aquote of the day(goodforall the semester): the camel’s ... · • sonochemistry, •...

Physical Chemistry EPM/01 1

Order of lectures is, perhaps, not the most logical one, but it is partially

dictated by my desire to synchronize the lectures and tutorials.

A quote of the day (good for all the semester):

The camel’s hump is an ugly lump

Which you well may see in the Zoo.

But uglier yet is the lump you get

From having too little to do.

Rudyard Kipling


�Subject:

Physical and chemical transformations of the matter (of

any kind) and the accompanying flows of energy.

�Method:

Mathematical, i.e., based on assuming or creating some

theoretical models based or tested on empirical

observations. This method leads to formulation of

hypotheses, theories and laws of nature related to their

specific subjects.

What is physical chemistry?


Major sub-topics:• Photochemistry,

• Sonochemistry,

• Electrochemistry,

• Thermochemistry,

• Chemical kinetics,

• Chemical thermodynamics (equilibria),

• Surface chemistry,

• Spectroscopy,

• Solid state chemistry,

• Quantum chemistry,

• Chemical Physics.

What is physical chemistry?

(2)


�Matter is everything possessing a feature known as inertia, due to its

mass (Newton).Chemical substance is a pure, isolated form of matter.The amount of matter is measured in moles.

�Energy is ability of a system (see later) to perform work (simplified).

(forms of energy, preservation of energy, energy conversion, energy carriers)

Basic concepts (1)

Mass-energy equivalence

E=mc2

In practice, relativistic (and quantum) effects are observed only in macro- or microscale and in the average observable scale are negligible.


Basic concepts (2)

Potential energy – depending on position of an object

in gravitation field in electric field

(of the Earth)

Ep=mgh Ep=q1q2/(4πε0r)

Kinetic energy – energy of movement:

Ek=½mv2

(translation, rotation, oscillation)


�Law of nature is a clearly defined piece of theory, usually

addressing a single phenomenon, i.e., a relation between different

observables involved in this phenomenon.

�Verbal formulation:Boyle’s Law:

At constant temperature, volume of a gas changes inversely

proportionally to its pressure.

�Mathematical Formula:

Basic concepts (3)

const.const.at VPPVPV ;P

P

V

V T 2211

1

2

2

1=====


System (definition)

System is a part of universe subject to

observation or being an object of theoretical

considerations, separated from the rest of

universe (surroundings) physically or just in our

imagination.

system + surroundings = universe

Introduction to

thermodynamics


The three types of systems:

� Open

�Closed

�Isolated

Introduction to

thermodynamics (2)


Ways of the energy transfer:

�work

�heat.

Types of work

Mechanical (change of volume, change of shape), change of the interface area,electrical.

Introduction to

thermodynamics (3)


HEAT

Introduction to

thermodynamics (4)


Endothermic process (systemlimited by a diathermic barrier)

Introduction to

thermodynamics (5)

before after


Endothermic process in an adiabatic system

Introduction to

thermodynamics (6)

before after


Physical properties of systems:� Extensive (additive), depending on the size of thesystem (number of components, their type and amount of each)

(e.g. mass or volume of a system)

� Intensive are not additive.

(e.g. temperature, pressure, density, molar quantities, refraction index)

Introduction to

thermodynamics (7)

∑= ii Xnx


State of a system:The following quantities permit complete

characterization of the state of a given system

P, V, T

Equation of state:

f(P,V,T)=0

For a system containing perfect gas only: pV=nRT

Introduction to

thermodynamics (8)


Thermochemistry (1)

• Enthalpy is heat exchanged at constant

pressure.

state property

• ∆∆∆∆H < 0 - exothermic process

• ∆∆∆∆H > 0 - endothermic process

initfinHHH −=∆

initfinHHH ∆−∆=∆


Thermochemistry (2)

A standard reaction enthalpy is the reaction

enthalpy when reactants in their standard

states are converted to products in their

standard states. Denoted as:

The standard state of a substance is its pure

form at pressure of 105 Pa.

0H∆


Thermochemistry (3)

• The standard molar enthalpy of formation

of a compound is the standard reaction

enthalpy per mole of compound for the

synthesis of the compound from its

elements in their most stable forms at 105 Pa

and specified temperature. Denoted as:

00 H ;H formf ∆∆


Thermochemistry (4)

• Standard enthalpies of formation of

elements are equal to ZERO (at 298K).

When several allotropic forms of an

element exist, the statement applies to the

most stable form of the element.

• Standard molar enthalpy of formation of a

compound is standard molar enthalpy of

this compound.


Thermochemistry (5)

ProductsReactants→

initfinrreactionHHHH ∆−∆=∆=∆

fin

0

Pri,f,i H =Hn ∆∆∑ init

0

Rei,f,i H =Hn ∆∆∑

0

Rei,f,i

0

Pri,f,i

0

r Hn-Hn=H ∑∑ ∆∆∆


Thermochemistry (6)

• The standard molar enthalpy of combustion is

the change in enthalpy per mole of the substance

(fuel) when it is burned (reacts with oxygen)

completely under standard conditions. Denoted as

• conditions superimposed on products in complete

combustion:

H →→→→ H2O(l); C →→→→ CO2(g); N →→→→ N2(g)

00 H ;H combc ∆∆


Thermochemistry (7)

Using standard molar enthalpies of combustion

one can calculate the reaction enthalpy as follows:

This is an exceptional formula, where a property

of the products is subtracted from that of the

reactants.

0

Pri,c,i

0

Rei,c,i

0

r Hn-Hn=H ∑∑ ∆∆∆


Hess law

A reaction enthalpy is the sum of the enthalpies of

any sequence of reactions (all at the same

temperature and pressure) into which the overall

reaction may be divided.

Hess's law results from a rigorous application of the statement

that enthalpy is a state property. It does not matter what was the

way to obtain the substance or into how many steps the overall

reaction was split.


Hess law (2)

Hess law may be applied to each and every state

quantity (not only to enthalpy).

Another wording:

If any reaction (target reaction) may be represented

as a linear combination of some other reactions

(partial reactions) then any state property of this

reaction is the same linear combination of the

respective state properties of the partial reactions.


Hess law (3)

Thermochemical equations:

1.State of matter (phase: solid, liquid, gaseous) and

any specific form of all reactants and products must

be indicated.

2.The heat released or absorbed, i.e., the ∆H must be

shown.


Hess law (4)

Rules of manipulation:

1.When the reaction is rewritten in reversed direction,

the sign of its ∆H is changed.

2.When stoichiometric coefficients in the reaction are

multiplied, the ∆H must be multiplied by the same

factor.

3.When two reactions are added, their enthalpy

changes must be added, too.


Hess law. Example

Target reaction:

Component reactions:

Step 1. Begin with the thermochemical equation having at

least one of the reactants or products on the correct side of

the arrow in the target reaction:

?H O(l);H+(g)2CO(g)O2+(g)HC :T 0

T22221

22 =∆→

mol]226.75[kJ/=H (g);HC(g)H+2C(s) :A 0

A222 ∆→

mol]-393.5[kJ/=H (g);CO(g)O+C(s) :B 0

B22 ∆→

mol]-285.9[kJ/=H O(l);H(g)O+(g)H :C 0

C2221

2 ∆→


Hess law. Example (2)

Hence, we start with reaction B and multiply both sides (as

well as its standard enthalpy) by a factor of 2:

Step 2. To add a reactant or product, add another reaction,

having the desired substance at the same side of the arrow

as in the overall reaction. To cancel a reactant or product

(which is not present in the target reaction), add another

reaction, having the desired substance at the opposite side of

the arrow than in the reaction obtained so far.

2B=D (g);2CO(g)2O+2C(s) 22 →



In our case, to complete the right side of reaction (T), we

add reaction (C) to reaction (D):

Step 3. If the reaction obtained in step 2 is identical with

reaction T, then the procedure is completed, else - repeat

step 2 (do not use reactions already used in it).

In our case, reaction (E) is not yet equal to reaction (T).

We repeat step 2 to eliminate carbon and hydrogen at the

left side of reaction (E), by adding reversed reaction (A) to

reaction (E) or subtracting (A) from (E).

C+2B=E O(l);H+2CO(g)O2+(g)H+2C(s) 22221

2 →



Our goal is achieved. Reaction (F) is equal to reaction (T).

The linear combination we found is T=2B+C–A. We

calculate the change in enthalpy in reaction (T) in the same

manner:

T=A-C+2B=F O(l);H+2CO(g)O2+(g)HC 22221

22 →

J/mol]-1299.65[k= ...

...=(-226.75)+(-285.5)+-393.52=H-H+H2=H 0

A

0

C

0

B

0

T ×∆∆∆∆

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