Exam 2-solutions - Version 462 Exam 2 sutcliffe (51045) 1...

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Unformatted text preview: Version 462 Exam 2 sutcliffe (51045) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points This is a BONUS question. Do this one last!! Yes, it requires you to use the quadratic equa- tion. The equilibrium constant for N 2 O 4 2 NO 2 has the numerical value of 0.0466. If the initial concentrations are [N 2 O 4 ] = 1.0 M, [NO 2 ] = 0.0 M, what are the final concentrations of [N 2 O 4 ] and [NO 2 ], respectively? 1. 0.8 M ; 0.2 M 2. 0.9 M ; 0.2 M correct 3. 1.2 M ; 0.4 M 4. 1.0 M ; 0.22 M 5. 0.8 M ; 0.1 M Explanation: K = 0 . 0466 [N 2 O 4 ] ini = 1 . 0 M [NO 2 ] ini = 0 . 0 M N 2 O 4 2 NO 2 ini, M 1 . . , M- x 2 x eq, M 1 .- x 2 x K = [NO 2 ] 2 [N 2 O 4 ] = 0 . 0466 (2 x ) 2 1 .- x = 0 . 0466 4 x 2 = 0 . 0466- . 0466 x 4 x 2 + 0 . 0466 x- . 0466 = 0 x = 0 . 102 [N 2 O 4 ] = 1- . 102 = 0 . 90 M [NO 2 ] = 2 x = 0 . 20 M 002 10.0 points What is the form of the equilibrium constant for the process H 2 O( ) + CO 2 (g) H 2 CO 3 (aq)? 1. K p = [H 2 CO 3 ] [H 2 O] [CO 2 ] 2. K c = [H 2 CO 3 ] 2 [H 2 O] [CO 2 ] 3. No choice is correct. 4. K c = [H 2 CO 3 ] [CO 2 ] correct 5. K p = P H 2 CO 3 P 2 H 2 O P 2 CO 2 Explanation: Only gases and solutions are included in K expressions. Gases measured in terms of pressure (in units of atmospheres) are used for K p ; solutions measured in terms of molar concentrations are used for K c . 003 10.0 points What is the H + ion concentration in a 0.50 mol/L solution of a weak base that has an ionization constant ( K b ) of 2.0 10 8 ? 1. 1 . 10 4 mol/L 2. 8 . 10 16 mol/L 3. 2 . 10 10 mol/L 4. 1 . 10 8 mol/L 5. 1 . 10 10 mol/L correct Explanation: 004 10.0 points Note: Youll need to solve a quadratic to do this question. Consider the following reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) . At equilibrium, [PCl 5 ] = 2.00 M and [PCl 3 ] = [Cl 2 ] = 1.00 M. If suddenly 1.00 M PCl 5 (g), PCl 3 (g), and Cl 2 (g) are each added, calculate the equilibrium concentration of PCl 3 (g). 1. 1.43 M Version 462 Exam 2 sutcliffe (51045) 2 2. 3.35 M 3. 2.18 M 4. 2.32 M 5. 1.95 M 6. 1.35 M correct Explanation: K = (1)(1) 2 = 0 . 5 after the addition each concentration in- creases by 1 M and then the reaction must shift to the left to reestablish equilibrium. K = (2- x ) 2 (3 + x ) = 0 . 5 which converts to 0 = x 2- 4 . 5 x + 2 . 5 The quadratic formula will yield two an- swers for x . x = 0 . 64929 or 3 . 85078 The 3.85078 answer is not physically possible and therefore x = 0 . 64929. [PCl 3 ] = 2- x = 1 . 35078 M 005 10.0 points The strength of an acid or base is a measure of 1. how many hydroxide or hydrogen ions it contains....
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This note was uploaded on 03/26/2012 for the course CHEM 302 taught by Professor Mccord during the Fall '10 term at University of Texas at Austin.

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Exam 2-solutions - Version 462 Exam 2 sutcliffe (51045) 1...

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