The decomposition of NH 3 to N 2 and H 2 was studied on two surfaces: Surface E a (kJ/mol) W 163 Os 197 Without a catalyst, the activation energy is 335 kJ/mol. a. Which surface is the better heterogeneous catalyst for the decomposition of NH 3 ? Why? b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form Rate = k [ NH 3 ] [ H 2 ] How can you explain the inverse dependence of the rate on the H 2 concentration?
The decomposition of NH 3 to N 2 and H 2 was studied on two surfaces: Surface E a (kJ/mol) W 163 Os 197 Without a catalyst, the activation energy is 335 kJ/mol. a. Which surface is the better heterogeneous catalyst for the decomposition of NH 3 ? Why? b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form Rate = k [ NH 3 ] [ H 2 ] How can you explain the inverse dependence of the rate on the H 2 concentration?
Solution Summary: The author explains that the substance that speeds up the rate of reaction without getting consumed itself in a chemical reaction is known as catalyst.
The decomposition of NH3 to N2 and H2 was studied on two surfaces:
Surface
Ea (kJ/mol)
W
163
Os
197
Without a catalyst, the activation energy is 335 kJ/mol.
a. Which surface is the better heterogeneous catalyst for the decomposition of NH3? Why?
b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction.
c. The decomposition reaction on the two surfaces obeys a rate law of the form
Rate
=
k
[
NH
3
]
[
H
2
]
How can you explain the inverse dependence of the rate on the H2 concentration?
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
For the compound: C8H17NO2
Use the following information to come up with a plausible structure:
8
This compound has "carboxylic acid amide" and ether functional groups.
The peaks at 1.2ppm are two signals that are overlapping one another.
One of the two signals is a doublet that represents 6 hydrogens; the
other signal is a quartet that represents 3 hydrogens.
Vnk the elements or compounds in the table below in decreasing order of their boiling points. That is, choose 1 next to the substance with the highest bolling
point, choose 2 next to the substance with the next highest boiling point, and so on.
substance
C
D
chemical symbol,
chemical formula
or Lewis structure.
CH,-N-CH,
CH,
H
H 10: H
C-C-H
H H H
Cale
H 10:
H-C-C-N-CH,
Bri
CH,
boiling point
(C)
Сен
(C) B
(Choose
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell