Question

The following data was collected at 20C for the reaction of bromphenol blue (HBPB^2-) and hydroxide ion (OH^-), shown in the equation.

HBPB^2- + OH^- $\rightleftharpoons$ BPB^3- + H₂O

The times required to consume a small but constant amount of HBPB^2- at varying initial HBPB^2- and OH^- concentrations were measured and are recorded below.

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	152

a) Determine the reaction order with respect to OH^-.

[OH-] = 1

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

[HBPB^2-] = 1

c) Write the rate law for the given equation.

R = k [HBPB^2-]¹[OH^-]1
------------------------------------------------------------
d) Assume that a small, constant amount of HBPB2- consumed in the experiment described corresponds to a HBPB2- concentration change of 7.22 x 10-7 M.

1. Calculate the rate constant (at room temperature).

2. What other information do you need to calculate activation energy?

3. Why is it better to transfer one group of solution 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.

4. For which part of the experiment would it be the most important to keep the flask in the constant-temperature water bath after mixing? What would happen if the temperature of that reaction mixture were not carefully controlled?

I need help with these questions.

Answer 1

The method you are using in your calculations is for an equation relating
(reaction rate) = k([HBPB]^x)([OH]^y)

but since it's time that you're given and not reaction rates, you have to substitute (amount of reactant consumed) / (time) for (reaction rate):

(amount of reactant consumed) / (time) = k([HBPB]^x)([OH]^y)

From your data and form a ratio using data for two reactions in which one of the reactants has the same concentration. For our purposes let's use determinations 1 and 2, where HBPB is constant.

(amount of HBPB consumed) / (time₁) k([HBPB]^x)([OH]₁^y)
---------------------------------------------------------- = -----------------------------
(amount of HBPB consumed) / (time₂) k([HBPB]^x)([OH]₂^y)
Since [HBPB] is constant in both the numerator and denominator, the amount of HBPB consumed is also constant for both the numerator and denominator since the reactions go to completion. Now we can cancel out like terms in the numerator and denominator and simplify:

(time₂) [OH]₁^y
------- = ----------
(time₁) [OH]₂^y

time₂ over time₁ is approximately 4 i.e 75s to 290s. [OH]₁ over [OH]₂ is also 4. Now we write out the equation again:

4 = 4^y

and we see y=1. Therefore the order of the reaction with respect to hydroxide is 1. You can repeat the same thing with your other determinations to find out the reaction order for HBPB, though this method will become more intuitive and you won't have to write out all the steps.

4 a)

the rate constant is given by

R = k [HBPB^2-]¹[OH^-]¹

some details are required like initial concentration

c) It is very important to monitor how long it takes for a certain reaction to occur. If that's the case, mixing the solutions from flasks allows you to mix the entire solutions all at once. This gives you a definite starting point for the reaction. If you add the second solution in smaller portions, the concentration of the second reactant keeps going up and down and up and down. You don't have a clear starting point in that case.