Energy

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Chapter 6

• Energy
• Thermodynamics

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6.1 Nature of Energy

• The ability to do work or produce heat.
• Conserved - can be converted from one form to
  another but can neither be created nor destroyed.
• Work is a force acting over a distance.
• Potential due to position or composition - can
  be converted to work.
• Kinetic due to motion of the object.
• KE 1/2 mv2
• (m mass, v velocity)

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Heat and Temperature

• Temperature reflects random motion of particles
  in a substance.
• Heat is the measure of energy content.
• Heat is energy transferred between objects
  because of temperature difference.
• State Function - property of a system that
  depends only on its present state.
• Independent of the path, or how you get from
  point A to B.

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The Universe

• Is divided into two halves, the system and the
  surroundings.
• The system is the part you are concerned with.
• The surroundings are the rest.
• Exothermic reactions release energy to the
  surroundings.
• Endothermic reactions absorb energy from the
  surroundings.

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Heat
Potential energy
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Heat
Potential energy
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Direction

• Every energy measurement has three parts.
• A unit ( Joules or calories).
• A number - how many.
• A sign to tell direction.
• Negative - exothermic
• Positive- endothermic

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Surroundings
System
Energy
DE lt0
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Surroundings
System
Energy
DE gt0
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Same rules for heat and work

• Heat given off is negative.
• Heat absorbed is positive.
• Work done by system on surroundings is negative.
• Work done on system by surroundings is positive.
• Thermodynamics - The study of energy and the
  changes it undergoes.

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• (YDVD)

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First Law of Thermodynamics

• The energy of the universe is constant.
• Law of conservation of energy.
• q heat
• w work
• DE q w
• Take the systems point of view to decide signs.

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What is work?

• Work is a force acting over a distance.
• work force ? distance
• since pressure force / area,
• work pressure ? volume
• Work can be calculated by multiplying pressure by
  the change in volume at constant pressure.
• units of Latm

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Work needs a sign

• If the volume of a gas increases, the system has
  done work on the surroundings.
• work is negative
• wsystem ?P?V
• Expanding work is negative.
• Contracting, surroundings do work on the system w
  is positive.
• 1 Latm 101.3 J

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Examples

• Calculate the ?E for a system undergoing an
  endothermic process in which 15.6 kJ of heat
  flows and where 1.4 kJ of work is done on the
  system
• What amount of work is done when 15L of gas is
  expanded to 25 L at 2.4 atm pressure?
• If 2.36 J of heat are absorbed by the gas above.
  What is the change in energy?

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6.2 Enthalpy

• Abbreviated H
• H E PV (thats the definition), at constant
  pressure.
• DH DE PDV
• the heat at constant pressure qp can be
  calculated from DE qp w qp - PDV
• qp DE P DV DH
• Where qP ?H at constant pressure.
• ?H energy flow as heat (at constant
  pressure).

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Examples

• When 1 mole of methane (CH4) is burned at
  constant pressure, 890 kJ of energy is released
  as heat. Calculate the DH for a process in which
  5.8 g sample of methane is burned at a constant
  pressure.
• Consider the following reaction
• 2H2(g) O2(g) ? 2H2O(l) DH-572kJ
• How much heat is evolved when 4.03 g of hydrogen
  is reacted with excess oxygen?

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Calorimetry

• Measuring heat. We use a calorimeter.
• The heat capacity for a material, C, is
  calculated.
• C heat absorbed/DT DH/ DT
• specific heat capacity
• heat capacity per gram J/Cg or J/Kg
• molar heat capacity
• heat capacity per mole J/Cmol or J/Kmol

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Calorimetry

• Constant pressure calorimeter (coffee cup
  calorimeter, used for solutions).
• heat specific heat x m x DT
• heat molar heat x moles x DT
• Make the units work and youve done the problem
  right.
• A coffee cup calorimeter measures DH.
• The specific heat of water is 1 cal/gºC (4.184
  J/gºC)
• Heat of reaction DH s x mass x DT

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Examples

• The specific heat of graphite is 0.71 J/gºC.
  Calculate the energy needed to raise the
  temperature of 75 kg of graphite from 294 K to
  348 K.
• A 46.2 g sample of copper is heated to 95.4ºC and
  then placed in a calorimeter containing 75.0 g of
  water at 19.6ºC. The final temperature of both
  the water and the copper is 21.8ºC. What is the
  specific heat of copper?

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Calorimetry

• Constant volume calorimeter is called a bomb
  calorimeter.
• Material is put in a container with pure oxygen.
  Wires are used to start the combustion. The
  container is put into a container of water.
• The heat capacity of the calorimeter is known and
  tested.
• Since DV 0, PDV 0, DE q

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Bomb Calorimeter

• thermometer
• stirrer
• full of water
• ignition wire
• Steel bomb
• sample

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Properties

• Intensive properties - not related to the amount
  of substance.
• Ex. density, specific heat, temperature.
• Extensive property - does depend on the amount of
  stuff.
• Ex. heat capacity, mass, heat from a reaction.

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6.3 Hesss Law

• Enthalpy is a state function.
• The change in enthalpy is the same whether the
  reaction takes place in one step or a series of
  steps.
• We can add equations to to come up with the
  desired final product, and add the DH.
• Two rules
• If the reaction is reversed the sign of DH is
  changed.
• If the reaction is multiplied, so is DH.

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Rules

• 1. If a reaction is reversed, ?H is also
  reversed.
• N2(g) O2(g) ? 2NO(g) ?H 180 kJ
• 2NO(g) ? N2(g) O2(g) ?H ?180 kJ
• 2. If the coefficients of a reaction are
  multiplied by an integer, ?H is multiplied by
  that same integer.
• 6NO(g) ? 3N2(g) 3O2(g) ?H ?540 kJ

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Examples

• When using Hesss Law, work by adding the
  equations up to make it look like the answer.
• Make the other compounds cancel out.
• N2(g) 2O2(g) ? 2NO2(g) ?H1 68kJ
• Above reaction is carried out in two steps below
• N2(g) O2(g) ? 2NO(g) ?H2 180kJ
• 2NO(g) O2(g) ? 2NO2(g) ?H3 -112kJ

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2NO, O2
-112 kJ
180 kJ
H (kJ)
2NO2
68 kJ
N2
O2
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Example

• Given(BDVD)

DHº -1300. kJ
DHº -394 kJ
DHº -286 kJ
Calculate DHº for this reaction
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Example
Given
DHº 77.9kJ
DHº 495 kJ
DHº 435.9kJ
Calculate DHº for this reaction
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EXAMPLE

• P4(s) 6Cl2(g) ? 4PCl3(g) ?H -1225.6kJ
• P4(s) 5O2(g) ? P4O10(s) ?H -2967.3kJ
• PCl3(g) Cl2(g) ? PCl5(g) ?H - 84.2kJ
• PCl3(g) 1/2O2(g) ? Cl3PO(g) ?H -285.7 kJ
• Calculate the ?H for the reaction
• P4O10(s) 6PCl5(g) ? 10Cl3PO(g)

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6.4 Standard Enthalpies of Formation

• Standard States
• Compound
• For a gas, pressure is exactly 1 atmosphere.
• For a solution, concentration is exactly 1 molar.
• Pure substance (liquid or solid), it is the pure
  liquid or solid.
• Element
• The form N2(g), K(s) in which it exists at 1
  atm and 25C.

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Standard Enthalpy of Formation

• The enthalpy change that occurs in the formation
  of one mole of a compound for a reaction at
  standard conditions (25ºC, 1 atm , 1 M
  solutions).
• Symbol DHºf
• There is a table in Appendix 4 (pg A21) It is a
  table of standard heats of formation. The amount
  of heat needed to for 1 mole of a compound from
  its elements in their standard states.

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Standard Enthalpies of Formation

• Need to be able to write the equations.
• What is the equation for the formation of NO2 ?
• ½N2 (g) O2 (g) NO2 (g)
• Have to make one mole to meet the definition.
• Write the equation for the formation of methanol
  CH3OH.

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Since we can manipulate the equations

• We can use heats of formation to figure out the
  heat of reaction.
• Lets do it with this equation.
• C2H5OH(l) 3O2(g) 2CO2 (g) 3H2O(l)

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Thermite Reaction

• Using enthalpies of formation, calculate the
  standard change in enthalpy for the thermite
  reaction.
• 2Al(s) Fe2O3(s) --gt Al2O3(s) 2Fe(s)
• (YDVD)

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Its a gas!

• Methanol (CH3OH) is often used as a fuel in high
  performance engines in race cars. Using the data
  in table 6.2, compare the standard enthalpy of
  combustion per gram of methanol with that per
  gram of gasoline. Gasoline is actually a mixture
  of compounds, but assume for this problem that
  gasoline is pure liquid octane (C8H18)