Transcript Chapter 14: Thermodynamics

Chapter 16: Thermodynamics
Review of the Law
• 1st law of Thermodynamics is a restatement of
the law of conservation of energy
– Energy can never be created nor destroyed, it may
only change form
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This is true only within a closed system
The universe is a closed system
So the amount of energy in the universe is constant
By default, the amount of heat energy is constant, but
may exist in potential and kinetic heat energy.
Review of the Law
• Consider the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H₂O(g) + energy
– The combustion of Methane produces a quantity
of energy released as heat.
– Potential energy stored in the bonds of CH4 and O2
is released as heat as they react to form water and
CO2.
– Potential energy has been converted to thermal
energy, but the amount of energy in the system
itself has not changed.
Review of the Law
• The 1st law of Thermodynamics allows us to
answer the following questions:
– How much energy is involved in the change?
– Does energy flow into or out of the system?
– What form does the energy finally assume?
• It DOES NOT, however, allow us to account for
“why” a process occurs in a given direction.
This is the main question to be addressed in
this chapter.
Spontaneity
• A process is considered spontaneous if it
occurs without outside intervention
– This process may be fast or slow
– Thermodynamics allow us to determine the
direction of the process (chemical change), but
can tell us nothing about the speed of the process.
• Speed of the process is dependent on rate of reaction.
∆[A]
• Remember that rate of reaction =
∆t
Spontaneity
• We can think of spontaneity in terms of a few
simple scenarios:
– A ball spontaneously rolls down a hill, but not
back up. Why?
– Heat flows from a hot object to a cooler one,
never the opposite. Why?
– Wood burns spontaneously to form CO2 and
Water, but this is not reversible.
Spontaneity and Entropy
• After many years of observation, one common
characteristic is present in all spontaneous
processes; Entropy.
– Entropy (S) is the driving force for all spontaneous
processes.
– Most accurately defined as a measurement of
randomness or disorder in a system.
• Represents the multitude of possible arrangements
available to a system that exists in a given state.
Spontaneity and Entropy
• These two concepts combined together
present a level of statistical probability that is
beyond the scope of this course, however, we
can simplify it greatly using a simple
metaphor.
– Think of your room, is it clean? Is it orderly?
– Think of your clothes and personal items as gas
molecules. If you expend no energy to confine
these things to a certain area, what happens?
High Entropy, low energy.
Low entropy, high energy
In other words
• Without work (energy), there is no order
(entropy), it takes work to keep things in their
place in a non-random state.
• This is true of all things in nature. Think of the
balance between different aspects of an
ecosystem. Each part exerts work (energy) on
another, thus keeping that other part in order.
– Predator prey interactions, survival of the fittest, etc…
– What happens when humans introduce outside forces
into ecosystems?
• Imbalance occurs, energy dynamics are un-equalized,
entropy increases in certain areas.
Positional Probability of Gases
• The most likely state of a gas in a container is
the one that requires the less energy to
obtain, as well as the one with the most
microstates.
Positional Probability
• Also explained in changes of state.
– In a solid state, molecules of a substance are very
close together, with very few positional
microstates
– In a liquid state, molecules of a substance are
further apart, with more positional microstates
– In a gas state, molecules of a substance are very
far apart, with very many positional microstates.
Entropy and Positional Probability
(entropy)
Sgas > Sliquid > Ssolid
• Gases posses the most Entropy (S) as a result of
the most available microstates, it takes energy
(from somewhere at some point in time) to
produce a liquid or a solid; a gas is always the most
likely state of matter when talking about entropy.
Positional Probability
• With this all in mind,
– Which has the higher positional entropy?
• Solid Carbon Dioxide or Gaseous Carbon Dioxide?
• Nitrogen Gas at 1atm or at 1.0 x 10-2atm?
– Predict the sign (negative or positive) of the
following processes (∆S=Sfinal – Sinitial)
• Solid sugar dissolved in water?
• Iodine Vapor condenses on a cold surface to form
crystals.
2nd Law of Thermodynamics
• In any spontaneous process there is always an
increase in the entropy of the universe
∆Suniverse = ∆Ssystem + ∆Ssurroundings
• This would indicate that there is a decrease of
energy in the universe, but remember that
there is constant work being put into
individual systems that keep everything
constant.
2nd Law of Thermodynamics
• In any spontaneous process there is always an
increase in the entropy of the universe
∆Suniverse = ∆Ssystem + ∆Ssurroundings
• If ∆Suniverse is positive, the process is spontaneous
in the direction it is written.
• If ∆Suniverse is negative, the process is spontaneous
in the opposite direction.
• If ∆Suniverse is zero, the process has no tendency to
occur and is at equilibrium.
Another Example from Biology
• In living cells, large molecules are made from
simpler ones, is this process consistent with
the 2nd law?
– Making large molecules from simpler ones gives a
negative value for ∆Ssystem
– Remember that ∆Suniverse = ∆Ssystem + ∆Ssurroundings
– Only if the value for ∆Ssurroundings is positive and
larger can this be spontaneous process, and in
cells it is.
Effect of Temperature on Spontaneity
• To illustrate this, lets consider the following
process:
H2O(l) →H2O(g)
• Consider the water to be the system and
everything else as the surroundings.
• Since we are moving from a liquid to a gas,
entropy (Ssystem) increases and is positive.
Effect of Temperature on Spontaneity
H2O(l) →H2O(g)
• Lets assume this process absorbs 50 joules (J)
of heat energy to the surroundings; it takes
heat to evaporate water.
– ∆Ssurroundings is negative(endothermic).
– Thus, ∆ Ssystem and ∆Ssurroundings are in opposition.
• Since we don’t have actual values, how will be
determine whether the process is spontaneous?
Effect of Temperature on Spontaneity
• ∆Ssurroundings depends on the temperature at
which the heat is transferred.
• The sign of ∆Ssurroundings (-/+) depends on the
direction of the heat flow.
Therefore:
heat=magnitude of entropy
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡(𝐽)
change=
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
∆S and Enthalpy
• Exothermic process:
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡(𝐽)
∆Ssurroundings =+
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
• Endothermic process:
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡(𝐽)
∆Ssurroundings =− 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
• In terms of Enthalpy (H)
∆Ssurr= -
∆𝑯
𝑻
– The negative sign is necessary because ∆H is
determined in terms of the reaction system, not the
surroundings. So a negative ∆H indicates a exothermic
reaction because the reaction system is losing heat,
but then ∆Ssurr would be positive …and vice versa.
Sample Problem
• Calculate the Entropy change in the reaction
system of:
a) Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ
b) Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g) ∆H = 778 kJ
both at 25°C and 1atm
– use the equation: ∆Ssurr= -
∆𝑯
𝑻
Sample Problem
a) Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ
• ∆Ssurr= -
−𝟏𝟐𝟓 𝒌𝑱
𝟐𝟗𝟖 𝑲
∆Ssurr= 419 J/K - exothermic, surroundings absorb
heat
a) Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g) ∆H = 778 kJ
• ∆Ssurr= -
𝟕𝟕𝟖 𝒌𝑱
𝟐𝟗𝟖 𝑲
∆Ssurr= - 𝟐. 𝟔𝟏 𝐤𝐉/𝐊 - endothermic, surroundings
lose heat
Effect of ∆H, ∆S and T on
Spontaneity
∆H
∆S
Low Temp
High Temp
Example
-
+
Spontaneous
(∆G < 0)
Spontaneous
(∆G < 0)
N20→N2 + O2
-
Nonspontaneous
(∆G > 0)
Nonspontaneous
(∆G > 0)
O2→O3
-
Spontaneous
(∆G < 0)
Nonspontaneous
(∆G > 0)
H2O(l) → H2O(s)
+
Nonspontaneous
(∆G > 0)
Spontaneous
(∆G < 0)
H2O(l) → H2O(g)
+
-
+
Section II:
Entropy Changes in Chemical
Reactions
• Key definitions:
– ∆Hrxn - standard change in enthalpy for a reaction.
“standard heat of reaction”
– ∆Hf-Standard Enthalpy of formation
“heat of formation”
• Compounds to elements: negative ∆Hf
• Elements to compounds: positive ∆Hf
– ∆Srxn = Standard entropy change for a reaction
∆Srxn = Sproducts + Sreactants
Section II:
Entropy Changes in Chemical
Reactions
• Key definitions:
– ∆Hrxn = ∆H1 + ∆H2
Reactants → elements
Elements → products
Reactants → products
∆H1= -∑∆Hf (reactants)
∆H2= ∑∆Hf (products)
∆Hrxn= ∆H1 + ∆H2
Section II:
Entropy Changes in Chemical
Reactions
• Key definitions:
– Third Law of thermodynamics
• The entropy of a perfect crystal at absolute zero is
“zero”.
• A crystal at absolute zero has only one possible way to
arrange its molecules or atoms. The positional
probability is 1, and therefore there is no level of
disorder.
Section II:
Entropy Changes in Chemical
Reactions
• Key definitions:
– Third Law of thermodynamics
• The entropy of a perfect crystal at absolute zero is
“zero”.
• A crystal at absolute zero has only one possible way to
arrange its molecules or atoms. The positional
probability is 1, and therefore there is no level of
disorder.
Entropy is related to
complexity of structure
• Many substances can
exist in different forms in
the same state. These are
called allotropes.
• Example is carbon.
– Graphite is carbon
structure as sheets
– Diamonds are carbon
structred in threedimensional crystals.
• Which has greater
entropy?
Entropy is related to
molecular complexity
• In general, with
increasing molecular
complexity, entropy
increases.
• At the same time,
dissolving a solid into an
aqueous solution
increases entropy
S°(J/mol*K)
NO(g)
NO2(g)
N2O4(g)
210.8
240.1
304.4
S°(J/mol*K)
KClO3(s)
143.1
KClO3(aq)
265.7
Entropy Changes in Chemical
Reactions
• The major premise of this section is
calculating the standard entropy change for a
reaction (∆Srxn )
– Simply stated, we can calculate this by subtracting
the entropy of products from reactants:
• ∆Srxn = ∑npS(products) - ∑nrS(reactants)
– Where n is the mole coefficients, p and r represent “products”
and “reactants” respectively.
– S is the entropy of each portion of the reaction.
Entropy Changes in Chemical
Reactions
• ∆Srxn = ∑npS(products) - ∑nrS(reactants)
– YOU WILL HAVE TO FIND THE VALUES FOR “S” IN
THE APPENDIX IN THE BACK OF YOUR BOOKS. ON
THE EXAM, THEY SHOULD PROVIDE A TABLE OR
GIVE YOU THE VALUES IN THE QUESTIONS.
– Make sure you pay close attention to the state of
matter when extracting the values for S.
Example
• Calculate the ∆Srxn for the equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
∆Srxn = ∑npS(products) - ∑nrS(reactants)
Reactant or product
NH3(g)
O2(g)
NO(g)
H2O(g)
S (J/mol*K)
192.8
205.2
210.8
188.8
Example
• Calculate the ∆Srxn for the equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
• ∆Srxn = [4mol(210.8 J/K*mol) + 6mol (188.8 J/K*mol)][4mol (192.8 J/K*mol) + 5mol (205.2 J/K*mol)]
= 1976 J/K – 1797.2 J/K
= 178.8 J/K
Example #2
• Calculate ∆Srxn for the equation:
H2S(g) + O2(g) → H2O(g) + SO2(g)
Step 1: Balance the equation
Step 2: extract values of “S” for each reactant
and product
Step 3: insert into equation:
∆Srxn = ∑npS(products) - ∑nrS(reactants)
Gibbs Free Energy (G)
• Free energy (G)-energy available to do work
• Josiah Gibbs-a guy they named free energy
after.
G= H – TS
Change in free energy for a constant
temperature process:
∆G = ∆H - T∆S
Gibbs Free Energy (G)
• ∆G<0-a reaction is accompanied by a release
of unusable energy. Reaction is spontaneous
• ∆G>0-a reaction is not spontaneous as
written, it is spontaneous in the opposite
direction
• ∆G=0-equilibrium
– Don’t forget…
Effect of ∆H, ∆S and T on
Spontaneity
∆H
∆S
Low Temp
High Temp
Example
-
+
Spontaneous
(∆G < 0)
Spontaneous
(∆G < 0)
N20→N2 + O2
-
Nonspontaneous
(∆G > 0)
Nonspontaneous
(∆G > 0)
O2→O3
-
Spontaneous
(∆G < 0)
Nonspontaneous
(∆G > 0)
H2O(l) → H2O(s)
+
Nonspontaneous
(∆G > 0)
Spontaneous
(∆G < 0)
H2O(l) → H2O(g)
+
-
+
Standard free energy of reaction
• ∆G°rxn-free energy
change for a reaction
when it occurs under
standard state
conditions, when
reactants in their
standard states are
converted to products
in their standard states.
Standard free energy of reaction
• ∆G°rxn = ∑np∆G°f (products) - ∑nr∆G°f (reactants)
• “Same” equation as standard enthalpy change
– Where p and r are mole coefficients, ∆Gf is the standard free
energy of formation
– ∆G°f --The free energy change that occurs when 1
mole of the compound is synthesized from its
elements in their standard states.
– ∆Gf – found in table in the appendices of your book
Difference between ∆G and ∆G°
• Under non-standard state conditions we must
use ∆G to predict direction of reaction.
– Remember the table of effects of H, S, and T on
∆G.
• Under equilibrium, ∆G° tells us whether
products or reactants are favored.
– Negative ∆G° favors product formation
– Positive ∆G° favors reactant formation.
Examples
• Calculate standard free energy changes for the
following reactions at 25°C
– CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
– 2MgO(s) → 2Mg(s) + O2(g)
Examples
– CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• -818.0 kJ/mol
– 2MgO(s) → 2Mg(s) + O2(g)
• 1139 kJ/mol
Phase Transitions
• At the temperature at which a phase
transition occurs, the system is at equilibrium,
so ∆G=0. Therefore;
∆G = ∆H - T∆S
0 = ∆H - T∆S
∆H
∆S =
T
Phase Transition Example
• When 1 mole of ice melts at 0°C, Calculate the
change in entropy (S).
– ∆H is represented by the molar heat of fusion, T is
the dual melting and freezing point of 273°K
So:
−6010 J/mol
∆Sice →water =
= -22.0 J/K * mol
273 K
And
6010 J/mol
∆Swater→ice =
= 22.0 J/K * mol
273 K
Free energy and chemical Equilibrium
• During the course of a chemical reaction, not
all the reactants and products will be at their
standard states. Under this condition, the
relationship between ∆G and ∆G° can be
described as:
∆G = ∆G° + RTlnQ
Where Q is the equilibrium (reaction) quotient.
Free energy and chemical Equilibrium
∆G = ∆G° + RTlnQ
• At equilibrium, ∆G = 0, and Q = K, so under
equilibrium conditions:
∆G = ∆G° + RTlnK
0= ∆G° + RTlnK
∆G°= - RTlnK
Key equations for this Chapter
Chapter 16 Homework
• Pg 831, #’s 55-65 odd, 67, 71, 75, 77, 83, 85,
88