Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. F e 2 O 3 ( g ) + 2 A I 2 O 3 ( s ) → 2 F e ( s ) △ r H ° = − 852 kJ mol -1 1.00 of granular F e 2 O 2 and 2.00 molof AI are mixed at room temperature ( 25 ° C ) , and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat capacity over a broad temperature range is about 0.8 J g − 1 ° C − 1 (The melting point of iron is 1530 °C.) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the meting point of iron.
Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. F e 2 O 3 ( g ) + 2 A I 2 O 3 ( s ) → 2 F e ( s ) △ r H ° = − 852 kJ mol -1 1.00 of granular F e 2 O 2 and 2.00 molof AI are mixed at room temperature ( 25 ° C ) , and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat capacity over a broad temperature range is about 0.8 J g − 1 ° C − 1 (The melting point of iron is 1530 °C.) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the meting point of iron.
Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic.
F
e
2
O
3
(
g
)
+
2
A
I
2
O
3
(
s
)
→
2
F
e
(
s
)
△
r
H
°
=
−
852
kJ mol
-1
1.00 of granular
F
e
2
O
2
and 2.00 molof AI are mixed at room temperature
(
25
°
C
)
, and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat capacity over a broad temperature range is about
0.8
J g
−
1
°
C
−
1
(The melting point of iron is 1530 °C.) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the meting point of iron.
3. Consider the compounds below and determine if they are aromatic, antiaromatic, or
non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I
electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly
drawn and you should be able to tell that the bonding electrons and lone pair electrons
should reside in which hybridized atomic orbital 2. You should consider ring strain-
flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti-
aromaticity)
H H
N
N:
NH2
N
Aromaticity
(Circle)
Aromatic Aromatic Aromatic Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic TT
electrons
Me
H
Me
Aromaticity
(Circle)
Aromatic Aromatic Aromatic
Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic πT
electrons
H
HH…
Chapter 7 Solutions
Selected Solutions Manual For General Chemistry: Principles And Modern Applications
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