Reaction Rate Data Mixture Time Rate [1] [BrO3] [H*] 1000/t (sec) 103 1 9.72 .002 M .008 M .02 M 2 47 21.3 .004M .008M .02M 3 50 20 .002M . 016 M 02M 4 24 41.7 .002M . 008M .04M 107 9.35 001GM 0032M .032M -o)(8.0) = k[ ][8,0, ]*[ #* ]P (0.0401 (2.0) : rate 1 = =.0032 25 = k [ ]m[BrO, ]"[ H ]P 125) rate 2 = O 32 Solve for m, the order of the reaction with respect to the I, to two decimal places and then round off to the nearest integer. (2 decimal places) (nearest integer) m = m = Apply the same approach to find the value of n, the order of the reaction with respect to the BrO, ion. Show your set-up. (2 decimal places) n = (nearest integer) n = Apply the method once again to find p, the order with respect to the H* ion. Show your set-up.

Chemistry
10th Edition
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you help me figure out the questions below? I found the missing values in the chart but I need help with questions below the chart

(o.10) (5.0) .02
(o.10) (10.0)
=.04
25
(0.040)(5.0)
(25)
(0.040)(10.0)
25
=0.008
=0.016
Reaction Rate Data
(25)
Mixture
Time
Rate
[1]
[BrO3]
[H*]
(sec)
103
1000/t
1
9.72
.002 M
.008 M
.02 M
47
21.3
.004M
.008M
.02M
3
50
20
.002M
.016M
.02M
. 008M
. 0032M
24
41.7
.002M
.04M
5
107
9.35
001GM
• 032M
olo)(も.0)
= k[ ]m[8rOs ]"[ H* ]P
(0.040) (2.0)
: rate 1 =
%3D
=.0032
25
= k [ ]m[BrO, 1 [ H ]P
125)
rate 2 =
FO 32
Solve for m, the order of the reaction with respect to the I, to two decimal places and then round off
to the nearest integer.
(2 decimal places)
(nearest integer)
m =
m =
Apply the same approach to find the value of n, the order of the reaction with respect to the BrO3
ion. Show your set-up.
tee your Ke
(2 decimal places)
(nearest integer)
n =
n =
Apply the method once again to find p, the order with respect to the Ht ion. Show your set-up.
Rate
(2 decimal places) p =
(nearest integer)
%3D
Having found m, n, andp (nearest integers, the rate law is written as:
rate =
Transcribed Image Text:(o.10) (5.0) .02 (o.10) (10.0) =.04 25 (0.040)(5.0) (25) (0.040)(10.0) 25 =0.008 =0.016 Reaction Rate Data (25) Mixture Time Rate [1] [BrO3] [H*] (sec) 103 1000/t 1 9.72 .002 M .008 M .02 M 47 21.3 .004M .008M .02M 3 50 20 .002M .016M .02M . 008M . 0032M 24 41.7 .002M .04M 5 107 9.35 001GM • 032M olo)(も.0) = k[ ]m[8rOs ]"[ H* ]P (0.040) (2.0) : rate 1 = %3D =.0032 25 = k [ ]m[BrO, 1 [ H ]P 125) rate 2 = FO 32 Solve for m, the order of the reaction with respect to the I, to two decimal places and then round off to the nearest integer. (2 decimal places) (nearest integer) m = m = Apply the same approach to find the value of n, the order of the reaction with respect to the BrO3 ion. Show your set-up. tee your Ke (2 decimal places) (nearest integer) n = n = Apply the method once again to find p, the order with respect to the Ht ion. Show your set-up. Rate (2 decimal places) p = (nearest integer) %3D Having found m, n, andp (nearest integers, the rate law is written as: rate =
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