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Unformatted text preview: Version 408 Exam 1 laude (51635) 1 This printout should have 30 questions. Multiplechoice 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 is5,400 J....
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 Spring '11
 Cathy

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