Chapter 6 Thermochemistry

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Chapter 6 Thermochemistry

• 6.1 The Nature of Energy

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The Nature of Energy

• Energy- the capacity to do work or produce heat
• Law of conservation of energy- energy can be
  converted but not created or destroyed. Energy of
  universe is constant

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Types of Energy

• potential energy- (PE) due to position or
  composition
• ex. attractive or repulsive forces
• kinetic energy- (KE) due to motion of the object
• KE ½mv2 depends on mass and volume

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Types of Energy

• (a) PEA gt PEB
• (b)
• ball A has rolled down the hill
• has lost PE to friction and PE in ball B

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Transfer of Energy

• Temperature- measure of the average kinetic
  energy of the particles. It reflects random
  motion of particles in substance
• Two Ways to Transfer Energy
• Heat- (q) transfer of energy between two objects
  because of a temperature difference
• Work- (w) force acting over a distance

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Pathway

• the specific conditions of energy transfer
• energy change is independent of pathway because
  it is a state function
• State function a property of a system that
  depends only on its present state
• work and heat depend on pathway so are not state
  functions
• state function- depends only on current
  conditions, not past or future

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Transfer of EnergyChemical Energy

• System - part of the universe you are focused on
• Surroundings- everything else in the universe
• usually
• system what is inside the container. Reactants
  or products of a chemical reaction
• surroundings container ,room, etc.

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Transfer of Energy

• Exothermic
• energy is produced in reaction
• flows out of system
• container feels hot to the touch
• Endothermic
• energy is consumed by the reaction
• flows into the system
• container feels cold to the touch

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Transfer of Energy
CH4(g) 2O2 (g) CO2 2H2O(g)
energy (heat)
Combustion of Methane Gas is exothermic
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Transfer of Energy
N2(g) 2O2 (g) energy (heat) CO2
2H2O(g)
Reaction between nitrogen and oxygen is
endothermic
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Transfer of Energy

• the energy comes from the potential difference
  between the reactants and products
• energy produced (or absorbed) by reaction must
  equal the energy absorbed (or produced) by
  surroundings
• usually the molecules with higher potential
  energy have weaker bonds than molecules with
  lower potential energy

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Thermodynamics

• Thermodynamics- study of energy and its
  interconversions (transfers)
• First Law of Thermodynamics Energy of universe
  is constant
• (Law of conservation of energy)

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Internal Energy

• (E) sum of potential and kinetic energy in system
  
• can be changed by work, heat, or both
• E PE KE
• ?E q w

E of a system can be changed by flow of heat,
work or both
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Signs

• Signs are very important in thermodynamic
  quantities
• Signs will always reflect the systems point of
  view unless otherwise stated

?E q w
change in internal energy heat work
Exothermic - - -
Endothermic
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Signs
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Work

• common types of work
• Expansion- work done by gas
• Compression- work done on a gas

P is external pressure not internal like we
normally refer to
expansion ?V -w
compression -?V w
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Work
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Example 1

• Find the ?E for endothermic process where 15.6 kJ
  of heat flows and 1.4 kJ of work is done on
  system
• Since it is endothermic, q is and w is

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Example 2

• Calculate the work of expansion of a gas from 46
  L to 64 L at a constant pressure of 15 atm.
• Since it is an expansion, ?V is and w is -

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Example 3

• A balloon was inflated from 4.00 x 106 L to 4.50
  x 106 L by the addition of 1.3 x 108 J of heat.
  Assuming the pressure is 1.0 atm, find the ?E in
  Joules.
• (1 Latm101.3 J)
• Since it is an expansion, ?V is and w is -

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6.2 Enthalpy and Calorimetry
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Enthalpy

• Enthalpy is a property of a system
• definition H E PV
• since E, P and V are all state functions, then H
  is too
• for the following, the process is at constant P
  and the only type of work allowed is PV work
• ?E qP w qP - P?V ? qP ? E P?V
• H E PV ? ?H ?EP?V
• so, qP ?H at constant P

At constant P, ?H of a System is equal to the
Energy flow as heat
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Enthalpy

• For reactions carried out at constant
• P, the flow of heat is a measure of
• change in H of a system

• heat of reaction and change in enthalpy are used
  interchangeably for a reaction at constant P
• ?H Hproducts - Hreactants
• endo ?H exo - ?H

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Calorimetry

• science of measuring
• heat
• calorimeter- device used to experimentally find
  the heat associated with a chemical reaction
• substances respond differently when heated ( to
  raise T for two substances by 1 degree, they
  require different amount of heat)

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Heat Capacity

• (C) how much heat it takes to raise a substances
  T by one C or K
• the amount of energy depends on the amount of
  substance

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Heat Capacity

• Specific heat capacity
• (s) heat capacity per gram
• in J/Cg or J/Kg
• Molar heat capacity
• heat capacity per mole
• in J/Cmol or J/Kmol

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Constant-Pressure Calorimetry

• uses simplest calorimeter (like coffee-cup
  calorimeter) since it is open to air
• used to find changes in enthalpy (heats of
  reaction) for reactions occurring in a solution
  since qP ?H
• heat of reaction is an extensive property, so we
  usually write them per mole so they are easier to
  use

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Constant-Pressure Calorimetry

• when 2 reactants are mixed and T increases, the
  chemical reaction must be releasing heat so is
  exothermic
• the released energy from the reaction increases
  the motion of molecules, which in turn increases
  the T

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Constant-Pressure Calorimetry

• If we assume that the calorimeter did not leak
  energy or absorb any itself (that all the energy
  was used to increase the T), we can find the
  energy released by the reaction
• E released by rxn E absorbed by soln
• ?H qP sP x m x ?T

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Constant-Volume Calorimetry

• uses a bomb calorimeter
• weighed reactants are placed inside the rigid,
  steel container and ignited
• water surrounds the reactant container so the T
  of it and other parts are measured before and
  after reaction

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Constant-Volume Calorimetry

• Here, the ?V 0 so -P?V w 0
• ?E q w qV for constant volume
• E released by rxn ?T x Ccalorimeter

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Example 1

• When 1 mol of CH4 is burned at constant P, 890 kJ
  of heat is released. Find ?H for burning of 5.8 g
  of CH4 at constant P.
• 890 kJ is released per mole of CH4

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Example 2

• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0C
  is mixed with 1.00 L of 1.00 M Na2SO4 solution at
  25.0C in a coffee-cup calorimeter, solid BaSO4
  forms and the T increases to 28.1C. The specific
  heat capacity of the solution is 4.18 J/gC and
  the density is 1.0 g/mL. Find the enthalpy change
  per mole of BaSO4 formed.

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Example 2

• Write the net ionic equation for the reaction
• Ba2 (aq) SO42- (aq) ? BaSO4(s)
• Is the energy released or absorbed? What does
  that mean about ?H and q?
• exothermic -?H and qP
• How can we calculate ?H or heat?
• heat q sP x m x ?T
• How can we find the m?
• use density and volume

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Example 2

• Find the mass
• Find the change in T
• Calculate the heat created

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Example 2

• since it is a one-to-one ratio and the moles of
  reactants are the same, there is no limiting
  reactant
• 1.0 mol of solid BaSO4 is made so
• ?H -2.6x104 J/mol -26 kJ/mol

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Example 3

• Compare the energy released in the combustion of
  H2 and CH4 carried out in a bomb calorimeter with
  a heat capacity of 11.3 kJ/C. The combustion of
  1.50 g of methane produced a T change of 7.3C
  while the combustion of 1.15 g of hydrogen
  produced a T change of 14.3C. Find the energy of
  combustion per gram for each.

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Example 3

• methane CH4
• hydrogen H2
• The energy released by H2 is about 2.5 times the
  energy released by CH4

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6.3 Hess Law
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Hess Law

• since H is a state function, the change in H is
  independent of pathway
• Hess Law- when going from a set of reactants to
  a set of products, the ?H is the same whether it
  happens in one step or a series of steps

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Example 1
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Example 1

• N2(g) 2O2(g) ? 2NO2(g) ?H 68 kJ
• OR
• N2(g) O2(g) ? 2NO(g) ?H 180 kJ
• 2NO(g) O2(g) ? 2NO2(g)?H -112 kJ
• N2(g) 2O2(g) ? 2NO2(g) ?H 68 kJ

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Rules

• If a reaction is reversed, the sign of ?H must be
  reversed as well.
• because the sign tells us the direction of heat
  flow at constant P
• The magnitude of ?H is directly proportional to
  quantities of reactants and products in reaction.
  
• If coefficients are multiplied by an integer,
  the ?H must be multiplied in the same way.
• because ?H is an extensive property

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Example 2

• Using the enthalpies of combustion for graphite
  (-394 kJ/mol) and diamond (-396 kJ/mol), find the
  ?H for the conversion of graphite to diamond.
• Cgraphite (s) ? Cdiamond (s) ?H?

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Example 2

• (1) Cgraphite(s) O2(g) ? CO2(g) ?H-394kJ/mol
• (2) Cdiamond(s) O2(g) ? CO2(g) ?H-396kJ/mol
• to get the desired equation, we must reverse 2nd
  equation
• (1) Cgraphite(s) O2(g) ? CO2(g) ?H-394kJ/mol
• -(2) CO2(g) ? Cdiamond(s) O2(g) ?H396kJ/mol
• Cgraphite (s) ? Cdiamond (s) ?H-394 396
• ?H2 kJ/mol

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Example 3

• Find ?H for the synthesis of B2H6, diborane
• 2B(s) 3H2(g) ? B2H6(g)
• Given
• 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• B2H6(g) 3O2(g) ? B2O3(s) 3H2O(g)
  ?H2-2035kJ
• H2(g) ½O2(g) ? H2O(l) ?H3-286kJ
• H2O(l) ? H2O(g) ?H444 kJ

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Example 3

• Need 3 H2 (g) so 3 x (3)
• Need 3 H2O to cancel so 3 x (4)
• (1) 2B(s) 3/2O2(g) ? B2O3(s) ?H1-1273kJ
• -(2) B2O3(s) 3H2O(g) ? B2H6(g) 3O2(g)
  -?H2-(-2035kJ)
• 3x(3) 3H2(g) 3/2O2(g) ? 3H2O(l) 3?H33(-286kJ)
• 3x(4) 3H2O(l) ? 3H2O(g) 3?H43(44 kJ)
• 2B(s) 3H2(g) ? B2H6(g)
• ?H -1273 -(-2035) 3(-286) 3(44) 36kJ
• 
  

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6.4 Standard Enthalpies of Formation
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Standard Enthalpy of Formation

• ?Hf
• change in enthalpy that accompanies the formation
  of one mole of a compound from its elements in
  standard states
• means that the process happened under standard
  conditions so we can compare more easily

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Standard States

• For a COMPOUND
• for gas P 1 atm
• pure liquid or solid state
• in solution concentration is 1 M
• For an ELEMENT
• form that it exists in at 1 atm and 25C
• O O2(g) K K(s) Br Br2(l)

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Writing Formation Equations

• always write equation where 1 mole of compound is
  formed (even if you must use non-integer
  coefficients)
• NO2(g)
• ½N2(g) O2(g) ? NO2(g)
• ?Hf 34 kJ/mol
• CH3OH(l)
• C(s) H2(g) O2(g) ?CH3OH(l)
• ?Hf -239 kJ/mol

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Using Standard Enthalpies of Formation

• where
• n number of moles of products/reactants
• ? means sum of
• ?Hf is the standard enthalpy of formation for
  reactants or products
• ?Hf for any element in standard state is zero so
  elements are not included in the summation

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Using Standard Enthalpies of Formation

• since ?H is a state function, we can use any
  pathway to calculate it
• one convenient pathway is to break reactants into
  elements and then recombine them into products

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Using Standard Enthalpies of Formation
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Using Standard Enthalpies of Formation
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Example 1

• Calculate the standard enthalpy change for the
  reaction that occurs when ammonia is burned in
  air to make nitrogen dioxide and water
• 4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(l)
• break them apart into elements and then recombine
  them into products

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Example 1
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Example 1

• can be solved using Hess Law
• 4NH3(g) ? 2N2(g) 6H2(g) -4?HfNH3
• 7O2(g) ? 7O2(g) 0
• 2N2(g) 4O2(g) ? 4NO2(g) 4 ?HfNO2
• 6H2(g) 3O2(g) ? 6H2O(l) 6 ?HfH2O

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Example 1

• can also be solved using enthalpy of formation
  equation
• values are in Appendix 4

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Example 2

• Calculate the standard enthalpy change for the
  following reaction
• 2Al(s) Fe2O3(s) ? Al2O3(s) 2Fe(s)

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Example 3

• Compare the standard enthalpy of combustion per
  gram of methanol with per gram of gasoline (it is
  C8H18).
• Write equations
• 2CH3OH(l) 3O2(g) ? 2CO2(g) 4H2O(l)
• 2C8H18(l) 25O2(g)?16CO2(g) 18H2O(l)

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Example 3

• Calculate the enthalpy of combustion per mole

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Example 3

• Convert to per gram using molar mass
• so octane is about 2x more effective