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John Alexander University of Cincinnati Cincinnati, OH 45221
The Right Shift? A Problem in Chemical Equilibrium João C. M. Paiva* Department of Chemistry, University of Porto, R. Campo Alegre 687, 4169-007 Porto, Portugal;
[email protected] Victor M. S. Gil Department of Chemistry, University of Coimbra, Portugal António Ferrer Correia Department of Chemistry, University of Aveiro, Portugal
The Problem The understanding of chemical equilibrium is sometimes clouded by uncritical application of concepts and mechanization (particularly in the case of numerical calculations) (1, 2). With this in mind, we devised the following problem appropriate to a physical chemistry course or to an honors general chemistry course. At this level, a surprising result is obtained, which can be explored to promote critical thinking and to acknowledge the importance of reference states when Kc and Kp are being used. Here is the problem. Consider a generic chemical reaction A(s) + 2B(g) C(s) for which Kp = 1.0 at 298 K and Kp = 0.36 at 600 K. 1. From the effect of temperature on the value of Kp do you expect the reaction to be exothermic or endothermic? 2. Apply the van’t Hoff equation, ln Kp = ᎑∆H°/RT + ∆S°/R, to find that ∆H° = ᎑5.0 kJ/mol (of A or C). 3. Using Kc = Kp(RT )᎑∆ng, show that Kc,600 K/Kc,298 K = 1.5 and compare with Kp,600 K/Kp,298 K = 0.36/1.0 = 0.36. How can the former ratio be larger than 1 whereas the latter is smaller than 1? 4a. Taking account only the change of gas pressure with temperature, show that p B600 K /p B298 K = 2.0. 4b. In reality, the equilibrium position will also shift. Show that p B e,600 K /p B e,298 K = 1.7. 5. On the basis of 3 and 4, conclude that, on going from 298 K to 600 K at constant volume, the equilibrium is shifted to the right. 6. Comment on this conclusion, which seems strange considering that the reaction is exothermic. Would you expect the same result if Kp varied more drastically with T (a larger magnitude of ∆H°)?
obtained from Kp,600 K /Kp,298 K together with the expression for Kp. Step 5 considers either the result Kc,600 K /Kc,298 K > 1 or the fact that pBe,600 K /pBe,298 K < pB600 K /pB298 K to conclude that the equilibrium shifts to the right on going from 298 K to 600 K. Finally, in step 6 such a conclusion conflicts with the generic statement that equilibria shift to the left when the temperature is increased if the reaction is exothermic. This same step offers a rationalization for such a deviation from the expected result (small absolute values for ∆H °). It has been stressed (3) that if ∆H ° is small, Kp and Kc can vary in opposite ways with temperature (except if ∆ng = 0). This stems from the use of two different reference states for the equilibrium constants: 1 atm for Kp and 1 mol dm᎑3 for Kc. Should the reference state for Kc be the same as for Kp (1.0 atm is equivalent to 0.041 mol dm᎑3 at 298 K and to 0.020 mol dm᎑3 at 600 K), then Kc = Kp for any temperature. For example, in the expression for Kc, the value to be used instead of 0.041 corresponding to the concentration of B at 298 K (pB = 1.0 atm) would be 0.041 atm/0.041 atm = 1.0 (activity of B); hence Kc = 1.0. Similarly, for 600 K ( pB = 1.7 atm), instead of 0.034 it would be 0.034 atm/0.020 atm = 1.7 (activity of B); hence Kc = 0.35 (which is the same as Kp, considering the number of significant figures used). Otherwise, by referring ∆H ° to a reference state of 1 atm, the way Kc varies with T is given by (see, e.g., ref 4 ) d(ln Kc)/dT = (∆H ° – RT∆ng)/RT 2 to be compared with (van’t Hoff equation) d(ln KP)/dT = ∆H °/RT 2 The same discussion would apply should we use the SI unit for pressure (bar instead of atm), in spite of small changes in the equilibrium constants (5). Acnowledgments
Discussion Step 1 addresses a basic piece of knowledge. The answer is confirmed quantitatively in 2 by making use of the van’t Hoff equation, which is given to show that no emphasis is placed on memory. A similar approach is adopted in step 3 to calculate Kc from Kp. The comparison required in 3 can be made at different levels: (i) the ratios are not the same, (ii) the ratio Kc,600 K /Kc,298 K is larger than 1, whereas Kc,600 K /Kc,298 K is smaller than 1, and (iii) Kc increases with temperature whereas Kp decreases. In step 4 the ratio pB600 K /pB298 K is calculated using the ideal gas equation and the ratio pBe,600 K /pBe,298 K is to be
We thank the referees for their helpful suggestions. Literature Cited 1. Gil, Victor M. S.; Paiva, J. C. Chem. Educator 1999, 4 (4), 128–130. 2. Huddle, Benjamin P. J. Chem. Educ. 1998, 75, 1175. 3. Allsop, E.; George, N. Educ. Chem. 1984, 3, 54. 4. Glasstone, S. Thermodynamics for Chemists; Van Nostrand: New York, 1952; p 289. 5. Treptow, R. S. J. Chem. Educ. 1999, 76, 212.
JChemEd.chem.wisc.edu • Vol. 79 No. 5 May 2002 • Journal of Chemical Education
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