706 3 1 1 k 5967 10 5 0706 3 01 1 169 10 3 Rate II 3458 10 5 k706 3 005 1 k

# 706 3 1 1 k 5967 10 5 0706 3 01 1 169 10 3 rate ii

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k[0.706] 3 [.1] 1 ; k = 5.967 10 5 0.706 3 0.1 1 = 1.69 10 3 Rate II= 3.458 10 5 == k[.706] 3 [0.05] 1 ; k = 3.458 10 5 0.706 3 0.1 1 = 9.83 10 4 Rate III= 1.109 10 5 == k[.353] 3 [0.1] 1 ; k = 1.109 10 5 0.353 3 0.1 1 = 2.52 10 3 Rate IV= 0.112 10 3 == k[0.706] 3 [0.1] 1 ; k = 0.112 10 3 0.706 3 0.1 1 = 3.18 10 3 30 o C POST-LAB QUESTIONS: 1) Calculate the rate constant and write the rate law expression for the catalyzed decomposition of hydrogen peroxide. Explain how you determined the order of the reaction in H 2 O 2 and KI. The Rate expression can be expressed as: Rate=k[H 2 O 2 ] x [I - ] y Substituting the experimental data for I, and II, we have Rate I= 5.967 10 5 = k[0.706] x [.1] y Rate II= 3.458 10 5 = k[0.706] x [.05] y Reactions I and II were chosen because the concentration of H 2 O 2 is held constant while the concentration of I - is varied. Divide Rate I by Rate II Rate I= 5.967 10 5 = ¿ k[0.706] x [.1] y Rate II= 3.458 10 5 = ¿ k[0.706] x [.05] y
RateI RateII = 1.726 = 2 y ; ylog ( 2 ) = log ( 1.726 ) ; y = log ( 1.726 ) log ( 2 ) = 0.787 1 Substituting the experimental data for I, and III, we have Rate I= 5.967 10 5 = k[0.706] x [.1] y Rate III= 1.109 10 5 = k[0.353] x [.1] y Divide Rate I by Rate I1 Rate I= 5.967 10 5 = k[0.706] x [.1] y Rate III= 1.109 10 5 = k[0.353] x [.1] y RateI RateII I = 5.38 = 2 x ; xlog ( 2 ) = log ( 5.38 ) ; x = log ( 5.38 ) log ( 2 ) = 2.69 3 The rate equation is: Rate=k[H 2 O 2 ] 3 [I - ] 1 The rate is third order in H 2 O 2 and first order I - . The reaction is forth order overall. 2) The following mechanism has been proposed for this reaction: H 2 O 2 + I → IO H 2 O H 2 O 2 + IO → I + H 2 O + O 2 If this mechanism is correct, which step must be the rate-determining step? Explain. Answer: No it is not valid for observed rate law, because here in mechanism order with respect to H 2 O 2 to 1 but observed order is 3. 3) Use the Arrhenius equation (shown below) to determine the activation energy, E a , for this reaction. ln k 1 k 2 = E a R ( 1 T 2 1 T 1 ) k 1 = 1.69 10 3 T 1 =21.8 R= 8.314
k 2 = 3.18 10 3 T 2 = 30.0 ln 1.69 10 3 3.18 10 3 = E a 8.314 ( 1 30.0 1 21.8 ) (− 0.144 8.314 ) 0.013 = E a E a =− 92.09 KJ/mol

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• Spring '14
• BOBBYBURKES
• Chemistry