15. Figure 12.1 p. 562 Definition of Rate. 2NO 2 2NO + O 2
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20. At .90 M the rate is - (.98 - .76) = -0.22 = 5.5x 10 -4 M• s -1 (0-400) -400
21. At .45 M the rate is - (.52 - .31) = -0.22 = 2.7 x 10 -4 (1000-1800) -800
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41. Figure 12.4 A Plot of In (N 2 O 5 ) Versus Time (1st order)
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43. Figure 12.5: A Plot of (N 2 O 5 ) vs. Time for the Decomposition Reaction of N 2 O 5
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51. Z7e 546 Figure 12.7 A Plot of [A] Versus t for a Zero-Order Reaction
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53. Fig. 12.8 Decomposition Reaction N 2 O 2N 2 + O 2 Even though [N 2 O] is twice as great in (b) as in (a) the reaction occurs on a saturated platinum surface, so rate is zero order
55. Zero Order Graph Time Concentration A]/ t = slope t -k = A] t 1/2 = [A 0 ]/2k
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62. Use for Online HW pp Use for online HW Hint: on one problem you will have to first determine the order , then use the half-life equation (5 th line) to solve for k, then use the Integrated rate law (2 nd line) to solve for time (t).
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64. Figure 12.9 A Molecular Representation of the Elementary Steps in the Reaction of NO 2 and CO Overall: NO 2 + CO NO + CO 2 Step 1: NO 2 + NO 2 NO 3 + NO (k 1 ) Step 2: NO 3 + CO NO 2 + CO 2 (k 2 ) NO 3 is an intermediate