Oxygen Partial Pressure Grid Lines

Transcription

1 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. his is because of the ordinate being the G = R ln ( eqm ) O for oxidation reactions. hus for constant O values, G vs. is represented by straight lines with R ln O slope and G = 0 intercept. If these lines are superimposed on Ellingham diagram, constant oxygen partial pressures can be read from the oxygen partial pressure scale provided on Ellingham diagrams. herefore, along any straight line that passes through the origin of Ellingham diagram, there is a constant oxygen partial pressure and it can be read from the O scale. he intersections of the constant oxygen partial pressure lines and the copper-cuprous oxide equilibrium line give the equilibrium oxygen pressures for this reaction at various temperatures. For example, at 700 o C, the oxygen pressure in equilibrium with CuO and Cu is atm. 0 O 1 atm. /\G o CuO 0(K)

2 Figure Superposition of equilibrium oxygen pressure grid lines on Ellingham diagram. Cu-O-CuO equilibrium at 700 o C is used to illustrate the use of (O) scale he Oxides of Carbon Carbon is extensively used in materials processing. here are two oxides of carbon, and, both are gaseous substances. herefore, it is important to know how these oxides are distributed in an environment containing these gases. Ellingham lines for these oxides yield, another line that represent the equilibrium between and. / C(s) + O(g) = (g) G = (J) - C(s) + O(g) = (g) G = (J) (g) + O(g) = (g) G = (J) ( (cal)) hen, on the Ellingham diagram three lines are represented as shown in figure. G / K

3 he point of intercept of three lines represent --O equilibrium when and are at 1 atmosphere pressures / Ratio Grid Lines Oxygen partial pressure in a system can also be controlled using the --O equilibrium. From the reaction among these three species: G = - R ln O G = R ln O - R ln R ln O = G - R ln Since, G = (J) or ( (cal)) R ln = ( R ln O ) (cal) Since, R ln O is the ordinate of the Ellingham diagram, constant / ratios fall on straight lines on Ellingham diagram. hese straight lines have intercepts at G = J ( calories) (point C) and slopes of: R ln J ( R ln calories) herefore, a / ratio scale is provided on Ellingham diagrams. Along a straight line that passes through cal. at zero Kelvin on Ellingham diagram, ()/() ratio is constant and can be read from the / ratio scale.

4 0 / <1 O /\G o / >1 0 K Example: Equilibrium in - -O mixture

5 H/HO Ratio Grid Lines Ellingham diagram takes advantage of the relationship between equilibrium partial pressures of H and HO to develop H/HO ratio scale. From H-HO-O equilibrium, that is H(g) + O(g) = HO(g) G= ( ) = (cal) G = - R ln HO H O G = R ln O - R ln HO H H R ln = G - R ln O H O Since, G = (J) or ( (cal)) R ln = (6. - R ln H O ) (cal) HO Since, R ln O is the ordinate of the Ellingham diagram, constant H/HO ratios fall on straight lines on Ellingham diagram. hese straight lines have intercepts at G o = J ( calories) (point H) and H H slopes of: ( R ln J (6. - R ln cal). HO HO herefore, an H/HO ratio scale is provided on Ellingham diagrams. Along a straight line that passes through cal. at zero Kelvin on Ellingham diagram, (H)/(HO) ratio is constant and can be read from the H/HO ratio scale.

6 0 H /H O <1 O C /\G o H /H O >1 1 0 K 4.5. Effect of emperature on Equilibrium At equilibrium, G = - R ln K o G = -S From Gibbs-Helmholtz equation o G H o = - Replacing G = - R ln K, R ln K H o = - ln K H o or = R or from, ln K = -H/R + S/R

7 ln K or 1 H o = - R ln K slope>o /\H<0 1/ slope<0 /\H>0 hen, if H> 0, endothermic, then K if H< 0, exothermic, then K 1/, K increases with decreasing temperature. he variation of K with temperature can be deduced from consideration of Le Chatelier's principle; If heat is added to the system, then equilibrium is displaced in that direction which involves the absorption of heat Efffect of ressure on Equilibrium Although equilibrium constant is independent of pressure, Le Chatelier's principle gives that an increase in total pressure at constant temperature will shift the equilibrium in that direction which involves a decrease in the number of moles in the system. For a general equilibrium a A(g) + b B(g) = c C(g) + d D(g)

8 K = [(C)c (D)d]/[(A)a (B)b] K = [(XC )c (XD )d]/[(xa )a (XB )b] or K = Kx (c+d-a-b) K is independent of pressure, but it is made up of two terms; Kx and the pressure term. Depending on the values of a, b, c and d, change of pressure may have effect on Kx, qoutient of mole fractions. If c+d>a+b, increasing pressure decreases Kx. If c+d=a+b, pressure does not have any effect on Kx. If c+d<a+b, Kx is proportional with pressure Effect of Composition on Equilibrium Reaction Equilibria in Systems Containing Components in Condensed Solutions Consider the general equilibrium a A + b B = c C + d D occuring at the temperature and pressure. If none of the above components are in their standard states, then G = G + R ln a a c C a A a a d D b B Consider the oxidation of a solid M according to: M(s) + O (g) = MO (s) When all the components are at their standard states; G = G = A + B his is one of the lines on Ellingham diagram. When only M is in solid solution with a M, G is now given by G = G - R ln a M since a M < 1; G > G When only MO is in solution, then G = G + R ln a MO a MO < 1; G < G

9 a M <1 ΔG o ΔG a MO <1 o C