Exam 1-solutions-1 - Version 408 Exam 1 laude (51635) 1...

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Unformatted text preview: Version 408 Exam 1 laude (51635) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Constants R = 8 . 314 J K 1 mol 1 Definitions Q [C] c [D] d [A] a [B] b G H- TS K w [H + ][OH ] pH - log [H + ] pOH - log [OH ] Equations G = H- T S ln parenleftbigg P 2 P 1 parenrightbigg = H vap R parenleftbigg 1 T 1- 1 T 2 parenrightbigg = MRT T b = iK b m T f = iK f m P i = P i i P = summationdisplay i P i q = mc T q = m H G =- RT ln K K = e G RT ln K 2 K 1 = H R parenleftbigg 1 T 1- 1 T 2 parenrightbigg Q = K [at equil.] [H + ] = ( K a C a ) 1 / 2 [OH ] = ( K b C b ) 1 / 2 [H + ] = K a ( C a /C b ) pH = p K a + log( C b /C a ) [OH ] = K b ( C b /C a ) pOH = p K b + log( C a /C b ) 001 6.0 points Which of the following equations is untrue for a solution with a pH of 3? 1. pOH =- log pK w 10 3 correct 2. pOH =- log K w 10 3 3. [H + ] = 10 3 4. [OH ] = K w 10 3 5. all of these equations are true Explanation: pOH =- log[OH ] K w = [OH ][H + ] pOH =- log K w [H + ] =- log K w 10 3 It should be clear that K w , not pK w belongs in the correct expression for pOH. 002 6.0 points Consider the reaction: 2 HI( g ) H 2 ( g ) + I 2 ( g ) If we start out with pure HI and the equi- librium hydrogen gas concentration is 0.233 M at 730 K and at this temperature Kc = 0.12, what is the correct expression for the equilibrium concentration of HI(g)? 1. [HI] = parenleftbigg . 233 . 233 . 12 parenrightbigg 1 / 2 correct 2. [HI] = parenleftbigg . 233 . 233 . 12 parenrightbigg 3. [HI] = parenleftbigg . 233 . 12 parenrightbigg 1 / 2 4. [HI] = (0 . 233 . 233 . 12) 1 / 2 5. [HI] = (0 . 233 . 233 . 12) Explanation: Since we started from pure HI, the equilib- rium concentrations of hydrogen and iodine gas must be equal. 003 6.0 points Version 408 Exam 1 laude (51635) 2 10 g of water at 25 C is cooled to ice at- 50 C. What is overall H sys for this process? c ice = 2 J / g C c water = 4 J / g C H fusion = 340 J / g 1.-5,400 J correct 2. 1,400 J 3. 2,000 J 4.-2,000 J 5. 5,400 J 6.-1,400 J Explanation: Cooling water at 25 C to water at 0 C, H = m c water T = 10 4 (- 25) =- 1 , 000 J. Changing water to ice at 0 C (the freez- ing point of water), H = m (- H fus ) = 10 (- 340) =- 3 , 400 J. Finally, cooling ice from 0 C to- 50 C, H = m c ice T = 10 2 (- 50) =- 1000 J. Adding everything together, the H sys for the entire process is-5,400 J....
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Exam 1-solutions-1 - Version 408 Exam 1 laude (51635) 1...

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