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Unformatted text preview: Version 127 – Exam 2 – Mccord – (52450) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. McCord CH302 1pm This exam is only for McCord’s MWF 1pm CH302 class. 001 10.0 points The balanced equation for a particular reac tion has 4 moles of gaseous reactants and 2 moles of gaseous products. For this reaction at room temperature, which of the following would be true? 1. K c equals K p 2. K c is greater than K p correct 3. K c is unrelated to K p 4. K c is less than K p Explanation: n gas products = 2 n gas reactants = 4 K c and K p are related by the equation K p = K c ( RT ) Δ n where Δ n = n gas products − n gas reactants = 2 − 4 = − 2 Since Δ n is a negative number, the value of K p will be smaller than the value of K c ; this can also be stated as K c being greater than K p . 002 10.0 points Solid NH 4 CO 2 NH 2 (ammonium carba mate) dissociates completely into ammonia and carbon dioxide. When pure ammonium carbamate is put into an evacuated container and allowed to come to equilibrium with the gaseous products, the total pressure at 45 ◦ C is found to be 0.6 atm. What is the value for K p for this reaction at 45 ◦ C? 1. K p = 0 . 032 correct 2. K p = 0 . 08 3. K p = 0 . 004 4. K p = 0 . 216 5. K p = 0 . 06 Explanation: P total = 0 . 6 atm NH 4 CO 2 NH 2 (g) ⇀ ↽ 2 NH 3 (g)+ CO 2 (g) eq, P 2 x x 2 x + x = 0 . 6 atm x = 0 . 2 atm Thus P NH 3 = 0 . 4 atm and P CO 2 = 0 . 2 atm, so K p = P 2 NH 3 · P CO 2 = (0 . 4) 2 (0 . 2) = 0 . 032 003 10.0 points Rank following acids from most to least acidic: hydrocyanic acid (HCN) K a = 6 . 2 × 10 − 10 hypoiodous acid (HOI) K a = 2 × 10 − 11 chlorous acid (HClO 2 ) K a = 1 . 2 × 10 − 2 acetic acid (CH 3 COOH) K a = 1 . 8 × 10 − 5 1. CH 3 COOH > HCN > HOI > HClO 2 2. HClO 2 > CH 3 COOH > HCN > HOI correct 3. HCN > HOI > HClO 2 > CH 3 COOH 4. HOI > HClO 2 > CH 3 COOH > HCN Explanation: Explanation: The value of K a is directly proportional to the acidity (strength) of the acid. By ranking from the largest (1 . 2 × 10 − 2 ) to the smallest (2 × 10 − 11 ) Ka, one is ranking from the most to least acidic. 004 10.0 points Calculate the pH of the solution resulting Version 127 – Exam 2 – Mccord – (52450) 2 from the addition of 40.0 mL of 0.200 M HClO 4 to 60.0 mL of 0.150 M NaOH. 1. 1.78 2. 2.00 3. 12.00 correct 4. 12.22 5. 7.00 Explanation: V HClO 4 = 40.0 mL [HClO 4 ] = 0.200 M V NaOH = 60.0 mL [NaOH] = 0.150 M Here it’s important to find out which of these two species (HClO 4 and NaOH) is in ex cess. The one that is in excess will determine the pH of this solution. From the formulas of the two compounds, you can expect that they will react in a onetoone fashion....
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This note was uploaded on 06/15/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
 Spring '07
 Holcombe

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