 ( ) hen, G > 0. his implies that, MO2 formation is impossible. herefore, M is stable. Stability regions for M and MO2 are indicated in the figure. M-MO2-O2 coexist together on the line Relative Stabilities of Oxides Most of the lines on Ellingham diagram are almost parallel to each other. Consider two lines for M/MO and N/NO equilibria. Stability regions for M, N, MO and NO are indicated in figure. G o ( MO, NO) M MO N ( M, NO ) ( M, N ) NO he oxide whose region of stability is larger is more stable. From the figure, it can be said that NO is relatively more stable than MO. he element whose oxide is more stable is more reactive. he element of the less stable oxide is more stable in elemental form. M is more stable than N.

13 Oxidation-Reduction Reactions: MO can be reduced by N Ellingham lines are not always parallel to each other, sometimes they intercept each other. Consider the two lines for X/XO and Y/YO equilibria. X, Y, XO and YO stability regions are indicated in figure. G o ( XO, YO ) ( Y, XO ) X XO ( X, YO ) ( X, Y ) Y YO E In this case relative stability changes with temperature. Below E, YO is more stable than XO, but above this temperature XO becomes more stable. At the temperature E, all the components, X, Y, XO and YO could coexist. Another important information that could be seen here is, no stability region for Y and XO together below E and for X and YO together above E. his indicates that if Y and XO are brought together at a temperature below E, they are not stable together; therefore Y will be oxidized while XO is being reduced.

Oxygen Partial Pressure Grid Lines

Oxygen Partial Pressure Grid Lines 4.4.4. Oxygen artial ressure Grid Lines When values of G for oxides are displayed, the Ellingham diagram offers a simple and useful way to estimate equilibrium oxygen pressures as a function of temperature.