Test1-04vA-soln - Name Chemistry 1AA3 Instructors M Austen...

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Name: Page 1 of 16 Continued on next page. Chemistry 1AA3 Term Test February 6 th , 2004 Version A Solutions Instructors: M. Austen, P. Lock Duration: 100 minutes This test contains 16 numbered pages. There are 21 multiple-choice questions appearing on pages numbered 3 to 13, and a periodic table on page 16. Pages 14 and 15 are for rough work. You may tear off the last page to view the periodic table and the first page to view the data below. Question 1 identifies the version of the test that you are writing. Be sure to answer this question correctly. Questions 2 to 16 are each worth 2 marks, questions 17 – 21 are each worth 3 marks; the total marks available are 45. There is no additional penalty for incorrect answers. You are responsible for ensuring that your copy of the question paper is complete. Bring any discrepancy to the attention of your invigilator. ANSWER ALL QUESTIONS ON THE ANSWER SHEET PROVIDED. No work written on the question sheets will be marked. Instructions for entering multiple-choice answers are given on page 2. Select one answer in each question from the answers (A) through (E). Your answers to all questions must be entered onto THE ANSWER SHEET IN PENCIL . Enter one and only one answer for each question. Only Casio FX 991 electronic calculators may be used; but they must NOT be transferred between students. Use of periodic tables, other than the table on page 16, is not allowed. Some general data are provided on this page. Other data appears with the questions. R = 8.314 J/(K mol) = 0.08206 L atm/(K mol) N A = 6.022 × 10 23 mol -1 1 atm = 760 mm Hg = 101.325 kPa 0 ° C = 273.15 K K w = 1.0 x 10 -14 K b (NH 3 ) = 1.8 × 10 5 K a (HPO 4 2 ) = 4.8 × 10 13 K a (CH 3 NH 3 + ) = 2.3 × 10 11 K a (CH 3 COOH) = 1.8 × 10 5
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Name: Page 3 of 16 Continued on next page. 1. You are writing VERSION A of this test. Please choose answer A(1) for question 1. (A) bubble 1 Questions 2 through 16 are worth two (2) marks each. 2. Given the equilibrium constants shown below, at the same temperature, calculate the equilibrium constant, K c , for the reaction: N 2 (g) + O 2 (g) + Br 2 (g) 2NOBr(g). 2NO(g) N 2 (g) + O 2 (g) K c = 2.1 × 10 30 reverse NO(g) + ½ Br 2 (g) NOBr(g) K c = 1.4 double (A) 6.7 × 10 31 (B) 9.3 × 10 31 (C) 1.3 × 10 30 (D) 2.8 (E) 4.1 × 10 30 3. A self-contained breathing device uses the following chemical equilibrium: 4 KO 2 (s) + 2 CO 2 (g) 2 K 2 CO 3 (s) + 3 O 2 (g), K p = 28.5. In order to simulate normal breathing conditions, the equilibrium partial pressure of O 2 (g) must equal 0.20 atm. What is the equilibrium partial pressure of CO 2 (g) needed for these conditions? (A) 7.0 × 10 3 atm (B) 0.015 atm (C) 0.017 atm (D) 0.022 atm (E) 0.48 atm N 2 (g) + O 2 (g) 2NO(g) K c1 = 1 / (2.1 × 10 30 ) = 4.7 6 × 10 -31 2NO(g) + Br 2 (g) 2NOBr(g) K c2 = (1.4) 2 = 1.9 6 SUM of reactions N 2 (g) + O 2 (g) + Br 2 (g) 2NOBr(g) K c = K c1 * K c2 K c = (4.7 6 × 10 -31 )(1.9 6 ) = 9.3 × 10 -31 K p = 28.5 = ( ) ( ) ( ) 2 3 2 2 2 3 20 . 0 x CO P O P = 017 . 0 5 . 28 10 0 . 8 3 = × = x P(CO 2 ) = 0.017 atm
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