Thursday, 13 April 2017

Difference Between Electromotive force and Potential Difference

Electromotive force and potential difference are not same. They have following differences below:

Electromotive force

The potential difference between the two terminals of a cell is called electromotive force in an open circuit.
Electromotive force transmits current both inside and outside the cell.
Electromotive force emf is the cause.
Electromotive force is always greater than potential difference.
Electromotive force creates potential difference entire the circuit.
Electromotive force does not depend on the resistance of the circuit.
Electromotive force remains constant.
The part of the circuit where electrical energy is created from any other energy then that part contains the source of Electromotive force.

Potential difference

Bringing a unit positive charge from one point to another point in a circuit is called potential difference between two points.
Potential difference current transfers between any two points in the circuit.
Potential difference is the result.
Potential difference is always less than electromotive force.
Potential difference takes place between any two points in the circuit.
Potential difference of two points depends on the resistance of those points.
It does not remain constant.
Potential difference exists in the part of the circuit where electrical potential energy is transformed into another form of energy.

Friday, 7 April 2017

FSC Part 2 Date Sheet Karachi Board (BIEK)

Tuesday, 21 March 2017

Tuesday, 14 March 2017

Monday, 13 March 2017

Tuesday, 6 December 2016

SULPHURIC ACID (H2SO4)

On industrial scale, sulphuric acid can be prepared by the following two methods.
Contact process
Lead Chamber process

CONTACT PROCESS

Now a days, sulphuric acid is prepared by contact process all over the world.
Preparation of sulphuric acid by contact process is based upon the catalytic oxidation of SO2 to SO3.

DETAILS OF CONTACT PROCESS :

Following steps are involved in the preparation of H2SO4.
PREPARATION OF SO2.
PURIFICATION OF SO2.
OXIDATION OF SO2.
ABSORPTION OF SO3.
DILUTION OF OLEUM.

PREPARATION OF SO2

SO2 is obtained by burning sulphur or by heating iron pyrite (FeS2) in pyrite burner.

S + O2 → SO2

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

PURIFICATION OF SO2

SO2 contains a number of impurities such as dust particles, Arsenous oxide, vapours, sulphur etc. These impurities must be removed otherwise catalyst loses its efficiency (catalyst poisoning).

DUST CHAMBER:

SO2 is first passed through the dust chamber where steam is spread over the gas to remove dust particles, which settle down. Fe(OH)3 also sprayed over to remove oxides of Arsenic.

WASHING TOWER:

SO2 is then passed through a washing tower after cooling. Here it is sprayed by water to remove any other soluble impurities.

DRYING TOWER:

The gas is now dried by passing through drying tower where conc. H2SO4 (dehydrating agent) is sprayed. H2SO4 removes moisture from SO2.

ARSENIC PURIFIER:

Arsenic oxide is a poison for the catalyst. It is removed when the gas is passed over ferric hydroxide.

As2O3 + 2Fe(OH)3 → 2FeAsO3 + 3H2O.

In order to remove traces of As2O3, it is passed through a test box, where a strong beam of light is thrown against the gas. If there is no scattering of light in the box, it indicates that gas is free from As2O3.

OXIDATION OF SO2 TO SO3

CONTACT TOWER:
Oxidation of SO2 is carried out in contact tower where V2O5 is filled in different pipes. SO2 here reacts with air (O2) to produce SO3. Under above conditions 98% SO2 is converted into SO3.

2SO2 + O2 → 2SO3 + 45Kcal

CONDITIONS NECESSARY FOR MAXIMUM YIELD OF SO3:

Oxidation of SO2 is a reversible and exothermic process in which volume of product is less than the volumes of reactants. In order to obtain maximum amount of SO3, according to Le-Chatelier’s Principle following conditions are necessary.

CONCENTRATION:
Excess of O2.

TEMPEATURE:

A decrease in temperature favours reaction in forward direction. Optimum temperature for this process is 450oC to 500oC.

PRESSURE:

Since volumes of reactants are greater than the product (3:2), therefore, according to Le-Chatelier’s Principle a high pressure is favourable. Optimum pressure is about 1.5 to 1.7 atmosphere.

USE OF CATALYST:

At low temperature, rate of reaction decreases. To increase rate of reaction a catalyst vanadium pentaoxide (V2O5) is used.

ABSORPTION OF SO3 IN H2SO4

SO3 is not directly passed in water, because a dense fog of minute particles of H2SO4 is produced. It is therefore, dissolved in conc.H2SO4 to form pyrosulphuric acid (oleum).

SO3 + H2SO4 → H2S2O7 (OLEUM)

DILUTION OF OLEUM

Oleum is now diluted with water to form H2SO4 of required concentration.

H2S2O7 + H2O → 2H2SO4

THANKS FOR YOUR VIST TO
LOGICAL EDUCATION 4U

Wednesday, 30 November 2016

Ostwald process

The Ostwald process was developed by a man named Wilhelm Ostwald, after years of researching. It was created in 1902, patented in 1902, he then later was awarded the Nobel-peace prize for his work in 1909. Wilhelm Ostwald was born in Riga, Russian Empire to mother Elisabeth Leuckel and father Gottfried Wilhelm Ostwald (www.nobelprize.org, 1966).
This process was and still is a very important process because it is any easy way to create nitric acid in only two steps. Nitric acid, being used in so many things we don't really think about such as fertilizer and explosives...is at high demand. So the process the Wilhelm created is still being used today because it is reliable and the easiest way to create nitric acid for the high demand it is needed for.
The balanced equation for this reaction is different for each step:

Step 1:

4NH3 (g) + 5O2 (gas) → 4NO (gas) + 6H2O (gas)

In this step, ammonia is heated with oxygen. This yields nitric oxide (NO) and water as products.

Step 2:

2NO (gas) + O2 (gas) → 2NO2 (gas)

In this step the nitric oxide (NO) that was created in the first is combined with oxygen again to create nitrogen dioxide (NO2).

Step 3:

3NO2 (gas) + H2O (liquid) → 2HNO3 (aqueous solution) + NO (gas)

In this step, the nitrogen dioxide is absorbed with water to create nitric acid (HNO3) as an aqueous solution. This also yields nitric oxide (NO).

Step 4:

4NO2 (gas) + O2 (gas) + 2H2O (liquid) → 4HNO3 (aqueous solution)

For this last step, the nitric oxide created along with the nitric acid is recycled and combined with oxygen and water to create a higher concentration of nitric acid (HNO3).

This reaction is exothermic because during the process, it releases heat. (www.pem-news.de, n.d.)
Picture
In reference to LeChatelier's principle , since this reaction is exothermic...conditions that would favor the forward reaction and shift the equilibrium to the right would be decreasing the temperature, increase the concentration , and increasing the pressure and volume.
By decreasing the temperature, because this reaction is exothermic, the equilibrium will shift to the left away from the added energy. So for favorable conditions, we would want to decrease the temperature. By increasing the concentration, the equilibrium will shift away from the added product or reactant. Since we are constantly adding water and oxygen for this process as reactants, than increasing the concentration would create favorable conditions for the equilibrium. As for pressure and volume, the equilibrium would shift towards the side of the reaction with the last number of moles to help ease the pressure. So, taking step one's balanced equation as an example...the right side has less moles, so with increasing the pressure it would shift to the right. Which is favorable for the forward reaction.
Because of it's reaction when combined with organic compounds, most industrial probably don't use the most favorable conditions during the Ostwald process while creating because it will produce an unsafe concentration and corrosive tendencies within the nitric acid. This would be a safety hazard, which is actually why nitric acid is used in explosives.
The catalyst that is used for this reaction is a platinum gauze. It would be heated, however sometimes in substitute..a copper wire/rod can serve as a proper catalyst for this process (www.digipac.ca, n.d.).

Monday, 28 November 2016

HSC Part 1 Pre Engineering Result 2016 BIEK Karachi

HSC Part 1 Pre Engineering Result 2016 Karachi - BIEK Inter FA FSC ICOM Results - BIEK Karachi is one of the most popular educational boards of the country. BIEK Karachi is the regulatory body that aims to manage the high level intermediate education system in the affiliated region. Pre-engineering group of Intermediate is for those students who aim to build their future in Engineering. It serves as the building block of the student’s career. The students of Pre Engineering studies and gives examination of courses like Chemistry, Physics, and Mathematics. These tough courses are taught in colleges or high schools to regular students, while the private candidates prepare for examinations on their own.

Wednesday, 26 October 2016

DOWN'S PROCESS

INTRODUCTION:

J.C Down developed a process for the manufacture of sodium metal known as DOWN's Process.

PRINCIPLE:

The process based upon the principle of electrolysis of Aqeous sodium chloride solution

NaCl → 2Na + Cl2

RAW MATERIALS:

Fused sodium chloride is used as the electrolyte in the process.
As the sodium chloride is has the melting point of 801 degree celcius some amount of calcium chloride is added to lower the melting point to about 600 degree celcius which makes the process feasible.

PREPARATION OF SODIUM THROUGH DOWN'S PROCESS

Working:

When an electric current is passed through the molten mixture of NaCl and CaCl2, NaCl decomposes in to Na+ and Cl- ion. Na+ ions migrate towards cathode while Cl- ions towards the anode. The molten sodium collects in the cathode compartment where it rises to the top and is tapped off by a pipe. Chlorine is collected at the anode.

ELECTROCHEMICAL CHANGES

At cathode

Na+-ions migrate to cathode where they are reduced to Na.

2Na+ + 2e → 2Na (Reduction)

At anode

Cl--ions migrate to anode and oxidised to form chlorine gas.

2Cl- → Cl2 + 2e- (Oxidation)

Overall Reaction

2Na+ + 2e- → 2Na

2Cl- → Cl2 + 2e-

__________________

2Na+ + 2Cl- → 2Na + Cl2

During electrolysis calcium is also obtained at cathode but sodium and calcium are separated from each other due difference in density. Density of Na is 0.67gm/cc and the density of Ca is much higher than that of Na i.e. 2.54gm/cc. That's why they do not mix with each other.

LOGICAL EDUCATION 4U

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