Beng 130 Lecture 12 & 13

Beng 130 Lecture 12 & 13 - Equilibrium: Mass Action...

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Free Energy and Chemical Equilibria BENG 130 Lecture ¡ When equilibrium is establish A ' B ¢ That is rate forward = rate reverse or K f [A] = K b [B] ¢ Rearranging this equation can be rearrange to K f /K b = [B] / [A] ¢ The mass action expression : K eq = [B] / [A] ¡ For any generic chemical process at equilibrium ¡ aA + bB ' pP + qQ ¢ A mass action expression can be written: ¢ This is also referred to as the Law of Mass Action Law of Mass Action Equilibrium: Mass Action Expression k eq A > @ a •B > @ b P > @ p •Q > @ q ¡ N 2 (g) + 3H 2 (g) #'# 2NH 3 (g) (not 100 % process) ¡ As soon as NH 3 is form, it back reacts and form N 2 and H 2 . Law of Mass Action: k eq NH 3 > @ 2 N 2 > @ •H 2 > @ 3 K c Example: Law of Mass Action Write the expressions for K c and K p for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous: a) 3NO (g) ' N 2 O (g) + NO 2 (g) b) CH 4 (g) + 2H 2 S (g) ' CS 2 (g) + 4H 2 (g) c) Ni(CO) 4 (g) ' Ni (s) + 4CO (g) d) Fe 2 O 3 (s) + 3H 2 (g) ' 2Fe (s) + 3H 2 O (g) b) K p = P CS 2 p H 2 +, 4 p H 2 S 2 p CH 4 ; K c = CS 2 >@ 2 4 H 2 S 2 • CH 4 d) K p = p H 2 O + , 3 p H 2 + , 3 ;K c = H 2 O > @ 3 H 2 > @ 3
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Which is favored: Reactant or Product? For K eq > 1, the ratio [Product] [Product] x > [Reactant] > [Reactant] y For K eq = 1, the ratio [Product] x = [Reactant] = [Reactant] y For K eq < 1, the ratio [Product] x < [Reactant] < y For K eq > 1, at equilibrium Product is favored Product is favored. For K eq = 1, at equilibrium Product and Reactant are equal . For K eq < 1, at equilibrium Reactant is favored. Meaning of K eq k eq A >@ A •B b P p •Q q Product x Reactant y The Reverse of a Reaction; K eq relationship What is the equilibrium constant if a reaction is written in the reverse direction? instead of: COCl 2(g ) ' CO (g) + Cl 2 (g) (i) it is : CO (g) + Cl 2 (g) ' COCl ) (ii) What is the relationship in K eq between (i) and (ii) K eq (i) [CO] [Cl 2 ] [COCl 2 ] K eq (ii) [COCl 2 ] [CO] [Cl 2 ] [CO] [Cl 2 ] [COCl 2 ] [COCl 2 ] [CO] [Cl 2 ] ª# ¬# «# º# ¼# »# 0 1 K eq (i)= K eq (ii) -1 K forward 1 K reverse Effects of K eq and Variation of a Chemical Equation Consider the variation of a chemical reaction: A ' B K eq = [B] / [A] = [B] / [A] B ' A K rev rev = [A] / [B] = [A] / [B] K rev rev = 1 / K eq eq 1/2 A ' 1/2 B K’ = [B] = [B] 1/2 1/2 / [A] 1/2 1/2 K’ =( K eq ) 1/2 2B 2B ' 2A 2A K” = [B] = [B] 2 / [A] / [A] 2 K ” = ( K eq ) 2 Note: K rev rev = 1 / K eq eq or or K eq eq = 1 / K rev rev K’ K eq ) 1/2 or K eq = K’ 2 K ” = ( K eq eq ) 2 or or K eq eq = ( K ” ) 1/2 Example 1: K eq relationship The equilibrium constant for the reaction: 2SO 3(g) ' 2SO 2 (g) + O 2 (g) K c = 2.4 •10 -3 at 700°C. a) Calculate K c for: 2SO 2 (g) + O 2 (g) '# 2SO 3(g) b) Does the equilibrium favor SO 2 and O 2 , or does it favor SO 3 at this temperature ? a) K rev = 1 Kc p SO 3 + , 2 p SO 2 +, 2 p O 2 1 2.4 •10 0 3 416.7 b) K rev >> 1; Product Favored
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Adding Chemical Reactions How is K eq related to a series of chemical equations added: 1. 2NO 2(g) ' N 2 O 4(g) K 1 2. N 2 O 4 (g) + O 2(g) ' 2NO 3(g) K 2 3. 2NO 2(g) + O 2(g) ' 2NO 3(g) K 3 How is K eq related to a series of chemical equations added: 1. 2NO 2(g) ' N 2 O 4(g) K 1 2. N 2 O 4 (g) + O 2(g) ' 2NO 3(g) K 2 3. 2NO 2(g) + O 2(g) ' 2NO 3(g) K 3 K 1 [N 2 O 4 ] [NO 2 ] 2 K 2 [NO 3 ] 2 [N 2 O 4 ] [O 2 ] K 1 K 2 2 O 4 ] [NO 2 ] 2 [NO 3 ] 2 [N 2 O 4 ] [O 2 ] K 1 •K 2 [NO 3 ] 2 [NO 2 ] 2 [O 2 ] K 3 Heterogeneous Systems ¡
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Beng 130 Lecture 12 &amp; 13 - Equilibrium: Mass Action...

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