Transcript Spontaniety

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Spontaneous reactions are reactions, that
once started, continues by itself without
further input of energy from the outside.
If a reaction is spontaneous under a given set
of conditions, then the reverse reaction is
considered non-spontaneous.
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Identify whether the reactions below are
spontaneous or not:
– H2O (s)

H2O (l)
– 2H2 (g) + O2 (g)

2H2O (l)
*requires a spark to begin
– 2H2O (l)  2H2 (g) + O2 (g)
*requires consistent electric current
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Almost all exothermic reactions are considered to
be spontaneous (at 25 *C and 1 atm).
– ΔH for a spontaneous reactions tends to be negative.
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However, some endothermic reactions at specific
temperatures may be considered spontaneous,
for example the melting of ice at 1 atm above 0
*C:
H2O (s)

H2O (l)
ΔH = 6.0 kJ
Endothermic reactions that are non-spontaneous
at room temperature often become spontaneous
when the temperature is raised.
The Randomness Factor: In general, nature tends
to move spontaneously from more ordered to
more random states (order to disorder).
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The randomness factor discussed is treated
quantitatively as entropy (S). Basically the
greater the disorder (more random the
distribution of molecules) the greater the
entropy.
2nd Law of Thermodynamics: In any
spontaneous process there is always an
increase in the entropy of the universe.
Entropy, like enthalpy is a state property so
that ΔS = Sfinal - Sinitial
◦ unit for S = J/mol K
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A liquid has higher entropy than the solid
from which it is formed.
A gas has a higher entropy than the liquid
from which it is formed.
Increasing the temperature of a substance
increases its entropy.
A completely ordered pure crystalline solid
has an entropy of 0 K (3rd law of
thermodynamics).
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Elements have nonzero standard entropies.
Standard molar entropies of pure substances are
always positive quantities.
Aqueous ions may have negative entropy values.
As a group, gases tend to have higher entropies
than liquids. An increase in the number of moles
of a gas also leads to a higher entropy and vice
versa.
As a molecule becomes more complex, the
higher the entropy (more ways for the atoms to
move about with respect to one another).
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Predict whether ΔS is positive or negative for
each of the following processes:
◦ Taking dry ice from a freezer where its temperature
is -80°C and allowing it to warm to room
temperature
◦ Dissolving bromine in hexane
◦ Condensing gaseous bromine to liquid bromine
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Which of the following
reactions results in the
largest increase in entropy?
(A) CO2(s) CO2(g)
(B) H2(g) + Cl2(g)  2HCl(g)
(C) KNO3(s)  KNO3(l)
(D) C(diamond)  C(graphite)
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To calculate the standard entropy change, ΔS,
use the following relation:
ΔS = ∑ ΔS products - ∑ ΔS reactants
The 2nd Law of Thermodynamics: In a
spontaneous process, there is a net increase
in entropy, taking into account both the
system and surroundings, ΔS > 0
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Calculate the entropy change at 25ºC in J/K
for
2SO2(g) + O2(g)  2SO3(g)
given the following entropy data:
◦ SO2(g): 248.1 J/mol-K
◦ O2(g): 205.3 J/mol-K
◦ SO3(g): 256.6 J/mol-K
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[2(256.6)] - [2(248.1) + 1(205.3)] = -188.3 J/K
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Calculate ΔS for the reduction of aluminum
oxide by hydrogen gas to produce aluminum
and water vapor. Use the following entropy
data:
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Aluminum oxide: 51 J/mol K
Hydrogen gas:
131 J/mol K
Aluminum solid: 28 J/mol K
Water vapor:
189 J/mol K
Ans: 179 J/K
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Gibbs free energy is a quantity that basically
allows us to put the two quantities, enthalpy and
entropy, together in such a way as to arrive at a
single function whose sign will determine
whether the reaction is spontaneous. The basic
definition of Gibbs free energy is:
ΔG = ΔH - TΔS
(The Gibbs-Helmholtz Equation)
– G = Gibbs Free Energy (J)
– H = Enthalpy (J/mol)
– T = Temperature (K)
– S = Entropy (J/mol K -- please be aware that the mole
in the unit tends to cancel out from the ΔS equation)
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The Gibb’s free energy equation combines all
the information that we have learned thus far.
But what does the Gibb’s free energy value
tell us about a reaction? It tells us the
following:
◦ If ΔG is negative, the reaction is spontaneous in the
forward direction.
◦ If ΔG is equal to zero, the reaction is at equilibrium.
◦ If ΔG is positive, then the reaction is
nonspontaneous in the forward direction, but the
reverse reaction will be spontaneous.
◦ For elements at standard state (pure elements at
25ºC and 1 atm are assigned a value of zero).
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Using example 4, calculate:
◦ ΔH°
 ΔH°f Al2O3 (s) = -1675.7 kJ/mol
 ΔH°f H2O (g) = -241.82 kJ/mol
◦ ΔG° at 25°C
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The standard free energy of formation, ΔGf,
can also be used to solve for the free energy
of a reaction (reference Appendix 1 in text):
ΔGf rxn = ∑ ΔGf products - ∑ ΔGf reactants
If it is a negative quantity then the compound
can be formed spontaneously from the
elements, like in the formation of water:
H2 (g)
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+
½ O2 (g)

H2O (l)
ΔGf = -237.2 kJ
Elements in their elemental state will have a
ΔGf = 0
ΔS+
ΔG =
ΔG = ??
negative
spontaneous
at high
temperatures
spontaneous
at all
temperatures
ΔH–
0
K=1
ΔH+
ΔG = ??
ΔG = positive
spontaneous
at low
temperatures
non-spontaneous
at all
temperatures
ΔS–
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At what temperature does ΔG° become zero
for the reaction
Fe2O3 (s) + 3H2 (g)  2Fe (s) + 3H2O (g)
ΔH°f
(kJ/mol)
ΔS°
(J/mol K)
Fe2O3 (s)
-825.5
90
3H2 (g)
--------
130.7
2Fe (s)
--------
27
3H2O (g)
-241.8
188.8
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Which of the following reactions has the
largest positive value of entropy per mole of
Cl2 ?
(A) H2(g) + Cl2(g) ---> 2 HCl(g)
(B) Cl2(g) + 1/2 O2(g) ---> Cl2O(g)
(C) Mg(s) + Cl2(g) ---> MgCl2(s)
(D) 2 NH4Cl (s) ---> N2(g) + 4 H2(g) + Cl2(g)
(E) Cl2(g) ---> 2 Cl (g)
For which of the following processes would
ΔS have a negative value?
I. 2 Fe2O3(s) ---> 4 Fe(s) + 3 O2(g)
II. Mg2+ + 2 OH¯ ---> Mg(OH)2(s)
III. H2(g) + C2H4(g) ---> 3 C2H6(g)
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III
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N2 (g) + 3 H2 (g) ---> 2 NH3 (g)
The reaction indicated above is thermodynamically
spontaneous at 298 K, but becomes nonspontaneous
at higher temperatures. Which of the following is true
at 298 K?
(A) ΔG, ΔH, and ΔS are all positive.
(B) ΔG, ΔH, and ΔS are all negative.
(C) ΔG and ΔH are negative, but ΔS is positive.
(D) ΔG and ΔS are negative, but ΔH is positive.
(E) ΔG and ΔH are positive, but ΔS is negative.
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H2O (s) ---> H2O (l) When ice melts at its
normal melting point, 298.16 K and 1
atmosphere, which of the following is true for
the process shown above?
(A) ΔH < 0, ΔS > 0, ΔG > 0
(B) ΔH < 0, ΔS < 0, ΔG > 0
(C) ΔH > 0, ΔS < 0, ΔG < 0
(D) ΔH > 0, ΔS > 0, ΔG > 0
(E) ΔH > 0, ΔS > 0, ΔG < 0
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Of the following reaction, which involves the
largest decrease in Entropy?
(A) CaCO3 (s) ---> CaO (s) + CO2 (g)
(B) 2 CO (g) + O2 (g) ---> 2 CO2 (g)
(C) Pb(NO3)3 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)
(D) C3H8 (g) + O2 (g) ---> 3 CO2 (g) + 4 H2O (g)
(E) 4 La (s) + 3 O2 (g) ---> 2 La2O3 (s)
Answer the following questions about nitrogen, hydrogen, and
ammonia.
 (a)
In the boxes below, draw the complete Lewis electron-dot
diagrams for N2 and NH3.
 (b)
Calculate the standard free-energy change, ΔG°, that occurs
when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to
the reaction represented below:
N2(g) + 3 H2(g)  2 NH3(g) ∆G˚298 = -34 kJ mol-1
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(c) Given that ΔH˚298 for the reaction is −92.2 kJ mol-1, which is
larger, the total bond dissociation energy of the reactants or the
total bond dissociation energy of the products? Explain.
(d) The value of the standard entropy change, ΔS˚298 , for the
reaction is −199 J mol-1K-1. Explain why the value of ΔS˚298 is
negative.
(e) Assume that ΔH° and ΔS° for the reaction are independent of
temperature.
◦ (i) Explain why there is a temperature above 298 K at which the algebraic
sign of the value of ΔG° changes.
◦ (ii) Theoretically, the best yields of ammonia should be achieved at low
temperatures and high pressures. Explain.
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Use principles of thermodynamics to answer the following
questions.
(a) The gas N2O4 decomposes to form the gas NO2.
◦ (i) Predict the sign of ∆H˚ for the reaction. Justify your answer.
◦ (ii) Predict the sign of ∆S˚ for the reaction. Justify your answer.
(b) One of the diagrams below best represents the
relationship between ∆G˚ and temperature for the reaction
given in part (a). Assume that ∆H˚ and ∆S˚ are independent
of temperature.
Draw a circle around the correct graph. Explain why you
chose that graph in terms of the relationship:
∆G˚ = ∆H˚ - T∆S˚.
2 Fe(s) + O2(g)  Fe2O3(s)
∆Hf˚ = -824 kJ mol–1
Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g
sample of Fe(s) is mixed with 11.5 L of O2(g) at 2.66 atm and 298 K.
(a) Calculate the number of moles of each of the following before the reaction
occurs.
(i) Fe(s)
(ii) O2(g)
(b) Identify the limiting reactant when the mixture is heated to produce Fe2O3.
Support your answer with calculations.
(c) Calculate the number of moles of Fe2O3 produced when the reaction proceeds to
completion.
(d) The standard free energy of formation, ∆Gf˚ of Fe2O3 is –740. kJ mol–1 at 298
K.
(i) Calculate the standard entropy of formation ∆Sf˚ of Fe2O3 at 298 K.
Include units with your answer.
(ii) Which is more responsible for the spontaneity of the formation reaction
at 298K, the standard enthalpy or the standard entropy?
The reaction represented below also produces iron(III) oxide. The value of ∆H˚ for
the reaction is –280 kJ per mol.
2 FeO(s) + O2(g)  Fe2O3(s)
(e) Calculate the standard enthalpy of formation, ∆Hf˚ of FeO(s).
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For the gaseous equilibrium represented below, it is
observed that greater amounts of PCl3 and Cl2 are
produced as the temperature is increased:
PCl5(g)  PCl3(g) + Cl2(g)
(a) What is the sign of ∆S° for the reaction? Explain.
(b) What change, if any, will occur in DG° for the
reaction as the temperature is increased? Explain
your reasoning in terms of thermodynamic principles.
(c) If He gas is added to the original reaction mixture
at constant volume and temperature, what will
happen to the partial pressure of Cl2? Explain.
(d) If the volume of the reaction mixture is decreased
at constant temperature to half the original volume,
what will happen to the number of moles of Cl2 in the
reaction vessel? Explain.
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
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The reaction represented above is spontaneous at 25°C.
Assume that all reactants and products are in their
standard state.
(a) Predict the sign of ∆S° for the reaction and justify
your prediction.
(b) What is the sign of ∆G° for the reaction? How would
the sign and magnitude of ∆G° be affected by an
increase in temperature to 50°C? Explain your answer.
(c) What must be the sign of ∆H° for the reaction at
25°C? How does the total bond energy of the reactants
compare to that of the products?
(d) When the reactants are place together in a
container, no change is observed even though the
reaction is known to be spontaneous. Explain this
observation.