## Friday, April 30, 2010

### Faraday's Second Law of Electrolysis:

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Statement:

When the same quantity of electricity is passed through different electrolytes, the masses of the elements liberated or deposited are in proportion to the chemical equivalents of these elements. Faraday's laws are very useful for the determination of electrochemical equivalents of different substances.

$Electrochemical equivalent\,(Z)= \frac{quantity\,of \, substance\,deposited\,(W)}{quantity\,of\,electricity\,passed(Q)}$

$Z\,=\,\frac{W}{Q}$
Chemical Equivalent:

The chemical equivalent of an element is numerically equal to its relative atomic mass in grams divided by its the valency of the ion.
Faraday's second law of electrolysis can also be stated as under:
"The mass of different substances liberated or deposited by the same quantity of electricity is proportional to the atomic masses divided by the valencies of their ions."

Explanation:

Take three solutions of electrolytes: AgNO3, CuSO4 and Al(NO3)3 in a series, pass some quantity of electricity through them for the same time. Now Ag Cu and Al metals collect at the cathode. Their masses are directly proportional to their equivalent masses.

According to Faraday, if 96,500 Coulombs (or 1 Faraday) is passed through these electrolytes,

we get $Ag\frac{108}{1}\,or 108g,Cu\frac{63.5}{2}\,or\,31.75g\,and Al\frac{27}{3}=9g$ which are the equivalent masses of

Ag, Cu and Al respectively.$Equivalent\,Mass\,=\frac{Atomic\,mass\,of \,element}{valency\,of\,element}$
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### Faraday's First Law of Electrolysis:

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Faraday's put forward his two laws of electrolysis in 1833.

Statement:

The mass of an elements which is deposited on an electrode during electrolysis is directly proportional to the quantity of electricity which passes through the electrolyte.

Explanation:

If W is the amount of substance which liberates or deposited at the electrode on passing the electricity through the electrolyte and the quantity of electricity is Q, then

$W\,\alpha \,Q$

or W = ZQ

Z is the electrochemical constant for a given substance.

As $Q\,=\,A\times \,t$

$\therefore$ We can write the statement of the first law of electrolysis mathematically as under:

$W\times A\times t$

or W= Zat

If 1 ampere electric current passes through the electrolyte for 1 second then W=Z It means that on passing the current of 1 ampere for 1 second the weight of the substance deposited is equal to the electrochemical constant. For doing the calculations of electrochemical problems, we must know the units too.

unit of charge (Q) = Coulomb (C)
unit of mass (m) = Kilogram (kg)
unit of current (A) = ampere (A)
unit of electrochemical equivalent (Z) = kg/C

Note:
Faraday's first law of electrolysis is written as:

W = ZAt

Here;

W= is actually mass and not weight, as mas is commonly called weight.
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### Mass Number and Atomic Number:

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Mass Number:

The mass number is the number of neutrons added to the number of protons. The mass number of the most common isotope can be obtained from the periodic table. If you take the decimal number on the periodic table and round it to the nearest whole number, you have the mass number.

For example:

The atomic weight of Iron(Fe) is 55.847.
When rounded it gives a mass number of 56.

The atomic number of Fe is 26. so most Fe atoms have 30 (56-26) neutrons. In addition, all neutral Fe atoms have 26 protons and 26 electrons. Atoms of the same element with a different number of neutrons are called isotopes. The most common isotope of an element is the one that is on the periodic table.

Atomic Number:

The atomic number of an element is what distinguishes it from all other elements. An atom's atomic number is the number of protons there are in the nucleus. Hydrogen's atomic number is 1. Helium's atomic number is 2. Any atom that has an atomic number of 1 is a hydrogen atom no matter how many electrons or neutrons the atom has.
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### Percent Composition, Empirical Formula,Molecular Weight Formula

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Sometimes we want to find out what the formula of a compound would be. To figure this out, we analyze the compound into amounts of the elements for a given amount of the compound. This is expressed as the percent composition which is the mass percentages of each different element in a compound. We must know the molecular weight of the compound in order to determine the molecular formula.

Say we have an element X in a compound. This element X is just part of the whole compound. We define the mass percentage of X as the parts of X per hundred parts of the total, by mass. That is:

Mass % X = (mass of X in the whole)/(mass of the whole) * 100%

Percent Composition Example:

Calculate the percent composition of Mg(NO3)2

1Mg = 1 x 24 = 24
2N = 2 x 14 = 28
6O = 6 x 16 = 96
148g/mole
%Mg = 24/148 x 100 = 16.2%
%N = 28/148 x 100 = 18.9%
%O = 96/148 x 100 = 64.0%

The percentage composition of a compound leads directly to its empirical formula. An empirical formula for a compound is the formula of a substance written with the lowest integer subscripts. For example, hydrogen peroxide has the molecular formula H2O2. The molecular formula tells us the precise number of atoms of different elements in the substance. The empirical number tells us ratio of numbers of atoms in the compound. The empirical formula of hydrogen peroxide is HO, while the molecular formula is H2O2. Compounds with different molecular formulas can have the same empirical formulas and such substances will have the same percentage composition. An example is acetylene, C2H2 and benzene, C6H6. In order to obtain the molecular formula of a substance, you need to know the percent composition and the molecular weight. The molecular weight allows us to choose the correct multiple of the empirical formula for the molecular formula.

Empirical Formula Example:

Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H

Divide each percent by that element's atomic weight. To get the answers to whole numbers, divide through by the smallest one.

Ca = 54.09/40 = 1.352 1.352/1.352 = 1
O = 43.18/16 = 2.699 2.699/1.352 = 2
H = 2.73/1 = 2.73 2.73/1.352 = 2

CaO2H2 =

Ca(OH)2

Calcium Hydroxide.

The molecular weight of a substance is the sum of the atomic weight of the atoms in a molecule. The atomic weight is the average atomic mass for a naturally occurring element. This means that molecular weight is the average mass of a molecule of a substance. Molecular weight is expressed in atomic mass units. For example, we might want to find the molecular weight of a molecule of water. We have 2 atoms of H, with each hydrogen atom weighing 1 amu. We multiply 2 H atoms by 1 amu a piece to get 2 amu. We add the 16 amu from one O atom to the 2 amu from the oxygen to get a total of 18 amu for one molecule of water.

H2O = 2 atoms of hydrogen and 1 molecule of oxygen

2 H * 1 amu = 2 amu
1 O * 16 amu = 16 amu

2 amu + 16 amu = 18 amu = molecular weight of water

The formula weight of a substance is the sum of the atomic weights of all atoms in a formula unit of the compound. Formula weight doesn't depend on whether or not the substance is a molecule.For example, sodium chloride, which is NaCl, has a formula weight of 58.44 amu. This results from having 22.99 amu from Na and 35.45 amu from Cl. You would use the formula weight for substances which are not molecules, such as ionic compounds.

Gram Formula Weight Examples:

Find the gram formula weight of H2SO4

1.
2H = 2 x 1 = 2
1S = 1 x 32 = 32
4O = 4 x 16 = 64
98g/mole

Find the gram formula weight of Na2CO3 . 10 H2O
2.
2Na = 2 x 23 = 46
1C = 1 x 12 =12
3O = 3 x 16 = 48
10H2 = 10 x 18 = 180
286g/mole.
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### Chemical & Physical Changes of Matter:

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A chemical change is a change in which one kind of matter is changed into a different type of matter. Some examples of chemical changes: the rusting of your car, setting your shoe on fire, digesting food, and the burning of magnesium metal in oxygen to form magnesium oxide. All of these materials combine chemically with another material , and cannot be separated by any physical means.

A physical change is a change in the form of matter but not in its identity. An example of a physical change would be the dissolving of one thing into another thing. For instance, dissolving sugar into water. The water and the sugar retain their chemical identities and can be separated by physical means. Another example is ice melting to water. Ice and water are both H2O. The identity of the matter is not changed, just the state that it is in.

### Compounds, Elements, Mixtures:

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John Dalton (1766-1844) provided us with the Atomic Theory. His theory says that all matter is composed of small particles called atoms.

1. All matter is composed of indivisible atoms. An atom is an extremely small particle of matter that retains its identity during chemical reactions.
2. An element is a type of matter composed of only one kind of atom, each atom of a given kind having the same properties.
3. A compound is a type of matter composed of atoms of two or more elements chemically combined in fixed proportions.
An element is a substance that is composed of only one kind of atom like aluminum, iron, or neon. Today, 109 elements are known and listed on the periodic table.

We now know that atoms are not indivisible , since they contain protons, neutrons, electrons and other subatomic particles. However, other than this mistake Dalton's ideas are essentially correct.

A compound is a substance of more than one element, chemically combined. A more scientific definition is that a compound is a type of matter composed of atoms of two or more elements chemically combined in fixed proportions. An example of a compound would be water. It is a compound that contains the elements hydrogen and oxygen fixed in the ratio 2 to 1. A compound has new properties unlike the elements which make it up. A compound has a chemical formula such as H2O.

A mixture is a material that can be separated by physical means into two or more substances. A classic example of a mixture lab would be one in which you were presented a mixture of sand, iron filings, and salt. You are told to separate these materials. How do you do that? Well, think about the various physical properties of each material. You use a magnet to separate out the iron filings. You then mix water with the sand and salt mixture. You swish the water, salt and sand around for a while and then filter it. The salt dissolved into the water, so the salt water solution passes through the filter while the sand gets left in the filter. Now we slowly heat up the salt water solution, and evaporate the water, and we are left with salt. In a mixture, the compounds are not in a definite proportion. For example, a teaspoon of salt in a liter of water is salt water, but so is a cup of salt in a liter of water.
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### Definition of Ideal Gas Law:

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The ideal gas law is a combination of all the gas laws. The ideal gas law can be expressed as PV = nRT.

1. P is the pressure in atm
2. V is the volume in liters
3. n is the number of moles
4. R is a constant
5. T is the temperature in Kelvin

The constant R is calculated from a theoretical gas called the ideal gas. The most commonly used form of R is .0821 L * atm / (K * mol). This R will allow the units to cancel so the equation will work out.

To find the volume of 2.00 moles gas that is at 1.00 atm of pressure and 235 Kelvin, use the ideal gas law equation.

(1.00 atm)(V) = (2.00 mol)(.0821 L * atm / (K * mol))(235 kelvin)

V = (38.587 L * atm) / (1.00 atm)

V = 38.6 L
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### Combined Law

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The combined gas law is a combination of Boyle's Law and Charles's Law; hence its name the combined gas law. In the combined gas law, the volume of gas is directly proportional to the absolute temperature and inversely proportional to the pressure.

This can be written as PV / T = constant. Since for a given amount of gas there is a constant then we can write P1V1 / T1 = P2V2 / T2.

1. P1 is the initial pressure
2. V1 is the initial volume
3. T1 is the initial temperature (in Kelvin)
4. P2 is the final pressure
5. V2 is the final volume
6. T2 is the final temperature (in Kelvin)

This equation is useful if you have the current volume, temperature, and pressure of a gas, and if you have two of the three final values of the gas.

For example:

If you have 4.0 liters of gas at STP, and you want to know the volume of the gas at 2.0 atm of pressure and 30o C, the equation can be setup as follows:

(1.0)(4.0) / 273 = (2.0)(V2) / 303
(V2)(2)(273) = (1)(4)(303)
V2 = 2.2

Therefore the new volume is 2.2 liters.
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### Charles's Law:

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Charles's Law can be stated as the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature.

V / T =constant,

1. V is the volume
2. T is the absolute temperature (measured in Kelvin)

Charles's Law can be rearranged into two other useful equations.

V1 / T1 = V2 / T2

1. V1 is the initial volume
2. T1 is the initial temperature
3. V2 is the final volume
4. T2 is the final temperature

V2 = V1 (T2 / T1)

1. V2 is the final volume
2. T2 is the final temperature
3. V1 is the initial volume
4. T1 is the initial temperature

Important:

Charles's Law only works when the pressure is constant.

Note:

Charles's Law is fairly accurate but gases tend to deviate from it at very high and low pressures.
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### Oxidation and Reduction:

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When forming compounds, it is important to know something about the way atoms will react with each other. One of the most important manners in which atoms and/or molecules react with each other is the oxidation/reduction reaction. Oxidation/Reduction reactions are the processes of losing and gaining electrons respectively. Just remember, "LEO the lion says GER:" Lose Electrons Oxidation, Gain Electrons Reduction. Oxidation numbers are assigned to atoms and compounds as a way to tell scientists where the electrons are in a reaction. It is often referred to as the "charge" on the atom or compound. The oxidation number is assigned according to a standard set of rules. They are as follows:

1. For single atoms in an ion, their oxidation number is equal to their charge.
2. An atom of a pure element has an oxidation number of zero.
3. Cl, Br, and I are always -1 in compounds except when they are combined with O or F.
4. H is normally +1 and O is normally -2.
5. Fluorine is always -1 in compounds.
6. The oxidation number of a compound is equal to the sum of the oxidation numbers for each atom in the compound.
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### Definition of Enthalpy:

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Enthalpy is an interesting concept: it is defined by its change rather than a single entity. A state property, the word enthalpy comes from the Greek "heat inside". If you have a chemical system that undergoes some kind of change but has a fixed volume, the heat output is equal to the change in internal energy (q = ΔE). We will define the enthalpy change, ΔH, of a system as being equal to its heat output at constant pressure:

dH = q at constant pressure
Where:

ΔH = change in enthalpy

We define enthalpy itself as:

H = E + PV

Where:

H = enthalpy
E = energy of the system
PV = pressure in atm times volume in liters

You will not need to be able to calculate the enthalpy directly; in chemistry, we are only interested in the change in enthalpy, or ΔH.

ΔH = Hfinal - Hinitial or ΔH = H(products) - H(reactants)

Tables of enthalpies are generally given as ΔH values.

Example Problem:

Calculate the ΔH value of the reaction:
HCl + NH3 → NH4Cl
(ΔH values for HCl is -92.30; NH3 is -80.29; NH4Cl is -314.4)

ΔH = ΔHproducts - ΔHreactants
ΔHproducts = -314.4
ΔHreactants = -92.30 + (-80.29) = -172.59
ΔH = -314.4 - 172.59 = 141.8

We can also represent enthalpy change with the equation:

ΔH = ΔE + P ΔV

Where:

ΔV is the change in volume, in liters
P is the constant pressure

If you recall, work is defined as P ΔV, so enthalpy changes are simply a reflection of the amount of energy change (energy going in or out, endothermic or exothermic), and the amount of work being done by the reaction. For example, if ΔE = -100 kJ in a certain combustion reaction, but 10 kJ of work needs to be done to make room for the products, the change in enthalpy is:

ΔH = -100 kJ + 10 kJ = -90 kJ

This is an exothermic reaction (which is expected with combustion), and 90 kJ of energy is released to the environment. Basically, you get warmer. Notice the convention used here -- a negative value represents energy coming out of the system.

You can also determine ΔH for a reaction based on bond dissociation energies. Breaking bonds requires energy while forming bonds releases energy. In a given equation, you must determine what kinds of bonds are broken and what kind of bonds are formed. Use this information to calculate the amount of energy used to break bonds and the amount used to form bonds. If you subtract the amount to break bonds from the amount to form bonds, you will have the ΔH for the reaction.

Example Problem:

Calculate ΔH for the reaction:
N2 + 3H2 → 2NH3

(The bond dissociation energy for N-N is 163 kJ/mol; H-H is 436 kJ/mol; N-H is 391 kJ/mol)

ΔH = ΔHproducts - ΔHreactants
To use the bond dissociation energies, we must determine how many bonds
are in the products and the reactants. In NH3 there are 3 N-H bonds so in 2 NH3
there are 6 N-H bonds. In N2 there is 1 N-N bond and in 3H2 there are 3 H-H bonds.
ΔHproducts = 6(391) = 2346
ΔHreactants = 163 + 3(436) = 1471
ΔH = 2346 - 1471 = 875
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### Entropy:

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Entropy is a measure of the disorder of a system. Take your room as an example. Left to itself,
your room will increase in entropy (i.e., get messier) if no work (cleaning up) is done to contain the disorder. Work must be done to keep the entropy of the system low. Entropy comes from the second law of thermodynamics, which states that all systems tend to reach a state of equilibrium. The significance of entropy is that when a spontaneous change occurs in a system, it will always be found that if the total entropy change for everything involved is calculated, a positive value will be obtained. Simply, all spontaneous changes in an isolated chemical system occur with an increase in entropy. Entropy, like temperature, pressure, and enthalpy, is also a state property and is represented in the literature by the symbol "S". Like enthalpy, you can calculate the change of S (ΔS).

Δ S = S final - S initial or Δ S = S (products) - S (reactants)

Where:

ΔS is change in entropy

Sfinal and Sinitial are the final and initial entropies, respectively.
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### Heat and Wrok:

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Heat and work are both forms of energy. They are also related forms, in that one can be transformed into the other. Heat energy (such as steam engines) can be used to do work (such as pushing a train down the track). Work can be transformed into heat, such as might be experienced by rubbing your hands together to warm them up.

Work and heat can both be described using the same unit of measure. Sometimes the calorie is the unit of measure, and refers to the amount of heat required to raise one (1) gram of water one (1) degree Celsius. Heat energy is measured in kilocalories, or 1000 calories. Typically, we use the SI units of Joules (J) and kilojoules (kJ). One calorie of heat is equivalent to 4.187 J. You will also encounter the term specific heat, the heat required to raise one (1) gram of a material one (1) degree Celsius. Specific heat, given by the symbol "C", is generally defined as:

$C=\frac{q}{M\triangle\,T}$

Where:

C = specific heat in cal/g-°C
q = heat added in calories,
m = mass in grams
ΔT = rise in temperature of the material in °C.

The value of C for water is 1.00 cal/g-°C.

The values for specific heat that are reported in the literature are usually
listed at a specific pressure and/or volume, and you need to pay attention to
these settings when using values from textbooks in problems or computer models.

Example Problem:

If a 2.34 g substance at 22°C with a specific heat of 3.88 cal/g-°C is heated with 124 cal of energy, what is the new temperature of the substance?

$\bigtriangleup T=\frac{q}{MC}$

$\bigtriangleup T=\frac{124}{(2.34)(3.88)}=17.7\,^{\circ}C$

new T = 22 + 13.7 = 35.7°C

Two other common heat variables are the heat of fusion and the heat of vaporization.
Heat of fusion is the heat required to melt a substance at its melting temperature, while
the heat of vaporization is the heat required to evaporate the substance at its boiling point.

Chemical work is primarily related to that of expansion. In physics, work is defined as:

$w\,=d\,\times \,f$

Where:

w = work, in joules (N×m) (or calories, but we are using primarily SI units)
d = distance in meters
f = opposing force in Newtons (kg*m/s2)

In chemical reactions, work is generally defined as :

w = distance × (area × pressure)

The value of distance times area is actually the volume. If we imagine a reaction taking place in a container of some volume, we measure work by pressure times the change in volume.

w = ΔV × P

Where:

ΔV is the change in volume, in liters

If ΔV=0, then no work is done.

Example Problem:

Calculate the work that must be done at standard temperature and pressure
(STP is 0°C and 1 atm) to make room for the products of the octane combustion:

$2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O$

Knowing that 25 moles of gas are replaced by 34 moles of gas in this reaction, we can
calculate a net increase of 9 moles of gas. Knowing the molar volume of an ideal gas at
STP (22.4 L/mol), the change in volume and the work of expansion can be calculated
dV = 9 moles ∗ 22.4 L/mol = 202 L
The external pressure is 1.0 atm (standard pressure), so the work required is:
w = dV ∗ P = 202 L ∗ 1.00 atm = 202 l-atm
Using the conversion factor of 1 L-atm = 101 J, the amount of work in joules is:
w = 202 L-atm ∗ 101 j/L-atm = 2000 J, or 2kJ of energy.
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### Definition of pH:

What is of interest in this reading, however, is the acid-base
nature of a substance like water. Water actually behaves both like
an acid and a base. The acidity or basicity of a substance is defined
most typically by the pH value, defined as below:

$pH=-log[H^{+}]$

At equilibrium, the concentration of $[H^{+}]$ is 1.00 × 10-7,
so we can calculate the pH of water at equilibrium as:

$pH=-log[H^{+}]$ = -log[1.00 × 10-7] = 7.00

Solutions with a pH of seven (7) are said to be neutral,
while those with pH values below seven (7) are defined as acidic
and those above pH of seven (7) as being basic.

pOH gives us another way to measure the acidity of a solution.
It is just the opposite of pH. A high pOH means the solution is acidic
while a low pOH means the solution is basic.

$pOH=-log[H^{-}]$

$pH+pOH=14.00\,at\,25^{\circ }C$

### WATER:

We typically talk about acid-base reactions in
aqueous-phase environments -- that is, in the presence of water.
The most fundamental acid-base reaction is the dissociation of water.

$H_{2}O\,\Leftrightarrow H^{+}+OH^{-}$

In this reaction, water breaks apart to form a hydrogen
ion $[H^{+}]$ and a hydroxide ion $[OH^{-}]$ In pure water,
we can define a special equilibrium
constant (Kw) as follows:

$Kw=[H^{+}][H^{-}]=1.00\times 10^{-14}$

Where Kw is the equilibrium constant for water at 25° C (unitless)
$[H^{+}]$ is the molar concentration of hydrogen

$[OH^{-}]$ is the molar concentration of hydroxide

An equilibrium constant less than one (1) suggests that the reaction
prefers to stay on the side of the reactants -- in this case, water likes
to stay as water. Because water hardly ionizes, it is a very poor conductor of electricity.

### The Mole:

Given the equation above, we can tell the number of moles of
reactants and products. A mole simply represents Avogadro's
number (6.022 x 1023) of molecules. A mole is similar to a
term like a dozen. If you have a dozen carrots, you have twelve
of them. Similarly, if you have a mole of carrots, you have
(6.022 x 1023) carrots. In the equation above there are no numbers
in front of the terms, so each coefficient is assumed to be one (1).
Thus, you have the same number of moles of

AgNO3, NaCl, AgCl, NaNO3.

Converting between moles and grams of a substance
is often important. This conversion can be easily done when
the atomic and/or molecular mass of the substance(s) are known.
Given the atomic or molecular mass of a substance, that mass in
grams makes a mole of the substance.

For example,

calcium has an atomic mass of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.

### Empirical Formula and Molecular Formula:

While the empirical formula is the simplest form of a compound,
the molecular formula is the form of the term as it would appear
in a chemical equation. The empirical formula and the molecular
formula can be the same, or the molecular formula can be any
positive integer multiple of the empirical formula.

Examples of empirical formulas:

AgBr, Na2S, C6H10O5.

Examples of molecular formulas:

P2, C2O4, C6H14S2, H2, C3H9.

One can calculate the empirical formula from the masses or
percentage composition of any compound.
We have already discussed percent composition in the section above.
If we only have mass, all we are doing is essentially eliminating
the step of converting from percentage to mass.

Example:

Calculate the empirical formula for a compound that has
43.7 g P (phosphorus) and 56.3 grams of oxygen.
First we convert to moles:

$\frac{43.7\,grams\,P}{1}\times \frac{1\,mol}{30.97\,grams}=1.41\,moles$

$\frac{56.3\,grams\,O}{1}\times \frac{1\,mol}{16.00\,grams}=3.52\,moles$

Next we divide the moles to try to get an even ratio:

$Phosphorus:\frac{1.41}{1.41}=1.00$
$Oxygen:\,\frac{3.52}{1.41}=2.50$

When we divide, we did not get whole numbers so we must multiply by two (2).

Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number.

Example:

If we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necessary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2).

## Thursday, April 29, 2010

### Electronegativity:

Definition:

Electronegativity is defined as the relative tendency of atom in a molecule to attract shared pair of electrons to itself.

Explanation:

An American chemist, Linus Pauling, calculated the electronegativities of different elements taking fluorine as a standard. He devised a scale which ranged from O to 4. Fluorine has the highest value of electronegativity of 4.

Group Trends in Electronegativity:

It increases on going across a period from left to right because new shells are being added to each atom and the power of a nucleus to attract electrons decreases.

Periodic Trends in Electronegativity:

It increases on going across a period from left to right because the nuclear charge increases within the same shell holding the electrons tightly.

Electronegativity values help to determine the nature and type of bond. The strength of a bond between two atoms increases with the difference in their electronegativities. If the difference is zero, the bond is non-polar (pure) covalent. If the difference is 1.7, the bond is polar covalent (partially ionic). If the difference is more than 1.7, the bond is ionic.

### Ionization Energy:

Definition:

"The energy required to remove the first outermost electron is called the first ionization energy. The energy required to remove the outermost electron from a monovalent positive ion is called the second ionization energy and so on. The higher the ionization and third ionization energies are always larger than the first. The higher the ionization energy, the more difficult it is to remove an electron. It is measured in KJ/mole or ev/atom.

KJ=Kilo Joule and e.v. = electron volt

Dependence of Ionization Energy:

1. Nuclear Charge: The larger the nuclear charge, the greater the I.E.
2. Shielding Effect: The greater the shielding effect, the less the I.E.
3. Radius: The greater the distance between the nucleus and outermost electrons, the less the I.E.
4. Sub-levels: Electrons in the P sub-level have a lower I.E. than a sub-level electrons.
Group Trends in I.E.

I.E. decreases down the group from top to bottom.

Periodic Trends in I.E.

I.E. increases going across a group from left to right.

### Definitation of Alpha Particles, Beta Particles and Gamma Particles

$(a)\,\,\,\alpha - Particles:$

They are positively charged particles emitted from certain radioactive nuclei. Each alpha particle consists of two protons and two neutrons and is identical to the nucleus of helium atom.

$(b)\,\,\,\beta - Particles:$

They are fast moving electrons emitted from certain radioactive nuclei. A beta particle is formed when a neutron decomposes.

$(c)\,\,\,\gamma-Particles:$

They are high energy electromagnetic radiations emitted from certain radioactive nuclei. Gamma rays have no mass or electrical charge.

### The Discovery of Electrons

The credit of the discovery of electrons goes to the British physicist J.J. Thomson in 1897. He used a glass tube called the discharge tube for his experiment. The tube is fitted with two metal electrodes and a vacuum pump.The two electrodes are connected to a high voltage source. The tube is evacuated. On increasing a high voltage across the tube, a beam of bluish light emerges from the negative electrode, the cathode. These rays travel to anode in straight lines. These rays were called cathode rays. These rays were deflected to the positive plate in electric and magnetic fields. From this Thomson concluded that the cathode rays were made up of very small negatively charged particles, which he named electrons. Whatever the nature of electrodes or gas may be, electrons are always produced. It proves that electrons are the fundamental particles of all atoms.

Properties of Cathode Rays:

• They produce shadows if an opaque object comes in their way. It proves that they travel in straight lines.
• They can revolve a light paddle wheel. It shows that they are material particles.
• On striking the walls of the tube, they glow or produce fluorescence.
• They deflect towards positive pole in the electric and magnetic fields. It shows that they are negatively charged.
• They can produce mechanical pressure. It shows that they have kinetic energy.
• They raise the temperature of the body on which they fall.
• Their charge to mass (e/m) ration is constant. Its value is fixed which is the same for all electrons.
• Cathode rays are emitted irrespective of the nature of electrodes or the gas present in the discharge tube. It shows they are the essential constituents of all matter.

### Law of Conservation of Mass and Landolt Experiment

Law of Conservation of Mass:
Antoine Lavoisier, a French chemist, stated the law of conservation of mass in 1785.

Statement:

"In any chemical reaction, the initial weights of reactants is equal to the final weights of products"OR
"Mass is neither created nor destroyed during a chemical reaction but it only changes from one form to another"

Example:
1. Water forms by the union of hydrogen and oxygen. If we weigh the reactants (hydrogen and oxygen) before the chemical reaction, we find the weight of the product (water) equal to the combined weight of reactants.
2. The weight of iron increases on rusting. The increase in weight is equal to the weight of oxygen added to iron.
Landolt Experiment:
H. Landolt was German Chemist. He proved the law of conservation of mass by using an H-shaped glass tube. he filled silver nitrate in limb A and hydrochloric acid in limb B. The tube was sealed and weighed before the chemical reaction. The reactants were mixed by inverting and shaking the tube. A white precipitate of silver chloride was formed along with nitric acid. The tube was weighed again. He found that there was no change in weight during the following chemical reaction.

$AgNO_{3}+HCl(aq)\rightarrow AgCl_{s}+HNO_{3}(aq)$
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## Wednesday, April 28, 2010

### Balancing Equations Using Matrices

This is a fundamental skill in chemistry, as you might have noticed from the short reading in stoichiometry! Balancing equations means writing chemical equations such that the amount of stuff you start with in the reaction equals the amount of stuff you end up with as a product. In other words, if I start baking bread with 10 pounds of flour, I should end up with 10 pounds of bread, unless some is lost onto the floor or if some of it goes up in smoke!

A simple example goes a long way. We can form water by combing hydrogen gas (H2) and oxygen (O2) in the presence of electricity. The reaction looks like this:

H2 + O2 ---> H2O

If you do some of the gram molecular weight calculations you will find this:

2 grams of hydrogen + 32 grams of oxygen = 18 grams of water

What this says is that you start with 34 grams of stuff and end up with 18 grams of stuff. You've lost 16 grams of stuff, and in this reaction that just doesn't happen! Where did the 16 grams go?

They're not lost, we just haven't balanced the equation! You might have also noticed that there are two oxygen s on the left and only one on the right! We need to get things in the correct proportions for this reaction to be balanced. The balanced reaction looks like this:

2 H2 + O2 ---> 2 H2O

This says that we need two hydrogen molecules to combine with one oxygen molecule to form two new water molecules. If we do the math:

(2 x 2 grams of hydrogen) + 32 grams of oxygen = (2 x 18 grams of water)

we now have 36 grams of stuff on the left and 36 grams on the right. We also now have 4 hydrogens on the left, four hydrogens on the right, two oxygens on the left, and two oxygens on the right. We can say that this equation is mass balanced. In your studies of chemistry, you will also need to be concerned with charge balancing, but we'll let your profs help you with that!

Balancing equations is an art, but if you have a calculator that can handle what is known as a "matrix", you have a foolproof way of balancing equations! A matrix is a group of numbers, arranged in rows and columns, like this:

$\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$

This is called a "2 by 2" or "2 x 2" matrix, because it has two rows (going across) and two columns (going down). In this application, you will have to do three matrix operations:

1. Multiply two matrices
2. Find the determinant of a matrix
3. Find the inverse of a matrix

Fortunately, graphing calculators make this particularly easy! To help you understand a little of what you are doing, let's explain finding the determinant. The determinant is a single number generated by cross-multiplying the terms in the matrix. You must have a square matrix (n X n) to be able to find the determinant. The equation for finding the determinant is:

$det\begin{pmatrix} a & b \\ c & d \end{pmatrix}=a\bullet d-c\bullet b$
$det\begin{pmatrix} 8 & 3 \\ 4 & 2 \end{pmatrix}=8\bullet 2-4\bullet 3=16-12=4$

The example below the equation shows a sample calculation for a 2 x 2 matrix. Notice that you are cross multiplying the opposite terms, then subtracting out the other set of opposite set of multiplied terms. Pretty easy.

Here is how this is done on the TI-82 Graphical Calculator. These instructions are SPECIFIC to the TI-82:

1. Turn the calculator on (yep, common sense, but want to make sure that's done!)
2. Hit the "MATRIX" button
3. Use right arrow key to scroll over to "EDIT"
4. Type "1" for Matrix A
5. Type "2" for number of rows, the ENTER
6. Type "2" for number of columns, then ENTER
7. Type each of the numbers, following each with ENTER: 8-ENTER-3-ENTER-4-ENTER-2-ENTER
8. Type the blue "2nd" button then "QUIT" (above the MODE button)
9. Type MATRIX, then scroll to MATH
10. Hit "1" for "det"
11. Hit MATRIX then "1". You should see "det [A]" in the window.
12. Hit the ENTER key, you should see the result "4"

Now we are ready to talk about balancing equations. Let's choose a simple reaction:

a Cr + b O2 ---> c Cr2O3

We have two different elements, Cr and O, so we will need two different equations. We are trying to calculate the values of a, b, and c, the coefficients of the reaction. The two equations look like this:

Cr: 1a + 0b = 2c
O: 0a + 2b = 3 c

The "1a" means that there is one Cr behind the "a" coefficient, zero Cr's behind the "b" coefficient, and 2 Cr's behind the c coefficient. We use the same technique for oxygen. We now have two matrices (called Matrix A and Matrix B):

$A=\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$
$B=\begin{pmatrix} 2 \\ 3 \end{pmatrix}$

To obtain the values of a, b, and c, do these steps:

* To calculate a and b, multiply the inverse of Matrix A times Matrix B times the determinant of Matrix A: [A]-1 x [B] x det[A]
* To calculate c, calculate the determinant of Matrix A.

Here is how to do the first step:

1. Enter Matrix A as a 2x2 matrix, using the technique described above
2. Enter Matrix B as a 2x1 matrix, using the same technique
3. Once this is done, you can hit "2nd" then "QUIT" to get a blank screen.
4. Call up Matrix A by typing MATRIX 1 (you should see [A])
5. Hit "x-1" to set up the inverse
6. Hit the "x" for multiply.
7. Hit MATRIX 2 to call up Matrix B
8. Hit the "x" for multiply
9. Hit MATRIX, scroll to MATH, then 1 to do a determinant
10. Hit MATRIX 1 to call up Matrix A. You should now have: [A]-1 * [B] *det [A]
11. Hit the ENTER key to get your result of [[ 4 ] [3 ] ]. The coefficient a is equal to 4, and the coefficent b is equal to 3!

To do the second part, calculating c:

1. Hit MATRIX, scroll to MATH, select 1 for the determinant
2. Hit MATRIX 1 to call up Matrix A, hit the ENTER key to get the result for c as 2

The balanced equation is now:

4 Cr + 3 O2 ---> 2 Cr2O3

### Nuclear Chemistry:

Nuclear chemistry deals with the nuclei of atoms breaking apart. Atoms are continually undergoing decay. When studying nuclear chemistry, there is a typical format used to represent specific isotopes.

mass number
238 $\rightarrow$ element
92
charge or atomic number

Nuclear equations are typically written in the format shown below. There are 5 different types of radioactive decay.

1. Alpha decay follows the form:

$^{X}_{Z}A\rightarrow ^{4}_{2}He+^{X}_{Z}^{-}_{-}^{4}_{2}B$

Where A is the parent isotope (the atom being broken apart) B is the daughter isotope or the isotope formed. When an element is broken down in alpha decay it looses two neutrons and two (2) protons. This means that the name of the element will change as well, moving back two (2) places on the periodic table. Alpha decay is not very penetrating because the He atoms capture electrons before traveling very far. However it is very damaging because the alpha particles can knock atoms off of molecules. Alpha decay is the most common in elements with an atomic number greater than 83.

2. Beta negative decay follows the form:

$^{X}_{Z}A\rightarrow _{-}_{1}^{0}e^{-1}+_{Z}_{+}_{1}^{X}B$

The beta emission increases the atomic number by one (1) by adding one (1) proton. At the same time, one (1) neutron is lost so the mass of the daughter isotope is the same as the parent isotope. Beta negative decay is more penetrating than alpha decay because the particles are smaller, but less penetrating than gamma decay. Beta electrons can penetrate through about one (1) cm of flesh before they are brought to a halt because of electrostatic forces. Beta decay is most common in elements with a high neutron to proton ratio.\

3. Gamma decay follows the form:

$^{X}_{Z}A\rightarrow _{0}^{0}\gamma +_{Z}^{X}B$

In gamma emission, neither the atomic number or the mass number is changed. A high energy gamma ray is given off when the parent isotope falls into a lower energy state. Gamma radiation is the most penetrating of all. These photons can pass through the body and cause damage by ionizing all the molecules in their way.

4. Positron emission (also called Beta positive decay) follows the form:

$_{Z}^{X}A\rightarrow ^{O}_{1}e+_{Z}_{-}_{1}^{X}B$

In this reaction a positron is emitted. A positron is exactly like an electron in mass and charge force except with a positive charge. It is formed when a proton breaks into a neutron with mass and neutral charge and this positron with no mass and the positive charge. Positron emission is most common in lighter elements with a low neutron to proton ratio.

5. Electron capture follows the form:

$_{Z}^{X}A+_{-}_{1}^{0}e\rightarrow _{Z}_{-}_{1}^{X}B$
electron particle

In this reaction a nucleus captures one (1) of its own atom's inner shell electrons which reduces the atomic number by one. This captured electron joins with a proton in the nucleus to form a neutron. Electron capture is common in larger elements with a low neutron to proton ratio.

All elements with an atomic number over 83 are considered radioactive. Radioactivity can be measured using a Geiger counter, a cylinder containing a low-pressure gas and two (2) electrodes. Radiation ionizes the atoms in the cylinder and allows current to flow between the electrodes.

All radioactive elements disintegrate according to their specific half life. The half life of a radioactive substance is the time required for half of the initial number of nuclei to disintegrate. The decay rate expresses the speed at which a substance disintegrates. The following equation represents the relationship between the number of nuclei remaining, N, the number of nuclei initially present, N0, the rate of decay, k, and the amount of time, t.

$In \frac{N}{N_{0}}=-kt$

The relationship between the half-life of a radioactive substance and k, the rate at which it decays can also be found.

$t\tfrac{1}{2}=\frac{.693}{k}$

### Isotopes and it's uses

Isotopes of Hydrogen:

1. Protium: It has on e proton and no neutron in its nucleus. Its atomic number and mass number both are 1. It is denoted by $_{1}^{1}H\,or\,P$
2. Deuterium: It has one proton and one neutron in the nucleus. Its atomic number is 1 and mass number is 2. It is denoted by $_{1}^{2}H\,or\,D$
3. Tritium: It has one proton and two neutrons in the nucleus. Its atomic number is 1 and mass number is 3. It is represented by $_{1}^{3}H\,or\,T$
Isotopes of Oxygen:

Oxygen has also got three isotopes. The atomic number of oxygen is 8. It means all the three isotopes of oxygen have 8 protons but they have 8, 9 and 10 neutrons in their nuclei having mass numbers 16, 17 and 18 respectively. They are represented as:

$_{8}^{16}O,\,\,^{17}_{8}O,\,^{18}_{8}O$

Isotopes of Carbon:

Carbon has three isotopes. the atomic number of carbon (C) is 6. However the mass numbers of C are 12, 13 and 14. It means all of them have 6 protons but the number of neutrons are 6, 7 and 8 respectively.

$_{6}^{12}C,\,\,^{13}_{6}C,\,^{14}_{6}C$

Isotopes of Uranium:

There are three common isotopes of uranium. Their atomic numbers are the same, i.e. 92 but their mass numbers are 234, 235 and 236 respectively. They are represented as

$_{92}^{234}U,\,\,^{235}_{92}U,\,^{236}_{92}U$

Isotopes of Chlorine:

Chlorine has two isotopes. Their masses are 35 and 37. Both the isotopes of chlorine have atomic number 17 it means that each chlorine atom has 17 protons. Cl-35 has18 neutrons and Cl-37 has 20 neutrons. They are are represented as:

$^{35}_{17}Cl,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,^{37}_{17}Cl,$

Uses of Isotopes:

1. They are used as tracers in physical, chemical and biological researches.
2. They are used in the diagnosis and treatment of diseases like cancer.