5.112 Principles Of Chemical Science Worksheet With Answers - Problem Set #4 Solutions - 2011 Page 4
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16Problem set #4 solutions for 5.112, Fall 2011
Page 4 of 16
b) Write 5 reactions which when summed will result in the reaction: SrO Sr + O.
Four of these reactions are related to the energies for the processes given in the table below.
IE
of Sr
549 kJ/mol
1
IE
of Sr
1064 kJ/mol
2
EA of O
141 kJ/mol
-
-844 kJ/mol
EA of O
Then, calculate the bond dissociation energy of CaO into its neutral atoms, Ca and O.
Five possible reactions are:
Reaction
ΔE
→ Sr
ΔE
+1
-1
-1
Sr
+ e
= -(IE
of Sr) = -549 kJ mol
A
1
→ Sr
ΔE
+2
-1
+1
-1
Sr
+ e
=-(IE
of Sr)= -1064 kJ mol
B
2
→ O + e
ΔE
-1
-1
-1
O
= +141 kJ mol
c
→ O
ΔE
-2
-1
-1
-1
O
+ e
= -844 kJ mol
D
SrO → Sr
ΔE
+2
-2
-1
+ O
= 3610 kJ mol
coulomb
CaO ⇒ Ca + O
We can now calculate the energy required to dissociate into neutral atoms (which is called the
dissociation energy);
ΔE=ΔE
+ ΔE
+ ΔE
+ ΔE
+ ΔE
A
B
C
D
Coulomb
(
)
(
)
(
)
(
)
(
)
-1
-1
-1
-1
-1
= −549 kJ mol
+ −1064 kJ mol
+ 141 kJ mol
+ −844 kJ mol
+ 3610 kJ mol
-1
= 1294 kJ mol
-1
The dissociation energy is 1294 kJ mol
.
Question 7 (out of 3 points)
Because of its particle nature, light can exert pressure on a surface. Suppose a 1.5-W laser is
focused on a circular area 0.15 mm in radius. The wavelength of the laser light is 569 nm.
Assume that the photons travel perpendicular to the surface and are perfectly reflected, and
their momenta change sign after reflection from the wall. Using results from the kinetic
theory of gases, calculate the pressure exerted by the light on the surface. The momentum p of
a photon is related to its energy through E = pc.
The fall of photons on a surface exerts a pressure. Use the deBroglie relation to compute the
momentum of the photons:
-34
h
6.626 x 10
J s
-27
-1
p =
=
= 1.1645 x 10
kg m s
λ
-9
569 x 10
m
Each photon bounces away from the wall with momentum of the same magnitude but opposite
sign. The change in momentum per collision is:
(
)
(
)
−27
-27
-1
-27
-1
-1
Δp = p
− p
= 1.1645 x 10
− -1.1645 x 10
= 2.33 x 10
kg m s
kg m s
kg m s
2
1
-1
The number of photons colliding per second equals the power of the laser (1.5 watt = 1.5 J s
)
divided by the energy delivered per photon. The energy transported by one 550 nm photon is:
(
)
(
)
-34
8
-1
6.626 x 10
J s
2.9979 x 10
m s
hc
-19
E =
=
(
)
=3.491 x 10
J
λ
-9
569 x 10
m
-1
Clearly, it requires lots of such photons to transport 1.5 J s
:
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