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1aa3-2012-tut3-a

# 1aa3-2012-tut3-a - CHEMISTRY 1AA3 Week of TUTORIAL PROBLEM...

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CHEMISTRY 1AA3 Week of JANUARY 23, 2012 TUTORIAL PROBLEM SET 3 QUESTIONS ______________________________________________________________________________ Page 1 of 5 1. What are the overall orders of the reactions to which the following rate laws apply? (a) rate = k[H 2 ][I 2 ] (b) rate = k[NO 2 ] 2 [H 2 ] (c) rate = k[Cl 2 ] (3/2) [O 2 ] SOLUTION: (a) 2 (b) 3 (c) 2.5 Note: orders of reactions are usually integers, but can be fractional values such as 1.5 or 0.5 in rare cases 2. The equation for the reaction of NO(g) with O 2 (g) at 660K is: 2 NO(g) + O 2 (g) 2 NO 2 (g) The reaction rate data below was measured by observing the disappearance of NO: Concentration (mol.L -1 ) Initial Rxn Rate Expt. No. [NO] [O 2 ] (mol L -1 s -1 ) 1 0.020 0.020 2.5 x 10 -4 2 0.020 0.040 5.0 x 10 -4 3 0.060 0.040 4.5 x 10 -3 (a) Derive the rate law for the reaction in terms of the disappearance of NO(g). (b) What is the overall order of the reaction? (c) What is the numerical value of the specific rate constant k at 660K? Give the units of k. (d) If the concentrations of NO and O 2 had been expressed in terms of partial pressure p(N0) and p(O 2 ), both in units of atmospheres, what would the units of k be now? SOLUTION: (a) The general form of the rate law is: v o = k[NO] m [O 2 ] n From experiments 1 & 2, as [O 2 ] is doubled, rate of reaction also increases by a factor of two. Thus, rate is proportional to [O 2 ] 1 .

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2 From experiments 2 and 3, as [NO] is tripled (while [O 2 ] is held constant), rate of reaction changes by a factor of 9: (4.5 x 10 -3 /5.0 x 10 -4 =) 9; i.e., reaction rate is proportional to [NO] 2 .
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1aa3-2012-tut3-a - CHEMISTRY 1AA3 Week of TUTORIAL PROBLEM...

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