Final_rev2-solutions-1

Final_rev2-solutions-1 - Version 208 – Final rev2 –...

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Unformatted text preview: Version 208 – Final rev2 – laude – (51635) 1 This print-out should have 60 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 5.0 points Consider the irreversible reaction A + 2 B → 3 C . Which of the following correctly expresses the rate of change of [B] ? 1. Δ[B] Δ t =- 1 2 (rate of rxn) 2. Δ[B] Δ t = +(rate of rxn) 3. Δ[B] Δ t = + 1 2 (rate of rxn) 4. Δ[B] Δ t =- (rate of rxn) 5. Δ[B] Δ t =- 2 (rate of rxn) correct 6. Δ[B] Δ t = +2 (rate of rxn) Explanation: As B is a reactant and consequently disap- pears during the reaction, its rate of change will be negative and the inverse of the coeffi- cient is used when writing the rate: (rate of rxn) =- 1 2 Δ[B] Δ t- 2 (rate of rxn) = Δ[B] Δ t 002 5.0 points Which of the following statements is/are always true concerning K w ? I) It gets larger as the temperature in- creases II) It equals 10 − 14 III) K w = [H + ][OH − ] 1. I only 2. II, III 3. I, II, III 4. III only 5. II only 6. I, II 7. I, III correct Explanation: All equilibrium processes are temperature dependent, and because auto-protolysis is en- dothermic, K w increases as temperature in- crease. Thus statement I is true, but state- ment II is only true at room temperature. Statement III is the definition of K w . 003 5.0 points Consider the potential energy diagram shown below. A B Reaction progress Energy(kJ) 250 300 350 What is the activation energy E a for the reaction A → B? 1. 350 kJ 2. 250 kJ 3. 150 kJ 4. 100 kJ correct 5.- 50 kJ Explanation: We need enough energy to ‘get to the top of the hill’ in order to fall down to the products. So the energy must raise from 250 kJ (at A) up to the ‘top of the hill’ at 350 kJ: 350 kJ- 250 kJ = 100 kJ Version 208 – Final rev2 – laude – (51635) 2 004 5.0 points Which of the following pairs of solutions would result in a buffer upon mixing? 1. 1 L of 0 . 5 M CH 3 NH 2 ; 1 L of 0 . 5 M HF 2. 1 L of 2 M (CH 3 ) 2 NH 2 Cl; 1 L of 1 M HCl 3. 2 L of 0 . 5 M NaHCOO; 1 L of 1 M LiOH 4. . 25 L of 1 M H 2 SO 3 ; 0 . 5 L of 1 M NH 3 5. . 5 L of 1 M Ca(OH) 2 ; 2 L of 1 M CH 3 COOH correct Explanation: A buffer prepared by a neutralization re- action requires either a weak acid and less strong base or a weak base and less strong acid. The only pair of substances that is us- able is the strong base Ca(OH) 2 and the weak acid CH 3 COOH. 005 5.0 points Consider the following reaction and its rate constant. A → B k = 0 . 103 M − 1 · min − 1 What will be the concentration of A after 1 hour if the reaction started with a concen- tration of 0.400 M ? 1. 0.115 M correct 2. 0.384 M 3. 0.308 M 4. 0.236 M 5. 0.152 M 6. 8 . 28 × 10 − 4 M 7. 0.0843 M 8. 0.361 M Explanation: 1 [A] t- 1 [A] = akt 1 [A] t = 1 [A] + akt = 1 . 400 + (0 . 103 M − 1 · min − 1 ) (60 min) = 8 . 68 [A] t = 0 . 115 M 006 5.0 points A fatty acid consists of a ten-carbon alkane with a carboxylic acid functional group....
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This note was uploaded on 06/14/2011 for the course MATH 305G taught by Professor Cathy during the Spring '11 term at University of Texas.

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Final_rev2-solutions-1 - Version 208 – Final rev2 –...

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