Chapter 14 Solutions - Chemistry 112 Lecture Problem...

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Chemistry 112 – Lecture Problem Solutions CHAPTER 14 Example 1 Solution : a) Use stoichiometry to relate the rate of N 2 to the rate of H 2 : - . ( ) × ( ) ( )=- . 1 2 mol N2 g 1 s 3 mol H2 g 1 mol N2 g 3 6 mol H21 s The sign stays the same because H 2 is on the same side of the reaction as N 2 . Therefore, H 2 reacts at a rate of -3.6 mol/sec. b) Again, use stoichiometry to relate the rate of N 2 to the rate of NH 3 : - . ( ) × ( ) ( )=- . 1 2 mol N2 g 1 s 2 mol NH3 g 1 mol N2 g 2 4 mol NH31 s This time, however, the sign must be flipped because the two substances are on separate sides of the reaction. Therefore, the rate of reaction in terms of NH 3 is 2.4 mol/sec. Example 2 Solution : a) First, write the rate law for the equation and fill in variables for the unknown values: = rate kNOxBr2y Notice that between trials 1 and 2, the constant k and the concentration of Br 2 are both held constant. We can divide the rate law of trial 1 by trial 2 and obtain a mathematical relationship: = rate1rate2 kNO1xBr2ykNO2xBr2y = rate1rate2 NO1NO2x = lnrate1rate2 x∙lnNO1NO2 / / = . . ln24 M s150 M s x∙ln0 10 M0 25 M Substitute the values in and solve for the unknown to obtain = x 2 . Since NO is held constant between trials 1 and 3, divide the rate law of trial 1 by the rate law of trial 3 to solve for y in the same manner that x was solved for to obtain = y 1 . Now, all of the rate law is known except the rate constant
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This note was uploaded on 03/10/2009 for the course CHEM 112 taught by Professor Vandersluys,lorschmid,kylem during the Spring '07 term at Pennsylvania State University, University Park.

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Chapter 14 Solutions - Chemistry 112 Lecture Problem...

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