19-Chemical Equilibria - 5.111 Lecture 19 CHEMICAL...

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5.111 Lecture # 19. CHEMICAL EQUILIBRIA [Chapter 9 from the “Chemical Principles” textbook, 4th edition, by Peter Atkins & Loretta Jones, Freeman, New York, 2008] Consider the chemical process aA+bB<—>cC+dD (1) where A, B, C, and D are reactants/products; and a, b, c, and d are their respective stoichiometric coefficients. . Reversibility. Forward and reverse reactions within process (1). If we mix A and B, with no C and D initially present (i.e., initially [C] = [D] = 0), then in the beginning the forward reaction will dominate, with the reverse reaction being insignificant. However, as the forward reaction progresses, it slows down because the reactants (A and B) become depleted; in contrast, the reverse reaction picks up due to build—up of C and D. Eventually, Rateforward = 1{actereverse and we say that the system is at equilibrium. Dynamic (as opposed to static) nature of this equilibrium. No net change in the composition of the reaction mixture. [Slide 19.1] The Law of Mass Action. At equilibrium in process (1): K=([C]C[D]d)/([A1“[B]b> (2) where K is the eguilibrium constant. Units of K. K is independent of [A], [B], [C], and [D]; however, K is dependent on some other process parameters, notably on the temperature. Experimental evidence for the law of mass action. K values have been measured and tabulated for numerous reactions at different temperatures. At a given temperature, they vary by many orders of magnitude depending on the reaction. Now consider the chemical process cC+dD<—>aA+bB (3) which is process (1) reversed. Its equilibrium constant, K ', is K’ = ([A]“[B]b)/ ([C]C[D]d) (4) 3260.... 535:33 5:323”. Thus K=<K9" (5) It was implicitly assumed that in chemical reactions (1) and (3), all reactants/products were in the same phase, i.e., all (g) or all (1). Such equilibria are called homogeneous. What if they are not, for instance, some are (l) and some are (s)? Heterogeneous equilibria. Since molar concentrations of a pure liquid or a pure solid are constant, they (in contrast to amounts) do not change during a reaction. Therefore, they are ignored in equilibrium constants calculations. For example, Ca(OH)2 (s) <—->~ Ca2+(aq) + 2 OH' (aq) K: [Ca2+][OH']2 (6) 01' Ni (S) + 4C0 (g) H Ni(C0)4(g) K=[Ni(CO)4]/[CO]4 (7) What does the value of an equilibrium constant tell us? 0 The extent of reaction: consider equation (2) above. [Slide 19.2] Reactants Products — _'-_~—-—. v--“> -_-_ _““ “ l — _ ‘—“-“ “1.. .Ili' HWIIIIIIIIIQ MWIIII ; , 35 WWI-II MW..- HWII: I I MW.- ml- 'II KMIIIIIIIII UWIIIIIIIII. qWIIIIIIIII WIIIIIIIII will-I ‘ I k EMill-III WIIII fill-l WWII-III... NWIIIIIIIII “W::=:: I M $IIIIIIIII 'I-IIIIIII K=1o3 K=10'3 o The direction of reaction: The reaction quotient, Q. Defined the same way as K in equation (2) above, except that [A], [B], [C], and [D] are at any stage of the reaction, not necessarily at equilibrium as for K. Comparing the values of Q and K [Slide 19.3]: if Q > K, then the reverse reaction tends to dominate; If Q < K, then the forward reaction tends to dominate. 0 Note that the value of K says nothing about the rate of reaching equilibrium. 0 It allows calculating the equilibrium composition of a reaction mixture. Consider the reaction: A <——> B (i.e., a = b = 1) having the equilibrium constant K under given conditions. If the initial concentrations are [A]0 and [B]o, then at equilibrium [A] + [B] = [A]o + [B]o and [B]/ [A] 2K Therefore, at equilibrium, [A]=([A]o+[B]o)/(1+K) and [B]=K([A]o+[B]o)/(1+K) How do chemical equilibria respond to changes in the experimental conditions? E:_5___:_um an p.333: 32:99:. 5.3 3 3:3 3:33am 3:339. 35 ouanOuu—u 3 3:3 anus—yo...“— Le Chatelier’s principle: VWien a stress is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the extent of the stress. For example, consider reaction (1) at equilibrium aA+bB<—>cC+dD What would be the effect of increasing/decreasing the concentration of a reactant or a product according to Le Chatelier’s principle? Explanation based on Q and K. Product removal as a process enhancement strategy. Other common stresses in Le Chatelier’s principle: 0 Pressure Consider the reaction N2 (g) + 3 H2 (g) <—> 2 NHz (g) The effect of increasing/decreasing the pressure. Explanation based on Q and K. 0 Temperature Endothermic and exothermic reactions. The effect of increasing/decreasing the temperature. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern