Exam 2-solutions

# Exam 2-solutions - Version 462 Exam 2 sutclie(51045 This...

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Version 462 – Exam 2 – sutclife – (51045) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore 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 ±or N 2 O 4 2 NO 2 has the numerical value o± 0.0466. I± the initial concentrations are [N 2 O 4 ] = 1.0 M, [NO 2 ] = 0.0 M, what are the Fnal concentrations o± [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 . 0 0 . 0 Δ, M - x 2 x eq, M 1 . 0 - x 2 x K = [NO 2 ] 2 [N 2 O 4 ] = 0 . 0466 (2 x ) 2 1 . 0 - x = 0 . 0466 4 x 2 = 0 . 0466 - 0 . 0466 x 4 x 2 + 0 . 0466 x - 0 . 0466 = 0 x = 0 . 102 [N 2 O 4 ] = 1 - 0 . 102 = 0 . 90 M [NO 2 ] = 2 x = 0 . 20 M 002 10.0 points What is the ±orm o± the equilibrium constant ±or 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 o± pressure (in units o± atmospheres) are used ±or K p ; solutions measured in terms o± molar concentrations are used ±or K c . 003 10.0 points What is the H + ion concentration in a 0.50 mol/L solution o± a weak base that has an ionization constant ( K b ) o± 2.0 × 10 8 ? 1. 1 . 0 × 10 4 mol/L 2. 8 . 0 × 10 16 mol/L 3. 2 . 0 × 10 10 mol/L 4. 1 . 0 × 10 8 mol/L 5. 1 . 0 × 10 10 mol/L correct Explanation: 004 10.0 points Note: You’ll need to solve a quadratic to do this question. Consider the ±ollowing 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. I± suddenly 1.00 M PCl 5 (g), PCl 3 (g), and Cl 2 (g) are each added, calculate the equilibrium concentration o± PCl 3 (g). 1. 1.43 M

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Version 462 – Exam 2 – sutclife – (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. 2. how vigorously it reacts with the other molecules. 3. how completely it dissociates in aqueous solution. correct 4. the strength oF its attraction to water molecules.
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Exam 2-solutions - Version 462 Exam 2 sutclie(51045 This...

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