Unformatted text preview: Order with respect to BrO3: first order (rou nd to th e nearest who le nu mb er) Reaction Order Determination for H+ Place yo ur plot relatin g to the reactio n o rd er fo r H+ here (cover this instruction bo x so that your g raph is an ap rop riate size). Properly label your graph (labels for axes, including units, and a title). Include a trendline and its equation and R2 value on your graph. This is done by rightclicking on one of the data points on your graph and cho sing "Ad Trendline" from the drop down menu. The first tab asks what type of trendline you wish to use, and the Options tab al ows you to include the trendline equation and R2 value. Order with respect to H+: secon d o rd er (rou nd to th e nearest who le nu mb er) Calculation of Rate Constant, k Determination of Activation Energy, Ea Type your calculation of "k" here: k=rate/[BrO3]^b[I]^i[H 3O +]^h= (3.0e8)/(0.0 80M )^1(0.0 20M )^1(0.020 M )^2= 4.6875 Type your calculation of Ea here: ln(k 1/k2) = Ea/R (1/T2  1/T 1)  > Ea = R ln(k1/k2)/(1/T 2  1/T1)Ea= (8.314 J/K.m ol)(ln(8.53125/17.76)/(1/298.26 7  1/273) = 19645.314 J/m ol = 19.645 kJ/m ol Type an exam ple calc ulation for determ ining the CV+ concentration from the absorbance data: A=[CV+]e where e=absorptivity and A=absorbance so, [C V+]=absorbance/absorptivity =(0.172/41761.71352)= 1.7241e6 M Results and Discus ion Part I 1. B ased on your d ata, write ou t the com plete rate law includ in g value and un its for th e rate constant. (1 pt) R ate= k[BrO3]^b[I]^i[H 3O]^h  > R ate= (17.76 L^2/(m ol^2*s) *[BrO3][I][H+]^2 2. T he literature values for the reactio n o rd ers are 1 fo r BrO 3 and I an d 2 fo r H+ C om pare the ac uracy of yo ur o rd ers to these literature values. (C alcu late the % er o r.) Discus your m ost likely so urces of er or. (2 p ts) BrO3 = (1.00.705)/(1.0) = 29.5% I = (1.00.6862) = 31.38% H + = (2.07532.0) = 3.765% Overal m y literature values w ere pret y ac urate, especial y for H +. One sorce of er or m ay have be n that the reaction of the tim er w as slow, w hich w ould have m ade the tim e in acurate. Anohter sorce of er or would have be n that the date w as transfered inac urately by other students in other groups. 3. H ow d o th e activation energ ies for the catalyzed an d u ncatalyzed reaction co mp are (in clude a % dif erence in your d iscu s io n)? Is this in lin e with wh at is exp ected? (2 pts) T he Ea for the uncatalyzed reaction w as 19.645 kJ/m ol and 43.59 kJ/m ol for the catalyzed rection. 43.59 19.645/43.59 = 54.932%. T his is not as I w ouldhave expected it to be because during the reaction, the catalys t al owed the reaction to proce d m ore quickly. The er or in this calc ulation m ust have stem ed from the tim es recorded for these trial runs. Part I 1. B ased on your d ata, write ou t the com plete rate law includ in g value and un its for th e rate constant. Th e literature valu es o f the o rd ers with resp ect to C V+ are and O H are 1 and 1 respectively. H ow do your valu es com pare? (2 p ts) R ate= (2.34 e2 m ol^1 sec^1)*[C V+][O H] M y values for the order of CV+ and OH  are som ewhat close to the literature values: source of er or: C V+ = (1.01.0)/1 = 0% OH  = (1.01.0)/1 = 0%OH  = (1....
View
Full Document
 Spring '08
 N.
 Reaction, Kinetics, Run, Rate equation, CV+

Click to edit the document details