1)The following data were collected at 20 degrees Celsius for the reaction of br
ID: 486998 • Letter: 1
Question
1)The following data were collected at 20 degrees Celsius for the reaction of bromphenol blue (HBPB2-) and hydroxide ions (OH-).
HBPB2- + OH- <-------> BPB3- + H2O
The times required to consume a small but constant amount of HBPB2- at varying initial HBPB2- and OH- concentrations were measured and are recorded below
a) Determine the reaction order with respect to OH-.
b) Determine the reaction order with respect to HBPB2-.
c) Write the rate law for the reaction of HBPB2- with OH-.
2) Assume that the small constant amount of HBPB2- consumed in the experiment described corresponds to an HBPB2- concentration change of 7.22 x 10-7 M.
a) Calculate the rate constant for this reaction at room temperature.
b) What other information do you need to calculate the activation energy for this reaction?
Determination [HBPB2-], M [OH-], M Time, s 1 7.22 x 10-6 1.00 75 2 7.22 x 10-6 0.25 290 3 3.63 x 10-6 1.00 152Explanation / Answer
•hold [A] in trial 1 and 2 constant, then rate2/rate1= k([B2]/[B1])y so 1x104/1x10-4= k (.2/.1)y and y = 0. This is a 0th order reaction in B.
•now hold [B] in trial 1 and 3 constant, then rate3/rate1= k([A3]/[A1])x so3x104/1x104=k(.3/.1)x
and x = 1. This is a 1st order reaction in A.
Let HBPB2- be A and OH be B
A)Reaction order with respect to OH
[OH-], M
Time, s
rate
rate3/rate1
Order of the reaction
1.00
75
0.0133or 1.33x102
6.57x103/1.33x102=
4.94x101
x=1
Hence 1st order reaction
0.25
290
8.62x104
1.00
152
6.57 x103
B)Reaction order with respect to HBPB2-
[HBPB2-], M
Time, s
rate
rate2/rate1
Order of the reaction
7.22 x 10-6
75
7.22 x 10-6/75=
0.096 x 10-6
0.025 x 10-6/0.096 x 10-6
=0.260 x 10-0
y=0
Hence ‘0’ order reaction
7.22 x 10-6
290
0.025 x 10-6
3.63 x 10-6
152
0.025 x 10-6
C) Rate = k[A]x[B]yreaction order = x + y=1+0=1
Rate = k [A]1[B]0= k [A]is 1st order in [A]and 0th order in [B]and 1st order for the reaction
.
2. HBPB2- concentration change of 7.22 x 10-7 M.
Rate constant
We have rate = k [HBPB2-]1
rate0.096 x 10-6 M/s /7.22 x 10-6M=k(0.0133M1s1)=k
b) As temperature increases, gas molecule velocity also increases (according to the kinetic theory of gas). This is also true for liquid and solid substances. The (translational) kinetic energy of a molecule is proportional to the velocity of the molecules (KE = 1/2 mv2). Therefore, when temperature increases, KE also increases; as temperature increases, more molecules have higher KE, and thus the fraction of molecules that have high enough KE to overcome the energy barrier also increases.
The fraction of molecules with energy equal to or greater than Ea is given by the exponential term eEaRT
in the Arrhenius equation:
k=AeEa/RT
[OH-], M
Time, s
rate
rate3/rate1
Order of the reaction
1.00
75
0.0133or 1.33x102
6.57x103/1.33x102=
4.94x101
x=1
Hence 1st order reaction
0.25
290
8.62x104
1.00
152
6.57 x103
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