Question

1. The following data were collected at 20 ^oC for the reaction of bromphenol blue, HBPB^2-, and hydroxide ions, OH^-, shown in the equation below

HBPB^2-   +    OH^- <---> BPB^3-    +    H₂O

a)What is the rate equation for the reaction?

Determination	[HBPB2-]0, M	[OH-]o, M	Rate, M/sec
1	7.22 x 10-6	1.00	9.627 x10-9
2	7.22 x 10-6	0.25	2.490 x10-9
3	3.63 x 10-6	1.00	4.750 x10-

The reaction was run three times with different initial concentrations of the reactants yielding the following rates:

b)Determine the reaction order with respect to HBPB^2-

c)Determine the reaction order with respect to OH^-

d)Calculate the rate constant for this reaction at 20 ^oC

e)What is the ‘rate law’ for this reaction?

f)What other information do you need to calculate the activation energy for this reaction?

2. Why is it better to transfer one group of solutions into one flask and a second group into a second flask and then mix them, instead of pipetting the solutions sequentially into the same flask?

3. What will be the concentration of [I^-] in the reaction flask after mixing (See Run #6): 1.0 ml starch + 2.0 mL 1.2 x10^-2 M Na₂S₂O₃ + 4.0 mL 0.20 M KI + 4.0 mL 0.20 M KNO₃ + 4.0 mL 0.20 M (NH₄)₂S₂O₈ + 4.0 mL 0.20 M (NH₄)₂SO₄ ?

4. What will be the concentration of [S₂O₈^2-] in the same reaction mixture as above?

Answer 1

Solution :-

Q1). Using the data of the concentration and rate for the different tirals we can calculate the order of each reactant

a) rate equation for the reaction is

rate = -[delta HBPB2-]/ delta T

b) Calculating the order of HBPB^2- using the data from trial 1 and 3 whre the concentration of OH- is constant

Rate 3 / rate 1 = ([HBPB2-]3/[HBPB2-]1)^m

4.750*10^-9 / 9.627*10^-9 = (3.63*10^-6 / 7.22*10^-6)^m

0.5 = 0.5^m

Log 0.5 = m * log 0.5

Log 0.5 / log 0.5 = m

1 = m

so the order of the HBPB2- is 1

c) now lets calculate the order with OH- using the data from trial 1 and 2

rate2 / rate 1 = ([OH-]2/[OH-]1)^n

2.490*10^-9/ 9.627*10^-9 = (0.25/1)^n

0.25 = 0.25^n

Log 0.25 = n * log 0.25

Log 0.25 / log 0.25 = n

1=n

So order with OH- is also 1

d) Calculating the rate constant

Rate= K [ HBPB2-][OH-]

9.627*10^-9 = K [7.22*10^-6] [1.00]

9.627*10^-9 / [7.22*10^-6] [1.00] = K

1.33*10^-3 = K

e) rate law for the reaction is as follows

Rate= K [ HBPB2-][OH-]

f) To calculate the activation energy of the reaction we need the information of the rate at the different temperature.

2. Why is it better to transfer one group of solutions into one flask and a second group into a second flask and then mix them, instead of pipetting the solutions sequentially into the same flask?

Solution :- It is beeter to put the two solutions in the didfferent flask and then add them because when we directly pipette out solutions in the same flask then we cannot determine the time neded for the reaction.

3. What will be the concentration of [I^-] in the reaction flask after mixing (See Run #6): 1.0 ml starch + 2.0 mL 1.2 x10^-2 M Na₂S₂O₃ + 4.0 mL 0.20 M KI + 4.0 mL 0.20 M KNO₃ + 4.0 mL 0.20 M (NH₄)₂S₂O₈ + 4.0 mL 0.20 M (NH₄)₂SO₄ ?

Solution :- total volume of solution = 1 ml + 2 ml + 4 ml + 4 ml +4ml + 4ml = 19 ml

Using the initial volume and molarity of the KI we can calculate the final concentration of the I- in the mixture

M1V1 = M2V2

M2 = M1V1/V2

     = 0.2 M * 4.0 ml / 19 ml

    = 0.0421 M

So the molarity of the I- in the mixture = 0.0421 M

4. What will be the concentration of [S₂O₈^2-] in the same reaction mixture as above?

Solution :-

M1V1 = M2V2

M2 = M1V1/V2

    = 0.2 M * 4.0 ml / 19ml

   = 0.0421 M

So the concentration of the S2O8^2- = 0.0421 M