kinetics in chemistry please help ! Studying the Kinetics of a Chemical Reaction
ID: 945198 • Letter: K
Question
kinetics in chemistry please help !
Studying the Kinetics of a Chemical Reaction: Iodine Clock Pre-lab Assignment The following data were collected at 20 degree C for the reaction of bromphenol blue, HBPB^2-, and hydroxide ions, OH^-, shown in the equation below What is the rate equation for the reaction? The reaction was run three times with different initial concentrations of the reactants yielding the following rates Determine the reaction order with respect to HBPB^2- Determine the reaction order with respect to OH^- Calculate the rate constant for this reaction at 20 degree C What is the 'rate law' for this reaction? What other information do you need to calculate the activation energy for this reaction?Explanation / Answer
a)
The rate of reaction, if we assume it as elementary rate law
should depend on the reactants
Rate = k*[HBP-2][OH-]
b)
In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.
Note that the generic formula goes as follows:
r = k [A]^a [B]^b
Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:
r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)
If we assume K1 and K2 are constant, then K1= K2 cancel each other
r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)
Then, order according to [A] and [B]
r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b
If we get two points in which A1 = A2, then we could get B, and vise versa for A...
From the data shown in YOUR table
choose point 1 and 3 so [OH-] cancels out
(9.627*10^-9)/(4.75*10^-9) = ((7.22*10^-6)/(3.63*10^-6))^a * (1/1)^b
2.0267 = 1.988^a * 1
a = 2.0267 /1.988
a = 1
1st order with respect to HBPB-2
b)
For OH-
chose point 1 and 2 so it cancels out
(9.627*10^-9)/(2.49*10^-9) = (7.22*10^-6)/(7.22*10^-6)^a * (1)/(0.25){ b
3.9 = 4^b
b = 1
1st order with respect to OH-
c)
rate for this reaction at 20°
Rate = k*[HBPB-2][OH-]
choose any point
9.627*10^-9 = k*(7.22*10^-6)(1)
k = (9.627*10^-9)/ (7.22*10^-6)
k = 0.0013333
e)
rate law:
Rate = k*[HBBP-2]^a[OH-]`^ b
Rate = 0.0013333*[HBBP-2][OH-]
f)
information for activtion energy required:
K = A*exp(-E/(RT))
we will need
A = the coefficient of collision
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