Chemistry 431Chemistry 431

L t 7Lecture 7Enthalpy

NC State University

MotivationThe enthalpy change ΔH 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 + PVm

For an ideal or perfect gas this becomes:

Hm = Um + RT

Usually when we write ΔH 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:

ΔH = ΔU + PΔV at const. T and P

g

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

ΔH = ΔU + ΔnRT 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) Δn = -1W i t thi l f th f lWe arrive at this value from the formula

Δn = 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

ΔH = CPΔT

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:

ΔHm = CP,mΔT

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.

ΔH = CPΔT ΔU = CVΔT

CP = ΔHΔT

= ∂H∂T

P

CV = ΔUΔT

= ∂U∂T

V

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:

ΔH C (T)dTT2

C (T) bT c

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

ΔH = 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 1ΔH = a T2 – T1 + b2 T2

2 – T12 + c 1

T1– 1

T2

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

ΔHfus ΔHvap

Solid ⇔ Liquid ⇔ GasFusion Vaporization

qΔHfreeze ΔHcondFreezing Condensation

ΔHsub

Solid ⇔ GasSublimation

Solid ⇔ GasΔHvap dep

Vapor 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:Δ H = Δ H

Δ forwardH = – Δ reverseH

Δ fusH = – Δ freezeHΔvapH = – ΔcondHΔsubH = – Δvap depH

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

ΔsubH Δvap 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. ΔsubH > 0

B Δ H < 0B. ΔcondH < 0

C. ΔfusH > 0fus

D. ΔvapH < 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.

ΔsubH = Δ fusH + ΔvapH

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 Δ H = Δ H Δ HA. ΔsubH = ΔfusH - ΔvapH

B. ΔvapH = ΔsubH - ΔfusH

C. ΔfusH = ΔsubH + ΔvapH

D. ΔvapH = ΔsubH + ΔfusH

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

The negative value of ΔH 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 ΔH 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, ΔrH , 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

Δ rH∅ = νHm

∅(products)Σ – νHm∅(reactants)Σp

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) ΔH = +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) ΔH = +738 kJMg+(g) Mg2+(g) + e-(g) ΔH = +1451 kJMg (g) Mg2 (g) + e (g) ΔH = +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) ΔH = -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) Δ H = 890 kJCH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔcH = -890 kJ

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

ΔHm= ΔUm + ΔνgasRTThe quantity Δνgas 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 Δνgas= -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 Δνgas for the reaction as written:

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

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

QuestionDetermine Δνgas for the reaction as written:

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

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

ΔHm= ΔUm + ΔνgasRTThe quantity Δνgas 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 Δνgas= -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. ΔνgasRT B. ΔUm + ΔνgasRTgC. ΔUm - ΔνgasRTD. Δνgas

QuestionWhat is the work term for expansion against the atmosphere?

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

A. ΔνgasRT B. ΔUm + ΔνgasRTgC. ΔUm - ΔνgasRTD. Δνgas

Hess’s LawWe often need a value of ΔH that is not in the thermochemicaltables. We can use the fact that ΔH 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 ΔH of fusion and ΔH ofexample of this using the sum of ΔH of fusion and ΔH of vaporization to obtain ΔH of sublimation.

H ’ l i f l t t t f thi tHess’s 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) ΔH = +1312 kJCl(g) + e-(g) Cl-(g) ΔH = -349 kJ

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

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

Question Consider the reactions:

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

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

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

Question Consider the reactions:

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

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

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

Question Consider the reactions:

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

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

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

Application of Hess’s LawppWe can use the property known as Hess’s law to obtain a standard enthalpy of combustion for propene from the py p ptwo reactions:C3H6(g) + H2(g) C3H8(g) ΔH = -124 kJC H (g) + 5O (g) 3CO (g) + 4H O(l) ΔH = -2220 kJC3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔH = -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) ΔH = -2344 kJand now we can subtract:H2(g) + 1/2O2(g) H2O(l) ΔH = -286 kJ2(g) 2(g) 2 ( )to obtain:C3H6(g) + 9/2O2(g) 3CO2(g) + 3H2O(l) ΔH = -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:

Δ rCP = νCP(products)Σ – νCP(reactants)Σand then substitute this into the expression:

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

Δ rH∅(T2) = Δ rH

∅(T1) + Δ rCPdTT1

T2

If the heat capacities are all constant of the temperaturerange then: Δ rH

∅(T2) = Δ rH∅(T1) + Δ rCPΔT

Standard Enthalpies of FormationThe standard enthalpy of formation ΔfH 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, let’s examine the formation of water.H (g) + 1/2 O (g) H O(l) ΔH = 286 kJH2(g) + 1/2 O2(g) H2O(l) ΔH = -286 kJ

Therefore, we say that ΔfH (H2O, l) = -286 kJ/mol. Although ΔfH for elements in their reference states is zero,ΔfH is not zero for formation of an element in a different phase:C(s, graphite) C(s, diamond) ΔfH = + 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 ΔfH 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 ΔfH 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. Hess’s lawB. the van der Waal’s 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. Hess’s lawB. the van der Waal’s equation of stateqC. ideal gas lawD. none of the above