?H

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Enthalpy

• ?H
• Every chemical reaction and change of physical
  state releases or absorbs heat.
• Goal to determine whether heat is absorbed or
  released during a chemical reaction, therefore
  determine if exothermic or endothermic.

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Thermochemistry

• Study of heat changes that accompany chemical
  reactions and phase changes
• Thermochemical equations are used to represent
  these reactions
• 4Fe(s) 3 O2(g) ? 2Fe2O3(s) 1625kJ
• 27kJ NH4NO3(s) ? NH4(aq) NO3-(aq)
• First equation exothermic- heat pack
• Second equation endothermic cold pack

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Thermochemistry Terms

• System specific part of the universe that
  contains reaction or process you want to study
• Surroundings everything in universe other than
  system
• Universe system surroundings

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Enthalpy

• Impossible to know the total heat content of a
  system because it depends upon many factors
• Chemists are interested in changes in energy
  during reactions
• For many reactions, energy lost or gained can be
  measured by calorimeter at constant pressure
• Enthalpy(H) heat content of a system at
  constant pressure

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Enthalpy

• We measure change in H (heat content)

?H H2 H1 Products
reactants

• Exothermic reaction
  a downhill change
• Endothermic reaction an uphill
  energy change

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Exothermic Reactions
reactants

• Is ?H positive or negative?
• Why?
• Products - Reactants!

Enthalpy (kJ)
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Endothermic Reactions

• ?H is a positive

Enthalpy (kJ)
reactants
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Why is there a change in Enthalpy?

• due to
• energy required to break the bonds in the
  reactants and
• energy produced by forming the chemical bonds in
  the products.

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• A balanced equation can represent the energy
  absorbed or released

C(cr) O2 ? CO2 energy (393.5kJ) ?Hr -393.5kJ

• Energy change for the reaction is called the
• Enthalpy of Reaction
• and is represented by ?Hr

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Standard States
?Hºf

• ?H change in enthalpy
• Standard state enthalpy at 298.15K
  (25 C degrees) and 101.325 kilopascals
  (pressure)
• (No longer STP!)
• f formation

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Elements in the Natural State
?Hºf 0
ELEMENTS ONLY
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ALL OTHER COMPOUNDS
?Hºf

• Enthalpy of formation
• found in standard tables
• Table C-13 pg921

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Enthalpy of Formation

• ? Hf represents the production of one mole from
  ...
• its free elements in their standard states.
  (units kJ/mole)

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Calculation of Enthalpy of Reaction

• ?Hr S?Hºf (products) - S?Hºf (reactants)
• S (Sigma) is the symbol used to indicate
  summation.
• It means to ADD all the values of ?H for all the
  products and subtract ...
• the sum of all the ? H of the reactants

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EXAMPLE 1

• Calculate the enthalpy change in the following
  chemical reaction
  carbon monoxide oxygen ? carbon
  dioxide
• 1) Write a balanced equation
  
• 2CO (g) O2 ? 2CO2

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2CO (g) O2 ? 2CO2

• 2) Calculate ? Hf products
• 2 moles CO2 x (-393.5 kJ) -787.0 kJ
• 3) Calculate ? Hf reactants
• 2 moles CO 1 mole O2 2(-110.5 kJ) 1(0
  kJ)
• 
  -221 kJ
• ?Hr S?Hf (products) - S?Hf (reactants)

• ? Hr (-787.0 kJ) - (-221. kJ) -566.0 kJ
• This means the 566.0 kJ are released in the
  reaction

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Practice

• Compute ?Hºr for the following reactions
• 2NO(g) O2(g) ? 2NO2(g)
• -114.14kJ
• __FeO(cr) O2(g) ? __ Fe2O3(cr)

• -560.4kJ

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Enthalpy

• Solving for change in enthalpy we get
• ?Hr S?Hºf (products) - S?Hºf (reactants)

• Knowing the enthalpy of formation of each
  reactant and product, we can calculate the amount
  of energy produced or absorbed and predict
  whether a reaction will be exothermic or
  endothermic.

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Exothermic Reactions
reactants
115 kJ
Enthalpy (kJ)
75 kJ

• ? Hr 75 kJ -115 kJ
• - 40 kJ
• Negative ? Hr for EXOTHERMIC

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Endothermic Reactions
247 kJ
Enthalpy (kJ)
122 kJ
reactants

• ? H 247 kJ - 122 kJ
• 125 kJ
• Positive ? Hr for ENDOTHERMIC

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Spontaneous or Nonspontaneous That is the
Question ?

• Suppose ? H is negative
  ? Hf products lt ? Hf reactants
• Its EXOTHERMIC
• When ? Hr is negative reaction tends to be
• SPONTANEOUS
• Spontaneous means that it will occur without any
  outside influence

• How about ? H positive?

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Hesss Law

• the enthalpy change for a reaction is the sum of
  the enthalpy changes for a series of reactions
  that add up to the overall reaction.

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Hesss Law

• N2 2O2 ? 2NO2 41kJ
• N2O4 ? 2NO2 35kJ
• N2 2O2 ? N2O4

Reverse the second reaction to get
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Hesss Law

• N2 2O2 ? 2NO2 ?H 41kJ
• 2NO2 ? N2O4 ?H 35kJ
• N2 2O2 ? N2O4

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Hesss Law

• N2 2O2 ? 2NO2 ?H 41kJ
• 2NO2 ? N2O4 ?H 35kJ
• N2 2O2 ? N2O4

Reversing the second reaction reverses the sign
of the enthalpy of the reaction
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Hesss Law

• N2 2O2 ? 2NO2 ?H 41kJ
• 2NO2 ? N2O4 ?H-35kJ
• N2 2O2 ? N2O4

Now add the two
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Hesss Law

• N2 2O2 ? 2NO2 ?H 41kJ
• 2NO2 ? N2O4 ?H-35kJ
• N2 2O2 ? N2O4 6kJ

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End
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Entropy

• (S) is the measure of the degree of disorder in a
  system.
• A spontaneous process is one that occurs in a
  system left to itself. No external action is
  needed to make it happen.

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• Increase in disorder
• Decrease in disorder

?S gt 0
?S lt 0
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GIBBS FREE ENERGY

• Gibbs free energy indicates whether a reaction
  will occur or not.

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• Exergonic reactions (spontaneous)
• Endergonic reactions (nonspontaneous)

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• Endothermic reactions occur spontaneously when T
  ? S is large.
• The thermodynamic definition of a system in
  equilibrium is when ? H and ? S have the same
  sign, and there is some temperature at which ? H
  and T ? S are numerically equal.

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• A spontaneous reaction proceeds toward
  equilibrium.
• Chemical potential energy, G, is least at
  equilibrium.
• Enthalpy, entropy, and free energy depend on
  temperature. ( We will only work at 298.15 K and
  100.00 kPa-standard states)

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GIBBS FREE ENERGY CALCULATIONS

• ? H values are relative free elements are
  considered to have
• change in enthalpy is found by

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• entropy changes and Gibbs free energy is computes
  as follows

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• 1. Organize the data you will use from the
  appropriate table.
• 2. Multiply each ? Gf0 by the number of moles
  from the balanced equation. (Always make sure
  the chemical equation is balanced). Substitute
  these values into the equation used to determine
  ? Gf0 .

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• The Gibbs free energy decreases in a spontaneous
  reaction because the system is changing to a more
  stable state.

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• To find 1. Multiply each
  by the number of moles from the balanced
  equation.2. Substitute these values into the
  equation.