KINE 508: Seadying che Kinetics of s Chemical Reaction section Pre-Laboratory As

Question

KINE 508: Seadying che Kinetics of s Chemical Reaction section Pre-Laboratory Assignment 1. The following data were collected at 20 C for the (c) reaction of bromphenol blue (HBPB2-) and hydroxide with OH ions (OH), shown in Equation 16. Write the rate law for the reaction of HBPB2 The times required to consume a small but con- stant amount of HBPB- at varying initial HBPB2-and OH-concentrations were measured and are recorded below determinaton HBPEM IOH LM Bime .2 Assume that the small, constant amount df 7.22 × 10-6 1.00 75 7.22 × 10-6 290 3.63 x106 00 152 HBPB2 consumed in the experiment described in Pre-Laboratory Assignment 1 corresponds to an HBPB2-concentration change of 7.22 × 10" M. 025 290HBPB concentration (a) Determine the reaction order with respect to at room temperature. temperature. (a) Calculate the rate constant for this reaction OH knev4H8P8 (clined),.arnaunr ¢H67B cno me (b) What other information do you need to cal- (b) Determine the reaction order with respect to culate the activation energy for this HBPB2

Explanation / Answer

Q1.

a and 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

then choose point 1 and 2 so HBPB cancels

(1/75)/(1/290) = (1)^a * (1/0.25)^b

b = ln((1/75)/(1/290) ) / (ln(1/0.25))

b = 0.98 = 1

then, choose point 1 and 3 so OH cancels

(1/75)/(1/152) =(7.22*10^-6)/(3.63*10^-6)^a * (1/1)^b

ln(2) / ln (2) = a

a = 1

then, this is

c)

find rate law:

Rate = k*[HBPB-2][OH-]

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