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  • Chemistry 431Chemistry 431

    L t 7Lecture 7Enthalpy

    NC State University

  • MotivationThe enthalpy change H is the change in energy at constantpressure When a change takes place in a system that ispressure. When a change takes place in a system that isopen to the atmosphere, the volume of the system changes,but the pressure remains constant. In any chemical reactionsthat involve the creation or consumption of molecules in the vapor or gas phase there is a work term associated with thecreation or consumption of the gas.p g

  • Molar EnthalpypyEnthalpy can be expressed as a molar quantity:

    H

    We can also express the relationship between enthalpy andi t l i t f l titi

    Hm = Hn

    internal energy in terms of molar quantities:

    Hm = Um + PVmFor an ideal or perfect gas this becomes:

    Hm = Um + RT

    Usually when we write H for a chemical or physical changewe refer to a molar quantity for which the units are kJ/mol.

  • Enthalpy for reactions involving gases

    If equivalents of gas are produced or consumed in aIf equivalents of gas are produced or consumed in a chemical reaction, the result is a change in pressure-volumework. This is reflected in the enthalpy as follows.

    which can be rewritten for an ideal gas:

    H = U + PV at const. T and P

    g

    The number of moles n is the number of moles created or

    H = U + nRT at const. T and P

    The number of moles n is the number of moles created orabsorbed during the chemical reaction. For example,

    CH2=CH2(g) + H2(g) CH3CH3(g) n = -1W i t thi l f th f lWe arrive at this value from the formula

    n = nproducts - nreactants = 1 - 2 = -1

  • The temperature dependence of the enthalpy change

    Based on the discussion the heat capacity from the lastp ylecture we can write the temperature dependence of theenthalpy change as:

    H C T

    Note that we can use tabulated values of enthalpy at 298 Kf

    H = CPT

    and calculate the value of the enthalpy at any temperatureof interest. We will see how to use this when we considerthe enthalpy change of chemical reactions (the standardpy g (enthalpy change). The basic physics of all temperature dependence is contained in the above equation or morefrequently in the equation below as molar quantity:frequently in the equation below as molar quantity:

    Hm = CP,mT

  • Another view of the heat capacity

    At this point it is worth noting that the expressions for thep g pheat capacity at constant volume and constant pressurecan be related to the temperature dependence of U and H,respectivelyrespectively.

    H = CPT U = CVT

    CP = HT= H

    TP

    CV = UT= U

    TV

    The heat capacity is the rate of change of the energy withtemperature The partial derivative is formal way of sayingtemperature. The partial derivative is formal way of sayingthis.

  • The heat capacity is also a function of temperature

    We have treated the heat capacity as a constant up to p y pthis point. That is a valid approximation under many circumstances, but only over a limited range of temperature.In the general case the temperature dependence of theIn the general case the temperature dependence of theenthalpy can be described as:

    H C (T)dTT2

    C (T) bT c

    The parameters a, b, and c are given in Tables. Actually,

    H = CP(T)dTT1

    , CP(T) = a + bT + cT2

    p , , g y,this expression is readily integrated in the general case togive:

    b 2 2 1 1H = a T2 T1 + b2 T22 T1

    2 + c 1T1 1T2

  • Enthalpy of physical changepy p y gA physical change is when one state of matter changes into

    th t t f tt f th b t Th diffanother state of matter of the same substance. The differencebetween physical and chemical changes is not always clear,however, phase transitions are obviously physical changes.p y p y g

    Hfus HvapSolid Liquid Gas

    Fusion Vaporizationq

    Hfreeze HcondFreezing Condensation

    HsubSolid Gas

    SublimationSolid Gas

    Hvap depVapor Deposition

  • Properties of Enthalpy as a State Function

    Th f t th t th l i t t f ti i f l f thThe fact that enthalpy is a state function is useful for theadditivity of enthalpies. Clearly the enthalpy of forwardand reverse processes must be related by:p y

    so that the phase changes are related by: H = H

    forwardH = reverseH

    fusH = freezeHvapH = condHsubH = vap depH

    Moreover, it should not matter how the system is transformedfrom the solid phase to the gas phase. The two processes of

    subH vap depH

    p g p pfusion (melting) and vaporization have the same net enthalpyas sublimation.

  • QuestionQuestionWhich is statement is false:Which is statement is false:A. subH > 0

    B H < 0B. condH < 0

    C. fusH > 0fus

    D. vapH < 0

  • Addivity of Enthalpiesy pBecause the enthalpy is a state function the same magnitudemust be obtained for direct conversion from solid to gas asmust be obtained for direct conversion from solid to gas asfor the indirect conversion solid to liquid and then liquid to gas.

    subH = fusH + vapH

    Of course, these enthalpies must be measured at the sametemperature. Otherwise an appropriate correction would need

    p

    to be applied as described in the section on the temperaturedependence of the enthalpy.

  • QuestionWhich statement is true?A H = H HA. subH = fusH - vapH

    B. vapH = subH - fusH

    C. fusH = subH + vapH

    D. vapH = subH + fusH

  • Chemical ChangegIn a chemical change the identity of substances is alteredduring the course of a reaction. One example is the g phydrogenation of ethene:

    CH =CH (g) + H (g) CH CH (g) H = -137 kJCH2=CH2(g) + H2(g) CH3CH3(g) H = -137 kJ

    The negative value of H signifies that the enthalpy of thet d b 137 kJ d if th ti t k lsystem decreases by 137 kJ and, if the reaction takes place

    at constant pressure, 137 kJ of heat is released into the surroundings, when 1 mol of CH2=CH2 combines with 1 mol g , 2 2of H2 at 25 oC.

  • Standard Enthalpy Changespy gThe reaction enthalpy depends on conditions (e.g. T and P).It is convenient to report and tabulate information under aIt is convenient to report and tabulate information under astandard set of conditions.

    Corrections can be made using heat capacity for variationsin the temperature. Corrections can also be made forvariations in the pressure. p

    When we write H in a thermochemical equation, we alwaysmean the change in enthalpy that occurs when the reactantsmean the change in enthalpy that occurs when the reactantschange into the products in their respective standard states.

  • Standard Reaction EnthalpypyThe standard reaction enthalpy, rH , is the difference between the standard molar enthalpies of the reactantsbetween the standard molar enthalpies of the reactantsand products, with each term weighted by the stoichiometric coefficient.

    The standard state is for reactants and products at 1 bar

    rH = Hm

    (products) Hm(reactants)p

    of pressure. The unit of energy used is kJ/mol.

    The temperature is not part of the standard state and it isThe temperature is not part of the standard state and it ispossible to speak of the standard state of oxygen gas at100 K, 200 K etc. It is conventional to report values at298 K and unless otherwise specified all data will bereported at that temperature.

  • Enthalpies of IonizationThe molar enthalpy of ionization is the enthalpy thataccompanies the removal of an electron from a gas phaseatom or ion:atom or ion:

    H(g) H+(g) + e-(g) H = +1312 kJ

    For ions that are in higher charge states we must considersuccessive ionizations to reach that charge state. For example, for Mg we have:p , g

    Mg(g) Mg+(g) + e-(g) H = +738 kJMg+(g) Mg2+(g) + e-(g) H = +1451 kJMg (g) Mg2 (g) + e (g) H = +1451 kJ

    We shall show that these are additive so that the overallthl l h i 2189 kJ f th tienthlalpy change is 2189 kJ for the reaction:

    Mg(g) Mg2+(g) + 2e-(g)

  • Electron Gain EnthalpyThe reverse of ionization is electron gain. The correspondingenthalpy is called the electron gain enthalpy. For example:

    Cl(g) + e-(g) Cl-(g) H = -349 kJ

    The sign can vary for electron gain. Sometimes, electrongain is endothermic.

    The combination of ionization and electron gain enthalpycan be used to determine the enthalpy of formation of salts.

    Other types of processes that are related include moleculardissociation reactions.

  • Enthalpies of CombustionpStandard enthalpies of combustion refer to the completecombination with oxygen to carbon dioxide and water.combination with oxygen to carbon dioxide and water.For example, for methane we have:

    CH (g) + 2O (g) CO (g) + 2H O(l) H = 890 kJCH4(g) + 2O2(g) CO2(g) + 2H2O(l) cH = -890 kJ

    Enthalpies of combustion are commonly measured in abomb calorimeter (a constant volume device). Thus,Um is measured. To convert from Um to Hm we needto use the relationship:to use the relationship:

    Hm= Um + gasRTThe quantity gas is the change in the stoichiometric coefficients of the gas phase species We see in thecoefficients of the gas phase species. We see in the above express that gas= -2. Note that H2O is a liquid.

  • QuestionFill in the missing stoichiometric coefficients for thecombustion reaction:combustion reaction:

    C5H12(g) + XO2(g) YCO2(g) + ZH2O(l)

    A. X=4, Y=8, Z=12B. X=8, Y=5, Z=6C. X=4, Y=10, Z=6D. X=2, Y=1, Z=6D. X 2, Y 1, Z 6

  • QuestionFill in the missing stoichiometric coefficients for thecombustion reaction:combustion reaction:

    C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l)

    A. X=4, Y=8, Z=12B. X=8, Y=5, Z=6C. X=4, Y=10, Z=6D. X=2, Y=1, Z=6D. X 2, Y 1, Z 6

  • QuestionDetermine gas for the reaction as written:

    C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l)

    A. gas = 3B. gas = 8gC. gas = -4D. gas = -3

  • QuestionDetermine gas for the reaction as written:

    C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l)

    A. gas = 3B. gas = 8gC. gas = -4D. gas = -3

    Hm= Um + gasRTThe quantity gas is the change in the stoichiometric coefficients of the gas phase species We see in thecoefficients of the gas phase species. We see in the above express that gas= -2. Note that H2O is a liquid.

  • QuestionWhat is the work term for expansion against the atmosphere?

    C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l)

    A. gasRT B. Um + gasRTgC. Um - gasRTD. gas

  • QuestionWhat is the work term for expansion against the atmosphere?

    C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l)

    A. gasRT B. Um + gasRTgC. Um - gasRTD. gas

  • Hesss LawWe often need a value of H that is not in the thermochemicaltables. We can use the fact that H is a state function to advantage by using sums and differences of known quantitiesto obtain the unknown. We have already seen a simple example of this using the sum of H of fusion and H ofexample of this using the sum of H of fusion and H of vaporization to obtain H of sublimation.

    H l i f l t t t f thi tHesss law is a formal statement of this property.

    The standard enthalpy of a reaction is the sum of the standardpyenthalpies of the reactions into which the overall reactionmay be divided.

  • Question Consider the reactions:

    H(g) H+(g) + e-(g) H = +1312 kJCl(g) + e-(g) Cl-(g) H = -349 kJ

    Which statement is true about the charge transfer f C f C ?from H to Cl to form H+ and Cl-?

    A. H = 963 kJB. H = 1661 kJC. H = 1312 kJD H = -349 kJD. H = -349 kJ

  • Question Consider the reactions:

    H(g) H+(g) + e-(g) H = +1312 kJCl(g) + e-(g) Cl-(g) H = -349 kJ

    Cl(g) + H(g) Cl-(g) + H+(g) H = 963 kJWhich statement is true about the charge transfer f C f C ?from H to Cl to form H+ and Cl-?

    A. H = 963 kJB. H = 1661 kJC. H = 1312 kJD H = -349 kJD. H = -349 kJ

  • Question Consider the reactions:

    H(g) H+(g) + e-(g) H = +1312 kJCl(g) + e-(g) Cl-(g) H = -349 kJ

    What further information do you need to calculate thef C ( ) C ( )?enthalpy for the reaction H2 + Cl2 2H+(aq) + 2Cl-(aq)?

    A. H of ionization and H of electron capture pB. H of formation, H of dissociation and H of solvationC. H of ionization and H of solvationD H of dissociation and H of solvationD. H of dissociation and H of solvation

  • Question Consider the reactions:

    H(g) H+(g) + e-(g) H = +1312 kJCl(g) + e-(g) Cl-(g) H = -349 kJ

    What further information do you need to calculate thef C ( ) C ( )?enthalpy for the reaction H2 + Cl2 2H+(aq) + 2Cl-(aq)?

    A. H of ionization and H of electron capture pB. H of formation, H of dissociation and H of solvationC. H of ionization and H of solvationD H of dissociation and H of solvationD. H of dissociation and H of solvation

  • Application of Hesss LawppWe can use the property known as Hesss law to obtain a standard enthalpy of combustion for propene from the py p ptwo reactions:C3H6(g) + H2(g) C3H8(g) H = -124 kJC H (g) + 5O (g) 3CO (g) + 4H O(l) H = -2220 kJC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) H = -2220 kJ

    If we add these two reactions we get:C H ( ) H ( ) 5O ( ) 3CO ( ) 4H O(l) H 2344 kJC3H6(g) + H2(g) + 5O2(g) 3CO2(g) + 4H2O(l) H = -2344 kJand now we can subtract:H2(g) + 1/2O2(g) H2O(l) H = -286 kJ2(g) 2(g) 2 ( )to obtain:C3H6(g) + 9/2O2(g) 3CO2(g) + 3H2O(l) H = -2058 kJ

  • Variation in Reaction Enthalpy with Temperature

    Since standard enthalpies are tabulated at 298 K we needSince standard enthalpies are tabulated at 298 K we needto determine the value of the entropy at the temperature ofthe reaction using heat capacity data. Although we have

    thi d i th l th l l ti fseen this procedure in the general case the calculation forchemical reactions is easier if you start by calculating theheat capacity difference between reactants and products:p y p

    and then substitute this into the expression:

    rCP = CP(products) CP(reactants)and then substitute this into the expression:

    If th h t iti ll t t f th t t

    rH(T2) = rH

    (T1) + rCPdTT1

    T2

    If the heat capacities are all constant of the temperaturerange then: rH(T2) = rH(T1) + rCPT

  • Standard Enthalpies of FormationThe standard enthalpy of formation fH is the enthalpy forformation of a substance from its elements in their standardstates. The reference state of an element is its most stable form at the temperature of interest. The enthalpy offormation of the elements is zero.formation of the elements is zero.

    For example, lets examine the formation of water.H (g) + 1/2 O (g) H O(l) H = 286 kJH2(g) + 1/2 O2(g) H2O(l) H = -286 kJ

    Therefore, we say that fH (H2O, l) = -286 kJ/mol. Although fH for elements in their reference states is zero,fH is not zero for formation of an element in a different phase:C(s, graphite) C(s, diamond) fH = + 1.895 kJ/molC(s, g ap te) C(s, d a o d) f 895 J/ o

  • Question Consider the formation of carbon dioxide at 298 K:

    C(s) + O2(g) CO2(g)

    How would you find the heat of formation of oxygen?How would you find the heat of formation of oxygen?A. Look up fH for C(s) and subtract it from that of CO2.B. Look it up the standard thermodynamic tables.C f f f O fC. The heat of formation of O2 is zero by definition.D. It is equal to the standard bond energy of two

    oxygen atoms.yg

  • Question Consider the formation of carbon dioxide at 298 K:

    C(s) + O2(g) CO2(g)

    How would you find the heat of formation of oxygen?How would you find the heat of formation of oxygen?A. Look up fH for C(s) and subtract it from that of CO2.B. Look it up the standard thermodynamic tables.C f f f O fC. The heat of formation of O2 is zero by definition.D. It is equal to the standard bond energy of two

    oxygen atoms.yg

  • Question Consider the formation of carbon dioxide at 150 K:

    C(s) + O2(g) CO2(g)

    How would you find the heat of formation of CO ?How would you find the heat of formation of CO2?

    Apply a correction to the enthalpy from:

    A. Hesss lawB. the van der Waals equation of stateqC. ideal gas lawD. none of the above

  • Question Consider the formation of carbon dioxide at 150 K:

    C(s) + O2(g) CO2(g)

    How would you find the heat of formation of CO ?How would you find the heat of formation of CO2?

    Apply a correction to the enthalpy from:

    A. Hesss lawB. the van der Waals equation of stateqC. ideal gas lawD. none of the above