Unit 8 Chapter 16 Thermochemistry

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Unit 8- Chapter 16Thermochemistry
Thermodynamics

• One of Mrs. Dorrs faaaaaaavorite topics!!!

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The first law of thermodynamics

• Recall the law of conservation of energy- energy
  can be neither created or destroyed.
• The 1st Law of Thermodynamics is simply a
  restatement- The energy of the universe is
  constant.
• We are going to look at the energy involved in
  chemical reactions and prove not only the 1st law
  of thermodynamics, but the 2nd as well

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Systems

• System ? surroundings open system
• matter
• System ? surroundings closed system
• (matter is trapped but it can
  exchange energy)
• matter
• System ? surroundings Isolated system
• energy (neither the matter or
  energy from the system can be
  exchanged)

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Boundaries of Systems

• Diathermic (closed system) boundary that permits
  heat transfer
• Adibatic (isolated system) prevents heat or
  matter transfer

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Work

• Work is defined as force acting over a distance
• W F d it is measured in Joules
  (J)
• note 1 Latm
  101.32 J
• Consider a frictionless piston system with a
  constant Pexternal see side board
• Total work
• W -?Pext dV ? -Pext?dV
• W -Pext (Vf Vi)
• Therefore W -Pext ?V

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• Example 1 Consider an ideal gas in a piston
  chamber expanding against constant Pext and at
  constant temperature. The gas has an initial
  volume of 2.0L, a final volume of 5.50L, an
  initial pressure equal to 8.00atm and an external
  pressure of 1.75atm. Determine a). the
  work and b). the final pressure.

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• Work
• Expansion ? W -Pext ?V
• Compression ? W Pext ?V
• If Pext 0 what happens to work?
• There is NO WORK under the condition of free
  expansion (Pext 0)
• The only other time there is NO WORK is when ?V
  0

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Example

• Consider a can of spray paint
• A. Can of spray discharged (can system)
• B. Can of spray discharged (spray system)
• A. no work being done b/c no ?
• B. work done on atmosphere/surroundings by the
  system/spray
• W -Pext ?V

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System Processes

• Irreversible ? No Equilibrium
• Reversible ? continuous Equilibrium state with
  the surroundings (small incremental changes)
• For Reversible Changes Pext Pint
• Wrev -?Pint dV
• So we can say Wrev -? nRT/V dV
• Then see side board for full derivation

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Heat- Its getting hot in here

• Heat, q if heat is flowing in/ gaining heat
• - if heat is flowing out/ losing heat
• Heat is the measure of thermal energy transfer
  that can be determined in the temperature change
  of the object (its a way to follow the energy
  change of a system)

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Equations with Heat

• 1. q m c ?T ?intensive property
• m is mass, c is specific heat of substance
• 2. q C ?T ?extensive property
• C is the heat capacity of substance
• 3. q IVt ?intensive property
• I is current, V is voltage, t is time

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Internal Energy The 1st Law of Thermodynamics

• Internal energy, E total energy of system
• Recall adiabatic isolated system no heat or
  matter transfer!
• Recall the 1st Law- For an isolated system, the
  total energy of a system remains constant,
  therefore ?E 0
• ?E q W (general)
• ?E W (adiabatic)

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Example

• A gas sample changes in volume from 2.0L to 4.0L
  against an external pressure of 1.50 atm, while
  absorbing 1,000J of heat, what is ?E of the
  system?

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State Functions

• State function- path independent
• Ex- P, V, T, ?E, ?H
• NOT State functions- q and W
• In-exact differentials ?dW W
• ?dq q
• Exact differentials ?dE ?E

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Enthalpy

• At constant pressure, the change in enthalpy ?H
  of the system is equal to the energy flow as heat
• H E PintV
• ?H ?E ?(PV)
• ?H ?E ?ngRT
• If no gases are involved, or if ?n 0, then ?E
  ?H
• See side of board for chain rule derivation

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Example

• A piston filled with 0.0400mol of an ideal gas
  expands reversibly from 50.00mL to 375.0mL _at_
  constant temperature of 37?C. As it does so, it
  absorbs 208J of heat. Calculate q, W, ?U, and
  ?H for this process.

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?H

• ?H ?Hproducts ?Hreactants
• If ?H is positive the reaction is endothermic
• If ?H is negative the reaction is exothermic

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Enthalpy Calorimetry

• Recall a calorimeter is used to experimentally
  determine the heat associated with a chemical
  reaction
• Calorimetry is the science of measuring heat
• Heat capacity C heat absorbed
• ______________________
• inc. in
  temperature
• Heat capacity can be given per gram specific
  heat capacity c ? J/?Cg or J/Kg
• Or it can be given per mole molar heat capacity
  C ?J/?Cmol or J/Kmol

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Constant Pressure Calorimetry

• Constant pressure!
• For these reactions ?H qp
• So we can use qp mc?T or qp C?T

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Example of Constant Pressure Calorimetry

• When 1.00L of 1.00M Ba(NO3)2 solution at 25.0?C
  is mixed with 1.00L of 1.00M Na2SO4 solution at
  the same temperature in a calorimeter, the white
  solid BaSO4 forms and the temperature of the
  mixture increases to 28.1?C. Assuming that the
  calorimeter absorbs only a negligible quantity of
  heat, that the specific heat capacity of the
  solution is 4.18J/?Cg, and that the density of
  the final solution formed is 1.0g/mL, calculate
  the enthalpy change per mole of BaSO4 formed.

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Constant Volume Calorimetry

• Calorimetry experiments can also be done at
  constant volume
• For a constant volume procses, the ?V 0, so
  work, w 0 also
• ?E q W q qv

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Constant Volume Example

• Consider the combustion of octane, C8H18. A
  0.5269g sample of octane is placed in a bomb
  calorimeter known to have a heat capacity of
  11.3kJ/?C. The octane is ignited in the presence
  of excess oxygen, and the temperature increase of
  the calorimeter is 2.25?C. Determine the amount
  of energy released per mole of octane.

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Heat of Reactions

• Standard states ?H?
• For a Compound-
• The standard state of a gaseous substance is a
  pressure of exactly 1 atm
• For a liquid or solid the standard state is the
  pure liquid or solid
• For a substance present in solution, the standard
  state is a concentration of exactly 1M
• For an Element-
• The standard state of an element is the form in
  which the element exists under conditions of 1atm
  and 25?C. (ex- O2(g), Na(s), and Hg(l)

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• Standard enthalpy of formation, ?Hf? the change
  in enthalpy that accompanies the formation of one
  mole of a compound from its elements with all
  substances in their standard states
• The degree symbol ? indicates standard conditions
• ?H?rxn ??H?products - ??H?reactants
• DO NOT put pure elements in this eqn. ex-O2
• When a rxn is reversed, change the sign on ?H
• When the balanced eqn. is multiplied by an
  integer, the value of ?H must be multiplied by
  the same integer
• Use this equation for heat of fusion, heat of
  formation, heat of vaporization, heat of
  combustion, heat of synthesis ? any enthalpy
  change!

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Read Chapter 6, section 5 6, pages 267-281 by
Friday.
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Example

• Determine the standard enthalpy change for the
  combustion of methane.

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Hesss Law

• Recall enthalpy, ?H is independent of pathway (a
  state function)
• Hesss Law states in going from a particular set
  of reactants to a particular set of products, the
  change in enthalpy is the same whether the
  reaction takes place in one step or in a series
  of steps
• So ?H? ?npH?products - ?nr ?H?reactants

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Hesss Law Example

• Diborane (B2H6) is a highly reactive boron
  hydride, which was once considered as a possible
  rocket fuel for the U.S. space program.
  Calculate the ?H for the synthesis of diborane
  from its elements according to the equation
• 2B(s) 3H3(g)
• Using the following data
• 2B(s) 3/2O2(g) ? B2O3(s) ?H -1273 kJ
• B2H6(g) 3O2(g) ? B2O3(s) 3H2O(g) ?H -2035 kJ
• H2(g) ½ O2(g) ? H2O(l) ?H -286 kJ
• H2O(l) ? H2O(g) ?H 44 kJ

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The 2nd Law of Thermodynamics

• Recall that the 1st Law of Thermodynamics deals
  with energy change
• The 2nd Law of Thermodynamics states that in any
  spontaneous process there is always an increase
  in the entropy of the universe
• To reword the entropy of the universe is
  increasing
• Energy is conserved in the universe, but entropy
  is not

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Entropy

• Spontaneous process is one that occurs without
  outside intervention
• Note a spontaneous process irreversible
  process
• If it is a reversible process ?S 0
• Thermodynamics tells us only about the direction,
  not the speed.
• Entropy, S the thermodynamic function that
  measures randomness or disorder

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Entropy

• ?Suniv ?Ssys ?Ssurr
• If ?Suniv is , it is spontaneous in the
  direction written
• If ?Suniv is -, it is spontaneous in the opposite
  direction
• If ?Suniv 0, the process has no tendency to
  occur the system is _at_ equilibrium

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The Effect of Temperature on Spontaneity

• For Reversible Irreversible processes
• ?Ssurr qsurr / Tsurr
• Since _at_ constant P q ?H we can say
• ?Ssurr - ?H / T
• This means that if the rxn is exothermic, ?H -,
  but since heat flows into the surroundings,
  ?Ssurroundings is positive

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?Ssurr Example

• Antimony, the pure metal is recovered via
  different reactions, depending on the composition
  of the ore.
• Iron is used to reduce antimony in sulfide ores
• Sb2S3(s) 3Fe(s) ? 2Sb(s) 3FeS(s) ?H -125kJ
• Carbon is used as the reducing agent for oxide
  ores
• Sb4O6(s) 6C(s) ? 4Sb(s) 6CO(g) ?H 778kJ
• Calculate ?Ssurr for each of these reactions at
  25?C and 1atm.

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Summary of Entropy

• Recall ?Suniv ?Ssurr ?Ssys
• _______________________________________
• ?Ssys ?Ssurr ?Suniv Spontaneous?
• Yes
• - - - No
• - ? Yes, if ?Ssys is larger
  than ?Ssurr
• - ? Yes, if ?Ssurr is larger than
  ?Ssys

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Free Energy

• Free energy, G is defined by the following
  relationship ?GRxn ??Gproducts -
  ??Greactants
• And G H - TS
• For a process that occurs _at_ constant T, the
  change in free energy ?G is
• ?G? ?H? - T?S? This is called the
  Gibbs-Helmholtz Eqn.
• To relate it to spontaneity
• ?Suniv - ?G / T _at_ constant T P
• This means _at_ constant T P a process is
  spontaneous only if ?G is NEGATIVE.

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• We can re-phrase this to say a process (_at_
  constant T P) is spontaneous in the direction
  in which the free energy decreases (-?G means
  ?Suniv).
• We now have two variables that determine
  spontaneity (?G), ?S ?H
• Case Result
• ?S , ?H - Spontaneous _at_ all temps.
• ?S , ?H Spontaneous _at_ high temps.
• ?S -, ?H - Spontaneous _at_ low temps.
• ?S , ?H NOT Spontaneous _at_ ANY Temperature

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Example

• At what temperatures is the following process
  spontaneous _at_ 1atm?
• Br2(l) ? Br2(g)
• ?H? 31.0kJ/mol ?S? 93.0J/Kmol
• What is the normal boiling point of liquid Br2?

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Entropy the States of Matter

• Solids have low entropy, liquids have higher
  entropy, solutions have higher entropy, gases
  have the highest entropy
• If the of gaseous molecules in the products is
  greater than the of gaseous molecules in the
  reactants, the forward rxn favors entropy, its
• The reverse rxn is favored if the opposite is
  true
• Ex Predict the sign of ?S? for each rxn
• CaCO3(s) ? CaO(s) CO2(g)
• 2SO2(g) O2(g) ? 2SO3(g)

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Free Energy Chemical Reactions

• Standard free energy change, ?G?, the change in
  free energy that will occur if the reactants in
  their standard states are converted to the
  products in their standard states
• Note you can use Hesss Law with ?G?!
• ?G? ?npG?products - ?nr ?G?reactants
• ?G? 0 for the formation of an element in its
  standard state

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Pressure Free Energy

• Free energy is affected by Pressure
• G G? RTln(P)
• G is the free energy of the gas _at_ pressure P, G?
  is the free energy of the gas _at_ 1atm, and R is
  the universal gas constant, use 3.14 J/molK
• Also, ?G nRTln(Pf/Pi)
• ?G ?G? RTlnQ ? Q is the rxn quotient
• For 2A 3B ? 4C Q C4/ A2B3

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Pressure Example

• Calculate ?G at 25?C for this rxn where carbon
  monoxide gas at 5.0 atm and hydrogen gas at
  3.0atm are converted to liquid methanol
• CO(g) 2H2(g) ? CH3OH(l)

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Free Energy Equilibrium

• Equilibrium represents the lowest free energy
  value available
• _at_ equilibrium ?G 0 and Q K
• So, ?G 0 ?G ? RTln(K) or ?G ? -RTln(K)
• K is the equilibrium constant where
  eA fB ? gC hD
• Cg Dh
• K ------------
• Ae Bf

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?G ? and K yes, memorize these!!!

• ?G ? K
• --------------------------------------------------
  -------
• ?G ?0 K 1
• ?G ? lt 0 K gt 1
• ?G ? gt 0 K lt 1

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• Also, when ?G 0, then then ?H T?S or T ?H
  / ?S

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Equilibrium Example

• The overall reaction for the corrosion (rusting)
  of iron by oxygen is
• 4Fe(s) 3O2(g) ? 2Fe2O3(s)
• Using the following data, calculate the
  equilibrium constant for this reaction at 25?C.
• Substance ?H?f (kJ/mol) S? (J/Kmol)
• Fe2O3(s) -826 90
• Fe(s) 0 27
• O2(g) 0 205

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