Chemistry Principles And Practice
3rd Edition
ISBN: 9781305295803
Author: David Reger; Scott Ball; Daniel Goode
Publisher: Cengage Learning
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Chapter 13, Problem 13.67QE
Interpretation Introduction
Interpretation:
An approximate energy-level diagram for the given catalyzed and uncatalyzed exothermic reactions has to be sketched.
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5
What would the complete ionic reaction be if aqueous solutions of potassium sulfate and barium acetate were mixed?
ed
of
Select one:
O a
2 K SO4 + Ba2 +2 C₂H3O21
K+SO4 + Ba2+ + 2 C2H3O21
K+SO42 + Ba2 +2 C2H3O2
BaSO4 +2 K+ + 2 C2H3O
estion
Ob.
O c.
Od.
2 K SO4 +Ba2 +2 C₂H₂O₂
BaSO4 + K+ + 2 C2H3O
BaSO4 + K + 2 C2H301
→Ba² +SO42 +2 KC2H3O
s page
(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the
reagents and product(s) for each step to receive full credit. The number of steps is provided.
(OC 4)
4 steps
4 steps
OH
b.
LTS
Solid:
AT=Te-Ti
Trial 1
Trial 2
Trial 3
Average
ΔΗ
Mass water, g
24.096
23.976
23.975
Moles of solid, mol
0.01763
001767
0101781
Temp. change, °C
2.9°C
11700
2.0°C
Heat of reaction, J
-292.37J -170.473
-193.26J
AH, kJ/mole
16.58K 9.647 kJ 10.85 kr
16.58K59.64701
KJ
mol
12.35k
Minimum AS,
J/mol K
41.582
mol-k
Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn &
Show your calculations for:
AH in J and then in kJ/mole for Trial 1:
qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J
qsin =
qrxn =
292.35 292.37J
AH in J = 292.375 0.2923kJ
0.01763m01
=1.65×107
AH in kJ/mol =
=
16.58K
0.01763mol
mol
qrx
Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con
Kelvin.)
AS=AHIT
(1.65×10(9.64×103) + (1.0
Jimai
Chapter 13 Solutions
Chemistry Principles And Practice
Ch. 13 - Prob. 13.1QECh. 13 - Prob. 13.2QECh. 13 - What is the difference between the integrated and...Ch. 13 - Prob. 13.4QECh. 13 - Explain why half-lives are not normally used to...Ch. 13 - Derive an expression for the half-life of a...Ch. 13 - Prob. 13.7QECh. 13 - Prob. 13.8QECh. 13 - Prob. 13.9QECh. 13 - Prob. 13.10QE
Ch. 13 - Prob. 13.11QECh. 13 - Prob. 13.12QECh. 13 - Prob. 13.13QECh. 13 - Prob. 13.14QECh. 13 - Prob. 13.15QECh. 13 - Prob. 13.16QECh. 13 - Prob. 13.17QECh. 13 - Prob. 13.18QECh. 13 - Prob. 13.19QECh. 13 - Prob. 13.20QECh. 13 - Prob. 13.21QECh. 13 - Prob. 13.22QECh. 13 - Nitrogen monoxide reacts with chlorine to form...Ch. 13 - Prob. 13.24QECh. 13 - Prob. 13.25QECh. 13 - Prob. 13.26QECh. 13 - Prob. 13.27QECh. 13 - Prob. 13.28QECh. 13 - Prob. 13.29QECh. 13 - Prob. 13.30QECh. 13 - Prob. 13.31QECh. 13 - Prob. 13.32QECh. 13 - Prob. 13.33QECh. 13 - Write a rate law for NO3(g) + O2(g) NO2(g) +...Ch. 13 - Prob. 13.35QECh. 13 - Prob. 13.36QECh. 13 - Prob. 13.37QECh. 13 - Rate data were obtained at 25 C for the following...Ch. 13 - Prob. 13.39QECh. 13 - Prob. 13.40QECh. 13 - Prob. 13.41QECh. 13 - Prob. 13.42QECh. 13 - Prob. 13.43QECh. 13 - Prob. 13.44QECh. 13 - Prob. 13.45QECh. 13 - Prob. 13.46QECh. 13 - Prob. 13.47QECh. 13 - Prob. 13.48QECh. 13 - When formic acid is heated, it decomposes to...Ch. 13 - Prob. 13.50QECh. 13 - The half-life of tritium, 3H, is 12.26 years....Ch. 13 - Prob. 13.52QECh. 13 - Prob. 13.53QECh. 13 - Prob. 13.54QECh. 13 - Prob. 13.55QECh. 13 - Prob. 13.56QECh. 13 - The decomposition of ozone is a second-order...Ch. 13 - Prob. 13.58QECh. 13 - Prob. 13.59QECh. 13 - Prob. 13.60QECh. 13 - A reaction rate doubles when the temperature...Ch. 13 - Prob. 13.62QECh. 13 - Prob. 13.63QECh. 13 - Prob. 13.64QECh. 13 - Prob. 13.65QECh. 13 - The activation energy for the decomposition of...Ch. 13 - Prob. 13.67QECh. 13 - Prob. 13.68QECh. 13 - Prob. 13.69QECh. 13 - Prob. 13.70QECh. 13 - Prob. 13.71QECh. 13 - Prob. 13.72QECh. 13 - Prob. 13.73QECh. 13 - Prob. 13.74QECh. 13 - Prob. 13.75QECh. 13 - Prob. 13.76QECh. 13 - Prob. 13.77QECh. 13 - Prob. 13.78QECh. 13 - Prob. 13.79QECh. 13 - Prob. 13.80QECh. 13 - The gas-phase reaction of nitrogen monoxide with...Ch. 13 - Prob. 13.82QECh. 13 - Prob. 13.83QECh. 13 - A catalyst reduces the activation energy of a...Ch. 13 - Prob. 13.85QECh. 13 - Prob. 13.86QECh. 13 - Prob. 13.87QECh. 13 - Prob. 13.88QECh. 13 - Prob. 13.89QECh. 13 - Prob. 13.90QECh. 13 - Prob. 13.91QECh. 13 - Prob. 13.92QECh. 13 - Prob. 13.93QECh. 13 - Prob. 13.94QECh. 13 - Prob. 13.95QECh. 13 - Prob. 13.96QECh. 13 - Prob. 13.98QECh. 13 - Prob. 13.99QE
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