# ANSWERS CHEM 1420 Test 1 - Your name 1420-002 HOUR TEST 1...

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Your name _______________________________________ 1420-002 HOUR TEST -- February 14, 2007 _____ 1. The K eq values apply to the following two equations at 25°C: H 2 ( g ) + I 2 ( g ) W 2HI( g ) K eq = 6.17 x 10 2 H 2 ( g ) + Br 2 ( g ) W 2HBr( g ) K eq = 1.88 x 10 19 Calculate K eq for the equation: I 2 ( g ) + 2HBr( g ) W Br 2 ( g ) + 2HI( g ) (A) 3.28 x 10 -17 . (B) 3.04 x 10 16 . (C) 1.56 x 10 22 . (D) 8.64 x 10 -23 . (E) 4.29 x 10 17 . (F) 8.05 x 10 21 . (G) 1.00. (H) 1.16 x 10 22 . Invert the second equation, then add the two equations H 2 ( g ) + I 2 ( g ) W 2HI( g ) K eq = 6.17 x 10 2 2HBr( g ) W H 2 ( g ) + Br 2 ( g ) K eq = 5.32 x 10 -20 K = (6.17 x 10 2 )(5.32 x 10 -20 ) = 3.28 x 10 -17 _____ 2. K eq = 1.03 x 10 5 for the reaction CO( g ) + H 2 O( g ) W CO 2 ( g ) + H 2 ( g ) . A 1-liter flask at equilibrium has 0.002 mole of CO, 0.0015 mole of H 2 O, and 0.01 mole of CO 2 . How many moles of H 2 exist in the flask? (A) 1.24. (B) 0.39. (C) 5.20. (D) 30.9 . (E) 0.0034. (F) 0.87. _____ 3. For the reaction I 2 ( g ) + Br 2 ( g ) W 2IBr( g ), a flask with an initial concentration of 1.0 M I 2 , 1.0 M Br 2 , and no IBr is allowed to reach equilibrium. After equilibrium is established, the concentration of I 2 is found to be 0.285 M. What is K eq ? (A) 12.4. (B) 25.2 . (C) 5.78. (D) 1.44. (E) 0.85. (F) 0.045. (G) 1.45 x 10 -6 . Set up table for I 2 ( g ) + Br 2 ( g ) W 2IBr( g ) Initial 1.0 1.0 0 Change Final 0.285 Fill in table for I 2 ( g ) + Br 2 ( g ) W 2IBr( g ) Initial 1.0 1.0 0 Change -0.715 -0.715 +1.430 Final 0.285 0.285 1.430

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K = (1.43) 2 /0.285)(0.285) = 25.2 _____ 4. For the exothermic equilibrium CO( g ) + O 2 ( g ) W CO 2 ( g ), if the temperature is increased, the reaction (A) shifts to the left . (B) shifts to the right. (C) does not change. _____ 5. For the reaction C(s) + CO 2 (g) W 2CO(g), at equilibrium at 850 / C the relative molar amounts of C, CO 2 , and CO in a 1.00-L flask are 0.050, 0.062, and 0.54 respectively. What is the value of the equilibrium constant K? (A) 14.25. (B) 0.030. (C) 0.15. (D) 4.70 . (E) 2.82. (F) 2.00. K = (CO) 2 /(CO 2 ) = (0.54) 2 /(0.062) = 4.70 _____ 6. What effect does a catalyst have on the value of the K eq for a chemical reaction? (A) no effect . (B) the ratio increases. (C) the ratio decreases. (D) the ratio increases with increasing temperature. (E) the ratio decreases with increasing temperature.
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