Introduction to Entropy
01
Second Law of Thermodynamics: Reactions
proceed in the direction that increases the entropy
of the system plus surroundings.
Introduction to Entropy
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A spontaneous process is one that proceeds on its own
without any continuous external influence.
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02
salt =
rusty heap of metal!
A nonspontaneous process takes place only in the presence
of a continuous external influence.
25°C
Introduction to Entropy
03
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The measure of molecular disorder in a system is
called the system’s entropy (S)
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Entropy has units of J/K
Introduction to Entropy
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ΔS = Sfinal – Sinitial
Positive ΔS indicates
Negative ΔS indicates
increased disorder
decreased disorder
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Introduction to Entropy
05
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To decide whether a process is spontaneous, both
enthalpy and entropy changes must be considered:
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Spontaneous process:
Nonspontaneous process:
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Increase in enthalpy (+ΔH)
Decrease in entropy (–ΔS)
Spontaneous
Situations leading to ΔG < 0:
ΔG = ΔH – TΔS
ΔH<0 , TΔS>0
ΔH<<0, TΔS slightly negative
ΔH slightly positive, TΔS>>0
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Gibbs Free Energy Change (ΔG): Uses enthalpy
and entropy to determine spontaneity of a reaction
ΔG = ΔH – TΔS
Decrease in enthalpy (–ΔH)
Increase in entropy (+ΔS)
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Introduction to Free Energy
ΔG < 0
spontaneous
ΔG = 0
equilibrium
ΔG > 0
nonspontaneous
Equilibrium
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Situations leading to ΔG = 0:
ΔG = ΔH – TΔS
ΔH and TΔS are equally negative
ΔH and TΔS are equally positive
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Nonspontaneous
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Situations leading to ΔG > 0:
Example
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ΔG = ΔH – TΔS
10
Predict whether ΔS° is likely to be positive or negative for
each of the following reactions. Using tabulated values,
calculate ΔS° for each:
a. 2 CO(g) + O 2(g) → 2 CO 2(g)
ΔH > 0, TΔS > 0
ΔH slightly negative, TΔS << 0
ΔH >> 0, TΔS is slightly positive
ΔS = negative
b. 2 NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
c. C2H4(g) + Br2(g) → CH2BrCH2Br(l)
ΔS = negative
d. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
Example
Which of the following reactions are spontaneous under
standard conditions at 25°C?
a.
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
ΔG° = –55.7 kJ spontaneous
b.
2 C(s) + 2 H 2(g) → C2H4(g) nonspontaneous
ΔG° = 68.1 kJ
c.
N2(g) + 3 H2(g) → 2 NH3(g)
spontaneous
ΔH° = –92 kJ; ΔS° = –199 J/K
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ΔS = positive
ΔS = positive
Example
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12
Equilibrium (ΔG° = 0): Estimate the temperature at which
the following reaction will be at equilibrium. Is the reaction
spontaneous at room temperature?
N2(g) + 3 H2(g) → 2 NH3(g)
ΔH° = –92.0 kJ
ΔS° = –199 J/K
ΔG = ΔH – TΔS = 0
T = ΔH° = -92.0kJ/ -0.199kJ/K = 463K = 190°C
ΔS°
ΔG° = (-92kJ) - (298K)(-0.199kJ/K) = -32.9kJ spontaneous!
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Example
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Benzene, C6H6, has an enthalpy of vaporization,
ΔHvap, equal to 30.8 kJ/mol and boils at 80.1°C.
What is the entropy of vaporization, ΔSvap, for
benzene?
ΔGvap = ΔHvap – TΔSvap
ΔGvap = 0 at boiling temp, T = 80.1 °C + 273.15 = 353.3K
ΔSvap = ΔHvap/T = (30.8kJ/mol)/353.3K = 0.0872kJ/Kmol
=87.2J/Kmol
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