EXAM1solutions - Version 395 – Exam 1 – flowers –...

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Unformatted text preview: Version 395 – Exam 1 – flowers – (53885) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Use 2 pencils. No programmable calcula- tors. No books. No notes 001 10.0 points Consider the equation 1 × 10 − 5 = [2 x ] 2 (0 . 1- 3 x ) 3 (0 . 7- x ) which describes an equilibrium. 0.1 and 0.7 are starting conditions for two reactants. By inspection, the solution for x must be 1. greater than 0.00 and less than 2.000. 2. less than 0.100. 3. greater than 0.10 and less than 0.700. 4. less than 0.0333. 5. greater than 0.00 and less than 0.0333. correct Explanation: 3 A + B ⇀ ↽ 2 C ini, M 0.1 0.7 Δ, M- 3 x- x +2 x eq, M . 1- 3 x . 7- x 2 x K = [C] 2 [A] 3 [B] Since there is no C initially, Q = 0 ( < K ) and A and B must react to produce 2 x amount of C. If [A] = 0.1, it is the limiting reactant and the reaction can produce at most 0.067 of C. 002 10.0 points The system H 2 (g) + I 2 (g) ⇀ ↽ 2 HI(g) is at equilibrium at a fixed temperature with a partial pressure of H 2 of 0.200 atm, a partial pressure of I 2 of 0.200 atm, and a partial pres- sure of HI of 0.100 atm. An additional 0 . 29 atm pressure of HI is admitted to the con- tainer, and it is allowed to come to equilib- rium again. What is the new partial pressure of HI? 1. 0.154 2. 0.132 3. 0.158 4. 0.168 5. 0.136 6. 0.13 7. 0.138 8. 0.15 9. 0.152 10. 0.156 Correct answer: 0 . 158 atm. Explanation: P I 2 = P H 2 = 0 . 2 atm P HI = 0 . 1 atm H 2 (g) + I 2 (g) ⇀ ↽ 2 HI(g) K p = ( P HI ) 2 P H 2 · P I 2 = (0 . 1 atm) 2 (0 . 2 atm) (0 . 2 atm) = 0 . 25 new P HI = (0 . 1 + 0 . 29) atm = 0 . 39 atm Adding the products shifts the equilibrium to the left. H 2 (g) + I 2 (g) ⇀ ↽ 2 HI(g) ini, atm 0.200 0.200 . 39 Δ, atm + x + x- 2 x eq, atm 0 . 200 + x . 200 + x . 39- 2 x (0 . 39- 2 x ) 2 (0 . 2 + x ) 2 = 0 . 25 . 39- 2 x . 2 + x = √ . 25 . 39- 2 x = (0 . 5)(0 . 2) + 0 . 5 x 2 . 5 x = 0 . 29 x = 0 . 116 P HI = 0 . 39 atm- (2) (0 . 116 atm) = 0 . 158 atm 003 10.0 points Version 395 – Exam 1 – flowers – (53885) 2 The volume of water increases as it freezes. 1. True correct 2. False Explanation: 004 10.0 points The first stage in the production of nitric acid by the Ostwald process is the reaction of am- monia gas with oxygen gas, producing nitric oxide gas (NO) and liquid water. The nitric oxide further reacts with oxygen to produce nitrogen dioxide gas. Finally, nitrogen dioxide gas, when dissolved in water, produces nitric acid and nitric oxide. What is the coefficient of oxygen in the second balanced equation? (Balance the equations with the smallest pos- sible whole number coefficients.) 1. 3 2. 11 3. 9 4. 2 5. 5 6. 10 7. 8 8. 1 correct 9. 4 10. 6 Explanation: The balanced equations are 4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O( ℓ ) 2 NO(g) + O 2 (g) → 2 NO 2 (g) 3 NO 2 (g) + H 2 O( ℓ ) → 2 HNO 3 (aq) + NO(g) 005 10.0 points A 2 . 54 g sample of a protein of molar mass...
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This note was uploaded on 02/04/2009 for the course CH 53885 taught by Professor Flowers during the Fall '08 term at University of Texas.

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EXAM1solutions - Version 395 – Exam 1 – flowers –...

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