Fill in the following table. Look below the table for instructions on how to cal

Question

Fill in the following table. Look below the table for instructions on how to calculate the values for each column of the table.

Slope=-0.341 Intercept=-1.807

A)Determine the value of q from the graph. Explain your answer.

B)What is the rate law for the reaction?

C)For each trial, calculate the rate constant. What is average value of the rate constant?

D)In this experiment, you assumed that [S2O82–] >> [S2O32–]. To find out if this assumption is correct, calculate the ratio of [S2O82–] / [S2O32–] for Trial 3. (In Trial 3, the [S2O82–] was lowest, and therefore the ratio [S2O82–] / [S2O32–] is the smallest).

E)In this experiment, you assumed that only a small amount of S2O82– was used during the time trial so that the concentration of this reactant did not change appreciably during the course of the reaction. Calculate how much S2O82– was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?

Really need help with these problems! Thank you :)

Fill in the following table. Look below the table for instructions on how to calculate the values for each column of the table.

a b c d e f g h i Trial t [I3-] / t (M/s) log([I3-] / t) V 0.2 M KI added (mL) [I–]0(M) log[I–]0 V of 0.2 M (NH4)S2O8 (mL) [S2O8]0 (M) log[S2O8]0 1 19s 0.0116 -1.94 25 0.077 -1.11 25 0.077 -1.11 2 40s 0.0058 -2.24 25 0.077 -1.11 12.5 0.038 -1.42 3 85s 0.0144 -1.84 25 0.077 -1.11 6.25 0.019 -1.72 4 41s 0.0563 -1.25 12.5 0.038 -1.42 25 0.077 -1.11

Slope=-0.341 Intercept=-1.807

Explanation / Answer

a) for a reaction your ra = dC/t = KCn using Log we get logC/t =logk nlogC as it have a relation liner we have

Y = mx + b where b = logK and m = n or order of reaction in our graph we have than b = -1.807 and m = 0.341

as b = logK =) K = 10b =) K = 10-1.807 K = 1.6x10-3

b) n = 0.341 as n is near to 0 we can says than the rate is of order 0 for I3-  

the rate law is C/t = K

C) Kt = dC/dt =) K = dC/dt/t   

for trial 1 dC/dt = 0.0116 t = 19s ) K = dC/dt/t =  0.0116/19s = 6.1x10-4

for trial 2 dC/dt = 0.0058 t = 40s ) K = dC/dt/t =  0.0058/40s = 1.5x10-4

for trial 3 dC/dt = 0.0144 t = 85s ) K = dC/dt/t =  0.0144/85s = 1.7x10-4

for trial 4 dC/dt = 0.0563 t = 41s ) K = dC/dt/t = 0.0563/41s = 1.4x10-3

3I-   + S2O8-2 -> [S2O3-2 ] + I3-   

Kt = Ln [S2O8-2 ]o / [S2O8-2 ]f as [S2O8-2 ]f = [S2O8-2 ]o - [S2O3-2 ]

Kt = Ln[S2O8-2 ]o/[S2O8-2 ]o - [S2O3-2 ] ekt = [S2O8-2 ]o/[S2O8-2 ]o - [S2O3-2 ] =

ekt[S2O8-2 ]o - ekt[S2O3-2 ] = [S2O8-2 ]o =

[S2O3-2 ] = [S2O8-2 ]o (1-ekt)/ ekt  for [S2O8-2 ]o = 0.077M t = 19s and K = 1.6x10-3s-1

[S2O3-2 ] = 0.077M o (-1 + e1.6x10-3s-1 19s)/ e1.6x10-3s-1 19s = 0.03M =) [S2O8-2 ]f = 0.077M- 0.030M = 0.047M

the radio [S2O3-2 ]/ [S2O8-2 ] = 0.030/0.047 = 0.64 or 64%

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