Exam 1-solutions-1

# Exam 1-solutions-1 - Version 408 – Exam 1 – laude...

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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 Δ G = Δ H- T Δ S ln parenleftbigg P 2 P 1 parenrightbigg = Δ H ◦ vap R parenleftbigg 1 T 1- 1 T 2 parenrightbigg Π = MRT Δ T b = iK b m Δ T f = iK f m P i = P ◦ i χ i P = summationdisplay i P i q = mc Δ T q = m Δ H Δ G ◦ =- RT ln K K = e − Δ G ◦ RT ln K 2 K 1 = Δ 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 Δ H sys for this process? c ice = 2 J / g · ◦ C c water = 4 J / g · ◦ C Δ 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, Δ H = m · c water · Δ T = 10 · 4 · (- 25) =- 1 , 000 J. Changing water to ice at 0 ◦ C (the freez- ing point of water), Δ H = m · (- Δ H fus ) = 10 · (- 340) =- 3 , 400 J. Finally, cooling ice from 0 ◦ C to- 50 ◦ C, Δ H = m · c ice · Δ T = 10 · 2 · (- 50) =- 1000 J. Adding everything together, the Δ H sys for the entire process is-5,400 J....
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Exam 1-solutions-1 - Version 408 – Exam 1 – laude...

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