chem 2880 - kinetics thermodynamics i - enthalpy · chem 2880 - kinetics 14 in both plots, the...

CHEM 2880 - Kinetics

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Thermodynamics I - Enthalpy

Tinoco Chapter 2

Secondary Reference: J.B. Fenn, Engines, Energy and Entropy,

Global View Publishing, Pittsburgh, 2003.


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Thermodynamics

• An essential foundation for understanding physicaland biological sciences.

• Relationships and interconversions between variousforms of energy (mechanical, electrical, chemical,heat, etc)

• An understanding of the maximum efficiency withwhich we can transform one form of energy intoanother

• Preferred direction by which a system evolves, i.e.will a conversion (reaction) go or not

• Understanding of equilibrium

• It is not based on the ideas of molecules or atoms. Alinkage between these and thermo can be achievedusing statistical methods.

• It does not tell us about the rate of a process (howfast). Domain of kinetics.


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Surroundings, boundaries, system

When considering energy relationships it is important todefine your point of reference.


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Types of Systems

Open: both mass and Closed: energy can be energy may leave and enter exchanged no matter can

enter or leave

Isolated: neither mass nor energy can enter or leave.


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Energy Transfer

Energy can be transferred between the system and thesurroundings as heat (q) or work (w).

This leads to a change in the internal energy (E or U) ofthe system.

Heat• the energy transfer that occurs when two bodies at

different temperatures come in contact with eachother - the hotter body tends to cool while the coolerone warms until thermal equilibrium is achieved(they are both at the same temperature)

• heat transfer depends on the heat capacity (C) of thebodies involved - C (J K ) reflects how much energy-1

is required to heat up an object or substance by 1°Cor 1 K (CG = molar heat capacity, J K mol )-1 -1

Note: the equation q = C)T applies only when C isindependent of T, which is usually true over a small )T• sign-convention: heat is positive if it flows into the

system (from the surroundings) and negative if it


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flows out of the system (into the surroundings)Work• an energy transfer between system and surroundings

that is not heat• any energy transfer that has or could have as its sole

effect the raising of a weight (Fenn, p.6)• sign-convention: work is positive if it is done by the

surroundings on the system and negative if it is doneby the system on the surroundings

• can be mechanical, electrical, gravitational etc.• common type discussed in thermodynamics is work

of increasing or decreasing volume

See Tinoco pp. 18-24 for more examples of work.


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How do you tell if an energy transfer is heat or work?

• often depends on how your define the system andsurroundings

• one method that can differentiate between the twotypes of energy transfer is considering what happensif a thermal insulator is placed between the systemand the surroundings (Fenn, p.6)• if the energy transfers as heat, then this will

likely effect the energy transfer• if there is no effect, then the transfer is likely

work


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Example: Consider a battery connected to a hotplateheating a water bath. (Fenn, p.7-8)

• If the system is defined as the battery alone and thehotplate and water bath are considered part of thesurroundings, then the energy transfer is work• surrounding the battery with a thermal insulator

would not effect the heating of the water• If the system is defined as the battery and hotplate

and the water bath is part of the surroundings, thenthe energy transfer is heat• placing a thermal insulator between the hotplate

and the water bath would effect the heating ofthe water

• If the system is defined as the battery, hotplate andwater bath, then no energy transfer takes place(neither work nor heat).

There are however some exceptions to this method fordistinguishing work and heat. e.g. Radiation• a mirrored surface can reflect electromagnetic

radiation, preventing any energy transfer, howeverthis radiation can represent a heat (thermal radiation)or a work (radio waves) energy exchange


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The First Law of Thermodynamics

Energy is conserved

• Energy can be transferred between the system and surroundings and it can change form, but the totalenergy of the system plus surroundings remainsconstant

• if the only forms of energy exchanged are q and w

• if other forms of energy are involved then terms mustbe added to this equation for each form of energy

• for an isolated system )U = 0• for any change in a system, )U depends only on the

initial and final state and not on the process by whichthe change occurred

Sign Conventionwork done on the system by the surroundings +work done by the system on the surroundings -

heat absorbed by the system from the surroundings +(endothermic)heat absorbed by the surroundings from the system -(exothermic)


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State Variables and Functions

• state variables depend only on the state of the systemand not how it arrived at that state• e.g. P, V, T, n

• generally only a few state variables are required tocompletely describe a system - other variables can bedetermined from these few• e.g. for a liquid if P, V, T and chemical

composition are specified then density, surfacetension, refractive index etc can be determined

• q, w, and C (among others) depend on the path takenand are not state variables

• state functions depend only on the state of the systemand not how it arrived at that state• e.g. U, H, S, G

The enthalpy (H) of the system is defined as:


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• changes in state functions depend only on the initialand final states and not on the path taken from onestate to another• U (E) and is a state function, q and w are not• in taking a system from an initial state (A) to

some new final state (B), a number of differentpaths are generally available. Although theinternal energy change will be the same for allpaths, the amounts of heat and work willgenerally be different for different paths.

• e.g. if glucose is combusted or metabolized, thesame amount of energy is released

• if a system undergoes a long series of steps andreturns to its initial state (a cyclic process), thechange in U or any other state variable is zero


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Intensive vs Extensive

• intensive state properties are independent of the sizeof the system - they do not change when a system isdivided e.g. T, P

• extensive state properties are proportional to the sizeof the system - they do change when a system isdivided e.g. mass, V, n, U

• an extensive property can be converted to anintensive property by describing it per unit ofmaterial (m, n, or V)

Extensive Intensiveenergy (J) energy/mol (J mol )-1

heat capacity (J K ) molar heat capacity(J K mol )-1 -1 -1

mass (kg) density (kg mol )-1

and volume (m ) 3


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Equations of State

• link state variables, most frequently P, V and T

Solids and Liquids

• solids and liquids don’t change V much with P or Tso a first approximation for an equation of state of asolid or liquid is

V�constant

• find density at one temperature and you can use thatat any temperature to describe the system

• liquids and solids do change V slightly with P and T- Tinoco Fig 2.3 shows example for liquid water


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In both plots, the volume of water changes by ~3% overthe temperature and pressure ranges shown• the change in V with P at constant T (293 K) is linear

• 45.9 x 10 atm is the isothermal compressibility of-9 -1

water at 293 K - fractional decrease in V of water foran increase in P of 1 atm

• the change in V with T at constant P (1 atm) is notlinear, it has a minimum at 277 K


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Gases

• V does change significantly with P and T• first approximation is the ideal gas law

PV = nRT

• this law assumes ideal gas behaviour – gas moleculesare perfectly elastic, hard spheres of negligiblevolume, with no attractive nor repulsive forcesbetween them, moving in a completely chaoticmanner with perfectly elastic collisions betweenmolecules and with the walls of the container

• the ideal gas law applies to all gases at low pressures• at higher pressures it is a good approximation for

most gases, accurate to within ±10% at roomtemperature and atmospheric pressure

• other, more accurate equations of state for gasesattempt to account for non-ideal behaviour andcontain parameters relating to individual gases• the van der Waals gas equation

accounts for attractive forces between molecules(a) and the intrinsic volume of the molecules (b)


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• in this case, changing P from 0 to 1000 atm causes a1000x decrease in V (compared to 3% decrease forliquid water)


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Mixtures• for mixtures, the amount (m or n) of each component

must be specified• for gases the situation is particularly simple as the

ideal gas law applies to each component individuallyand the total pressure is equal to the sum of thepartial pressures of each component

• liquids and solids are more complicated e.g. thevolume of a mixture of liquids is not necessarilyequal to the sum of the volumes of the components l


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Paths

• changes in state functions are independent of the pathtaken so it is often convenient to pick a path forwhich the energy change(s) are easy to calculate

1 1 1 2 2 2• a change from P , V , T to P , V , T might be easiestto calculate considering a series of path steps whereone state variable is held constant in each step

1 1 1 2 1, 2 2 2• first P , V , T to P , V, T then to P , V , T etc.

• there are several types of named paths:

sysConstant P Isobaric )P = 0

sysConstant T Isothermal )T = 0Constant V Isochoric )V = 0No heat transferred Adiabatic q = 0final state = initial state Cyclic no change in any

state variable orfunction

• in biological systems, the paths of most interest willusually be isobaric and isothermal rather thanisochoric


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Dependence of U and H on P, V and T

Liquids and Solids

• considering changes in q and w for the types of pathsdiscussed on the previous page:

Isobaric

P• q and C are not state variables, so C must be used to

Pcalculate q

• )V at constant P will be negligible, so no work isdone


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Isochoric

V V• here, C must be used to calculate q

• V is constant, so w = 0

Isothermal

• again, )V will be negligible, and w = 0

So for any change of state for a pure solid or pure liquid,)U can be calculated using a convenient path and thensumming the appropriate q and w values.

e.g. a isobaric path might be combined with anisothermal path, or an isochoric path might becombined with an isothermal path


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For solids and liquids, the volume does not change

P V P Vappreciably with pressure so C .C and )U .)U

The enthalpy change for the same processes can also becalculated:

)H = )U +)(PV)

but any volume change with a change in pressure ortemperature in a solid or liquid is negligible and )(PV) .0, thus for processes involving constant pressure ortemperature

)H = )U


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Tinoco Example 2.9

Calculate )U and )H in joules for heating 1 mole ofliquid water from 0 °C and 1 atm to 100 °C and 10 atm. The volume for the water is essentially independent ofpressure; it can be calculated from the average density ofwater, 0.98 g ml .-1

The overall reaction is:

2 2H O (l) (0 °C, 1 atm) 6 H O (l) (100 °C, 10 atm)

We can do this in two steps, one isobaric and oneisothermal:

2 2H O (l) (0 °C, 1 atm) 6 H O (l) (100 °C, 1 atm)

2 2H O (l) (100 °C, 1 atm) 6 H O (l) (100 °C, 10 atm)

)U = )U(isobaric) + )U(isothermal)

P P T T)U = q + w + q + w

T TFor a liquid, q and w are zero and since there is little

Pchange in volume, w is also zero.


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Because there is a change in pressure, )H�)U.


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Gases

A similar analysis can be done to determine the energychanges for gases undergoing changes in P, V and T. Themajor difference is that changes in V and thus w aresignificant for gases.

Isobaric

P• w is not zero

Isochoric• no change


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Isothermal

T ex• w is not zero, but for a reversible path where P = P

By comparing two different paths for the same overallprocess, one isothermal + isobaric, the other isothermal +isochoric:

For any change in P, V and T for gases:

Relationships between q, )U and )H

For any system undergoing an isochoric process:

Vq = )U

For any system undergoing an isobaric process:

Pq = )H


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Phase Changes

Phase Change Name

gas 6 liquid or solid Condensation

solid 6 liquid Fusion, melting

liquid 6 solid Freezing

liquid 6 gas Vaporization

solid 6 gas Sublimation

We are interested in phase changes at constant T and P.

P exw = -P )V

Pq = )H

P P ex)U = q + w = )H - P )V

)H values for phase changes for many substances havebeen tabulated

For phase changes involving only solids and liquids, )Vis often negligible and the work term can be ignored.

For any phase change involving the gas phase, thevolume of the other phase (solid or liquid) is negligible

gascompared to that of the gas and )V = V .


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If )H or )U are known at one temperature and need to becalculated at another, appropriate paths can be chosen.

Example: We know )H for the vaporization of water at100 °C and need to calculate it at the temperature ofhuman skin 35 °C. (See reaction scheme Tinoco p. 45)

Follow the path:

2 21 H O (l) (35 °C, 1 atm) 6 H O (l) (100 °C, 1 atm)

2 22 H O (l) (100 °C, 1 atm) 6 H O (g) (100 °C, 1 atm)

2 23 H O (g) (100 °C, 1 atm) 6 H O (g) (35 °C, 1 atm)

2 2H O (l) (35 °C, 1 atm) 6 H O (g) (35 °C, 1 atm)

Step 1 is an isobaric heating of liquid water

1 P P 1)H = q = C (l))T

Step 2 is the vaporization of water at 100 °C which wecan look up in our standard tables

Step 3 is the isobaric cooling of water vapour

2 P P 2 2 1)H = q = C (g))T Note that )T = -)T


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Thus for the overall reaction

P 1 P 2)H(35 °C) = )H(100 °C) + C (l))T + C (g))T

P 1 P= )H(100 °C) + (C (l))T - C (g)))T

or more generalized:

This equation does not apply only to phase changes. Itcan be used to determine the enthalpy change for anyreaction at another temperature.


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Chemical Reactions

The energy derived from chemical reactions is of greatimportance in biological systems.

Enthalpy ()H) Changes

For a chemical reaction:

A B C Dn A + n B 6 n C + n D

the enthalpy change is:

If a reaction gives off heat, an exothermic reaction:

products reagents3H < 3H and )H is negative.

If a reaction absorbs heat, an endothermic reaction:

products reagents3H > 3H and )H is positive.

The enthalpy changes for many reactions have beentabulated and these values can be used to determineenthalpy changes for other reactions by addition andsubtraction - Hess’ Law.


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Example Tinoco p. 48

Given the following enthalpies of reaction at 25 °C,calculate the heat of reaction for the oxidation of aqueousglycine to aqueous urea.

Combustion (oxidation) of two moles of solid glycine -1163.5 kJ mol-1

Hydrolysis of solid urea 133.3 kJ mol-1

Dissolution of solid glycine 15.69 kJ mol-1

Dissolution of solid urea 13.93 kJ mol-1

The corresponding reactions are:

2 2 21) 2NH CH COOH (s) + 3O (g, 1 atm) -1163.5 kJ mol-1

2 2 36 4CO (g, 1 atm) + 2H O (l) + 2NH (g, 1 atm)

2 2 22) H NCONH (s) + H O (l) 133.3 kJ mol-1

2 36 CO (g, 1 atm) + 2NH (g, 1 atm)

2 2 23) NH CH COOH (s) + 4H O (l) 15.69 kJ mol-1

2 26 NH CH COOH (aq)

2 2 24) H NCONH (s) + 4H O (l) 13.93 kJ mol-1

2 26 H NCONH (aq)


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The overall reaction we are looking for is:

2 2 22NH CH COOH (aq) + 3O (g, 1 atm)

2 2 2 26 H NCONH (aq) + 3CO (g, 1 atm) + 3H O (l)

So we must add, subtract and multiply the reactions wehave been given to achieve our overall reaction.

We need to reverse reaction 3 (changing the sign of )H)and multiply it by 2 and leave reaction 4 as is to get theaqueous glycine and urea in the correct places

2 2-2x3) 2NH CH COOH (aq) -31.38 kJ mol-1

2 2 26 2NH CH COOH (s) + 4H O (l)


2 26 H NCONH (aq)

we want the solid forms of glycine and urea to cancel, soreaction 1 stays as is and reaction 2 needs to be reversed.



2 3-2) CO (g, 1 atm) + 2NH (g, 1 atm) -133.3 kJ mol-1

2 2 26 H NCONH (s) + H O (l)


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2 22x-3) 2NH CH COOH (aq) -31.38 kJ mol-1

2 2 26 2NH CH COOH (s) + 4H O (l)


2 26 H NCONH (aq)



2 3-2) CO (g, 1 atm) + 2NH (g, 1 atm) -133.3 kJ mol-1

2 2 26 H NCONH (s) + H O (l)

2 2 22NH CH COOH (aq) + 3O (g, 1 atm)

2 2 2 26 H NCONH (aq) + 3CO (g, 1 atm) + 3H O (l)

)H = -31.38 + 13.93 -1163.5 -133.3 kJ mol-1

)H = -1314.2 kJ mol -1


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Energy ()U) Changes

)U = )H + )(PV)

At constant pressure:

if only solids and liquids are involved, )V.0 and

)U = )H

If gases are involved, then only the )V of the gases isimportant and

)(PV) = )nRT)U = )H + )nRT

where )n = 3n(gaseous products) - 3n(gaseous reagents)


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Standard Enthalpies of Formation

A convention adopted for enthalpies of reaction is that theabsolute enthalpy of all elements in their most stable stateat 1 atm pressure is zero. These are called standard statesand are designated by superscript zero. i.e. H°

The standard enthalpy of a compound is defined as theenthalpy of formation of 1 mole of the compound at 1 atmpressure from the elements in their standard states. This

fis the standard enthalpy of formation )H °

f)H ° values for thousands of compounds have beendetermined and tabulated. Note that standard statespecifies 1 atm pressure but not a specific temperature,

fthough most )H ° values are determined at 25 °C.

The )H° of a reaction can be determined using the tables

fof )H ° using

f f)H° = 3)H ° (products) - 3)H ° (reagents)


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Bond Energies

Bond dissociation energies can be used to approximate reaction enthalpies by determining a reaction path thatinvolves break and forming individual bonds.

Bond dissociation energies are the energies required tobreak bonds:

A-B (g) 6 A (g) + B (g)

Note that these reaction always take place in the gasphase at 1 atm pressure and 25 °C.

To determine a change in enthalpy for a reaction, thedifference between the total bond energies of the productsand reactants must be determined

)H° = 3BE(reagents) - 3BE(products)

Notice that unlike similar equations already discussed,this is “reagents - products” not “products - reagents”. This is due to bond energies being defined as the energyto break bonds not the energy to form them.

chem 2880 - kinetics thermodynamics i - enthalpy · chem 2880 - kinetics 14 in both plots, the...

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