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### Week7

Course: C 112, Fall 2009
School: Allegheny
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7 Week Solutions, p 1 Chem. 112 S09 Dr. Guldan Chemistry 112 Week 7 Solutions: 9.45 (a) The balanced equation is Cl2 (g) ( 2 Cl(g). The initial concentration of Cl2 (g) is Concentration (mol L-1 ) initial change equilibrium KC = 0.0020 mol Cl 2 = 0.0010 mol L-1. 2.0 L 2 Cl(g) 0 +2 x +2 x Cl2 (g) ( 0.0010 -x 0.0010 - x [Cl]2 (2 x ) 2 = = 1.2 10-7 [Cl 2 ] (0.0010 - x ) 4 x 2 = (1.2 10-7 ) (0.0010 x) - 4 x 2 +...

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7 Week Solutions, p 1 Chem. 112 S09 Dr. Guldan Chemistry 112 Week 7 Solutions: 9.45 (a) The balanced equation is Cl2 (g) ( 2 Cl(g). The initial concentration of Cl2 (g) is Concentration (mol L-1 ) initial change equilibrium KC = 0.0020 mol Cl 2 = 0.0010 mol L-1. 2.0 L 2 Cl(g) 0 +2 x +2 x Cl2 (g) ( 0.0010 -x 0.0010 - x [Cl]2 (2 x ) 2 = = 1.2 10-7 [Cl 2 ] (0.0010 - x ) 4 x 2 = (1.2 10-7 ) (0.0010 x) - 4 x 2 + (1.2 10-7 ) x - (1.2 10-10 ) = 0 x= x= - : (1.2 10-7 ) (1.2 10-7 ) 2 - 4(4) (-1.2 10-10 ) 2 4 4.4 10-5 -(1.2 10-7 ) 8 -6 x = -: 5.5 10 or+ 5.5 10-6 The negative answer is not meaningful, so we choose x = + 5.5 10-6 mol L-1. The concentration of Cl2 is essentially unchanged because 0.0010 - 5.5 10-6 0.0010. The concentration of Cl atoms is 2 (5.5 10-6 ) = 1.1 10-5 mol L-1. (b)The balanced equation is: F2 (g) ( 2 F(g). The problem is worked in an identical fashion to (a) but the equilibrium constant is now 1.2 10-4. The initial concentration of F2 (g) is Concentration (mol L-1 ) initial change F2 (g) 0.0010 -x 0.0020 mol F2 = 0.0010 mol L-1. 2.0 L ( 2 F(g) 0 +2 x equilibrium 0.0010 - x +2 x Week 7 Solutions, p 2 Chem. 112 S09 Dr. Guldan [F]2 (2 x) 2 KC = = = 1.2 10-4 [F2 ] (0.0010 - x ) 4 x 2 = (1.2 10-4 ) (0.0010 x) - 4 x 2 + (1.2 10-4 ) x (1.2 10-7 ) = 0 - (1.2 10-4 ) 2 - 4(4) (- 1.2 10-7 ) 2.4 -4 -3 -(1.2 10 ) 1.4 10 x= 8 -4 x = - 10 or + 10-4 1.9 1.6 x= The negative answer is not meaningful, so we choose x = +1.6 10-4 mol L-1. = The concentration of F2 is 0.0010 - 1.6 10-4 8 10-4 mol L-1. The 4 concentration of F atoms is 2 (1.6 10-=) 3.2 10-4 mol L-1. (1.2 10-4 ) - (c) Cl2 is more stable. This can be seen even without the aid of the calculation from the larger equilibrium constant for the dissociation for F2 compared to Cl2 . 9.47 Concentration (bar) initial change final K= pH2 e pBr2 pHBr 2 2 HBr(g) 1.2 10-3 -2 x 1.2 10-3 2 x - ( H 2 (g) + Br2 (g) 0 +x +x 0 +x +x Week 7 Solutions, p 3 Chem. 112 S09 Dr. Guldan 7.7 10-11 = 7.7 10 = -11 ( x) ( x ) x2 = - (1.2 10-3 2 x) 2 (1.2 10-3 - x2 (1.2 10-3 2 x) 2 - 2 x)2 x = 8.8 10-6 -3 (1.2 10 - 2 x ) - x = (8.8 10-6 ) (1.2 10-3 -6 2 x) x + 2(8.8 10 ) x (8.8 10-6 ) (1.2 10-3 ) = x 1.1 10-8 pH2 = pBr2 = 1.1 10-8 bar; the pressure of HBr is essentially unaffected by the formation of Br2 and H 2 . 9.49 1.0 g PCl5 -1 208.22 g mol PCl5 (a)Concentration of PCl5 initially = = 0.019 mol L-1. 0.250 L Concentration (mol L-1 ) PCl5 (g) initial change final 0.019 -x 0.019 - x ( PCl3 (g) 0 +x +x + Cl2 0 +x +x [PCl2 ][Cl 2 ] ( x) ( x) x2 KC = = = [PCl5 ] (0.019 - x ) (0.019 - x) x2 = 1.1 10-2 (0.019 - x) x 2 = (1.1 10-2 ) (0.019 x ) - x 2 + (1.1 10-2 ) x 2.1 10-4 = 0 - x= x= (1.1 10-2 ) - -(1.1 10-2 ) 2.1 (1.1 10-2 ) 2 - (4)(1)(- 2.1 10-4 ) 2.1 0.031 = + 0.010 or - 0.021 The negative root is not meaningful, so we choose x = 0.010 mol L-1. Week 7 Solutions, p 4 Chem. 112 S09 Dr. Guldan [PCl3 ] = [Cl 2 ] = 0.010 mol -1 ; [PCl5 ] = 0.009 mol -1. L L (b)The percentage decomposition is given by 0.010 ( 100 = 53%. 0.019 0.400 mol = 0.200 mol L-1. 2.00 L NH 3 (g) 0.200 +x 0.200 + x + H 2S(g) 0 +x +x 9.51 Starting concentration of NH 3 = Concentration (mol L-1 ) NH 4 HS(s) ( initial change final 1.6 10-4 (0.200 x) ( x) = + x 2 + 0.200 x - 1.6 10-4 0 = -- -- -- K C = [NH 3 ][H 2S] = (0.200 + x) ( x ) ( +0.200) 2 - (4) (1) ( - 1.6 10-4 ) 2.1 -0.200 0.2016 x= = + 0.0008 or - 0.2008 2.1 x= -(+ 0.200) The negative root is not meaningful, so we choose x = 8 10-4 mol L-1 (note that in order to get this number we have had to ignore our normal significant figure conventions). [NH 3 ] = + 0.200 mol -1 + 8 -4 mol -1 = 0.200 mol -1 L 10 L L [H 2S] = 8 10-4 mol L-1 Alternatively, we could have assumed that x << 0.2, the 0.200 x = 1.6 10-4 , x =p 8.0 10-4. 9.53 The initial concentrations of N 2 and O 2 are equal at 0.114 mol L-1 because the vessel has a volume of 1.00 L. Concentrations (mol L-1 ) N 2 (g) initial 0.114 + O 2 (g) 0.114 ( 2 NO(g) 0 change final KC = -x 0.114 x -x 0.114 x +2 x +2 x Week 7 Solutions, p 5 Chem. 112 S09 Dr. Guldan [NO]2 (2 x) 2 (2 x) 2 = = [N 2 ][O 2 ] (0.114 - x) (0.114 - x) (0.114 - x ) 2 (2 x ) 2 (0.114 - x) 2 (2 x) 2 (0.114 - x) 2 (2 x ) (0.114 - x) the concentrations of N 2 and O 2 1.00 10-5 = 1.00 10-5= 3.16 10-3 = 2 x = (3.16 10-3 ) (0.114 x) - 2.00316 x = 3.60 10-4 x = 1.8 10-4 [NO] = 2 x = 1.8 10-4 =p 2 3.6 10-4 mol L-1 ; remain essentially unchanged at 0.114 mol L-1. 9.55 The initial concentrations of H 2 and I 2 are [H 2 ] = 0.400 mol 1.60 mol = 0.133 mol -1 ; [I 2 ] = L = 0.533 mol -1 L 3.00 L 3.00 L + I 2 (g) 0.533 -x 0.533 - x ( 2 HI(g) 0 +2 x +2 x Concentrations (mol L-1 ) H 2 (g) initial change final 0.133 -x 0.133 - x At equilibrium, 60.0% of the H 2 had reacted, so 40.0% of the H 2 remains: (0.400)(0.133 mol -1 ) = 0.133 mol -1 - x L L x = 0.133 mol -1 - (0.400) (0.133 mol -1 ) L L x = 0.080 mol L-1 L-1 L-1 L-1 At equilibrium: [ H 2 ] = 0.133 mol - 0.080 mol = 0.053 mol Week 7 Solutions, p 6 Chem. 112 S09 Dr. Guldan [I 2 ] = 0.533 mol -1 - 0.080 mol -1 = 0.453 mol -1 L L L [HI] = 2 0.080 mol -1 = 0.16 mol -1 L L KC = [HI]2 0.162 = = 1.1 [H 2 ][I 2 ] (0.053) (0.453) 9.59 [BrCl]2 KC = [Cl 2 ][Br2 ] 0.031 = [Br2 ] = (0.145) 2 (0.495)[Br2 ] (0.145) 2 = 1.4 mol L-1 (0.495) (0.031) 9.61 We can calculate changes according to the reaction stoichiometry: Amount (mol) initial change final CO(g) 2.00 -x 2.00 - x + 3 H 2 (g) ( 3.00 -3 x 3.00 - 3 x CH 4 (g) + H 2 O(g) 0 +x 0 +x +x 0.478 According to the stoichiometry, 0.478 mol = x; therefore, at equilibrium, there are 2.00 mol 0.478 mol = 1.52 mol CO, 3.00 3(0.478 mol) = 1.57 mol H 2 , and 0.478 mol H 2 O. To employ the equilibrium expression, we need either concentrations or pressures; because K C is given, we will choose to express these as concentrations. This calculation is easy because V = 10.0 L: [CO] = 0.152 mol -1 ; [H 2 ] = 0.157 mol -1 ; [CH 4 ] = 0.0478 mol -1 ; L L L [H 2 O] = 0.0478 mol L-1 KC = [CH 4 ][H 2O] (0.0478) (0.0478) = = 3.88 [CO][H 2 ]3 (0.152) (0.157)3 9.65 (a)The initial concentrations are Week 7 Solutions, p 7 Chem. 112 S09 Dr. Guldan 1.50 mol 3.00 mol = 3.00 mol -1 ; [PCl3 ] = L = 6.00 mol -1 ; L 0.500 L 0.500 L 0.500 mol [Cl2 ] = = 1.00 mol L-1. 0.500 L [PCl5 ] = First calculate Q : Q= [PCl5 ] 3.00 = = 0.500 [PCl3 ][Cl 2 ] (6.00) (1.00) Because Q e K , the reaction is not at equilibrium. (b)Because Q < K C , the reaction will proceed to form products. (c) Concentrations L-1 (mol ) PCl3 (g) initial change final KC = [PCl5 ] [PCl3 ][Cl 2 ] 3.00 + x 3.00 + x = 2 (6.00 - x) (1.00 - x ) x - 7 x + 6.00 6.00 -x 6.00 - x + Cl2 (g) 1.00 -x 1.00 - x ( PCl5 (g) 3.00 +x 3.00 + x 0.56 = (0.56) ( x 2 - 7 x + 6.00) = 3.00 + x 0.56 x 2 - 3.92 x + 3.36 = 3.00 + x 0.56 x 2 - 4.92 x + 0.36 = 0 x= -(- 4.92) ( -4.92) 2 - (4) (0.56) (0.36) + 4.92 4.48 = (2) (0.56) 1.12 x = 9.2 or 0.07 Because the root 9.2 is larger than the amount of PCl3 or Cl2 available, it is physically meaningless and can be discarded. Thus, x = 0.071 mol L-1 , giving [PCl5 ] = 3.00 mol -1 + 0.07 mol -1 = 3.07 mol -1 L L L [PCl3 ] = 6.00 mol -1 - 0.07 mol -1 = 5.93 mol -1 L L L [Cl2 ] = 1.00 mol -1 - 0.07 mol -1 = 0.93 mol -1 L L L The number can be checked by substituting them back into the equilibrium constant expression: Week 7 Solutions, p 8 Chem. 112 S09 Dr. Guldan KC = [PCl5 ] [PCl3 ][Cl 2 ] (3.07) ? = 0.56 (5.93) (0.93) ? 0.56 = 0.56 9.73 (a)According to Le Chatelier's principle, an increase in the partial pressure of CO 2 will result in creation of reactants, which will decrease the H 2 partial pressure. (b)According to Le Chatelier's principle, if the CO pressure is reduced, the reaction will shift to form more CO, which will decrease the pressure of CO 2 . (c)According to Le Chatelier's principle, if the concentration of CO is increased, the reaction will proceed to form more products, which will result in a higher pressure of H 2 . (d)The equilibrium constant for the reaction is unchanged because it is unaffected by any change in concentration. 9.75 (a)According to Le Chatelier's principle, increasing the concentration of NO will cause the reaction to form reactants in order to reduce the concentration of NO; the amount of water will decrease. (b)For the same reason as in (a), the amount of O 2 will increase. (c)According to Le Chatelier's principle, removing water will cause the reaction to shift toward products, resulting in the formation of more NO. (d)According to Le Chatelier's principle, removing a reactant will cause the reaction to shift in the direction to replace the removed substance; the of NH 3 should increase. (e)According to Le Chatelier's principle, adding ammonia will shift the reaction to the right, but the equilibrium constant, which is a constant, will not be affected. (f)According to Le Chatelier's principle, removing NO will cause the formation of more products; the amount of NH 3 will decrease. amount Week 7 Solutions, p 9 Chem. 112 S09 Dr. Guldan (g)According to Le Chatelier's principle, adding reactants will promote the formation of products; the amount of oxygen will decrease. 9.79 (a)If the pressure of NO (a product) is increased, the reaction will shift to form more reactants; the pressure of NH 3 should increase. (b)If the pressure of NH 3 (a reactant) is decreased, then the reaction will shift to form more reactants; the pressure of O 2 should increase. 9.81 If a reaction is exothermic, raising the temperature will tend to shift the reaction toward reactants, whereas if the reaction is endothermic, a shift toward products will be observed. For the specific examples given, (a) and (b) are endothermic (the values for (b) can be calculated, but we know that it requires energy to break an X--X bond, so those processes will all be endothermic) and raising the temperature should favor the formation of products; (c) and (d) are exothermic and raising the temperature should favor the formation of reactants. 9.87 Because we want the equilibrium constant at two temperatures, we will r need to calculate H and S r for each reaction: (a) NH 4 Cl ( NH3 (g) + HCl(g) H r = H f (NH 3 , g) + H f (HCl, g) - nH f (NH 4 Cl, s) H = (- 46.11 kJ -1 ) + (-92.31 kJ -1 ) - (-314.43 kJ -1 ) mol mol mol r H S r r = 176.01 kJ mol -1 = S (NH 3 , g) +p (HCl, g) - S (NH 4 Cl, s) S S = 192.45 J -1 -1 + 186....

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East Los Angeles College - M - 298
MATH298: Warm-up excercises, week 4Set 2008/02/11 ; for self-control (solutions provided)norm 1. Determine the eigenvalues j and normalised eigenvectors vj for each of the following matrices A. Verify the results by substituting into the equation A
East Los Angeles College - M - 298
MATH298: Warm-up excercises week 5 (2008/02/18 ) - solutions1. Calculate first and second partial derivatives of the following functions, and check that the two mixed derivatives are equal to each other. (a) z(x, y) = x sin y. Solution: z = x sin y
George Mason - CS - 818
4/1/08Secure Sensor Network Routing: A Clean-Slate ApproachBryan Parno, Mark Luk, Evan Gaustad, and Adrian Perrig Carnegie Melon UniversityOverview Designa highly secure, highly available node-to-node sensor network routing protocol Security
UCLA - C - 160
Bioinformatics &amp; GenomicsChemistry 160 / 260 Winter 2002 Chris Lee Boyer Hall (MBI) 601Course Goals Understand conceptual foundations of genomics and bioinformatics, via examples of what has been done; Apply these principles towards inventing ne
UCLA - C - 160
Summation: Principles of Bioinformatics Review the key ingredients of the Recipe for Bioinformatics. Use the Human Genome results as examples for understanding the importance of these ingredients in future genomics and bioinformatics problems. Int
Ohio State - POLISCI - 585
Political Science 585: Summer 2007ChristensonPOLITICAL SCIENCE 585: TECHNIQUES IN POLITICAL ANALYSISSummer Quarter 2007 Mondays &amp; Wednesdays 10:30am 12:18pm Derby Hall 0125Instructor: Dino P. Christenson Office: Derby Hall 2160 Office hours: M
Ohio State - POLISCI - 585
Christenson PS 585: TECHNIQUES IN POLITICAL ANALYSIS Important political science articles that use elementary statistics* Ansolabehere, Stephen, Shanto Iyengar, Adam Simon, and Nicholas Valentino. &quot;Does Attack Advertising Demobilize the El
Ohio State - POLISCI - 585
Christenson PS 585: TECHNIQUES IN POLITICAL ANALYSIS A Short Guide to College Writing* Writing well is one of the most important skills to learn during your college career. However, becoming a good writer is often easier said than done. This guide is
UCSD - MAE - 210
MAE210A: Fluid Mechanics IFall Quarter 2005 http:/maecourses.ucsd.edu/mae210aSolutions Homework 11 Use sufces. [u ( u)]i = =ijk uj klmum xl um xl(il jm im jl )uj= uj =ui uj uj xi xj 1 ( uu) (u )ui xi 21 Finally, u ( u) = ( 2 uu