Explosive chemical reactions

Introduction :
Chemical reactions that releases energy are known as exothermic reactions.

Case I: When the reaction proceeds slowly, released energy will be dissipated smoothly and there will be few noticeable effects other than an increase in temperature.

Case II: On the other hand, when the reaction proceeds very rapidly, the energy will not be dissipated smoothly. A huge amount of energy can be deposited into a relatively small volume of atmosphere, then manifest itself by a rapid expansion of hot gases, which in turn can create a shock wave or propel fragments outwards at high speed.
There are three primary fields of application for these chemcial explosions: propellants, explosives and pyrotechnics.

Propellants works when they create a high gas pressure for moving projectiles or rockets and for similar uses.

Explosives works when they create a disruption of solid or liquid bodies, as in construction, mining or warfare.

Pyrotechnics works when they have effects that are mainly sound and light, but include many other varied applications, mainly on a small scale.

Fireworks are used as an application for entertainment, a show of light, noise and motion.

Some more examples:

Black Powder: Black powder was invented as a pyrotechnic substance, then it was applied as a propellant in firearms, and finally used in engineering and mining. The history of black powder and firearms relates to Cannon.

Smokeless Powder: Vielle discovered how to make a propellant from cellulose nitrate in 1886. The work was started with low-nitrogen guncotton, or pyrocotton, with 11%-12% of nitrogen, and plasticized it with ether and alcohol. Pyrocotton will dissolve completely in this solvent. This gel was rolled out into sheets and then the sheets were broken up into powder, after this the powder formed into grains, and these grains, mixed with various additives to control the rate of burning, chemical properties and stability in storage, made a propellant called smokeless powder that could replace gunpowder, and was more powerful.

Aromatic Explosives: One of the first aromatic explosives was picric acid, or trinitrophenol, C6H2(NO2)3OH. This particular explosive was first prepared in 1771 by Woulfe as a dye, and was also used in medicine, long before it was first employed as an explosive in 1830. Name was so given because of it's extremely sharp or bitter taste, and also in Greek word it means pikros, "sharp." It forms pale yellow crystals of density 1.76 g/cc, melting at 122°C and exploding above 300°C. It is too sensitive to heat to be poured into shells, and must be press-loaded, meanwhile another effect of it is corrosion of metals, forming sensitive picrates. The most famous aromatic explosive is trinitrotoluene, called TNT for short. TNT is deficient in oxygen, so makes a cloud of black smoke. It is a popular bursting charge for shells and bombs, replacing picric acid after World War I.

Atomic number 81

 Introduction :

• Atomic number 81 belong to P-block elements.
• Atomic number 81 is Thallium and chemical symbol is ‘Tl’ from periodic table.
• Thallium belongs to Group13 and period 6.
• General electronic configuration of p-block element is [Rare gas] nS² np^{1 to 6}
• Thallium has atomic number 81 and mass number 204.383 The data is obtained from the periodic table.
• Atomic number 81 was found in iron pyrites, crookesite, hutchinsonite, and lorandite.  It is obtained in the by-product of zinc and lead smelting.
• Electronic configuration of Thallium:
• 1S², 2S², 2P⁶, 3S², 3P⁶, 3d¹⁰, 4S², 4P⁶, 4d¹⁰, 4f¹⁴, 5s², 5p⁶, 5d¹⁰, 6S², 6P¹
• Electron per energy level: 2, 8, 18, 32, 18,3
• Number of Electrons (with no charge): 81
• Number of Neutrons (most common/stable nuclide): 123
• Number of Protons: 81
• Oxidation States: 3,1
• Crystal Structure of Atomic number 81 is Hexagonal.
• Density (293 K) of thallium is 11.85 g/cm³.
• In Greek thallos mean green twig, representation for bright green line in its spectrum.
• Thallium is a Soft gray metal that looks like lead.
• Sir William Crookes discovered Thallium in the year 1861 in England.
• Thallium belongs to metal group.

Thallium is very soft and malleable and at room temperature, it can be cut with a knife.  Thallium has a metallic luster, but by exposing to air, it quickly diminishes with a bluish-gray tinge that resembles lead. It is preserved by keeping it under oil.

Image of Thallium metal: Appearance of thallium metal is silvery white colour.
                                            


Properties of Atomic number 81

• Atomic radius and ionic radius of Group 13 elements: Atomic radius and ionic radius increases down the group from boron to thallium

Elements	Boron	Aluminum	Gallium	Indium	Thallium
Atomic radius (pm)	85	121	135	155	190
Ionic radius (pm)	41	53.5	76	94	102.5

• Ionization potential of Thallium:

      First Ionization potential: 6.1083 eV.
      Second Ionization potential: 20.428 eV.
      Third Ionization potential: 29.829 eV.

• Oxidation states of Atomic number 81: Group 13 elements exhibit oxidation state of +3.  Thallium exhibit oxidation state of +1 and +3.  It is exhibited when ns², np¹ electrons are involved in bonding.

                   Tl       : [Xe] 4f¹⁴, 5d¹⁰, 6s², 6p¹
                            Tl¹⁺     : [Xe] 4f¹⁴, 5d¹⁰, 6s², 6p⁰
                   Tl³⁺     :  [Xe] 4f¹⁴, 5d¹⁰, 6s⁰, 6p⁰

• Inert pair effect of Thallium: In Inert pair effect, the outermost s electrons to remain no ionized or unshared in compounds of post-transition metals (or p-block elements). The term inert pair effect is frequently used in relation to the increasing stability of oxidation states that are 2 less than the group valence for the elements of groups 13, 14, 15 and 16. The term "inert pair" was first proposed by Nevil Sidgwick proposed the term "inert pair" in 1927. As an example in group 13 the Tl has+1 oxidation state and it is the most stable one and Tl^III compounds are comparatively less. The stability of Group 13 elements is given in the order,

                                        Al^I < Ga^I < In^I < Tl^I.

• Melting point (M.P.) and boiling points (B.P.) of Group 13 elements: Melting point depends on the size of the atom.  Smaller the atomic size, higher is the meting point. Boiling point decreases from boron to thallium. 

Elements	Boron	Aluminum	Gallium	Indium	Thallium
M.P.  (0C)	4275	2740	2475	2350	1745
B.P.  (0C)	2300	933.25	302.9	429.75	577

• Isotopes of Thallium:  There are 25 isotopes in thallium.  Atomic masses ranges from 184 to 210. Stable isotopes are only ²⁰³Tl and ²⁰⁵Tl .  ²⁰⁴Tl is the most stable radioisotope which is having a half-life of 3.78 years.

Uses of Thallium

• Thallium sulphate is odorless and tasteless and was once widely used as rat poison and ant killer. Since 1972 it is prohibited.
• To treat ringworm, other skin infections and to reduce the night sweating of tuberculosis patients, thallium salts were used. However it is limited due to their narrow therapeutic index.

Chemical reaction of Thallium

• Chemical reaction of thallium with air: When thallium metal is heated to red hot in the presence of air, thallium (1) oxide which is poisonous is formed.

                 2Tl _(s) + O_{2 (g)} → Tl₂O _(s)

• Chemical reaction of thallium with water: When Thallium metal is exposed to moist air, it tarnishes slowly and then it dissolves in water to form thallium (1) hydroxide which is poisonous is formed.

                2Tl _(s) + 2H₂O _(l) → 2Tl (OH) _(aq) + H_{2 (g)} ↑

• Chemical reaction of thallium with halogens: Thallium metal reacts rapidly with halogen to form dihalides.  Thallium (111) fluoride, thallium (111) chloride and thallium (111) bromide are formed.  All these are poisonous.

                                        2Tl(s) + 3F₂ (g) → 2TlF₃(s)
                                        2Tl(s) + 3Cl₂ (g) → 2TlCl₃(s)
                                        2Tl(s) + 3Br₂ (l) → 2TlBr₃(s)

• Reaction of thallium with acids: Thallium reacts with sulphuric acid and hydrochloric acid slowly

Calculating percent yield

The predicted yield is determined by the masses used in a reaction and the mole ratios in the balanced equation. This predicted yield is the "ideal". It is not always possible to get this amount of product. Reactions are not always simple. There often are competing reactions. 

 For example, if you burn carbon in air you can get carbon dioxide and carbon monoxide formed. The two reactions occur simultaneously. Some carbon atoms end up in CO and others end up in CO2. The typical calculation in a starting class assumes that there is only one path for the reactants. This is an over simplification.You know for example from real life that food is not always converted to energy. If you eat a cookie, some of it could end up stored as "fat" Ugh!

Chemists, like all other people, aren't perfect. When a chemist does a synthesis, she will end up creating less product than expected because of spills, incomplete reactions, incomplete separations, or a dozen other reasons. The percent yield is a way of measuring how successful a reaction has been.

To compute the percent yield, figure out how much product you should have made by using basic stoichiometry. (Note: this may involve a limiting reagent problem.) Then simply divide the amount of stuff you did form by the expected amount and multiply by 100%. If you get a number > 100%, you've made a serious error someplace.

Obviously, you want a high percent yield: if you have a ten step synthesis where the product from one reaction ends up as the reactants for the next and each synthesis has 90% yield, you'll end up with only ~35% yield for the overall reaction.

Example: You burn 10.0 grams of methane in an excess of oxygen and form 19.8 grams of water. What was your percent yield?

Solution : First, you need to find out how much product you would expect to make using basic stoichiometry. The reaction of methane with oxygen is shown below

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)

You start with 10.0 grams of methane, which has a molecular weight of 16.04 g/mole, so you have 10.0 g/16.04 g/mole = 0.623 moles of methane.

The ratio between methane and water is 2 water for every 1 methane, so you expect to form

0.623 moles CH4 * 2 moles H2O/1 mole CH4 1.25 moles H2O

Now convert back to grams: the mole weight of water is 18.02 g/mole, so you should form

1.25 moles H2O * 18.02 g/mole = 22.5 g water

If the reaction had gone perfectly. You only formed 19.8 however, so your percent yield is

% yield = mass created/mass expected * 100%

% yield = 19.8 g/22.5 g = 88.0%

Standard gas volume

Introduction :
Each gas is characterized by the property of volume.  Volume is a space occupied by a gas. Volume is dependent on the temperature.  More the temperature more would be the volume.  Standard gas volume is defined as volume occupied by a gas at standard temperature and pressure.

 What are the figures for standard pressure and temperature?  The figures of standard pressure and temperature are adopted at 1 atmosphere and 273 degree Kelvin.  For the gas loss, the value of the units of pressure and temperature are always taken as atmosphere and degree kelvin.  The volume occupied by a gas at these standard conditions is called as standard volume.

Equations to find standard gas volume

Boyle established the relationship between pressure and the volume.Charles established the relation between pressure and temperature.
From both of these equations the universal gas law equation was derived and from the same ideal gas law equation was derived.
The ideal gas law equation is PV =nRT, here 'P' is pressure, 'V' is volume, 'n' is number of moles, 'R' is the gas constant and 'T' is temperature.  If the values for standard pressure and temperature are substituted at 1 atmosphere and 273 degree kelvin, the volume occupied by 1 mole of a gas comes to 22.4 liters.
Thus 22.4 liters is a standard volume of 1 mole of any gas.  This equation was derived by Avogadro. He quantified the volume occupied by 1 mole of gas at standard conditions

Illustration to show standard gas volume

The standard volume has been useful to find many other things in practical chemistry.Let us consider following problems.

Problem 1:
                   Find volume occupied by 64 grams of oxygen at STP
                                                       Or
                       Find the standard volume of 64 gms of oxygen.

Answer:  64 no. of moles of oxygen is = 64 / 32 = 2.
               If 1 mole occupies 22.4 litres, 2 moles would occupy 44.8 liters.

Problem 2:
                   Find the moles of the gas if the standard volume is 100 liters.

Answer: 1 mole occupies 22.4 liters,
             so 100 liters is occupied by 100 / 22.4 = 4.464 moles.

Percent volume concentration

Introduction :
Concentration of a solution depends upon the amount of solute present in the solution.  The concentration of the solution is expressed in many ways.  One of the important way of expressing the concentration is percent volume concentration.  Other means of expressing the concentration are mass percent which is nothing but the mass of the solute present in the solution expressed as percentage that is (mass of the solute/mass of solution) x 100.  Another way of expressing the concentration is by molarity it is the mole of solute present in a liter of the solvent it is expressed as mol/litre.  Yet another way of expressing the concentration is by molefraction is the fraction of moles that is present in the whole of the solution.  That is if two component A and B are present in the solution.  the mole fraction of A is the number of moles of A/moles of A + moles of B.

Concentration as volume percent:

In most of the liquid samples the concentration is expressed by volume percent
volume present = (volume of the solute/volume of the solution) x 100.  This is the easiest method of expressing the concentration of a liquid solution.
For example if you have 10%volume/volume of antifreeze that is ethylene glylcol in water it means that there is 10mL of ethylene glycol in 100mL of the soluiton or 10L of ethylene glycol in 100L of the solution

Problems on percent volume concentration:

Find out the volume percentage of ethanol when 10mL of ethanol is mixed in 190mL of the water?
Soluiton:

Volume of the solute that is ethanol is 10mL

volume of the solvent that is water is 190mL

total volume of the solution = volume of the solute + volume of the solvent = 10mL + 190Ml= 200mL
So volume percentage =(volume of solute  / volume of solution) x 100
= (10 / 200) x 100 = 5% v/V

Iupac nomenclature for organic compounds

IUPAC Rules for Nomenclature:

(i).Root words:

C1 -Meth, C2 -Eth, C3-Prop ,C4-Buta, C5- Penta, C6- Hexa, C7- Hepta,C8- Octa,C9-Nona,C10-Deca

(ii).Primary suffix:

                               Alkane   - C-C-       -ane
                               Alkene     -C=C-       -ene
                             alkyne       -C=_C-      -yne
1.Longest sum rule:
the longest continuous chain of carbon atoms is considered for naming the carbon compound, the prefix in the name of the compound depends on the number of carbons present in the compound.

2.Least Sum Rule:
In numbering the carbon atoms in the parent chain start at the end which  results in the use of lowest number for the substituent carbon atoms.

3.When identical substituent are present on the same carbon atom,the position of the substituent is repeated

4.if the carbon chain contain 2 or more than 2 identical substituent,then di- tri- tetra- words are used to indicate the number of substituents.

5.If the number of different groups are attached to the parent the chain ,then the naming is done in two ways:
(i). according to the IUPAC rules
(ii) the name is given to the compound following the increase in order of complexity of the group

6.when a carbon compound  has carbon chain having the same no.of carbon atoms,they having more number of branches is selected as the parent chain

7 If a carbon chain contains substituents which are at equal distance either from left are right side counting is done in such a way that the least number is given to substituens

8 When the main carbon chain in a carbon compound carries a branch which again contains a substituent,the position of the substituent in the branch is written in the brackets.

IUPAC Rules for naming Poly functional compounds:

A molecule can contain more than one functional group called as polyfunctional group,the functional group which specifies its class is called the principle functionl group the other functional groups are refered to as ' substituents'

Nomenclature Priority for determining the principle functional group.Highest priority at the top.
suffixes for functional groups:
Caboxylic acid     -COOH   (  -oic acid )
sulphonic acid    -SO3H  (sulphonic acid)
Ester                     - COOR ( -alkyl -  oate)
Acid halide           -COX  ( -oyl chloride)
amide                  -CO NH2 ( -amide)
nitrile                  -CN (-nitrile)
aldehyde         -CHO  (-al)
ketone             -CO-   (-one)
alcohol          -OH ( -ol)
amine             - NH2  (-amine)
ethers            -O-  (ether)
alkene          -C=C- (ene)
alkyne         -C=_C-   (-yne)


1.Identify principal functional group.this gives class name of  the structure.

2. Number the longest chain containing the principal functional group from the end closer to it.

3.Write the parent name corresponding to the no.of cabons in the longest chain.

4 Arrange the substituents names with position numbers in alphabetical order.

5.Prefix substituent name with the parent name.

6.The following functional groups are always named as substituents ,their names prefixed with the parent name.
ex:Cl - chloro; Br - bromo ;I -iodo ;F-flouro ;CN - cyano ; R - alkyl, OR - alkoxy ; NH2 -amine;NO2 -nitro  etc..

IUPAC Nomenclature rules with example:

7.Identify double/triple bonds. Number them with the number of the carbon atom at the head of the bond (i.e the carbon atom with the lesser number that it is attached to). For example a double bond between carbon atoms 3 and 4 is numbered as 3-ene. Multiple bonds of one type (double/triple) are named with a prefix (di-, tri-, etc.). If both types of bonds exist, then use "ene" before "yne" e.g. "6 13 diene 19 yne". If all bonds are single, use "ane" without any numbers or prefixes.

8. Arrange everything like in  this way: Group of side chains and secondary functional groups with numbers made in step 3 + prefix of parent hydrocarbon chain (eth, meth) + double/triple bonds with numbers (or "ane") + primary functional group suffix with numbers.
      Wherever it says "with numbers", it is understood that between the word and the numbers, you use the prefix(di-, tri-)

   9. Add punctuation:
         1. Put commas between numbers (2 5 5 becomes 2,5,5)
         2. Put a hyphen between a number and a letter (2 5 5 trimethylhexane becomes 2,5,5-trimethylhexane)
         3. Successive words are merged into one wordform (trimethyl hexane becomes trimethylhexane)
            NOTE: IUPAC uses one-word names throughout. This is why all parts are connected.

Ex:  CH3-CH(OH)-CH2-CH(CH3)-COOH

         5       4             3         2               1

From priority order -COOH  have higher rank ,therefore -COOH is the principal functional group.the structure is named as a carboxylic acid.
Longest chain : 5 carbons
so,parent name = pentanoic acid
Order of substituent 4-hydroxy -2-methyl
therefore,Name of the compound : 4-Hydroxy-2-methyl pentanoic acid.

Sound wave absorption

 Introduction :                                            
Sound waves travel in air as progressive longitudinal waves. Elasticity and inertia of the air  enable the sound wave to propagate with certain velocity. Sound cannot be transmitted through vacuum. It can travel through any solid, liquid and gas. All the frequencies of the vibrating bodies can not produce the sensation of hearing.

Sound  is a type of energy propagated in longitudinal waves. The source of sound could be any body which is vibrating. Some of the  examples are  a tuning fork which is excited, the wire of a stringed instrument being plucked , a bell which is struck with a hard thing,  etc.  When a sound wave is incident on any surface, a part of  the incident energy is always absorbed. Absorbption of  sound energy  vary  with different substances.

Thick screens or curtains, mats, carpets, wood, card boards are some of the examples of sound absorbers. Human bodies are very good absorbers of sound.  Best absorbers are those which absorb sound completely. Open windows and doors are therefore perfect absorbers. The characteristic absorption of a surface can be different at different frequencies.

Absorption coefficient of sound wave:

Absorption coefficient :  The absorption coefficient of a surface is defined as the ratio of the sound energy absorbed by the surface to the sound energy absorbed by an open window of equal area in the same time.

If Es  and Ew  are the amounts of sound energies absorbed by a given surface and an open window of the same area during same time, then

              The absorption coefficient        a     =    `(Es)/(Ew)`

Thus if  'a'  is the absorption coefficient of a surface and s is the surface area, the sound energy absorbed by it is given by  A  =  as.  Absorption coefficient 'a' has no unit but the SI unit of 'as' is metric sabin.

The absorption coefficients of certainsubstnces at a frequency of 512 Hz are given below.

S.No          Substance             Absorption Coefficient

1.               Marble                         0.01

2.               Glass                            0.028

3.               Carpet                          0.2

4.               Heavy Curtains             0.52

5.                Fibre glass                   0.69

6.              Open window                1.00