properties of exponentials - nc state universityfranzen/public_html/ch201/lecture/lec… · ·...

Thermodynamics

Heat Capacity

Calorimetry

Enthalpy

Thermodynamic cycles

Adiabatic processes

NC State University

Motivation

The enthalpy change DH is the change in energy at constant

pressure. When a change takes place in a system that is

open to the atmosphere, the volume of the system changes,

but the pressure remains constant. In any chemical reactions

that involve the creation or consumption of molecules in the

vapor or gas phase there is a work term associated with the

creation or consumption of the gas.

Molar Enthalpy

Enthalpy can be expressed as a molar quantity:

We can also express the relationship between enthalpy and

internal energy in terms of molar quantities:

For an ideal or perfect gas this becomes:

Usually when we write DH for a chemical or physical change

we refer to a molar quantity for which the units are kJ/mol.

Hm = Hn

Hm = Um + PVm

Hm = Um + RT

Enthalpy for reactions

involving gases If equivalents of gas are produced or consumed in a

chemical reaction, the result is a change in pressure-volume

work. This is reflected in the enthalpy as follows.

which can be rewritten for an ideal gas:

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

absorbed during the chemical reaction. For example,

CH2=CH2(g) + H2(g) CH3CH3(g) Dn = -1

We arrive at this value from the formula

Dn = nproducts - nreactants = 1 - 2 = -1

DH = DU + PDV at const. T and P

DH = DU + DnRT at const. T and P

The measurement of heat

We must carefully distinguish between heat and temperature.

When we add heat to the system its temperature increases.

We can use measurement of the temperature to determine

how much heat has been added. However, we need to know

the heat capacity of the system in order to do this.

The heat capacity is called C. If we perform a heat exchange

at constant volume then we designate the heat capacity as CV.

If the process occurs at constant pressure we call the heat

capacity CP.

Heat capacity =Heat supplied

Temperature rise

CV,P =qV,P

DT

Molar and Specific Heat

Capacities

We use molar heat capacities for pure substances. As the

name implies the units are J/mol-K for the molar heat

capacity. We write the molar heat capacity at constant

volume as CV,m.

For mixtures we cannot use a molar heat capacity and so

we use the specific heat capacity, which is the heat capacity

per gram of material with units of J/g-K.

Heat Capacity for

a Diatomic Molecule

For a diatomic molecule there is contribution from rotations

as well as translations. This means that as heat is added to

the system the rotational levels can be populated in addition

to an increase in molecular speed. The kinetic theory of

gases considers only the speed. An approximate rule is that

we obtain a contribution to the heat capacity, CV of 1/2nR for

each degree of freedom. We saw that for a monatomic gas

the heat capacity was CV = 3/2nR. A diatomic gas has two

rotational degrees of freedom and so the heat capacity is

approximately CV = 5/2nR. What does this say about CP?

Well, the relationship between CP and CV holds for all gases

so CP = 7/2nR for a diatomic “ideal” gas.

The temperature dependence

of the enthalpy change Based on the discussion the heat capacity from the last

lecture we can write the temperature dependence of the

enthalpy change as:

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

and calculate the value of the enthalpy at any temperature

of interest. We will see how to use this when we consider

the enthalpy change of chemical reactions (the standard

enthalpy change). The basic physics of all temperature

dependence is contained in the above equation or more

frequently in the equation below as molar quantity:

DH = CPDT

DHm = CP,mDT

Another view of

the heat capacity At this point it is worth noting that the expressions for the

heat capacity at constant volume and constant pressure

can be related to the temperature dependence of U and H,

respectively.

The heat capacity is the rate of change of the energy with

temperature. The partial derivative is formal way of saying

this.

DH = CPDT DU = CVDT

CP = DHDT

= HT

P

CV = DUDT

= UT

V

Variation in Reaction Enthalpy

with Temperature Since standard enthalpies are tabulated at 298 K we need

to determine the value of the entropy at the temperature of

the reaction using heat capacity data. Although we have

seen this procedure in the general case the calculation for

chemical reactions is easier if you start by calculating the

heat capacity difference between reactants and products:

and then substitute this into the expression:

If the heat capacities are all constant of the temperature

range then:

DrH(T2) = DrH

(T1) + DrCPdT

T1

T2

DrH(T2) = DrH

(T1) + DrCPDT

DrCP = CP(products) – CP(reactants)

Calorimetry The science of heat measurement is called calorimetry.

A calorimeter consists of a container in a heat bath.

A physical or chemical process occurs in the container and

heat is added or removed from the heat bath. The

temperature increases or decreases as result. By knowing

the heat capacity of the bath we can measure the amount

of heat that has been added or removed from the system.

Energy in the

form of heat

flows into the bath.

Calorimetry In the studies of biological systems there are two important

types of calorimetry.

1. Differential scanning calorimetry (DSC)

2. Isothermal titration calorimetry (ITC)

In DSC the temperature is increased at a constant heating

rate and the heat capacity is measured. DSC is used for

determining the parameters associated with phase transitions

e.g. protein unfolding, denaturation, DNA hybridization etc.

In ITC the temperature is held constant while one component

is added to another. The heat of interaction (e.g. binding) is

measured using this method. ITC is widely used to determine

the enthalpy of binding, e.g. for protein-protein and

protein-drug interactions among other types of biological

applications.

A note on using current flow in

calorimetry measurements Energy is measured in Joules. Electrical energy is used

to delivery heat in calorimetry applications. In electrostatics

the units of energy are:

Joules = Coulombs * Volts

J = CV

Coulomb is a unit of charge and volt is a unit of potential.

A charge moving through a potential is a little like a waterfall.

The problem here is that we need a dynamic description

since “moving charge” is not “static”.

V

C

O


calorimetry measurements

As charge moves through the potential energy (heat) is

Released.

Power = Amperes * Volts

W = IV

Power has units of energy per unit time. So as the current

flows through a wire at a certain rate, heat energy is added

to the system at that rate.

V

C

O


calorimetry measurements

Over a period of time, t, the total energy added to the system

is:

Energy = Amperes * Volts * time

E = IVt

This equation is used in the text to describe a number of

calorimetry applications.

V

C

O

The standard state The standard state of a substance is its pure form a 1 bar of

pressure or 1 molar concentration. The standard state can

have any temperature, but the temperature should be specified.

For example, the standard enthalpy of vaporization of water

Is the enthalpy of vaporization at 373 K (the boiling point) and

1 bar of pressure.

If we lower the temperature below 373 K (for example to 372 K)

then water no longer boils. We can say that the reaction is

defined at the standard state of 1 bar (or 1 atm in practice) of

pressure. A reaction can take place at concentrations other

than the standard state, but there will be a dependence of

the enthalpy on pressure. Vaporization is a special case since

the atmosphere is always at 1 atm at sea level.

Enthalpy of physical change

A physical change is when one state of matter changes into

another state of matter of the same substance. The difference

between physical and chemical changes is not always clear,

however, phase transitions are obviously physical changes.

DH fus DHvap

Solid Liquid Gas

DH freeze DHcond

DHsub

Solid Gas

DHvap dep

Fusion

Freezing

Vaporization

Condensation

Sublimation

Vapor Deposition

Properties of Enthalpy as a

State Function

The fact that enthalpy is a state function is useful for the

additivity of enthalpies. Clearly the enthalpy of forward

and reverse processes must be related by:

so that the phase changes are related by:

Moreover, it should not matter how the system is transformed

from the solid phase to the gas phase. The two processes of

fusion (melting) and vaporization have the same net enthalpy

as sublimation.

D fusH = – D freezeH

DvapH = – DcondH

DsubH = – Dvap depH

D forw ardH = – D reverseH

Addivity of Enthalpies

Because the enthalpy is a state function the same magnitude

must be obtained for direct conversion from solid to gas as

for the indirect conversion solid to liquid and then liquid to gas.

Of course, these enthalpies must be measured at the same

temperature. Otherwise an appropriate correction would need

to be applied as described in the section on the temperature

dependence of the enthalpy.

DsubH = DfusH + DvapH

Chemical Change In a chemical change the identity of substances is altered

during the course of a reaction. One example is the

hydrogenation of ethene:

CH2=CH2(g) + H2(g) CH3CH3(g) DH = -137 kJ/mol

The negative value of DH signifies that the enthalpy of the

system decreases by 137 kJ/mol and, if the reaction takes

place at constant pressure, 137 kJ/mol of heat is released

into the surroundings, when 1 mol of CH2=CH2 combines

with 1 mol of H2 at 25 oC.

Standard Enthalpies of Formation

The standard enthalpy of formation DfHo is the enthalpy for

formation of a substance from its elements in their standard

states. The reference state of an element is its most

stable form at the temperature of interest. The enthalpy of

formation of the elements is zero.

For example, let’s examine the formation of water.

H2(g) + 1/2 O2(g) H2O(l) DHo = -286 kJ

Therefore, we say that DfHo (H2O, l) = -286 kJ/mol.

Although DfHo for elements in their reference states is zero,

DfHo is not zero for formation of an element in a different

phase:

C(s, graphite) C(s, diamond) DfHo = + 1.895 kJ/mol

Standard Enthalpy Changes

The reaction enthalpy depends on conditions (e.g. T and P).

It is convenient to report and tabulate information under a

standard set of conditions.

Corrections can be made using heat capacity for variations

in the temperature. Corrections can also be made for

variations in the pressure.

When we write DH in a thermochemical equation, we always

mean the change in enthalpy that occurs when the reactants

change into the products in their respective standard states.

Standard Reaction Enthalpy

The standard reaction enthalpy, DrHo, is the difference

between the standard molar enthalpies of the reactants

and products, with each term weighted by the

stoichiometric coefficient, ,

The standard state is for reactants and products at 1 bar

of pressure or 1 molar concentration. The unit of energy

used is kJ/mol.

The temperature is not part of the standard state and it is

possible to speak of the standard state of oxygen gas at

100 K, 200 K etc. It is conventional to report values at

298 K and unless otherwise specified all data will be

reported at that temperature.

Hess’s Law We often need a value of DH that is not in the thermochemical

tables. We can use the fact that DH is a state function to

advantage by using sums and differences of known quantities

to obtain the unknown. We have already seen a simple

example of this using the sum of DH of fusion and DH of

vaporization to obtain DH of sublimation.

Hess’s law is a formal statement of this property.

The standard enthalpy of a reaction is the sum of the standard

enthalpies of the reactions into which the overall reaction

may be divided.

Application of Hess’s Law

We can use the property known as Hess’s law to obtain

a standard enthalpy of combustion for propene from the

two reactions:

C3H6(g) + H2(g) C3H8(g) DH = -124 kJ

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) DH = -2220 kJ

If we add these two reactions we get:

C3H6(g) + H2(g) + 5O2(g) 3CO2(g) + 4H2O(l) DH = -2344 kJ

and now we can subtract:

H2(g) + 1/2O2(g) H2O(l) DH = -286 kJ

to obtain:

C3H6(g) + 9/2O2(g) 3CO2(g) + 3H2O(l) DH = -2058 kJ

Calculate DrxnHo for the rxn :

2 SO3(g) 2 SO2(g) + O2(g)

given the following :

a) S(s) + O2(g) SO2(g) DfHo = -297 kJ

b) S(s) + 3/2 O2(g) SO3(g) DfHo = -396 kJ

Understanding the

additivity of enthalpies

Note the trends in the heats of solution for these

common compounds.

Enthalpy of solution

SOLUTE ΔHo in kJ/mol

hydrochloric acid -74.84

ammonium nitrate +25.69

ammonia -30.50

potassium hydroxide -57.61

caesium hydroxide -71.55

sodium chloride +3.88

potassium chlorate +41.38

acetic acid -1.51

sodium hydroxide -44.51

Bond pm kJ/mol Bond pm kJ/mol

H—H 74 436 C—C 154 348

H—C 109 413 C=C 134 614

C—C 154 348 C—N 147 308

H—N 101 391 CºC 120 839

N—N 145 170 C—O 143 360

H—O 96 366 C—S 182 272

O—O 148 145 O—O 148 145

H—F 92 568 C—F 135 488

F—F 142 158 O=O 121 498

H—Cl 127 432 C—Cl 177 330

Cl-Cl 199 243 C—Br 194 288

H—Br 141 366 N—N 145 170

Br-Br 228 193 C—I 214 216

H—I 161 298 NºN 110 945

I—I 267 151

Tabulated bond enthalpies

Thermodynamic cycles

The energies and enthalpies of ionic solids are dominated

by Coulombic interactions. The lattice enthalpy can be

calculated from Coulombic interactions. Using the

calculated lattice enthalpy and experimental data one can

obtain the enthalpy of formation of an ionic solid by means

of the Born-Haber cycle. This is illustrated on the next slide

for a generic salt of a monovalent ion M+X-.

The reactions needed to obtain M+X- in the solid phase

to M+ + X- in the vapor phase are given below. The overall

process of separating the ions in an ionic solid into the

constituent ions is written as

M+X- (s) M+ (g) + X- (g)

Lattice enthalpy

Enthalpies of Ionization The molar enthalpy of ionization is the enthalpy that

accompanies the removal of an electron from a gas phase

atom or ion:

H(g) H+(g) + e-(g) DHo = +1312 kJ/mol

For ions that are in higher charge states we must consider

successive ionizations to reach that charge state. For

example, for Mg we have:

Mg(g) Mg+(g) + e-(g) DHo = +738 kJ/mol

Mg+(g) Mg2+(g) + e-(g) DHo = +1451 kJ/mol

We shall show that these are additive so that the overall

enthlalpy change is 2189 kJ/mol for the reaction:

Mg(g) Mg2+(g) + 2e-(g) DHo = +2189 kJ/mol

Electron Gain Enthalpy The reverse of ionization is electron gain. The corresponding

enthalpy is called the electron gain enthalpy. For example:

Cl(g) + e-(g) Cl-(g) DH = -349 kJ/mol

The sign can vary for electron gain. Sometimes, electron

gain is endothermic.

The combination of ionization and electron gain enthalpy

can be used to determine the enthalpy of formation of salts.

Other types of processes that are related include molecular

dissociation reactions.

The Born-Haber cycle

The thermodynamic cycle is illustrated. The scheme shows

that if one knows all of the energies except one (difficult to

measure) quantity, one can calculate it using Hess’ law.

M+ + X-

1/2 X2 (g) + M (s)

Enthalpyof Formation

M (s)

X (g)

Lattice Enthalpy

Enthalpyof Vaporization

Bond Dissociation

ElectronAffinity

IonizationPotential

M+X-

Calculate the enthalpy change for the following rxn:

__ NH3(g) + __ O2(g) __ NO(g) + __ H2O(g)

Using enthalpies of formation

Calculate the enthalpy change for the following rxn:

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Using enthalpies of formation

Solution: Look up the enthalpies of formation of each

molecule in the thermodynamic tables. Remember that

the enthalpy of formation of O2 is zero.

Enthalpies of Combustion

Standard enthalpies of combustion refer to the complete

combination with oxygen to carbon dioxide and water.

For example, for methane we have:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DcombHo = -890 kJ/mol

Enthalpies of combustion are commonly measured in a

bomb calorimeter (a constant volume device). Thus,

DUm is measured. To convert from DUm to DHm we need

to use the relationship:

DHm= DUm + DgasRT

The quantity Dgas is the change in the stoichiometric

coefficients of the gas phase species. We see in the

above express that Dgas= -2. Note that H2O is a liquid.

Enthalpy of formation from combustion

The enthalpy of combustion DcombHo for propane,

C3H8, is -2220 kJ. What is DfHo of C3H8?

Solution: Write down the balanced equation

Look up the enthalpies of formation of the CO2 and H2O.

Remember that the enthalpy of formation of O2 is zero.

The tabulated values of enthalpy are given at 298 K,

Keep in mind that H2O is in the liquid phase.


The enthalpy of combustion DcombHo for propane,

C3H8, is -2220 kJ. What is DfHo of C3H8?


The enthalpy of combustion, DcombHo for propane,

C3H8, is -2220 kJ. What is DHof of C3H8?


What is the enthalpy of combustion for propane,

C3H8 assuming that water is in the vapor phase?

Enthalpy of formation


Look up the enthalpies of formation of the CO2 and H2O.

Remember that the enthalpy of formation of O2 is zero.

What volume of gas is produced if 50 grams of

C3H8 is combusted at 373 K at 1 atm? You may

assume all molecules are in the vapor phase.

Expansion following combustion


C3H8 is combusted? You may assume all molecules

are in the vapor phase (T = 373 K).



Calculate the change in number of moles of gas, Dngas.

Then calculate DV.


C3H8 is combusted? You may assume all molecules

are in the vapor phase.



Calculate the change in number of moles of gas, Dngas.

Then calculate DV.

Dngas = S nproducts – S nreactants

Dngas = 3 + 4 – 1 – 5 = 1

How much work can be extracted for an expansion

against a constant pressure of 10 atm if 50 grams of

C3H8 is combusted?

You may assume all molecules are in the vapor

phase at 373 K.A

Work done following combustion

How much work can be extracted for an expansion

against a constant pressure of 10 atm if 50 grams of

C3H8 is combusted?

You may assume all molecules are in the vapor

phase.

Work done following combustion

We can consider the heat released upon combustion

as the energy available to do work in an engine. If

we consider hydrocarbon fuels (methane, ethane,

propane, butane, and octane) as well as ethanol, we

can compare them both on a per mole and per mass

basis.

The mole basis is obtained using the enthalpy of

combustion.

Comparing fuels on a mass basis

Fuel DcombHo

CH4 -882 C2H6 -1561 C3H8 -2219 C4H10 -2878 C8H18 -5430 C2H6O -1370


Fuel DcombHo Mm

CH4 -882 16 C2H6 -1561 30 C3H8 -2219 44 C4H10 -2878 58 C8H18 -5430 114 C2H6O -1370 46


Fuel DcombHo Mm q (kJ/gram)

CH4 -882 16 55.1 C2H6 -1561 30 52.0 C3H8 -2219 44 50.4 C4H10 -2878 58 49.6 C8H18 -5430 114 47.6 C2H6O -1370 46 29.8


Methane vs. Methanol as a fuel

On a per mole basis methane and methanol have

similar enthalpies of combustion. On a mass basis

methanol provides 22 kJ/g

while methane produces 55 kJ/g.

Methanol is convenient because it is a liquid.

Atmospheric methane: a greenhouse gas

Methane Capture:

An important ecological problem

On a per mole basis methane is converted to methanol with a

heat of reaction of -164.5 kJ/ mol.

The direct conversion shown above does not work

well because it is slow unless the temperature is so high that

one risks complete oxidation to CO2.

Methane Conversion via Syngas

The conversion of CH4 to syngas is the most common

industrial method for producing CH3OH from CH4.

The first step or production of syngas is called cracking. The

second step requires a Cu or Pd catalyst. The overall process

is only about 10% efficient.

Photochemical Methane Conversion

An alternative approach is to use light to activate water in

order to make a hydroxyl radical, which then reacts with

methane.

This process requires sufficient light energy to cleave the

HO-H bond in H2O.

Methane Monooxygenase

The combustion of a fuel leads to heating of the gas

produced in the reaction. The heating can be calculated

using the heat capacity (or specific heat).

For octane, cp = 255.7 J/mol/K.

What is the s (the specific heat)?

Heating of a fuel in an engine

Consider the fact that after combustion the octane fuel has

been converted into CO2 and H2O in the vapor phase. For

H2O vapor, cp = 33 J/mol/K. What is the final temperature if

12 microliters of octane are combusted?

Heating of a fuel in an engine

Focus on energy

The work done in the internal combustion

engine is called pressure volume work.

For a simple irreversible

stroke the work is:

work = P DV

In a 3.0 L 6 cylinder

engine DV = 0.5 L.

Assuming the initial

volume is 80 micro-

liters, what is P?

We can estimate the

pressure from the

amount of octane is injected and combusted

A typical fuel injector will inject 12 microliters of

octane fuel per stroke. What is the pressure of

the gas created by combustion in a volume of 80

microliters?

In our first examination of this problem we will

ignore the heating and simply calculate the

conversion from the volume of liquid fuel to the

volume of H2O and CO2 according to the reaction

stoichiometry.

From combustion to useful work

First convert from volume to moles:

The change in the number of moles is:

Therefore the pressure is:

The pressure is ~100 atm.

Calculating pressure

Solar Heating

In an equilibration two objects that are initially at different

temperatures come to equilibrium at a final temperature.

For example, in a solar water heater we can imagine that a

slab of black stone acts as the absorber. If the stone has a

mass of 20 kg and a temperature of 80 oC.

What is the final temperature of 1 L of water that is initially at

10 oC? (for H2O s1 = 4.18 J/g-oC and stone s2 = 0.5 J/g-oC )

Equilibration

Biological implications

Obviously, the food we eat release heat in our bodies. This

heat is used both to maintain body temperature and for

processes that build up our bodies (anabolic processes).

We often talk about calories in the food we eat. The calorie

on a cereal box is equal to 1000 calories in the chemical

nomenclature. 1 calorie = 4.184 Joules. Therefore, the

Intake required for an average man of 12 MJ/day is about

3000 calories per day (in the sense of diet).

Differential relationships

for enthalpy We have defined a relationship between the enthalpy and

Internal energy

H = U + PV

The infinitesimal change in the state function H results in

H + dH = U + dU + (P + dP)(V + dV)

Therefore

dH = dU + PdV + VdP

Now we substitute dU = dq + dw into this expression

dH = dq + dw + PdV + VdP

Since dw = - PdV

dH = dq + VdP

At constant pressure, dP = 0, and we have

dH = dqP

Path Functions

We have seen that work and heat are path functions. The

magnitude of the work and heat depends not just on the

final values of the T and P, but also on the path taken.

We can summarize the paths and their implications in the

table below.

Path Condition Result

Isothermal DT = 0 w = -q

Constant V DV = 0 w = 0, DU = CvDT

Constant P DP = 0 w = -PDV, qP = CpDT

Adiabatic q = 0 DU = w

State Functions At present we have introduced two state functions:

Internal Energy DU

Enthalpy DH

State functions do not depend on the path, only on the

value of the variables.

We can make the analogy with elevation. The potential

energy at an elevation h, which we call V(h) does not

depend on how we got to that elevation. If we compare

V(h1) in Raleigh to V(h2) on Mt. Mitchell the difference

V(h2) - V(h1) is the same regardless of whether we drive

to Mt. Mitchell through Statesville or Asheville. The work

we do to get (i.e. how much gas we use in a car!) is a path

function.

properties of exponentials - nc state universityfranzen/public_html/ch201/lecture/lec… · ·...

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