Chemistry
10th Edition
ISBN:9781305957404
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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Question
![Complete the following exercises
1) Write the general rate law equation for the reaction
Acetone (C3H,0(aq)) + Hag) + Iodine (12(aq)
→ C3H501(aq) + Haq) + Taq)
Rate
K CAcetore J [ Lz)
2) Determine the value of the exponents. Write the complete ratio comparing one reaction
mixture to another and show the cancellation of terms. Remember to compare the rates
for two mixtures for which only one of the concentrations varies.
a)
Rate I_= 1.9x10" M/s= k [IGM)" (020) [0.0010M)"
Rate „I = 38x106 M/s = K [o.8oM]o2079 Po.00/ DM]2
%3D
3.8x10-6
2.7089 =(66)*
%3D
%3D
log 2.0789. 1oq1-67% log[0807
0:3178 = x Co 300)
l0.80]x
0.3178
Exponent for Acetone, x =
1.1
3010
b)
To.4OM) (0.0010M)
C02OM)" 10.0010M)
lo4oM)-y log Lozom)
Rate = 8-0 Y[O
M/s= k(o.80
)
%3D
%3D
Rate I= 3.4ylo-t M/s= Rlo.60)^
%3D
log 2-1053 = ylog (ouoy log Lozom)
O-20m)
y=o.3237
l107
Exponent for H", y =
0.3010
c)
6.
2.
Rate I =
3.8x10 M/s= k Co8oM)" L020M)
%3D
k Coi8o Mje Co 20)y coo00Sm)
log (0.0010M)- log (o.000sm)
=2 (0-3010)
Rate
%3D
4.0x10~ M/=
(० ५ ०-११
0.0223
-0.07
0.0223
Exponent for I2, z =-0.07
0:3010](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b574d51-0fdb-495e-9b46-97198cf41098%2F40736922-3b52-4f67-acdf-f54c2e91931e%2Favouhp9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Complete the following exercises
1) Write the general rate law equation for the reaction
Acetone (C3H,0(aq)) + Hag) + Iodine (12(aq)
→ C3H501(aq) + Haq) + Taq)
Rate
K CAcetore J [ Lz)
2) Determine the value of the exponents. Write the complete ratio comparing one reaction
mixture to another and show the cancellation of terms. Remember to compare the rates
for two mixtures for which only one of the concentrations varies.
a)
Rate I_= 1.9x10" M/s= k [IGM)" (020) [0.0010M)"
Rate „I = 38x106 M/s = K [o.8oM]o2079 Po.00/ DM]2
%3D
3.8x10-6
2.7089 =(66)*
%3D
%3D
log 2.0789. 1oq1-67% log[0807
0:3178 = x Co 300)
l0.80]x
0.3178
Exponent for Acetone, x =
1.1
3010
b)
To.4OM) (0.0010M)
C02OM)" 10.0010M)
lo4oM)-y log Lozom)
Rate = 8-0 Y[O
M/s= k(o.80
)
%3D
%3D
Rate I= 3.4ylo-t M/s= Rlo.60)^
%3D
log 2-1053 = ylog (ouoy log Lozom)
O-20m)
y=o.3237
l107
Exponent for H", y =
0.3010
c)
6.
2.
Rate I =
3.8x10 M/s= k Co8oM)" L020M)
%3D
k Coi8o Mje Co 20)y coo00Sm)
log (0.0010M)- log (o.000sm)
=2 (0-3010)
Rate
%3D
4.0x10~ M/=
(० ५ ०-११
0.0223
-0.07
0.0223
Exponent for I2, z =-0.07
0:3010

Transcribed Image Text:3) Complete the rate law equation by substituting the values of the exponents x, y and z in
the four equations. Determine the value ofk for each reaction mixture. Show your work
and give the correct units for k.
-2
2.4x10
Mixture I: k =
3,8y10-6
2-4
0.80X0-20 x 0-00/0
9.9x10
-6
Mixture II: k =
=2.5
1-6x020x0-06 10
Mixture III: k=
?ox10-6
20 x 10
= 2-5
6.80x0 Y0 Ao.
S xio-
-6
Mixture IV: k=
X10
4.0x10
0.80 x0.20x0.00
ニ
3.
)
Average k =
%3D
4) Write the complete rate law by substituting the values of x, y, z and k into the general rate
law.
5) What is the overall order of the reaction?
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