1422-Exam-2-2009-Answers

1422-Exam-2-2009-Answers - ANSWER KEY CHEM 1422 Exam # 2...

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CHEM 1422 Exam # 2 (March 26, 2009) ANSWER KEY Equilibrium & Acids/Bases Put a big X this box if you want your graded exam put out in the public racks outside Prof. Stanley’s office after grading . If you don’t check it, Prof. Stanley will keep your exam and you will have to stop by to pick it up from him personally. 1. (5 pts) For which of the following reactions will increasing the overall pressure make more products ? Circle your answer and briefly discuss your reasoning in the space below. a) 3NO 2 ( g ) + H 2 O( l ) 2HNO 3 ( aq ) + NO( g ) b) 3H 2 S( g ) + 2Fe( s ) Fe 2 S 3 ( s ) + 3H 2 ( g ) c) MgO( s ) + C( s ) Mg( s ) + CO 2 ( g ) d) 2HCl( g ) Cl 2 ( g ) + H 2 ( g ) e) AgCl( s ) + 2NH 3 ( aq ) Ag(NH 3 ) 2 + ( aq ) + Cl ( aq ) 2. (5 pts) Consider the reaction: 2HRe(CO) 5 ( soln ) Re 2 (CO) 10 ( soln ) + H 2 ( g ) Initially, [Re 2 (CO) 10 ] = [H 2 ] = 1 M . What is the concentration of HRe(CO) 5 at equilibrium? K eq = 4 Circle your answer and show your work in the space below. a) 0 M b) 0.2 M c) 0.4 M d) 0.8 M e) 2 M Initial: 0 1 M 1 M 2HRe(CO) 5 ( soln ) Re 2 (CO) 10 ( soln ) + H 2 ( g ) K eq = 4 @ Eq: 2 x 1 – x 1 – x () ; ; ; . eq xx x K x x x M x −− == = = =− = = = 2 10 2 2 2 22 5 5 [Re (CO) ] [H ] 1 1 1 4 take square root of each side: [HRe(CO) ] 2 2 1 2 ; solve for x : 1 4 1 5 02 [HRe(CO) ] = 2x = 0.4 2 3. (5 pts) Which of the following salts will be the least soluble in H 2 O? Circle your answer and show your work or reasoning below. a) AgCl (K sp = 1 × 10 –10 ) b) AgF (K sp = 1 × 10 –6 ) c) AgI (K sp = 1 × 10 –16 ) d) AgBr (K sp = 1 × 10 –13 ) e) Ag 3 PO 4 (K sp = 1 × 10 –20 ) f) AgCN (K sp = 1 × 10 –12 ) You can directly compare salts that have the same cation-anion ratio (stoichiometry). So AgCl, AgF, AgI, AgBr and AgCN all have the same 1:1 ratios. Of these, AgI has the smallest K sp value, so of these, AgI will be the least soluble with a [Ag + ] equal to: [Ag + ][I ] = ( x )( x ) = x 2 = 1 × 10 16 , or x = [Ag + ] = 1 × 10 8 M. Ag 3 PO 4 , has a 3:1 ratio and needs to be calculated separately to compare it’s [Ag + ] to that of AgI: [Ag + ] 3 [PO4 3- ] = (3 x ) 3 ( x ) = (27 x 3 )( x ) = 27 x 4 = 1 × 10 20 ; x 4 = (1 × 10 20 )/27 ; x 4 = 3.7 × 10 22 ; x = 4.4 × 10 6 ; [Ag + ] = 3 x = 1.3 × 10 5 M . This is a higher concentration relative to the AgI, so AgI is the least soluble. There are 3 gas molecules on the reactant side and only one on the product side. Increasing the pressure will shift the equilibrium to the side with fewer gas phase molecules to reduce the “crowding”. The other rxns either have the same # of gas phase molecules on each side (b & d, won’t be affected by pressure), more gas phase molecules on the product side (c), or no gas phase molecules (e).
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CHEM 1422 - Exam # 2 (Equilibrium & Acids/Bases) 2 4. (5 pts) What is the Δ Gº value (in kJ/mol) that corresponds to a K eq = 0.09 at 500 K. R = 8.314 J/molK Circle your answer and show your work or reasoning below.
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This note was uploaded on 11/12/2011 for the course CHEM 1422 taught by Professor Staff during the Fall '08 term at LSU.

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1422-Exam-2-2009-Answers - ANSWER KEY CHEM 1422 Exam # 2...

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