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AU - CHEM 1040 - Week of 2/19 notes - Class Notes

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AU - CHEM 1040 - Week of 2/19 notes - Class Notes

School: Auburn University
Department: Chemistry
Course: Fundamental Chemistry II
Professor: Michael Squillacote
Term: Fall 2015
Tags: Chemistry
Name: Week of 2/19 notes
Description: Everything we covered in lecture, but explained more thoroughly. all the problems are answered
Uploaded: 02/25/2018
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background image Magnitude of K c K c  large: “lies to the right” or “favors products” o Ex: 2H 2  (g) + O 2  (g)   2H 2 O (g)                     K c  = 3 x 10 81  (at 25 C) K c  small: “lies to the left” or “favors reactants” o 2CO 2  (g)   2CO (g) + O 2  (g)                          K c  = 3 x 10 ­91   K c  in between: the equilibrium mixture will contain comparable amounts of both  reactants and products Problem 2:  For the equilibrium, N 2  (g) + 2O 2  (g)   2NO 2  (g),  K c  = 1.4 x 10 ­10  at 1000 K, R=0.082061 (L x atm)/ (K x mol) a. Write the equilibrium expression (K c ) b. Calculate the value for K P c. Does the equilibrium favor the formation of NO 2  or N 2  and O 2  at this temperature? (answers at the end of this doc) Manipulating the Equilibrium Constant 1. If the equation is reversed, invert the equilibrium constant a. A(g) + B(g)   2C (g)       K ca b. 2C(g)   A(g) + B(g)        K cb  = 1/K ca 2. If the coefficients in the reaction are multiplied by a factor, K c1  is raised to a power equal  to that same factor a. A(g) + B(g)   2C (g)       K ca b. 2A(g) + 2B(g)   4C (g)   K cb  = (K ca ) 2 3. The equilibrium constant for a net reaction made up of 2 or more reactions is the product  of the equilibrium constants for the individual reactions a. A(g) + B(g)   C (g) + D(g)         K ca b. C(g) + F(g)   G(g) + A(g)          K cb  Overall: B(g) + F(g)   D(g) + G(g)           K overall  = (K ca )(K cb ) Problem 3: 
The following reactions have the indicated equilibrium constants at 700K,
K c1  = 54.0 for H 2 (g) + I 2 (g)   2HI(g) K c2  = 1.04 x 10 ­4  for N 2 (g) + 3H 2 (g)   2NH 3 (g) What is the K c  for: a. 2HI(g)   H 2 (g) + I 2 (g)
background image b. NH 3 (g)   ½ N 2 (g) + 3/2 H 2 (g) c. 2NH 3 (g) + 3I 2 (g)   6HI(g) + N 2 (g) K, Q, and the Direction of a Reaction Consider: H 2 (g) + I 2 (g)   2HI(g)                   K c  = 54.3  0.243 mol of H 2 , 0.146 mol of I 2 , 1.98 mol of HI in a 1.00 L container, would more HI be formed or consumed for the reaction to reach an equilibrium? Q = [HI]     2             =      (1.98)     2                           = 110.      [H 2 ][I 2 ]               (0.243)(0.146) HI needs to be consumed since the reaction right now is favoring the right, so there needs to be a 
shift, or yield, left.
Q>K, the reaction proceeds from right to left Q<K, the reaction proceeds from left to right Q = K, the reaction is at equilibrium Heterogeneous Equilibria Equilibria in which the products and reactants are in different phases The concentrations of solids and liquids DO NOT appear in the equilibrium expression o PbCl 2 (s)   Pb 2+ (aq) + 2Cl ­ (aq) K c  = [Pb 2+ ][Cl ­ ] Le Chatlier’s Principle A system at equilibrium responds to a stress in such a way that it relieves that stress Stress: o Addition/removal of a reactant/product o Pressure change o Temperature change (alters the K value) Addition/Removal Effect N 2 (g) + 3H 2 (g)   2NH 3 (g) K c  = [NH 3 ]   2                 [N 2 ][H 2 ] 3 Removal of NH 3 : shifts right Addition of NH 3 : shifts left Addition of N 2 : shifts right Removal of H 2 : shifts left

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School: Auburn University
Department: Chemistry
Course: Fundamental Chemistry II
Professor: Michael Squillacote
Term: Fall 2015
Tags: Chemistry
Name: Week of 2/19 notes
Description: Everything we covered in lecture, but explained more thoroughly. all the problems are answered
Uploaded: 02/25/2018
4 Pages 46 Views 36 Unlocks
  • Better Grades Guarantee
  • 24/7 Homework help
  • Notes, Study Guides, Flashcards + More!
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