Thermochemistry: The study of the heat transfer that accompanies chemical porcesses

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Thermochemistry The study of the heat transfer
that accompanies chemical porcesses

• For this unit we will define energy as the
  capacity to do work or to produce heat
• You already know the Law of Conservation of
  energy Energy is neither created nor Destroyed
  it is only converted from one form to another
• This means that the energy of the universe is
  constant.
• There are two type of energy that we need to
  concern ourselves with.
• Potential Energy Energy due to position or
  composition, in the case of composition, the
  energy is stored in chemical bonds, when the
  bonds are broken the energy is released.
• Kinetic Energy energy of movement of particles,
  this depends on the mass of the particle and its
  velocity Ke ½ mv2
• The units of both energies are expressed in
  Joules (J) or Kilo Joules (kJ)

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Heat transfer of energy between two objects due
to a temperature difference

• Not all energy can be explained by the transfer
  of energy between kinetic and potential energy
• Some energy is lost due transfer of energy to the
  surface, this is frictional heating.
• Work is defined as force acting over a distance.
• Regardless of the pathway taken to transfer the
  energy, by work or by heat, the result is the
  same the total energy transferred will be
  constant.
• State function (property) a property of the
  system that is independent of the pathway it took
  to get to that state, only depends on its present
  state. In other words the value of a state
  function (Property) doesn't depend on how the
  system arrived at that state, only the present
  characteristics of that state.
• Energy is a state function but work and heat are
  not.

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Chemical Energy

• These same ideas apply to chemical processes.
• For our purposes, we will refer to the system as
  part of the universe that we are focusing on, the
  reactants and the products.
• The surroundings will be everything else, the
  container, the room, your hand, anything else.
• When a reaction involves the evolution of heat it
  is exothermic. Energy is flowing out of the
  system as heat.
• When there is an exothermic reaction, the
  reactants had greater potential energy than the
  products, because energy gained by the
  surroundings is equal to energy lost by the
  reaction system.
• When a reaction absorbs energy from the
  surroundings, it is endothermic. The heat flow is
  into the system.
• Energy that flows into the system is equal to the
  potential energy gained by the system , in this
  case the products have greater potential energy.

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Thermodynamics the study of the flow of energy
and its interconversions Based on the 1st Law
Energy of the universe is constant.

• The internal energy of a system is defined as the
  sum of the work and the heat of the system.
• ?E q w
• change in internal energy heat work
• If the energy flow is endothermic heat is flowing
  into the system and the sign on q is
• If the energy flow is exothermic, then heat is
  flowing out of the system and the sign for heat
  is q indicating that the systems energy is
  decreasing

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Work

• If the system does work on the surroundings,
  energy flows out of the system, the work is
  negative, -w
• If the surroundings do work on the system then
  energy is flowing into the system and work is
  positive, w
• Example Calculate ?E for a system undergoing an
  endothemric process in which 15.6 kJ of heat
  flows and where 1.4 kJ of work is done on the
  system.

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Answer

• ?E q w
• q 15.6 kJ because this is an endothermic
  process
• w 1.4 kJ because work is being done on the
  system
• ?E 15.6 kJ 1.4 kJ 17.0 kJ

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Work can be done by a gas produced in a chemical
reaction

• Work can be done by a gas through expansion, or
  work can be done to the gas by compression
• Work can also be expressed as w -P?V
• Check your book for how we got to this equation
• Using this , work can be if the change in
  volume is increasing or work can be negative if
  the volume is decreasing.
• Pressure is always the external pressure causing
  the compression or expansion.

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Example A balloon is being inflated to its full
extent by heating the air inside it. At the end,
the balloon changes volume from 4.0 x 106 L to
4.5 x 106 L by the addition of 1.3 x 108 J of
heat. Assuming the balloon expands at a constant
pressure, 1 atm, calculate ?E ( use 101.3j 1 atm
L to convert to Joules)

• ?E q w
• q 1.3 x 108 J
• w-P?V -(1atm)(4.5x106L- 4.00x106L) -
  5.0x105Latm which can be converted to Joules
  using 101.3J 1 L atm
• ( -5.0x105Latm )(101.3 J/1 Latm) -5.1 x
  107J
• Note that the work is negative because gas is
  expanding and doing work on the surroundings
• ?E q w (1.3x108J) (-5.1x107 J) 7.9 x
  107J
• There is a net increase in the internal energy of
  the gas in the balloon so the ?E is positive.

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Enthalpy ?H heat

• Enthalpy has no easily interpreted meaning except
  at constant pressure ?H q remember q is heat.
• Enthalpy heat of reaction ?Hrxn H products
  H reactants
• If ?H is positive, then H of reactants is less
  than H of products, heat is absorbing into the
  system, it is endothermic.
• If ?H is negative, heat of reactants is greater
  than heat of products and heat is flowing out of
  the system , it is exothermic.

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Example

• When 1 mol of methane is burned at constant
  pressure, 890 kJ of energy is released as heat.
  Calculate the ?Hcomb for a process in which 5.8 g
  sample is burned at constant pressure.
• At constant pressure, 890 kJ/mol of CH4 is
  produced as heat.
• ?H q -890kJ/mol
• (5.8 g CH4/16.0g/mol CH4) 0.36 mol CH4
• (0.36 mol CH4) (-890 kJ/mol CH4) -320 kJ

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Calorimetry the science of measuring heat

• We use the equation q s x m x ?T
• q heat s specific heat capacity m mass ?T
  change in temperature (Tf Ti)
• Example When we mix 50.0 ml of 1.0 mol HCl at 25
  C with 50.0 ml of 1.0 NaOH at the same
  temperature, the temperature increase to 31.9 C
  the solutions density is 1.0 g/ml. It can be
  assumed that the caloirmeter doesnt absorb any
  heat. What is the heat? Also known as change in
  enthalpy.

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Answer

• ?T 31.9-25 6.9
• m 100.0 ml (1.0 g/ml) 100 g
• s4.18 J/Cg from the table given
• q (6.9 C) (100 g) (4.18J/Cg) 2.9 x 10 3 J
• The net ionic Equation for this reaction was
  HOH-?H20
• So if you want to know how much heat is produced
  per mol of H then
• 50.0 ml x 1L x 1.0mol H 5.0 x 10-2 mol H
• 1000ml 1 L
• 2.9x 103 J/5.0 x10-2 mol H 5.8 x 104 J/mol

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Hess Law in going from a particular set of
reactants to products, the change in enthalpy is
the same whether the reaction takes place in one
step or many steps.

• Ex N2 2O2 ? 2NO2 ?H1 68 kJ
• This same reaction can be carried out in two
  steps
• N2 O2 ? 2NO ?H2 180 kJ
• 2NO O2 ? 2NO2 ?H3 -112 kJ
• N2 2O2 ? 2NO2 ?H2 ?H3 68 kJ

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Rules of Hess Law

• If a reaction is reversed, then the sign is
  reversed on the ?H
• The magnitude of ?H is directly proportional to
  the quantities of reactants, if the coefficients
  are multiplied by a number so is the ?H.
• Xe 2F2 ? XeF4 ?H -251 kJ
• XeF4 ? Xe 2F2 ?H 251 kJ
• If we double the reaction
• 2Xe 4F2 ? 2XeF4 ?H 2(-251 kJ) -502kJ

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Example

• Calculate ?H for the synthesis of diborane from
  its elements
• 2B (s) 3H2(g) ? B2H6 (g)
• Us the following data
• 2B(s) 3/2 O2 (g)? B2O3 (s) ?H -1273 kJ
• B2H6 (g) 3 O2 (g) ? B2O3 (s) 3H2O (g) ?H
  -2035 kJ
• H2 (g) ½ O2 (g) ? H2O (l) ?H -286 kJ
• H2O ? H2O (g) ?H 44 kJ
• A couple of rules to follow
• Figure out how to rearrange the equations so
  their sum equal the overall equation
• If you multiply an equation by a number then you
  must also multiply the ?H
• If you reverse the equation, reverse the sign

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Answer

• Check each others work

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Standard Enthalpies of Formation

• The change in enthalpy that accompanies the
  formation of one mole of a compoundfrom its
  elements with all the element sin their standard
  states.
• ?H a degree symbol on a thermodynamic function
  indicates that the process is carried out under
  standard conditions 1atm 1mol 25 C

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The enthalpy change for a given reaction can be
calculated by subtracting the enthalpies of
formation of the reactants from the enthalpies of
formation of the products

• ?Hreaction?np ?Hf prod.- ?nr?Hf react.
• Things to know
• When a reaction is reversed the sign changes but
  not the magnitude
• When the balanced equation for a reaction is
  multiplied by an integer so must the ?H for that
  reaction
• Elements in their standard states are not
  included, in other words, the ?H for an element
  in its standard state is 0

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