Exam 2-solutions - Version 120 Exam 2 Lyon (53565) This...

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Version 120 – Exam 2 – Lyon – (53565) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. BE SURE THAT YOU HAVE BUB- BLED IN YOUR NAME AND EID COR- RECTLY. WE WILL BE UNABLE TO ±IX VERSION NUMBER ERRORS WITH- OUT YOUR EXAM HARDCOPY AND IT WILL TAKE SOME TIME TO MAKE THE CORRECTION. IT MAY BE A WEEK OR MORE BE±ORE YOU WILL KNOW YOUR EXAM SCORE I± YOU DO NOT COMPLETE THE BUBBLE SHEET COR- RECTLY. YOU MUST ALSO KEEP YOUR COPY O± THE EXAM AND MAKE IT AVAILABLE I± REQUESTED ±OR ANY REASON THIS SEMESTER. 001 10.0 points IdentiFy the list in which all salts produce a basic aqueous solution. 1. NH 4 Cl, C 6 H 4 NH 3 NO 3 , ±eI 3 2. AlCl 3 , Zn(NO 3 ) 2 , KClO 4 3. AgNO 3 , NaCHO 2 , CrI 3 4. CH 3 NH 3 Cl, KNO 3 , NaBz (sodium ben- zoate) 5. KCH 3 COO, NaCN, K± correct Explanation: 002 10.0 points What happens to the concentration oF HI(g) when the total pressure on the equi- librium reaction 2 HCl(g) + I 2 (s) 2 HI(g) + Cl 2 (g) is increased (by compression)? 1. decreases correct 2. remains the same 3. Unable to determine 4. increases Explanation: Increasing the total pressure on the sys- tem by decreasing its volume will shiFt the equilibrium toward the side oF the reaction with Fewer numbers oF moles oF gaseous com- ponents. IF the total number oF moles oF gas is the same on the product and reactant sides oF the balanced chemical equation, then changing the pressure will have little or no eF- Fect on the equilibrium distribution oF species present. 003 10.0 points The equilibrium constant For the reaction 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g) is K . Calculate the equilibrium constant For the reaction 2 NO(g) + 3 H 2 O(g) 2 NH 3 (g) + 5 2 O 2 (g) . 1. - K 2 2. - 2 K 3. - K 4. 1 K correct 5. 1 K Explanation: 004 10.0 points Calculate the equilibrium constant For the re- action 2 TiCl 3 (s) + 2 HCl(g) 2 TiCl 4 (g) + H 2 (g) at 25 C, iF Δ G = +46 . 6 kJ. 1. 1 . 5 × 10 8 2. 3 . 8 × 10 9
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Version 120 – Exam 2 – Lyon – (53565) 2 3. 6 . 6 × 10 18 4. 6 . 7 × 10 9 correct 5. 1 . 5 × 10 19 Explanation: 005 10.0 points Consider the famous ammonia preparation 3 H 2 (g) + N 2 (g) 2 NH 3 (g) The equation K = [ x ] 2 [0 . 1 - 3 x ] 3 [0 . 7 - x ] is not a possible correct description of the equilibrium situation because 1. the 0.1 and 0.7 in the denominator are incompatible. 2. [0 . 7 - x ] in the denominator should be [0 . 7 - 3 x ]. 3. [ x ] in the numerator should be [2 x ]. cor- rect 4. the denominator and numerator should be inverted. 5. The equation is correct. Explanation: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) ini, M 0.1 0.7 0 Δ, M - 3 x - x +2 x eq, M 0 . 1 - 3 x 0 . 7 - x 2 x K = [NH 3 ] 2 [H 2 ] 3 [N 2 ] = (2 x ) 2 (0 . 1 - 3 x ) 3 (0 . 7 - x ) 006 10.0 points Which of Acid X [HX] = 10 3 M K a = 1 . 32 × 10 6 Acid Y [HY] = 10 3 M K a = 3 . 47 × 10 3 is the stronger acid? 1.
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This note was uploaded on 04/27/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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Exam 2-solutions - Version 120 Exam 2 Lyon (53565) This...

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