Chapter-15-HW

Chapter-15-HW - Homework #4 Chapter 15 Chemical Kinetics 8....

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1 Homework #4 Chapter 15 Chemical Kinetics 8. Arrhenius Equation RT a E Ae k Therefore k depends only on temperature. The rate of the reaction depends on all of these items (a-d). 14. a) mol Ls b) mol c) s 1 d) L mol s e) 2 2 L mol s 15. Rate has units of s mol L therefore the units of the rate constant must compensate for any missing/additional units. 1 2 1 2 s L mol 17. a) General Rate Law     y x Cl NO k Rate 2 Experiment [NO] o (M) [Cl 2 ] o (M) Initial Rate ( ±² ³´µ ) 1 0.10 0.10 0.18 2 0.10 0.20 0.36 3 0.20 0.20 1.45 From experiments 1 and 2 ([NO] o is kept constant) as the initial concentration of Cl 2 is doubled the initial rate of the reaction doubles ( ¶·¸¹ ¶·º» ¼½·¾ ) ; therefore, the reaction is first order with respect to Cl 2 . From experiments 2 and 3 ([Cl 2 ] o is kept constant) as the initial concentration of NO is doubled the initial rate of the reaction quadruples ( º·¿À ¶·¸¹ ¼Á·¾ ) ; therefore, the reaction is second order with respect to NO.     2 2 Rate k NO Cl You could solve this problem the long way. Find 2 equations where the concentration of NO is constant     min 0.18 0.10 0.10 xy mol L k M M     min 0.36 0.10 0.20 mol L k M M Divide the 2 equations by each other.         min min 0.10 0.10 0.18 0.36 0.10 0.20 0.50 0.50 1 mol L mol L y k M M k M M y     2 k x
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2 Find 2 equations where the concentration of Cl 2 is constant     min 0.36 0.10 0.20 x mol L k M M     min 1.45 0.20 0.20 x mol L k M M Divide the 2 equations by each other.             min min 0.10 0.20 0.36 1.45 0.20 0.20 0.25 0.50 log 0.25 log 0.50 2 x mol L x mol L x k M M k M M x x     2 2 Cl NO k Rate b)     2 2 Rate k NO Cl     2 2 2 min min 0.18 0.10 0.10 180 mol L L mol k M M k 18. a) General Rate Law     y x O S I k 2 8 2 Experiment [I - ] o (M) [S 2 O 8 2- ] o (M) Initial Rate ( ±² ³´µ ) 1 0.080 0.040 12.5×10 -6 2 0.040 0.040 6.25×10 -6 3 0.080 0.020 6.25×10 -6 4 0.032 0.040 5.00×10 -6 5 0.060 0.030 7.00×10 -6 From experiments 1 and 2 ([S 2 O 8 2- ] o is kept constant) as the initial concentration of I - is doubled the initial rate of the reaction doubles ( ¶·¸¹º¶» ¼½ ½¸·¹º¶» ¼½ ¾¿¸À ) ; therefore, the reaction is first order with respect to I - . From experiments 1 and 3 ([I - ] o is kept constant) as the initial concentration of S 2 O 8 2- is doubled the initial rate of the reaction doubles ( ¶·¸¹º¶» ¼½ ½¸·¹º¶» ¼½ ¾¿¸À ) ; therefore, the reaction is first order with respect to S 2 O 8 2- .     2 8 2 O S I k b) Run #1    2 28 5 s s 1.25 10 0.080 0.040 0.0039 mol L L mol Rate k I S O k M M k       Run #2
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3    2 28 6 s s 6.25 10 0.040 0.040 0.0039 mol L L mol Rate k I S O k M M k       Run #3    2 6 s s 6.25 10 0.080 0.020 0.0039 mol L L mol Rate k I S O k M M k     Run #4
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Chapter-15-HW - Homework #4 Chapter 15 Chemical Kinetics 8....

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