1. Name H.Farkhanda Sharif
Section N1
ID#15003231016( Bs.Bt)
Submitted to Ma’am Urooj Fatima
Assignment 2
Principle of Chemistry
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2. Content
Entropy
Spontaneous &non-spontaneous
Entropy of vaporization & fusion
Entropy according to second law of
thermodynamics
Molecular interpretation of entropy
Absolute Scale of entropy or a perfect crystal
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7. How to calculate the entropy
∆S=
∆𝑞
𝑡
Where delta (∆) = change
q = ℎ𝑒𝑎𝑡
t= 𝑡𝑒𝑚𝑝𝑟𝑎𝑡𝑢𝑟𝑒
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8. Examples:1 calculate the change in entropy of a
large block of ice when 50.J of energy is removed
reversibly from it as heat at 0℃ in
qrev =50.j
T= 0℃
=0+273=273K
∆S=
∆𝑞
𝑡
=50/273
∆S =-0.18j/k
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9. Spontaneous
“A spontaneous change is the change that have
tendency to occur without any external influence”
These process are not takes place necessarily
rapidly.
Spontaneous change is form orderly manner to
disorder.
It is also called natural process.
It is exothermic reaction.
It increase the entropy.
e.g Ice melting in lukewarm water
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11. Non-Spontaneous
“A non-spontaneous change occur with external
influence”
Non-spontaneous change form dis-orderly
manner to orderly manner.
It is not naturally process.
It is endothermic reaction.
It decrease the entropy.
e.g compression of gas
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13. Example: Calculate the change entropy of the following
reaction.
CH4(g) + H2(g)→ C2H6(g)
SC2H6=229.5J/mol.k
SCH4(g) =219.8J/mol.k
SH2(g) =130.6J/mol.k
∆S= ∑ ∆S(products) -∑ ∆S(reactants)
=(1 mol×229.5J/mol.k)-(1 mol×219.8J/mol.k+1
mol×130.6J/mol.k)
∆S =-120.9J/K
Therefore the standard entropy is -120.9J/K
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14. Entropy of vaporization &
fusion
Entropy of vaporization
“The change in entropy per mole of molecules
when a substance change from a liquid into
vapour”
∆𝑺𝒗𝒂𝒑 = ∆𝑯𝒗𝒂𝒑/𝑻𝒃
Entropy of fusion
“The difference in the molar entropies of the
liquid and solid phases of a substance at a
stated temperature”
∆𝑺𝒇𝒖𝒔 = ∆𝑯𝒇𝒖𝒔/𝑻𝒃
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15. Entropy according to second law
thermodynamics
“The entropy of an isolated system increases in
any spontaneous process”
Entropy is a state function. and only measure in
a reversible process.
State function: Entropy change depend upon only
on the initial and final states, but not on the how the
change occur.
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16. A Molecular interpretation of Entropy
Entropy is a measure of disorder, and it is possible
to imagine a perfectly orderly state of matter which
have neither positional nor temperature disorder.
Such a state represents a natural zero of
entropy, a state of perfect order; therefore, we
should be able to set up an absolute scale of
entropy just as we can set up an absolute scale
of temperature
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17. A Molecular interpretation of
Entropy
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The entropy of system indicates its degree of
disorder.
Which gas has more entropy?
A decrease in number of molecule leads to a
decrease in etropy.
18. A Molecular interpretation of
Entropy
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There are atomic mode of motion:
- Translation(the moving of molecule in space
from one point to another)
-Vibration (the shortening and lengthening of
bonds, including the change in bond angels.
-Rotation( the spinning of molecule about axis)
19. Absolute Scale of entropy or a perfect
crystal
That is, S approaches 0 as T approaches to 0.
The “perfect crystal”
In 1877, the Austrian physicist Ludwig
Boltzmann proposed a molecular definition of
entropy that enables us to calculate the absolute
entropy at any temperature.
S = k ln W
K=1.38064852 × 10-23 m2 kg s-2 K-1
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20. Example: Calculate the entropy of a tiny solid made
up of four diatomic molecules of a compound such
as carbon monoxide, CO, at T = 0
When the four molecules have formed a perfectly
ordered crystal in which all molecules are aligned
with their C atoms
Because there is only one way of
arranging the molecules in the perfect
crystal, W = 1.
Using S = k ln W
S = k ln 1
S =0
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21. Example:How can we calculate the entropy of 1.00
mol of CO
For 1.00 mol CO, corresponding to 6.02 ×1023 CO
molecules, each of which could be oriented in either of
two ways, there are 2 6.02 ×1023
A chance of only 1 in 2 6.02 ×1023 of drawing a given
microstate in a blind selection.
We canexpect the entropy of the solid to be high and
calculate by using ln xa = a ln x that
From S = k ln W
S = k ln W
S = (1.3807 1023 J.K-1) (6.02 ×1023 ) ln 2
S = 5.76 J.K-1
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