The structure of the Atom

Atom is considered the smallest particles of matter, according to Dalton's Atomic Theory,  but it was found by some scientists that atom has still subatomic particles called the electron, proton and neutron.    

Electron was the first to be discovered by J.J Thomson, a British scientist, while he was studying the cathode ray tube.   He found out that this particle is negatively charged.   In 1908 to 1917 R.A. Milikan, an American physicist, found the charge of the electron to be -1.6022 x 10^-19 C.  From these data he also calculated the mass of the electron to be 9.10 x 10^-28 g.   J.J. Thomson proposed the plum pudding model of the atom.  He pictured the small electrons to be embedded  in a uniform and positive charge sphere.

After the discovery of proton, in 1896 a French scientist Henri Becquerel (1852-1908) while studying a uranium mineral , he found out that it spontaneously emits high energy radiation.  This spontaneous emission of radiation is called radioactivity.  Consequently, any element that spontaneously emits radiation is said to be radioactive.

Further study about radioactivity, a British scientist Ernest Rutherford revealed three types of radiation: alpha (𝛂), beta (𝛃) and gamma (𝛄).  Rutherford showed that 𝛂 and 𝛃 rays are consist of fast-moving particles which were called the 𝛂 and 𝛃 particles.  𝛃 particles are high-speed electrons and can be considered the radioactive equivalent  of cathode rays.  They are attracted to a positive plate.  The 𝛂 particles  are much more massive than the 𝛃 particles  and have a positive charge.  They are therefore attracted to negative plate.  𝛃 particles have a charge of -1 while the 𝛂 particles has +2 charge.  Rutherford showed that 𝛂 particles combine with electrons to forms atoms of helium.  He concluded that an 𝛂 particle consist of the positively charged core of the helium atom. He further concluded that 𝛄 radiation is high-energy radiation similar to x-rays; it does not consist of particles and carries no charge and not affected by electrical charge.

The Discovery of Nucleus of an Atom

Plum Pudding Model of Thomson was accepted for how many years but it was proved wrong by Rutherford.  In 1910, Rutherford together with his team use 𝛂 particles to probe the structure of the atom.   They carry out series of experiments using very thin foils of gold and other metals as targets for alpha particles from a radioactive source.  Below is the diagram of their gold foil experiment:

The result of the experiment were the following:

1.  Majority of particles penetrated the foil either undeflected or with only a slight deflection.
2. They also noticed that every now and then an 𝛂 particle was scattered (or deflected) at a large angle.
3.  Some instances, 𝛂 particle actually bounced back in the direction from which it had come.

From the result of the experiment Rutherford devised a new model of the atomic structure of an atom, he concluded that:

1.  Most of the atom is an empty space, since most of the 𝛂 particles pass through the gold foil with little deflection or no deflection at all.
2.  That the atom's positive charges are all concentrated in the nucleus, a dense central core of the atom.  Whenever an alpha particle came close to the nucleus in the scattering experiment, it experienced a large repulsive force and therefore large deflection.  An when an 𝛂 particle travelling directly towards the nucleus, it would experience a strong repulsion that would completely reverse the direction of the moving particle.

The positive charge found in the nucleus of the atom is called proton.  They found out that the charge of the proton has the same but opposite in magnitude of that of electron and has a mass of 1.67262 x 10^-24 g, about 1840 times heavier than that of the opposite charge electron.  The model is shown below:

At this point of investigation, scientists perceived that the mass of the atom is concentrated in the nucleus of an atom, but the nucleus only occupies 1/10¹³ of the volume of an atom.  A typical atomic radius is about 100 picometer (pm), whereas the radius of an atomic nucleus is only about 5 x 10^-3 pm.  To understand how small the atomic nucleus is, imagine the size of sports stadium, if that is the size of the atom; the size of the nucleus is comparable as that of a marble.


The neutron

Rutherford's atomic model of  an atom left one problem unsolved.  They found out that the mass of the total electrons and protons inside the atom did not equalize with the total mass of the atom, they concluded that there is still subatomic particle inside the nucleus that is missing.  The problem was solved when an English physicist James Chadwick, bombarded a thin sheet of beryllium  with 𝛂 particles, a very high energy radiation similar to 𝛄 rays was emitted by the metal.  His experiment showed that the rays consist of electrically neutral particles having a mass slightly greater than that of protons.  Chadwick named this particles as neutron.

Although there were new findings of some subatomic particles of atom.  The three fundamental particles of atom:  the electron, proton and neutrons are still very important in the study of chemistry.

The Atomic Theory of Matter

As you learned from my previous post that matter is anything that occupies space and has mass.  It means what we can see around us are matter.  It can be living matter or nonliving matter.  Do living matters and nonliving matters have similarities in their composition? What are matters made up?  In the fifth century B.C., Democritus (460-370 B.C.) expressed the notion that matter is made up of tiny indivisible particles called atomos, which means indivisible.  But Plato and Aristotle did not agree with his idea that is why the idea faded for many centuries, during those times Aristotelean philosophy dominated the Western culture.

The idea of atoms reemerged in Europe during the seventeenth century when scientific investigation should experimental evidence about "atomism" and give rise to the modern atomic theory formulated by John Dalton, an English scientist and school teacher in 1808.  His assumptions are summarized as follows:

1.  Each element is composed of extremely small particles called atoms.

2.  All the atoms of the given element, having the same size, mass, and chemical properties.  Atoms of different elements are different  and have different properties.

3.  Compounds are formed when one atom combines with one or more than one atoms. In any compound the ratio of the number of atoms is either an integer or simple fraction.

4.  Atoms are not changed to another atoms when combined chemically, rearrangement of atoms only occurs.


This theory states that its the atom that is considered the building blocks of matter.  They are the smallest particles that carries the identity of the element.

The Daltons atomic theory also explains some simple laws of chemical combination, one of these was the law of constant composition or law of definite proportion.  This law was the basis of  assumption no 3.  This law states that different samples of the same compounds always contain its constituents elements in the same proportion by mass.  Another law that supports no. 3 assumption of Dalton was the law of multiple proportion.  This law states that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. 

The assumption no. 4 also supports another law; the law of conservation of mass, which states that matter can neither be created nor destroyed during chemical reaction.  During chemical reaction, atoms are not destroyed they are just rearranged forming new substances.

Physical and Chemical Properties of Matter

Matter is anything that occupies space and has mass.  We have different kinds of matter.  We can distinguish one from the other through their physical and chemical properties.

Physical Properties of Matter

Physical property can be measured and observed without changing the identity and composition of the substance.  Some examples of this are color, odor, melting point, boiling point, hardness and density.  We can identify the color of a particular substance without changing the identity of the substance.

Physical property can be classified as intensive and extensive properties.  Intensive properties are properties that do not depend on the amount of the sample being studied.  Example; temperature, melting point and density.  Density of water is the same even the amount is changed, let's say a glass of water has 1 g/mL density same with 1 gallon of water.  Extensive properties of the substances depend on the quantity of the substance.  Example are the mass, length and volume.  A glass of water will have different volume with that of a gallon of water.  The volume also will change if the amount of the substances will change.

Physical properties can also be classified as intrinsic and extrinsic properties.  Intrinsic properties are inherent characteristics of the substance and depend on the kind of the material itself.  Some examples are taste, odor, color, transparency, solubility, melting point, boiling point, viscosity, refractive index, hardness, crystalline structure and atomic or molecular diameter.  Extrinsic properties are qualities that describe the outside appearance of matter.  Examples are size, shape, length, and mass.


Chemical Properties

Chemical properties are properties that can be observed when substance undergo chemical change. For example, hydrogen gas burns in the presence of oxygen gas forming water. Meaning combustibility of the substance is an example of chemical properties and this can be observed when a chemical reaction occur.  Another chemical properties include toxicity, flammability, chemical stability, enthalpy of formation, reactivity with other chemicals.


Physical Change and Chemical Change

Change can be classified as physical change and chemical change.  During physical change the substance only undergoes change in the physical appearance, but not the composition.  Example is the evaporation of liquid  water to water vapor.  The change only undergoes change in state from liquid to gas state but still the composition is still the same water.  All changes in state are all physical change.

Chemical change is a change in which the substance undergoes change in its composition.  Meaning a new substance is formed during the process.  Example, combustion of carbon forms carbon dioxide.  Combustion is the reaction of substance with oxygen, meaning carbon reacts with oxygen forming carbon dioxide.  This is an example of chemical change because the reactants carbon and oxygen have their own unique properties and once they react with each other the new substance carbon dioxide has different properties different from its origin.


Let us answer the following:

1.  Which of the following properties are extensive and intensive?
     a.  length
     b.  volume
     c.  boiling point
     d.  color
     e.  density

2.  Identify the property if physical or chemical
     a.  oxygen gas supports combustion
     b.  fertilizer help to increase agricultural production
     c.  water boils below 100^oC on top of the mountain
     d.  lead is denser than aluminum
     e.  uranium is a radioactive element

3.  Classify if physical change or chemical change
     a. The helium gas inside a balloon leaks out of the container after few hours.
     b.  A flashlight beam slowly gets dimmer and finally goes out
     c.  dissolving sugar in water
     d.  boiling of water
     e.  heating of sugar


Answer's Key

1.  a.  length     -  extensive
     b.  volume   -  extensive
     c.  boiling point  -  intensive
     d.  color     -   intensive
     e.  density   - intensive

2.  a.  chemical properties
     b.  chemical properties
     c.  physical properties
     d.  physical properties
     e.  chemical properties

3.  a.  physical change
     b.  chemical change
     c.  physical change
     d.  physical change
     e.  chemical change

Classification of Matter

Before studying further some topics in chemistry it is very important to know the basic concepts in chemistry, the classification of matter.  Matter is classified according to physical state ( as a gas, liquid, and solid) and according to the composition (as an element, compound and mixture).


States of Matter

Matter can be in the form of solid, liquid and gas, and this is called the phases or states of matter. This three forms of matter can be distinguished from one another by determining the corresponding properties of each.

1.  Solid has definite shape and volume.  It is rigid.  The molecules in solid are very closed to each          other and have definite arrangement.  The molecules can vibrate slightly in their fixed position and      cannot be compressed.

2.  Liquid has definite volume but no definite shape.  It takes the shape of the container it occupies.        The molecules are quite distant from each other and can move rapidly allowing each to slide one        another, that is why it can be poured. Liquids are not compressible.

3. Gas has no definite shape and volume.  It conforms to the volume and shape of the container.  The     molecules are far apart form each other and can move freely.  Gases are compressible.

Classification based from the Composition

Matter can be classified as pure substance and mixture.  Pure substance is a matter that has distinct properties and definite composition and does not vary from sample to sample.  Example is water, water has definite composition of hydrogen and oxygen.  Mixture, on the other hand, is composed of two or more substances that are combined.  It does not have definite composition.  Example of this is seawater, seawater is composed of two substances the water and salt.  Seawater has no definite composition of water and salt.  The components of mixture can be separated by physical means.


Pure Substance

Pure substances are classified into two classification, the elements and compounds. Element is the simplest form of matter.  It is only composed of only one kind of atom.  Since simplest form of matter, it cannot be decomposed into simpler form.  Example of this are O, oxygen atom, C carbon atom and N, nitrogen atom.  See more examples in the table below.

Some Common Elements  and their Symbols

Element                   Symbol
Flourine                         F
Chlorine                        Cl
Hydrogen                      H
Iodine                            I
Nitrogen                        N
Oxygen                         O
Sulfur                            S
Phosphorus                   P
Helium                          He
Aluminum                     Al
Barium                          Ba
Calcium                        Ca
Copper                          Cu
Magnesium                   Mg
Manganese                    Mn
Mercury                        Hg
Platinum                       Pt
Iron                               Fe
Lead                              Pb    

Note: You can refer to the Periodic Table of Element for complete list of all the elements.

Compound is composed of two or more elements chemically combined.  It has a definite composition and can be separated by chemical means.  Example of this is pure water, it is consist of 11% hydrogen and 89% oxygen by mass.  No matter how much amount of water we have, it still has the same composition.  This fixed composition of compounds is called Law of Definite Proportion (or Law of Constant Composition) formulated by the French chemist Joseph Louis Proust (1754-1826).  Compound also has fixed properties, water boils always at 100^oC, and freezes at 0^oC.

Examples of Compounds

Compound                  Formula
Water                                H₂O
Sodium chloride              NaCl
Carbon dioxide                CO₂
Carbon monoxide            CO
Ammonia                         NH₃
Magnesium chloride        MgCl₂
Potassium oxide               K₂O
Sulfur dioxide                  SO₂
Sulfur trioxide                  SO₃
Carbon tetrachloride        CCl₄
Iron (II) oxide                  FeO
Iron (III) oxide                 Fe₂O₃


Mixture

Mixture is composed of two or more substances.  The composition may vary from sample to sample. Each substance in a mixture retains its own identity and hence its own properties.  Mixture can be homogeneous and heterogeneous.  Homogeneous mixture is a mixture containing only one phase, meaning has uniform appearance while heterogeneous mixture containing two or more phases.  You can distinguish the components of heterogeneous mixture.  Example:  seawater (composed of salt and water) is a homogeneous mixture while sand in water is an example of heterogeneous mixture.  In seawater you cannot see the salt since it is being dissolve in water, while in sand in water, sand and water can still be seen and can be identified.


Types of Mixtures

Mixtures have three types; these are the solution, colloid, and suspension.  Solution is considered homogeneous mixture while colloids and suspension are heterogeneous mixture.  The three mixtures differ in their particles size.

1. Solution has the smallest particle size, that is why the particles are dissolved in the solvent.  The components cannot be separated by filtration, the particles will just flow together with the solvent. Example:  sugar in water, salt in water, alcohol in water, vinegar (acetic acid in water)

2. Colloid has intermediate particles size, enough to scatter light which is called the tyndall effect.  It may appear homogeneous to the naked eye but it is a heterogeneous mixture. The particles are suspended in the medium that is why it can scatter light. Example:  milk, cooked cassava starch in water, mayonnaise

3. Suspension has larger particles size.  The particles settle down at the bottom of the container after sometime. It can easily be separated by ordinary physical means like decantation and filtration.
Example:  sand in water, raw starch in water, clay in water


Methods of Separating Mixtures

Mixtures can be separated by ordinary physical means.  Mixtures can be separated differently.  Proper methods should be used in separating a particular type of mixture.  Knowing the kind of mixture and their corresponding properties will help you in determining the type of method to use to separate the mixture.

1.  Filtration is a method used to separate larger particle size mixture.  Filter paper is used in the laboratory attached to the funnel.  This method is best to use in suspension mixtures.

2.  Decantation is a method in which the particles are allowed to settle at the bottom for a few minutes and transferring the liquid to a different container.

3.  Distillation is a method in which a homogeneous mixture is heated allowing the liquid to vaporize.  The vapor is allowed to pass through a condenser, converting vapor back to liquid, and collected in  a receiving container.   Example, recovering pure water in seawater.

4.  Evaporation is another method of separating a homogeneous mixture.  This is when the solid particle is the needed by-product. The mixture is heated using evaporating dish, allowing volatile components to evaporate leaving the desired component on the container.  Example, recovering salt in seawater.

5.  Chromatography is another method of separating colored mixtures like, dyes, ink, and coloring agents in food.  This is done in a piece of paper in which a spot of the mixture is placed in a paper and is allowed to stand upright the solvent.  As the solvent travels on the paper the components are carried by the solvent with different rates depending on the difference of their masses.

Gibbs Free Energy

You learned from my previous post that processes can be spontaneous and non-spontaneous. Spontaneous processes can be endothermic and exothermic.  Usually, an spontaneous endothermic process has an increase  in disorder and entropy of the system.  On the other hand, a spontaneous exothermic process has a decrease in the entropy of the system.

Since spontaneous processes that result in the decrease in entropy are always exothermic, therefore the spontaneity of the reaction seems to involve two thermodynamic quantities the enthalpy and entropy.   An American mathematician Josiah Willard Gibbs (1839-1903) find a way to use ∆H and ∆S to predict whether a given reaction occurring at constant temperature and pressure will be spontaneous. Gibbs developed the so-called Gibbs Free Energy  (or simply free-energy) in answer to the problem above.  Gibbs Free energy is defined as

G  =   H  -     TS

where T is the absolute temperature.  For a process occurring at constant temperature, the change in free energy of the system, ∆G, is given by the equation

∆G  =  ∆H  -   T∆S

To determine how the function G relates to reaction spontaneity, we learned from my previous post that 

∆S_univ  =   ∆S_sys    +    ∆S_surr    =    ∆S_sys   +    (-∆H_sys / T)

Multiplying both sides by -T gives

-T∆S_univ  =  ∆H_sys   -  T∆S_sys

Looking at the two equations above, we can conclude that the free-energy change in the process occurring at constant temperature and pressure, ∆G, is equal to  -T∆S_univ.    We learned that the value of ∆S for spontaneous processes is positive, and so the sign of ∆G can provide us with valuable information about the spontaneity of processes that occur at constant temperature and pressure.   If both T and P are constant, the relationship between the sign  of ∆G and the spontaneity of the reaction  is as follows:

1.  If ∆G is negative, the reaction is spontaneous in the forward direction.

2.  If ∆G is zero, the reaction is at equilibrium.

3.  If ∆G is positive, the reaction in the forward direction is nonspontaneous ; work must be supplied      from the surroundings to make it occur.  However the reverse reaction will be spontaneous.

Standard Free Energy Changes

The Standard Free-Energy of reaction ( ∆G^o_rxn) is the free-energy change for a reaction when it occurs at standard state conditions, when reactants in their standard state is converted to products in their standard states.    The standard state for gaseous substances is 1 atm pressure.  For solid substances, the standard state is the pure solid; for liquids, the pure liquid.  For substances in solution, the standard state is normally a concentration of 1 M.  The temperature chosen for purposes of tabulating data is 25^oC or 298K.  For the standard heats of formation, the free energies of elements in their standard states are set to zero.  For the standard free-energy of formation values of some substances click here.

We can calculate the Standard Free-Energy of Reaction ( ∆G^o_rxn)  using the equation below:

∆G^o_rxn =    ∑nG^o_f (products) - ∑mG^o_f(reactants)

where n and m are stoichiometric coefficients, G^o_f is the standard free-energy of formation of a compound, that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states.

Let us see for example the standard free-energy change of the combustion of graphite.

        C(graphite)   +   O₂(g)  →    CO₂(g)

The value of ∆G^o_f of O₂(g) and C(graphite) is zero, since both are in their stable allotropic form, and so the ∆G^o_rxn is equal to the ∆G^o_f of CO₂.

Sample Problems:

Calculate the free-energy changes of the following reactions at 25^oC or 298 K.

1.  CH₄(g)  +   2O₂(g)   →   CO₂(g)   +   2H₂O(l)
2.  2MgO(s)    →   2Mg(s)   +    O₂(g)

Solution:

Before calculating be sure to find the values of free-energy of formation of the substances in the given link here.   Then substitute the values in the equation.

1.  ∆G^o_rxn =  [∆G^o_f(CO₂)  +  2∆G^o_f(H₂O)]  -  [∆G^o_f(CH₄) +   2∆G^o_f (O₂)
                 =  [ (-394.4 kJ/mol) + (2)(-237.2 kJ/mol)]  -  [(-50.8 kJ/mol) +  (2)(0 kJ/mol)]
                 =  (-394.4 kJ/mol - 474.4 kJ/mol)  - (-50.8 kJ/mol)
                 =  -868.8 kJ/mol + 50.8 kJ/mol
                 =  -818.0 kJ/mol

2.  ∆G^o_rxn  = [2∆G^o_f(Mg)  +  ∆G^o_f(O₂)]  -   [2∆G^o_f(MgO)]
                   = [(2)(0 kJ/mol)  +  (0 kJ/mol)] -  [(2)(-569.6 kJ/mol)]
                   = 0 - (-1139.2 kJ/mol)
                   = 1139 kJ/mol

Free Energy and Temperature

We calculated the Free-Energy change at 25^oC or 298K from the concept above.  However, we can also calculate Free-Energy Change at other temperatures, the formula presented above will be used.

∆G  =  ∆H  -   T∆S

In order to predict the sign of ∆G, we need to know the value of ∆H and ∆S.  Below are possible outcomes of the ∆H and ∆S relationships:

 1.  If both ∆H and ∆S are positive, then ∆G will be negative when the T∆S term is greater in                     magnitude than ∆H.   This condition is met when T is large.

2.  If ∆H is positive and ∆S is negative, ∆G will always be positive, regardless of temperature.

3.  If ∆H is negative and ∆S is positive, then ∆G will be negative regardless of temperature.

4.  If ∆H is negative and ∆S is negative, then ∆G will be negative only when T∆S is smaller in                  magnitude than ∆H.  This condition is met when T is small.

The temperatures that will cause the ∆G to be negative will depend on the actual values of ∆H and ∆S of the system.  The table below summarizes the effect of temperature  on the spontaneity of reaction:

 Let us have an example, the melting of ice to liquid water at 1 atm pressure:

H₂O(s)   →   H₂O(l)       ∆H  > 0,  ∆S >  0

The process above is endothermic, which means the ∆H is positive, and since their is an increase in disorder of molecules the ∆S is also positive, and -T∆S is negative.  At temperatures below 0^oC the magnitude of ∆H is greater than that of -T∆S, and therefore the positive ∆H dominates, leading to positive value of ∆G.  The positive value of ∆G means that the melting of ice is not spontaneous at T < 0^oC; rather the reverse process is the spontaneous, the freezing of liquid water to ice.

Before we proceed to calculation, let us first differentiate ∆G and ∆G^o.  ∆G is used to predict the direction of the reaction for nonstandard condition, while ∆G^o is used to predict the direction of the reaction for standard condition ( 1 M concentration) at 25^oC  and 1 atm pressure. The sign of ∆G^o
tell us whether the products or the reactants are favored when the reacting system reaches equilibrium.  Therefore, a negative value of  ∆G^o indicates that the reaction favors products formation whereas a positive value of ∆G^o indicates that there will be more reactants than products at equilibrium.

Sample Problem 1.

a) Using the standard enthalpies of formation and standard entropies ( click the following : Standard Enthalpy; Standard Entropies), calculate ∆H^o and ∆S^o at 298 K for the following reactions:
CaCO₃(s)    →  CaO(s)   +   CO₂(g)
b)  Using the values obtained in a) estimate ∆G^o at 298 K

Solution:

a)  ∆H^o = [∆H^o_f(CaO)  +  ∆H^o_f(CO₂)] -  [∆H^o_f(CaCO₃)]
             = [(-635.6 kJ/mol) + (-393.5 kJ/mol)] - (-1206.9 kJ/mol)
             = -1029.1 kJ/mol + 1206.9 kJ/mol
             = +177.8 kJ/mol

     ∆S^o = [S^o(CaO)  +  S^o(CO₂]  - [S^o(CaCO₃)]
             = [(39.8 J/K.mol)  +  (213.6 J/K.mol)]  -  (92.9 J/K.mol)
             = (253.4 J/K.mol) - (92.9 J/K.mol)
             = 160.5 J/K.mol

b)   ∆G^o  = ∆H^o  -  T∆S^o
               = (177.8 kJ/mol)  - [(298 K)(160.5 J/K.mol)( 1 kJ / 1000 J)
               = 177.8 kJ/mol -  47.83 kJ/mol
               =  130.0 kJ/mol
 
Sample Problem 2

a) Using standard enthalpies of formation and standard entropies, calculate ∆H^o and ∆S^o at 298 K for     the reaction:  2SO₂(g)  +  O₂(g)  →   2SO₃(g),
b) using the values obtained in a) estimate the ∆G^o at 400 K.

Solution:

a) ∆H^o = 2∆H^o_f(SO₃) - [2∆H^o_f(SO₂) +  ∆H^o_f(O₂)]
            = [(2)(-395.2 kJ/mol)] - [(2)(-296.9 kJ/mol) +  (0 kJ/mol)
            = (-790.4 kJ/mol) - (-593.8 kJ/mol)
            = -790.4 kJ/mol + 593.8 kJ/mol
            = -196.6 kJ/mol

     ∆S^o =  2S^o(SO₃)  -  [2S^o(SO₂)  +  S^o(O₂)
             =  2(256.2 J/K.mol) - [2(248.5 J/K.mol) +  205.0 J/K.mol]
             =  512.4 J/K.mol - (497 J/K.mol + 205.0 J/K.mol)
             =  512.4 J/K.mol -702 J/K.mol
             =  -189.6 J/K.mol

b)  ∆G^o  = ∆H^o  -  T∆S^o
              = -196.6 kJ/mol - [(400 K)(-186.6 J/K.mol)( 1 kJ /1000J)
              = -196.6 kJ/mol + 74.64 kJ/mol
              = - 122.0 kJ/mol

Entropy Changes in a Chemical Reaction

There is no easy method of measuring ∆S for a reaction, but we can use experimental measurements of variation of heat capacity for many substances at any temperature.   Absolute entropies are based on the reference point of zero entropy for perfect crystalline solid at 0 K.  Entropies are usually tabulated as molar quantities, in units of joules per mole-Kelvin (J/mol-K).

The molar entropy values of substances in their standard states are known as standard molar entropies and are denoted S^o.  The standard state for any substance is defined as the pure substance at 1 atm pressure.


Sample Molar Entropies of Some Selected Substances at 298 K

Substance                                     S^o, J/mol-K

Gases
H₂(g)                                                 130.7
N₂(g)                                                 191.6
O₂(g)                                                 205.2
H₂O(g)                                              188.8
NH₃(g)                                              192.5
CH₃OH(g)                                        237.6
C₆H₆(g)                                            269.2

Liquids
H₂O (l)                                              69.9
CH₃OH(l)                                        126.8
C₆H₆(l)                                            172.8

Solids
Li(s)                                                   29.1
Na(s)                                                  51.3
K(s)                                                    64.7
Fe(s)                                                   27.3
FeCl₃(s)                                            142.3
NaCl(s)                                              72.3
Al(s)                                                   28.3
Al₂O₃(s)                                            51.0

Based from the sample of molar entropies above, observations about S^o are the following:

1.  Unlike enthalpies of formation, the standard molar entropies of elements at the reference                      temperature at 298 K are not zero.

2.  The standard molar entropies of gases are greater than those of liquids and solids, consistent with         the interpretation of experimental observation.

3.  Standard molar entropies generally increase with increasing molar mass.

4.  Standard molar entropies generally increase with an increasing number of atoms in the formula of      the substance.

In summary, the number of significance of the vibrational degrees of freedom of molecules increase with the increasing mass and increasing number of atoms.

The entropy change in a chemical reaction can be calculated by the sum of the entropies of the products less the sum of the entropies of the reactants:

∆S  = ∑nS^o(products) -  ∑mS^o(reactants)

The coefficient n and m  are the coefficient in the balanced chemical equation.

Example 1

Calculate the ∆S^o for the synthesis of ammonia from N₂(g) and H₂(g)  at 298 K.

N₂(g)    +     3H₂(g)    →     2NH₃(g)

Solution:

Using the formula above, substitute the value of the molar entropies of the substances given in the table above:

∆S  =  ∑nS^o(products) -  ∑mS^o(reactants)

       =  2S^o(NH₃)   -  [S^o(N₂)  +   3S^o(H₂)
       =  (2 mol) (192.5 J/mol-K)   -   [(1 mol) (191.6 J/mol-K)  +  (3 mol)(130.7 J/mol-K)]
       =  - 198.4 J/K

Example 2

Using the standard entropies in the table presented above, calculate the standard entropy change, ∆S^o, for the following reaction  at 298 K.

Al₂O₃(s)  +  3H₂(g)   →    2Al(s)    +   3H₂O(g)

Solution:

∆S  =  ∑nS^o(products) -  ∑mS^o(reactants)

       =  [2S^o(Al)  +   3S^o (H₂O)]  -   [S^o(Al₂O₃)  +   3S^o (H₂)]
       = [(2 mol)(28.3 J/mol-K)  +  (3 mol)(188.8 J/mol-K)]  -  [(1 mol)(51.0 J/mol-K) +(3 mol)                                    (130.7 J/mol-K)]
       =  +180.39 J/K

FOR MOLAR ENTROPIES OF SOME SUBSTANCES JUST CLICK HERE

The Molecular Interpretation of Entropy

We learned that entropy is a measure of randomness and disorder of a system. In this topic entropy will be discussed based on how the structure and behavior of molecules affect the entropy of the system so as the third law of thermodynamics.

The entropy of the system increases (∆S > 0) as the gas molecules  spread out in a larger volume.  On the other hand, phases changes from solid to liquid or from liquid to gas  also lead to an increased entropy of the system.

There are processes that lead to the decrease in the entropy of the system.  Example of this is condensing a gas or freezing of liquid, this result in the increase in the order of the system and lead to the decrease of the entropy (∆S < 0).  A reaction that leads to a decrease in the number of molecules generally leads to the decrease in entropy.  Consider the example:

2NO(g)  +  O₂(g)  →   2NO_2(g)

The entropy change decreases (∆S < 0) since three molecules of gas react to form two molecules of gas.  The formation of N-O bonds imposes more order on the system, the fact that the atoms of the system are more tied up in the products  than in the reactants  that leads to the decrease in the entropy of the system.  The formation of new bonds  decreases the number of degrees of freedom, or forms of motions, available to the atoms ; that is; the atoms are less free to move  in random motion because of the formation of new bonds.  In other words, the greater the number of degrees of freedom of a system, the greater its entropy.

The degrees of freedom of molecules  are associated with three different types of motion for the molecules:
1.  Translational motion is a motion in which the entire molecule can move in one direction only, as        in the movements of gas molecules.  The molecules in a gas have more translational motion than        those in the liquid, which in turn have more translational motion than the molecules in solid.

2.  Vibrational motion is another motion that exist within molecule, in which molecules move                periodically toward and away from one another.

3.  Rotational motion is a motion of molecules as though they were spinning like tops.

The greater the energy that is stored in translational, vibrational, or rotational motion, the greater the entropy.

If the thermal energy of a system is decreased by lowering the temperature, the energy stored in translational, vibrational, and rotational forms of motion decreases.  As less energy is stored, the entropy of the system decreases.  If the temperature is keep lowered, there will come a time in which the motions are shutdown, a point of perfect order.  This occurs when the temperature reaches the absolute zero ( 0 K).  The entropy of a pure crystalline substance at absolute zero is zero, this is called the third law of thermodynamics.

But if the temperature is increased the reverse occurs. Entropy increases when the temperature is increased.  Just like when we increase the temperature in a solid substance, the substance will soon melt and the atoms or molecules  are free to move about the entire volume of the substance.    The added degrees of freedom of the individual molecules greatly increase the entropy of the substance.

Generally entropy increases when temperature is increased.  In other words, the entropies of the phases of a given substance follow the order S_solid <  S_liquid  < S_gas.

In general, the entropy is expected  to increase for processes in which:
1.  Liquids or solutions are formed from solids.
2.  Gases are formed either from solid or liquids.
3.  The number of molecules of gas increases during a chemical reaction.

Below are the examples:
Choose the sample of matter that has greater entropy in each pair:
1.  1 mol of NaCl(s) or 1 mol of HCl(g) at 25^oC
2.  2 mol of HCl(g) or 1 mol of HCl(g) at 25^oC
3.  1 mol of HCl(g) or 1 mol of Ar(g) at 25^oC
4.  1 mol of N₂(s) at 24K or 1 mol of N₂(g) at 298 K

Answer:
1.  1 mol of HCl since gaseous substance are more disordered than solid substance
2.  2 mol of HCl since it contains more molecules
3.  1 mol of HCl since it has capacity to stored more energy than Ar.  The structure of HCl molecules       can vibrate and rotate.
4.   Gaseous nitrogen sample because it is more disordered than the solid nitrogen

Entropy and Second Law of Thermodynamics

We learned from my previous post that energy is conserved during the process, and this is called the first law of thermodynamics.  Energy is just transformed from one form to another form of energy.  Second law of thermodynamics on the other hand, expresses the notion that there is an inherent direction in which processes occurs and usually expressed in terms of entropy.

Spontaneous Processes

Processes can be spontaneous and nonspontaneous.  Spontaneous process is a process that occurs without any ongoing outside intervention while nonspontaneous process needs outside intervention to occur.  Below are some examples of spontaneous processes:

- A waterfall runs downhill, but never up, spontaneously.
- A lump of sugar spontaneously dissolves in a cup of coffee.
- Water freezes spontaneously below 0^oC, and ice melts spontaneously above 0^oC.
- Heat flows from a hotter object to a colder one.
- A shiny nail left outdoors will eventually rust.

An spontaneous process has a difinite direction in which it occurs.  The spontaneity of the process can depend on temperature.  For example, the endothermic process of melting ice under atmospheric pressure.  When T > 0^oC, ice melts spontaneously; the reverse process, liquid water turning ice at this temperature is not spontaneous.  However when T < 0^oC, liquid water converts into ice spontaneously and at this temperature converting ice into water is not spontaneous.  When T = 0^oC, the normal boiling point of water, the solid and liquid phases are in equilibrium.

Example:
Predict whether the following processes are spontaneous, nonspontaneous and at equilibrium.
a.  When a piece of metal heated to 150^oC  is added to water  at 40^oC, the water gets hotter.
b.  Water at room temperature  decomposes into H₂(g) and O₂(g).

Answer:
a.  The process is spontaneous, heat is always transferred from a hotter object to a colder one.
b.  The process is nonspontaneous, at room temperature the reverse occurs.

Reversible and Irreversible Processes

A reversible process is a process in which the state of a system can change.  The change in the system can be restored to its original state by exactly reversing the change. Example is the interconversion of ice and water, at T = 0^oC the process is at equilibrium, now decreasing the temperature, liquid water will turn into ice and increasing the temperature will melt the ice into water.

An irreversible process is one that cannot be simply reversed to restore the system and its
surroundings to their original state.  When a system changes by an irreversible process, it must take a different path ( with a different value of q and w) to get back to its original state.

There are two very important concepts regarding reversible and irreversible processes:
1.  Whenever a chemical system is at equilibrium, reactants and products can inter-convert reversibly.
2.  In any spontaneous process, the path between reactants and products is irreversible

Entropy

Entropy is open described as the measure of  disorder or randomness of the system.  The more disorder or random the system, the larger its entropy.  Like internal energy and enthalpy, entropy is also a state function.  The change in the entropy of the system, ∆S = S_final   -     S_initial, depends only on the initial and final states of the system and not on the particular pathway by which the system changes.  A positive value of  ∆S indicates that the final state is more disordered than the initial state.  Example when gas expands to a large volume , its entropy increases.  A negative value of ∆S indicates that the final state is more disordered or less random than initial state.

Example 1
Predict whether the entropy increases or decreases for the following processes:
a.  freezing ethanol
b. evaporating a beaker of liquid bromine at room temperature
c.  dissolving glucose in water
d.  cooling a nitrogen gas from 80^oC to 20^oC.


Answer: 
a.  Entropy decreases,the ethanol molecules are held together rigidly.
b.  Entropy increases, bromine gas molecules occupy more positions in nearly empty space.
c.  Entropy increases, glucose is dissolved in water and so the more disorder the molecules.
d.  Entropy decreases, decreasing temperature decreases the molecular motion of molecules.


Example 2
By considering the disorder of the products, predict whether the ∆S is positive or negative for the following processes:
a.  H₂O(l)  →   H₂O(g)
b.  Ag⁺(aq)   +    Cl^-(aq)  →    AgCl(s)
c.  4Fe(s)  +  3O₂(g)   →   2Fe₂O₃(s)
d.  CO₂(s)   →     CO₂(g)

Answer:
a.  Entropy is positive there is a large increase in volume.
b.  Entropy is negative, ions are confined in more ordered positions.
c.  Entropy is negative, molecules are converted to solid, solid is more ordered than gas.
d.  Entropy is positive molecules becomes more disordered in gas form.

Second Law of Thermodynamics

Second law of thermodynamics relates the spontaneity of reaction to entropy.  The entropy of the universe increases in a spontaneous process and remain unchanged in an equilibrium process. We must both consider the entropy of the system and the surroundings.  The total change in the entropy, called the change in the entropy of the universe and denoted ∆S_univ, is the sum of the changes in the entropy of the system, ∆S_sys, and of the surrounding, ∆S_surr.

                                                      ∆S_univ  = ∆S_sys     +   ∆S_surr


In terms of ∆S_univ, the second law can be expressed as follows:  In any reversible process, ∆S_univ = 0, whereas in any irreversible (spontaneous) process, ∆S_univ >  0.


Reversible process  (Equilibrium)     :     ∆S_univ  =  ∆S_sys   +    ∆S_surr   =  0
Irreversible process (Spontaneous)   :     ∆S_univ  =  ∆S_sys    +   ∆S_surr   >  0

For a spontaneous process, the second law says that ∆S_univ must be greater than zero, but it does not restrict ∆S_sys and ∆S_surr.  Thus, it is possible that ∆S_sys and ∆S_surr to be positive or negative provided the sum of two quantities must be greater than zero.  For an equilibrium process, the ∆S_univ must be equal to zero, this means that the value of ∆S_sys and ∆S_surr must have the same magnitude but opposite in sign.

A special case of the second law concerns the entropy change in an isolated system, one that doesn't exchange energy or matter with its surroundings.  In any process that occurs in an isolated system leaves the surroundings completely unchanged.  Therefore, ∆S_surr = 0 for such process.  Thus for a special case of an isolated system, second law becomes: 

Reversible process (Equilibrium)         :    ∆S_univ  =  0
Irreversible process (Spontaneous)      ;    ∆S_univ  >  0

FOR ENTROPY AND SECOND LAW OF THERMODYNAMICS TEST QUESTIONS CLICK HERE

Hess Law

Hess Law is another way of manipulating the enthalpy change of the reaction.  It states that if a reaction is carried out in a series of steps, ∆H for the reaction will equal the sum of the enthalpy changes for the individual steps.  The overall enthalpy change  for the process is independent on the number of steps or the particular nature of the path.  Therefore we can calculate ∆H for any process provided we know the route of the step.

Let us try sample problems in order to understand Hess Law.

Sample Problem 1.

The enthalpy of combustion of C to CO₂ is -393.5 kJ/mol C, and the enthalpy of combustion of CO to CO₂ is -283.0 kJ/mol CO:

                             C(s)   +    O₂(g)   →   CO₂(g)                 ∆H = - 393.5 kJ/mol
                         CO(g)    +    1/2O₂(g)  →  CO₂(g)               ∆H =  -283.0 kJ/mol

Using these data, calculate the enthalpy of C to CO:

                          C(s)  +   1/2 O2(g)  →    CO(g) 

Solution:
 
 In the problem above, we have to analyze how the equation is being manipulated to arrive the net equation. Since the net equation show that C is in the reactant side, meaning the first equation is as is.  We also notice that CO is in the product side and therefore the second reaction is reversed.

                     C(s)   +    O₂(g)   →   CO₂(g)                          ∆H = -393.5 kJ/mol
                                   CO₂(g)  →    CO(s)  +   1/2O₂(g)       ∆H =   283.0 kJ/mol
                     C(s)  +   1/2O₂(g)  →  CO(s)                             ∆H = -110.5 kJ/mol             

CO₂ is cancelled out and one-half of O₂.

Sample Problem 2:

Carbon occurs in two forms, graphite and diamond.  The enthalpy of combustion of graphite is -393.5 kJ/mol and that of diamond is -395.4 kJ/mol. 

                   C(graphite)   +  O₂(g)  →   CO₂(g)                  ∆H =  -393.5 kJ/mol
                   C(diamond)  +  O₂(g)  →   CO₂(g)                  ∆H =  -395.4 kJ/mol

Calculate the ∆H for the conversion of graphite to diamond.

Solution:

Since the reactant is graphite the first equation is as is, and since diamond is in the product side, therefore the second equation is reversed and so the ∆H value will be the opposite.
            
                       C(graphite)   +  O₂(g)  →   CO₂(g)                              ∆H =  -393.5 kJ/mol
                                             CO₂(g) →   C(diamond)  +  O₂(g)        ∆H =  +395.4 kJ/mol 
                                  C(graphite)       →   C(diamond)                       ∆H =    +  1.9 kJ/mol

O₂ and CO₂ are cancelled out because both can be found in the reactant and product side.
                  

FOR ENTHALPY AND FIRST LAW OF THERMODYNAMICS TEST QUESTIONS CLICK HERE

                                   

Enthalpies of Formation

Elements combine to form compounds, compounds on the other hand also combine with another compound to form a different compounds, these are just some of the reactions that occur in our environment and inside the laboratory.  All these reactions can either absorb or release heat, and therefore has the ∆H.

A particular important process used for tabulating thermochemical data is the formation of compound from its constituent elements.  The enthalpy change associated with this process is called enthalpy of formation (or heat of formation) and is symbolized as ∆H_f, where the subscript f indicates that the substance is formed from its elements.

The value of any enthalpy change is dependent on the conditions of  temperature , pressure, and state of the reactants and products.  That is why , in order to compare the enthalpies of different reactions, a standard  condition is set which is called the standard state.  The standard state of a substance is set at 1 atm pressure, and the temperature is at 25^oC (298 K). And therefore, the standard enthalpy of reaction is defined as the enthalpy change when all reactants and products are in their standard states, and is denoted as ∆H^o, where the superscript "^o" denotes standard-state condition.

The standard enthalpy of formation of a compound is the change in enthalpy for the reaction that forms 1 mol of the compound from its elements, with all substances in their standard states.  By convention, the standard enthalpy of formation of most stable form of any element is zero, because there is no formation reaction needed when the element is already in the standard state. The standard enthalpy of formation is symbolized as ∆H^o_f.  Example is the O₂, molecular oxygen is more stable form than any other form of oxygen O, O₃, and therefore the ∆H^o_f = 0.

Knowing the  standard enthalpy of formation can help us calculate the standard enthalpy of reaction, ∆H^o_rxn.  It can be calculated by the formula below:

∆H^o_rxn = ⅀m∆H^o_f(products) - ⅀n∆H^o_f(reactants)

where m and n is the coefficient in the balanced equation and the symbol ⅀, means the sum of the standard enthalpy of products or the reactants.

Click HERE for the List of Enthalpy of Formation of Some Substances

Let us have sample problems:

Sample Problem 1

The thermite reaction  involves aluminum and iron (III) oxide

2Al(s)  +  Fe₂O₃(s)  →  Al₂O₃(s)  +   2Fe(l)

This reaction is highly exothermic and the liquid iron formed is used to weld metals.  Calculate the heat released in kilojoules per mol of Al reacted with Fe₂O₃.  The ∆H^o_f of Fe(l) is 12.40 kJ/mol.

Solution:
Click the link above for the ethalpy of formation of the given substances.

                     ∆H^o_rxn = [∆H^o_f(Al₂O₃) + 2∆H^o_f(Fe)]  - [2∆H^o_f(Al) + ∆H^o_f(Fe₂O₃)]
                                 = [( -1669.9 kJ/mol) + 2(12.40 kJ/mol)]  -  [2(0) + (-822.2kJ/mo)]
                                 = [-1669.9 kJ/mol + 24.8 kJ/mol] - [-822.2 kJ/mol]
                                 = -1645.1 + 822.1 kJ/mol
                                 = -822.8 kJ/mol

Sample Problem 2

The standard enthalpy change for the reaction

CaCO₃(s)  →  CaO(s)  +  CO₂(g)

is 178.1 kJ/mol.  Calculate  the standard enthalpy of formation of CaCO3(s).  Use the value of enthalpy of substances from the link above.

Solution:

                

                ∆H^o_rxn  =  [∆H^o_f(CaO) +  ∆H^o_f(CO₂)] - [∆H^o_f(CaCO₃)]
       178.1 kJ/mol   =  [-635.5 kJ/mol + (-393.5 kJ/mol)] -  [∆H^o_f(CaCO₃)]
    [∆H^o_f(CaCO₃)]  = -1029 kJ/mol - 178.1 kJ/mol
                              =  -1207.1 kJ/mol

FOR ENTHALPY AND FIRST LAW OF THERMODYNAMICS TEST QUESTIONS CLICK HERE