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Energy Changes in Chemical Reactions

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Title: Energy Changes in Chemical Reactions


1
Energy Changes in Chemical Reactions
  • Thermochemistry

2
Thermodynamics and Thermochemistry
  • Thermodynamics the study of heat and its
    transformations.
  • Thermochemistry the part of thermodynamics that
    deals with changes in heat that take place during
    chemical reactions as in most
  • chemical reactions,
  • energy is absorbed
  • or released.

3
What is Internal Energy?
  • Internal Energy (E)- the combined kinetic and
    potential energy of all particles in a system.
  • The total internal energy of a system is not
    usually known. Energy changes (DE) can be
    measured.

4
Examples of Potential and Kinetic that Contribute
to Internal Energy
interactions between particles of matter down to
the subatomic level (p, n, and e-) but we are
generally concerned about interactions at the
molecular level.
  • Kinetic Energy
  • Rotational nature of molecules
  • Vibrational nature of molecules
  • Potential Energy-
  • Intermolecular forces of attraction or repulsion
  • Intramolecular forces of attraction or repulsion

5
System
  • What is the system? The object or substance
    undergoing a change of physical state or
    reaction. What we are studying.
  • Surroundings are everything around the system
    that can exchange energy with the system.

6
Types of Systems
  • Open system can exchange mass and energy (usually
    in the form of heat) with the surroundings.
  • Closed system can exchange energy but not mass.
  • Isolated system no transfer of energy or mass.

7
State Functions
  • State functions are path independent. They
    depend only on present state and are independent
    of history of the system.
  • The change in a state function does not depend on
    how the process is carried out.

8
State Functions
9
Total Internal Energy (E) in a System
  • Etotal mgh ½ mv2
  • OR
  • Potential Energy Kinetic Energy
  • Internal energy of the system sum of kinetic
    and potential energies making up a substance down
    to the subatomic level. We often dont know the
    total energy of a system.
  • Chemists arent so interested in total energy in
    a system- theyre more interested in the energy
    changes that happen during chemical reactions.

10
First Law of Thermodynamics
  • The total energy in the universe is constant (Law
    of Conservation of Energy).
  • You cant get something for nothing
  • There are essentially two ways to change the
    energy of a system- heat(q) and work(w).
  • DE q w

11
Energy Change DE
  • Energy change in a system can be determined
  • DE Energy can be transferred as heat work
  • DE Ef Ei q w
  • Where work force X distance
  • OR
  • pressure X volume change
  • w p DV
  • Note Expansion volume -pDV. Work is being
    done by the system on the surroundings.

12
Work in Chemical Reactions
  • Often times in chemical reactions, there isnt a
    volume change or a force applied over a distance.
  • Therefore, for the chemist, heat changes are the
    primary concern when considering the energy
    changes that exist in a chemical reaction.
  • DE q
  • (for many, not all, chemical reactions)

13
Focus on Heat
  • Heat thermal energy (motion of atoms)
  • Something is hot because it has thermal energy
    and its atoms are moving rapidly.
  • The more energy, the faster they move.
  • The total thermal energy of a substance is the
    sum of the individual thermal energies of the
    atoms and molecules making up a substance.

14
Chemists are Interested in Heat Transfer (q)
during chemical reactions
  • Heat is a form of energy transfer.
  • Heat transfers from hot object to cooler object.
  • qgt0 Heat is transferred from surroundings to
    system. ENDOTHERMIC
  • qlt0 Heat is transferred from system to
    surroundings. EXOTHERMIC

15
Remember the Trick to Temperature
  • A measure of heat and relates to the average
    kinetic energy of atoms in a sample.
  • The higher the temperature, the greater the
    thermal motion. Thermal energy depends upon
    sample size.
  • Heat ? Temperature
  • Thermometers display changes in thermal energy.

16
Exothermic Process
  • By convention in an exothermic process, heat is
    transferred FROM the system TO the surroundings.
    Heat is a product.
  • 2H2 (g) O2 (g) ? 2H2O (l) energy
  • Combustion reactions are exothermic!
  • Chemical (potential) energy is converted to heat
    (kinetic energy).

17
Endothermic Reaction
  • Heat is transferred TO the system FROM the
    surroundings. Heat is a reactant.
  • Energy 2HgO (s) ? 2Hg(l) O2 (g)
  • Thermal energy (kinetic) is converted to chemical
    energy (potential).

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19
Specific Heat and Heat Capacity
  • Specific heat the heat needed to raise the
    temperature of one gram of a substance by 1.0 0
    C. Units J/g0C. Intensive property- not
    dependent on how much is present.
  • Heat Capacity the measure of an overall effect
    of heat transfer on the temperature of an object.
    Units J/g . Extensive property (depends on
    mass).

20
Specific Heats of Various Substances
21
Side Note Specific Heat of Water
  • The specific heat of water is one of the highest
    known specific heats.
  • Thats why temperature fluctuations are smaller
    near large bodies of water like Puget Sound than
    someplace like the desert in Arizona.

22
Calorimetry
  • The laboratory technique used to measure the heat
    released or absorbed during a chemical or
    physical change.
  • In a calorimeter, there is no volume change in
    the system, therefore,
    DE q
  • q mcDT

23
Calorimetry ? heat absorbed mcDTliquid
mcDTcalorimeter
  • Calorimetry depends on the assumption that all
    the heat involved changes the contents of the
    calorimeter and its contents? no heat is gained
    or lost through the environment.
  • For many chemical reactions, a styrofoam cup is a
    convenient calorimeter because it has a very low
    heat capacity (for most purposes negligible) and
    it is a good insulator.

24
Sample Calculation- Specific Heat
  • A piece of iron with a mass of 72.4 grams is
    heated to 100.oC and plunged into 100.g of water
    that is initially at 10.0oC. Calculate the final
    temperature that is reached assuming no heat loss
    to the surroundings.
  • Cwater 4.18 J/goC
  • Ciron 0.449 J/goC
  • NOTE the heat gained by the cooler body the
    heat lost by the warmer body 0
  • Tfinal 16.5oC

25
Sample Calorimetry Problem
  • A 1.5886 g sample of glucose was ignited in a
    bomb calorimeter. The temperature increased by
    3.682oC. The heat capacity of the calorimeter
    was 3.52 kJ/oC, and the calorimeter contained
    1.000 kg of water. Find the total heat released
    and the molar heat of reaction.
  • Remember Heat energy goes into heating up the
    water and the calorimeter itself.
  • Q -28.35 kJ
  • -3211 kJ/mol

26
Heats of Reactions
  • We describe the energy changes in a chemical
    reaction in terms of the amount of heat that can
    be produced or must be supplied.
  • All chemical reactions are accompanied by a
    change in the potential energy of the bonds,
    hence, heat energy changes.
  • Endothermic reactions ? Heat supplied
  • Exothermic reactions ? Heat produced

27
The Enthalpy
  • Enthalpy (H) is a state function defined as
  • H E PV
  • DH DE PDV
  • From the first law of thermodynamics
  • DE q - PDV
  • Therefore, at constant volume, DH q

28
Enthalpy, ?H A State Function
  • We can easily run reactions in the lab at
    constant pressure.
  • Enthalpy heat transfer at constant pressure.
  • ?H Hfinal Hinitial Hproducts Hreactants
    qp
  • The enthalpy change for a reaction is best shown
    by writing the balanced equation for the
    reaction, with the enthalpy change along side it.
    It is vital to include state symbols in all
    thermochemical equations because changes of state
    have their own enthalpy changes associated with
    them. (think ice vs water)

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32
Endothermic Vs Exothermic Reactions
  • For exothermic processes, q is negative then ?H
    is negative.The products are more likely to be
    stable than the reactants.
  • For endothermic processes, q is positive then ?H
    is positive. The products are more likely to be
    less stable than the reactants.

33
Energy Diagram for Exothermic and Endothermic
Reactions
Exothermic
Endothermic
34
Five Properties of Enthalpy
  • Extensive property- like mass, does depend on
    amount of substance present.
  • A state function path independent.
  • ? H has unique values qp.
  • If you reverse a reaction, the values of ?H are
    indentical, the signs are opposite.
  • ? H applies to balanced reactions with correct
    stoichiometric coefficients.

35
2 H2 (g) O2 (g) ? 2H2O (g) DH -483.6 kJ
  • The negative sign indicates the reaction is
    exothermic.
  • This value of DH is for the production of 2 mol
    of water. If 4 moles were produced, the value
    for DH would be twice -483.6 kJ
  • If the reaction is reversed, the sign of DH would
    be reversed.
  • DH does depend on the state of matter. The
    enthalpy change would be different for the
    formation of liquid water instead of gaseous
    water.

36
Sample Problem
  • 2NaHCO3 (s) ? Na2CO3(s) H2O(l) CO2 (g)
  • DH 91.6 kJ/mol
  • NaHCO3(s) ? ½ Na2CO3 ½ H2O (l) ½ CO2 (g)
  • DH?

37
Heats of Reaction DH
38
Sample Problems- Heats of Reaction
  • What is the heat of reaction for the dissolving
    of 4 moles of potassium nitrate in water?
  • How much energy is required to form 2 moles of
    methanol (CH3OH) from carbon dioxide and water?
  • Which produces more energy? The combustion of
    methanol or methane (CH4)?

39
Hess Law
  • Hesss law states that if a reaction occurs in a
    series of steps, then the enthalpy change for the
    overall reaction is simply the sum of the
    enthalpy changes of the individual steps.
  • Pay attention to the direction of the reaction
    and the stiochiometry of the reaction.
  • Remember DH is a state function.

40
State Functions do not depend on path, only
starting and finishing points.
41
Sample Hesss Law Problem
  • Given
  • C(s) O2 (g) ? CO2 (g) DH -393.5kJ
  • H2(g) 1/2O2 (g) ? H2O (l) DH -285.8 kJ
  • C2H2(g) 5/2O2(g) ? 2CO2(g) H2O(l)
  • DH -1200.8 kJ
  • Find the enthalpy change for
  • 2C(s) H2(g) ? C2H2(g)
  • 128 kJ

42
Enthalpies of FormationHeats of Formation DHfo
kJ/mol
  • Enthalpies of reaction can be calculated from
    individual enthalpies of formation for the
    reactants and products.
  • Standard states used
  • Pressure 1atm for gases
  • 1 molar solutions for aqueous solutions
  • For pure substances- the standard state is the
    most stable form at 1 atm and 25oC.

43
Standard Enthalpy of Formation- DHfo
  • Change in enthalpy when 1 mol of a compound is
    formed from its elements when all the substances
    are in their standard states.
  • The DHfo of an element in its standard state is
    zero.
  • The DHfo can be determined from the tabulated
    DHfo of the individual reactants and products.
  • What is the formation reaction for CaCO3(s)?
  • Ca(s) C(graphite) 3/2 O2 (g) ? CaCO3(s)
  • DH -1207.1 kJ/mol

44
DHorxn ? nDHfo products -? mDHforeactants
  • In using this equation be sure to consider the
    number of moles of each, DHfo for the individual
    compounds refer to the formation of one mole.
  • The DHo a value assumes reactants and products
    are at the same temperature.
  • Since DHfo values are tabulated at 25oC, we can
    only calculate DHo at this temperature.

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46
Sample Problem Standard Enthalpy of Reaction
  • Calculate DHo for
  • 6H2O (g) 4NO (g) ? 5 O2 (g) 4NH3 (g)
  • 904.68kJ

47
Bond Enthalpies
  • The breaking of bonds endothermic process
  • The formation of bonds exothermic process
  • The amount of energy associated with the
    formation / breaking of a particular bond is to a
    large extent independent of the bonding in the
    rest of the molecule.

48
C-H bond
  • The energy bonding a carbon atom to a hydrogen
    atom is about 413 kJ/mol in both methane and
    ethanol.

Ethanol
49
DH ?Bonds Broken ?Bonds Formed
  • Approximate enthalpy changes for reactions may be
    calculated by considering the bonds being broken
    and the bonds being formed in a reaction.
  • As before with enthalpy calculations, the number
    of bonds broken or formed must be considered in
    the calculation.

50
Sample Problem Bond Enthalpies
  • Consider the formation of ammonia from nitrogen
    and hydrogen
  • N2 (g) 3H2 (g) ? 2NH3 (g)
  • N?N 945 kJ/mol H-H 436 kJ/mol N-H 391 kJ/mol
  • DH ?
  • DH -93 kJ/mol

51
Sample Problem Bond Enthalpies
  • The balanced equation for the complete combustion
    of one mole of ethene, C2H4, in oxygen is shown
    below
  • C2H4 (g) 3O2(g) ? 2CO2 (g) 2H2O (g)
  • CC 612 kJ/mol C-H 412kJ/mol O-H 463 kJ/mol
    CO 743 kJ/mol OO 496 kJ/mol
  • DH ?

52
Second Law of Thermodynamics
  • All processes that occur spontaneously move in
    the direction of an increase in entropy of the
    universe.
  • Things are getting all messed up!
  • Nature Loves Disorder.

53
Entropy measure of disorderDS J/mol K
  • Entropy is another consideration in energetics.
    It deals with the desire of the universe to move
    to a more disordered state.
  • A positive value for entropy indicates the
    products of the reaction are more disordered than
    the reactants.
  • Entropy of a system can be calculated much the
    same way that enthalpy is calculated.

54
DSo Standard Molar Entropy
  • Entropy of 1 mole of a substance in its standard
    state.
  • Pressure 1atm for gases
  • 1 molar solutions for aqueous solutions
  • For pure substances- the standard state is the
    most stable form at 1 atm and 25oC.
  • For a reaction
  • DSorxn ? DSo products -? DSoreactants

55
Sample Problem Entropy
  • Calculate DSo for the following and determine if
    the reactions are favorable.
  • H2(g) 1/2O2(g) ? H2O(g)
  • H2(g) 1/2O2(g) ? H2O(l)
  • CaCO3(s) H2SO4(l) ? CaSO4(s) H2O(g) CO2(g)
  • a-44.8 J/mol K b -163.6 J /mol K c259 J/mol
    K

56
Spontaneity of Chemical Reactions
Why does this chemical reaction occur on its own
at room temperature?
Fe(s) CuSO4(aq) ? FeSO4(aq) Cu(s)
57
Chemists Goal- to be able to predict spontaneity
of a reaction
  • Enthalpy DH or values favorable?

Entropy DS or values favorable?
58
Gibbs Free Energy Equation
  • Gibbs free energy (G) is a thermodynamic
    function that combines the enthalpy, entropy, and
    temperature or a reaction.
  • DG DH TDS
  • If DGgt0, the reaction is not spontaneous energy
    must be supplied to cause the reaction to occur.
  • If DGlt0, the reaction is spontaneous.
  • If DG0, the reaction is at equilibrium

59
Gibbs Free Energy Equation
  • Just like with DH and DS, if there is a DG
    associated with a reaction and the reaction is
    reversed, the sign on DG changes.
  • Also, check the chemical equation for the
    stoichiometric proportions. Multiply the value
    of DGo if necessary.
  • Units for DGo are kJ/mol

60
Standard Gibbs Free Energy Change DGo
  • Just like with enthalpy and entropy, the standard
    Gibbs free energy change is calculated
  • DGorxn ? DGo products -? DGoreactants
  • Remember DGo of an element in its standard state
    is zero.

61
Sample Problem Gibbs Free Energy Change
  • Calculate DGo for and determine if the reaction
    is spontaneous
  • 2NH4Cl(s) CaO(s) ? CaCl2(s) H2O(l) 2NH3(g)
  • C2H4 (g) H2O(g) ? C2H5OH(l)
  • Ca(s) 2H2SO4(l) ? CaSO4(s) SO2(g) 2H2O(l)

62
Gibbs Free Energy Equation Under Non-Standard
Conditions
  • Standard States
  • Pressure 1atm for gases
  • 1 molar solutions for aqueous solutions
  • For pure substances- the standard state is the
    most stable form at 1 atm and 25oC.
  • What about when the concentrations of aqueous
    solutions are not 1.0 M or the pressure for gases
    is not at 1 atm?
  • DG DG0 RT ln Q Where Q reaction quotient
    DGo 2.303 log Q

63
Problem DG under Non Standard Conditions
  • Calculate DG (non standard conditions-note the
    lack of 0 symbol) for the following at 500K
  • 2NO(g) O2(g) ? 2NO2(g)
  • 2.00M 0.500M 1.00M
  • Note the conditions are non-standard-
    concentrations are not 1.0M.
  • DG

64
When Q Keq
  • If the system is at equilibrium, then DG 0 and
    the previous equation becomes
  • DG DG0 RT ln Q
  • At equilibrium Q K and DG 0
  • DGo -RT ln K or 2.303 RT log K

65
Problem DG at Equilibrium
  • Calculate DGo for (standard temp 298K)
  • 2O3 (g) ? 3O2(g) Kp 4.17X1014
  • DGo
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