Report Kimfis Lapindo

8/3/2019 Report Kimfis Lapindo

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Group 2

Ahmad Farizzi Artyani Batari Thomy Dj Vollmer

Faculty of Engineering

University of Indonesia

Analytical Chemistry

PBL II: Lapido Mud

FlowAtomic and Molecular Spectroscopy


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Table of Contents

PBL Mind Map ............................................................................................................ 1

Table of Contents ...................................................................................................... 2

Problem Definition .................................................................................................. 3

Theory .................................................................................................................................... 3

Lapindo Mud Flow ................................................................................................ 3

Analysis Method ................................................................................................................. 8

Infrared spectrometry ................................................................................................... 10

Answer of Questions .................................................................................... 19References............................................................................................................. 6


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PROBLEM DEFINITION

- Lapindo flow muds analysis using atomic and molecular spectroscopy- Atomic Absorption Spectrometry (AAS) and IR spectroscopy

THEORY

LAPINDO FLOW MUD

The Sidoarjo mud flow or Lapindo mud, also informally abbreviated as Lusiis a mud volcano in the sub

district ofPorong, Sidoarjo in East Java, Indonesia that has been in eruption since May 2006. This biggest

mud volcano in the world was created by the blowout of a natural gas well drilled by PT Lapindo

Brantas, although company officials contend that it was caused by a distant earthquake.

At its peak Lusi was spewing up to 180,000 m of mud per day. In mid August 2011, mud was being

discharged at a rate of 10,000 cubic meters per day, with 15 bubbles around the gush point. This was a

significant decline from a year previous, when mud was being discharged at a rate of 100,000 cubic

meters per day with 320 bubbles around the gush point. It is expected that the flow will continue for the

next 25 to 30 years. Although the Sidoarjo mud flow has been contained by levees since November

2008, resultant flooding regularly disrupts local highways and villages, and further breakouts of mud are

still possible
http://en.wikipedia.org/wiki/Poronghttp://en.wikipedia.org/wiki/Sidoarjohttp://en.wikipedia.org/wiki/East_Javahttp://en.wikipedia.org/wiki/Indonesiahttp://en.wikipedia.org/wiki/Blowout_(well_drilling)http://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/Leveehttp://en.wikipedia.org/wiki/Leveehttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/PT_Lapindo_Brantashttp://en.wikipedia.org/wiki/Blowout_(well_drilling)http://en.wikipedia.org/wiki/Indonesiahttp://en.wikipedia.org/wiki/East_Javahttp://en.wikipedia.org/wiki/Sidoarjohttp://en.wikipedia.org/wiki/Porong


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The content of lapindo flow mud

Test Results

Parameter Result marksStandart Quality

(PP no. 18/1999)

Arsen 0,045 Mg/L 5 Mg/L

Barium 1,066 Mg/L 100 Mg/L

Boron 5,097 Mg/L 500 Mg/L

Pb 0,05 Mg/L 5 Mg/L

Hg 0,004 Mg/L 0,2 Mg/L

Si 0,02 Mg/L 20 Mg/L

Trichlorophenol 0,017 Mg/L 2 Mg/L (2,4,6 Trichlorophenol)

400 Mg/L (2,4,4 Trichlorophenol)
http://id.wikipedia.org/wiki/Arsenhttp://id.wikipedia.org/wiki/Bariumhttp://id.wikipedia.org/wiki/Boronhttp://id.wikipedia.org/wiki/Timbalhttp://id.wikipedia.org/wiki/Raksahttp://id.wikipedia.org/wiki/Raksahttp://id.wikipedia.org/wiki/Timbalhttp://id.wikipedia.org/wiki/Boronhttp://id.wikipedia.org/wiki/Bariumhttp://id.wikipedia.org/wiki/Arsen


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Effect of lapindo flow mud

- decrease the gas sources- water pipe ofSurabayas PDAM was broke because of the surface is going down- 600 hectare of field was submerged- 16 Housing residents, schools and another public place were flooded- PERTAMINAs gas pipe was explode because of the high pressure mud flow- 30 factories are flooded with lusi- Etc

Cause of lapindo flow mud


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Mud eruption chronology

On May 28, 2006, PT Lapindo Brantas targeted gas in the Kujung Formation carbonates in the Brantas

PSC area by drilling a borehole named the 'Banjar-Panji 1 exploration well'. In the first stage of drilling

the drill string first went through a thick clay seam (5001,300 m deep), then sands, shales, volcanic

debris and finally into permeable carbonate rocks. At this stage the borehole was surrounded by a steel

casing to help stabilize it. At 5:00 a.m. local time (UTC+7) a second stage of drilling began and the drill

string went deeper, to about 2,834 m (9,298 ft), this time without a protective casing, after which water,

steam and a small amount of gas erupted at a location about 200 m southwest of the well. Two further

eruptions occurred on the second and the third of June about 8001000 m northwest of the well, but

these stopped on June 5, 2006. During these eruptions, gas was released and local villagers observed

hot mud, thought to be at a temperature of around 60 C (140 F).

A magnitude of 6.3 earthquake occurred in Yogyakarta at ~06:00 local time 27 May 2006, approximately

250 kilometers South West from Sidoarjo. Seven minutes after the earthquake a mud loss problem in

the well was noted. After two major aftershocks, the well suffered a complete loss of circulation. A loss

of circulation is when drilling mud that is pumped down a shaft does not return to the surface but is lost

into some opening or a fault system. This mud loss problem was finally stopped when a loss circulation

material was pumped into the well, a standard practice in drilling an oil and gas well. A day later the well

suffered a kick, an influx of formation fluid into the well bore. The kick appears to have been killed

within three hours. The next day, 29 May 2006, steam, water and mud began erupting 200 meters away

from the well, a phenomenon that is now known as the Lusi mud volcano.

Hypotheses on the possible causes

The birth of Lusi was a major disaster for the general population that lives nearby, the loss of their

houses, belongings as well as their livelihood. For the scientific community, however, it was a chance to

study the evolving geological process of a mud volcano. In the past, mud vulcanologists could only study

existing or ancient mud volcanoes during dormant periods. This is a rare occasion and a unique

opportunity to conduct scientific experiments to further our understanding. Lusi also offered

opportunities to study the down hole condition of a mud volcano from the neighboring Banjar-Panji

exploration well lithologies.

To explain what triggered the mud volcano, three hypotheses have been suggested, though none has

won universal support:

Hydro-fracturing of the formation, hence a drilling related problem

From a model developed by geologists working in the UK, the drilling pipe penetrated the overpressured

limestone, causing entrainment of mud by water. The influx of water to the well bore caused a

hydrofracture, but the steam and water did not enter the borehole; they penetrated the


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surrounding overburden and pressuredstrata. The extra pressure formed fractures around the borehole

that propagated 12 km to the surface and emerged 200 m away from the well. The most likely cause of

these hydraulic fractures was the unprotected drill string in the second stage of drilling. Though steel

casing is used to protect the well bore in oil or gas exploration, this can only be applied in stages after

each new section of the hole is drilled; see drilling for oil.

The relatively small distance, around 600 feet (180 m), between the Lusi mud volcano and the well being

drilled by Lapindo (the Banjarpanji well) may not be a coincidence, as less than a day before the start of

the mud flow the well suffered a kick. Their analysis suggests that the well has a low resistance to a

kick.Similarly, a NE-SW crack in the surface in the drill site may be evidence of an underground blowout.

The well may have suffered an underground blowout that resulted in a surface breach. Also, a likely

contributor is the dissociation of methane hydrates. These hydrates are formed when fresh water used

in fracking seeps into methane beds and at depth and pressure form crystal like rock which contain up

to 205 C (401 F) in each molecule. These hydrates cannot be formed in salt water. They are endemic

wherever drilling for oil and gas have used fresh water as part of the slurry that pressurizes wells on

land, permafrost, and under the earth crust along the continental shelves.

Solution of it

- Stop the flow mud with snubbing unit, it is use for pushing down the borehole which left, then itwill cover up the mud flow hole.

- Sidetracking to avoid the borehole which is left- putting some sort of heavy metal object shaped like balls to stop the eruption. It costs a fortune

and as it is nature, my personal opinion is that is not going to work. Its science: there is

pressure, there is a leak, you cover the leak, the pressure will go somewhere else.

Unfortunately, the only thing is to let the mother nature to reach its equilibrium.
http://en.wikipedia.org/wiki/Stratumhttp://en.wikipedia.org/wiki/Stratum


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ANALYSIS METHOD

Atomic Absorption Spectrhophotometry (AAS)

Introduction

Atomic-absorption (AA) spectroscopy uses the absorption of light to measure the concentration of gas-

phase atoms. Since samples are usually liquids or solids, the analyte atoms or ions must be vaporized in

a flame or graphite furnace. The atoms absorb ultraviolet or visible light and make transitions to higher

electronic energy levels. The analyte concentration is determined from the amount of absorption.

Applying the Beer-Lambert law directly in AA spectroscopy is difficult due to variations in the

atomization efficiency from the sample matrix, and nonuniformity of concentration and path length of

analyte atoms (in graphite furnace AA). Concentration measurements are usually determined from aworking curve after calibrating the instrument with standards of known concentration.

Schematic of an atomic-absorption experiment

Instrumentation

Light source

The light source is usually a hollow-cathode lamp of the element that is being measured. Lasers are also

used in research instruments. Since lasers are intense enough to excite atoms to higher energy levels,

they allow AA and atomic fluorescence measurements in a single instrument. The disadvantage of these

narrow-band light sources is that only one element is measurable at a time.
http://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lamps.htm#hollow-cathodehttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lasers.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/afs.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa-expt.gifhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/afs.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lasers.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/sources/lamps.htm#hollow-cathodehttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htm


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Atomizer

AA spectroscopy requires that the analyte atoms be in the gas phase. Ions or atoms in a sample must

undergo desolvation and vaporization in a high-temperature source such as a flame or graphite furnace.

Flame AA can only analyze solutions, while graphite furnace AA can accept solutions, slurries, or solid

samples.

Flame AA uses a slot type burner to increase the path length, and therefore to increase the total

absorbance (see Beer-Lambert law). Sample solutions are usually aspirated with the gas flow into a

nebulizing/mixing chamber to form small droplets before entering the flame.

The graphite furnace has several advantages over a flame. It is a much more efficient atomizer than a

flame and it can directly accept very small absolute quantities of sample. It also provides a reducing

environment for easily oxidized elements. Samples are placed directly in the graphite furnace and the

furnace is electrically heated in several steps to dry the sample, ash organic matter, and vaporize the

analyte atoms.

Light separation and detection

AA spectrometers use monochromators and detectors for uv and visible light. The main purpose of the

monochromator is to isolate the absorption line from background light due to interferences. Simple

dedicated AA instruments often replace the monochromator with a bandpass interference filter.

Photomultiplier tubes are the most common detectors for AA spectroscopy.

Picture of a flame atomic-absorption spectrometer:
http://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/optics/detector/pmt.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa.jpghttp://elchem.kaist.ac.kr/vt/chem-ed/optics/detector/pmt.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htm


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Infrared spectrometry

Theory

Introduction

The term "infra red" covers the range of the electromagnetic spectrum between 0.78 and 1000 mm. In

the context of infra red spectroscopy, wavelength is measured in "wavenumbers", which have the units

cm-1.

wavenumber = 1 / wavelength in centimeters

It is useful to divide the infra red region into three sections; near, midand farinfra red;

Region Wavelength range (mm) Wavenumber range (cm-1

)

Near 0.78 - 2.5 12800 4000

Middle 2.5 50 4000 200

Far 50 -1000 200 10

The most useful I.R. region lies between 4000 - 670cm-1.

Theory of infra red absorption

IR radiation does not have enough energy to induce electronic transitions as seen with UV. Absorption

of IR is restricted to compounds with small energy differences in the possible vibrational and rotational

states.

For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a net change in the

dipole moment of the molecule. The alternating electrical field of the radiation (remember that

electromagnetic radation consists of an oscillating electrical field and an oscillating magnetic field,

perpendicular to each other) interacts with fluctuations in the dipole moment of the molecule. If the

frequency of the radiation matches the vibrational frequency of the molecule then radiation will be

absorbed, causing a change in the amplitude of molecular vibration.


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Molecular vibrations

The positions of atoms in a molecules are not fixed; they are subject to a number of different vibrations.

Vibrations fall into the two main catagories ofstretching and bending.

Stretching: Change in inter-atomic distance along bond axis

Bending: Change in angle between two bonds. There are four types of bend:

Rocking Scissoring Wagging Twisting

Vibrational coupling

In addition to the vibrations mentioned above, interaction between vibrations can occur (coupling) if the

vibrating bonds are joined to a single, central atom. Vibrational coupling is influenced by a number of

factors;

Strong coupling of stretching vibrations occurs when there is a common atom between the twovibrating bonds

Coupling of bending vibrations occurs when there is a common bond between vibrating groups Coupling between a stretching vibration and a bending vibration occurs if the stretching bond is

one side of an angle varied by bending vibration Coupling is greatest when the coupled groups have approximately equal energies No coupling is seen between groups separated by two or more bonds


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Instrumentation

Sources

An inert solid is electrically heated to a temperature in the range 1500-2000 K. The heated material will

then emit infra red radiation.

The Nernst gloweris a cylinder (1-2 mm diameter, approximately 20 mm long) of rare earth oxides.

Platinum wires are sealed to the ends, and a current passed through the cylinder. The Nernst glower can

reach temperatures of 2200 K.

The Globar source is a silicon carbide rod (5mm diameter, 50mm long) which is electrically heated to

about 1500 K. Water cooling of the electrical contacts is needed to prevent arcing. The spectral output is

comparable with the Nernst glower, execept at short wavelengths (less than 5 mm) where it's output

becomes larger.

The incandescent wire source is a tightly wound coil of nichrome wire, electrically heated to 1100 K. It

produces a lower intensity of radiation than the Nernst or Globar sources, but has a longer working life.

Detectors

There are three catagories of detector;

Thermal Pyroelectric Photoconducting

Thermocouples consist of a pair of junctions of different metals; for example, two pieces of bismuth

fused to either end of a piece of antimony. The potential difference (voltage) between the junctions

changes according to the difference in temperature between the junctions

Pyroelectric detectors are made from a single crystalline wafer of a pyroelectric material, such as

triglycerine sulphate. The properties of a pyroelectric material are such that when an electric field is

applied across it, electric polarisation occurs (this happens in any dielectric material). In a pyroelectric

material, when the field is removed, the polarisation persists. The degree of polarisation is temperature

dependant. So, by sandwiching the pyroelectric material between two electrodes, a temperature

dependant capacitor is made. The heating effect of incident IR radiation causes a change in the

capacitance of the material. Pyroelectric detectors have a fast response time. They are used in most

Fourier transform IR instruments.

Photoelectric detectors such as the mercury cadmium telluride detector comprise a film of

semiconducting material deposited on a glass surface, sealed in an evacuated envelope. Absorption of IRpromotes nonconducting valence electrons to a higher, conducting, state. The electrical resistance of

the semiconductor decreases. These detectors have better response characteristics than pyroelectric

detectors and are used in FT-IR instruments - particularly in GC - FT-IR.


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IR Spectroscopy Tutorial: How to analyze IR spectra

Mono-functional molecules there are so many IR bands that it is not feasible to assign every band in an

IR spectrum. Instead, look for tell-tale bands -- the region from 4000-1300 cm-1 is particularly useful for

determining the presence of specific functional groups.

3500-3300 cm-1 NH stretch 1&Mac251;, 2&Mac251; amines

3500-3200 cm-1 OH stretch alcohols, a broad, strong band

3100-3000 cm-1 CH stretch alkenes

3000-2850 cm-1 CH stretch alkanes

1760-1665 cm-1 C=O stretch ketones, aldehydes, esters

1680-1640 cm-1 C=C stretch alkenes

looking in the region from 4000-1300. Look at the CH stretching bands around 3000:

Indicates:

Are any or all to the right of 3000? alkyl groups (present in most organic molecules)

Are any or all to the left of 3000? a C=C bond or aromatic group in the molecule

Look for a carbonyl in the region 1760-1690. If there is such a band:

Indicates:

Is an OH band also present? a carboxylic acid group

Is a CO band also present? an ester

Is an aldehydic CH band also present? an aldehyde

Is an NH band also present? an amide

Are none of the above present? a ketone

(also check the exact position of the carbonyl band for clues as to the type of carbonyl compound it is)


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Look for a broad OH band in the region 3500-3200 cm-1. If there is such a band:

Indicates:

Is an OH band present? an alcohol or phenol

Look for a single or double sharp NH band in the region 3400-3250 cm-1. If there is such a band:

Indicates:

Are there two bands? a primary amine

Is there only one band? a secondary amine

Other structural features to check for:

Indicates:

Are there CO stretches? an ether (or an ester if there is a carbonyl band too)

Is there a C=C stretching band? an alkene

Are there aromatic stretching bands? an aromatic

Is there a CC band? an alkyne

Are there -NO2 bands? a nitro compound

If there is an absence of major functional group bands in the region 4000-1300 cm-1 (other than CH

stretches), the compound is probably a strict hydrocarbon.

Also check the region from 900-650 cm-1. Aromatics, alkyl halides, carboxylic acids, amines, and amides

show moderate or strong absorption bands (bending vibrations) in this region.


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Group Frequencies

Typical Infrared Absorption Frequencies

Stretching Vibrations Bending Vibrations

Functional Class Range (cm-1

)Inten

sityAssignment

Range

(cm-1

)

Inten

sityAssignment

Alkanes 2850-3000 Str CH3, CH2 & CH

2 or 3 bands

1350-

1470

1370-

1390

720-

725

med

med

wk

CH2 &

CH3 deformati

on

CH3 deformati

on

CH2 rocking

Alkenes 3020-31001630-1680

1900-2000

medvar

str

=C-H & =CH2 (usuallysharp)

C=C (symmetry

reduces intensity)

C=C asymmetric

stretch

880-995

780-

850

675-

730

strmed

med

=C-H & =CH2(out-of-plane

bending)

cis-RCH=CHR

Alkynes 3300

2100-2250

str

var

C-H (usually sharp)

CC (symmetry

reduces intensity)

600-

700

str C-H

deformation

Arenes 3030

1600 & 1500

var

med-

wk

C-H (may be several

bands)

C=C (in ring) (2

bands)

(3 if conjugated)

690-

900

str-

med

C-H bending &

ring puckering

Alcohols &

Phenols

3580-3650

3200-3550

970-1250

var

str

str

O-H (free), usually

sharp

O-H (H-bonded),usually broad

C-O

1330-

1430

650-770

med

var-

wk

O-H bending

(in-plane)

O-H bend(out-of-plane)

Amines 3400-3500 (dil. soln.)

3300-3400 (dil. soln.)

wk

wk

N-H (1-amines), 2

bands

N-H (2-amines)

1550-

1650

660-

med-

str

NH2 scissoring

(1-amines)

NH2 & N-H
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir2http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir3http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4chttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4chttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir3http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir2


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1000-1250 med C-N 900 var wagging

(shifts on H-

bonding)

Aldehydes &

Ketones

2690-2840(2 bands)

1720-1740

1710-1720

1690

1675

1745

1780

med

str

str

str

str

str

str

C-H (aldehyde C-H)

C=O (saturated

aldehyde)

C=O (saturated

ketone)

aryl ketone

, -unsaturation

cyclopentanone

cyclobutanone

1350-

1360

1400-

1450

1100

str

str

med

-CH3 bending

-CH2 bending

C-C-C bending

Carboxylic

Acids&Derivative

s

2500-3300 (acids)

overlap C-H

1705-1720 (acids)

1210-1320 (acids)

1785-1815 (

acyl halides)

1750 & 1820

(anhydrides)

1040-1100

1735-1750

(esters)

1000-1300

1630-

1695(amides)

str

str

med-

str

str

str

str

str

strstr

O-H (very broad)

C=O (H-bonded)

O-C (sometimes 2-

peaks)

C=O

C=O (2-bands)

O-C

C=O

O-C (2-bands)C=O (amide I band)

1395-

1440

1590-1650

1500-

1560

med

med

med

C-O-H bending

N-H (1-

amide) II bandN-H (2-

amide) II band

Nitriles

Isocyanates,Isothi

ocyanates,

Diimides, Azides

& Ketenes

2240-2260

2100-2270

med

med

CN (sharp)

-N=C=O, -N=C=S

-N=C=N-, -N3, C=C=O
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir4bhttp://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed/irspec1.htm#ir5


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INTERPRETING AN INFRA-RED SPECTRUM

Example :

The infra-red spectrum for a simple carboxylic acid

Ethanoic acid

Ethanoic acid has the structure:

You will see that it contains the following bonds:

carbon-oxygen double, C=O

carbon-oxygen single, C-O

oxygen-hydrogen, O-H

carbon-hydrogen, C-H

carbon-carbon single, C-C

The carbon-carbon bond has absorptions which occur over a wide range of wavenumbers in the


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fingerprint region - that makes it very difficult to pick out on an infra-red spectrum.

The carbon-oxygen single bond also has an absorbtion in the fingerprint region, varying between

1000 and 1300 cm-1depending on the molecule it is in. You have to be very wary about picking out

a particular trough as being due to a C-O bond.

The other bonds in ethanoic acid have easily recognised absorptions outside the fingerprint region.

The C-H bond (where the hydrogen is attached to a carbon which is singly-bonded to everything

else) absorbs somewhere in the range from 2853 - 2962 cm-1. Because that bond is present in most

organic compounds, that's not terribly useful! What it means is that you can ignore a trough just

under 3000 cm-1, because that is probably just due to C-H bonds.

The carbon-oxygen double bond, C=O, is one of the really useful absorptions, found in the range

1680 - 1750 cm-1. Its position varies slightly depending on what sort of compound it is in.

The other really useful bond is the O-H bond. This absorbs differently depending on its

environment. It is easily recognised in an acid because it produces a very broad trough in the range2500 - 3300 cm-1.

The infra-red spectrum for ethanoic acid looks like this:

The possible absorption due to the C-O single bond is queried because it lies in the fingerprint

region. You couldn't be sure that this trough wasn't caused by something else.


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SOLUTION

Assignment 1 :

1. Are the above-ground basins built to containment of the mud a sustainable solution? How doyou think to solve this as a long term solution?

Answer:

No, because it only can make the lapindo mud not expand to another area and its not a best or

permanent way, that because the above-ground basins can collapse.

Geologist: just wait until it reaches the equilibrium by itself. Equilibrium will happen if the

ground runs out the overpressure gas or when theres ground erosion so the ground will cover

up the hole of overpressure gas.

2. What do you think will happen if releasing the mud into an aquatic environment?Answer:

It would dangerous for aquatic biota. Because the heavy metals or another dangerous metals

will poisoning the aquatic biota. Besides that it will contaminated water. And we all know that

we need water in our life every day and every activity.

3. Will you describe any organic and inorganic might be found in the mud?Answer:

Test Results

Parameter Result marksStandart Quality

(PP no. 18/1999)

Arsenic 0,045 Mg/L 5 Mg/L

Ba 1,066 Mg/L 100 Mg/L

B 5,097 Mg/L 500 Mg/L

Pb 0,05 Mg/L 5 Mg/L

Hg 0,004 Mg/L 0,2 Mg/L

FreeSi 0,02 Mg/L 20 Mg/L

Trichlorophenol 0,017 Mg/L 2 Mg/L (2,4,6 Trichlorophenol)

400 Mg/L (2,4,4 Trichlorophenol)
http://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianidahttp://id.wikipedia.org/wiki/Sianida


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4. What analysis methods can be used for analyzing heavy metals and organic compounds inpolluted soil such as lapindo mud?

Answer:

The best analysis method for analyzing heavy metals and organic compounds in polluted soils

(such as Lapido Muds) are the Atomic and Molecular Spectroscopy, which uses the absorption of

light to measure the concentration of gas-phase atoms. What makes it accurate is, that the

liquids or solids have to be vaporized and then analyzed in a flame or graphite furnace.

Example:

- Atomic-Absorption for Atomic Spectroscopy

- IR Spectra for Molecular Spectroscopy

Assigment 2:

1. What you know about Atomic Absorption Specthrophotometry (AAS)? Can you describe thefundamental theory and instrumentation in this method of analysis?

Answer:

Atomic-absorption (AA) spectroscopy uses the absorption of light to measure the concentration

of gas-phase atoms. Since samples are usually liquids or solids, the analyte atoms or ions must

be vaporized in a flame or graphite furnace. The atoms absorb ultraviolet or visible light and

make transitions to higher electronic energy levels. The analyte concentration is determined

from the amount of absorption.

Applying the Beer-Lambert law directly in AA spectroscopy is difficult due to variations in the

atomization efficiency from the sample matrix, and nonuniformity of concentration and pathlength of analyte atoms (in graphite furnace AA). Concentration measurements are usually

determined from a working curve after calibrating the instrument with standards of known

concentration.

2. How do you describe the procedure for performing analysis of heavy metal in soil/mud usingAAS?

Answer:

In order to analyze a sample for its atomic constituents, it has to be atomized. The atomizersmost commonly used nowadays are flames and electrothermal (graphite tube) atomizers. The

atoms should then be irradiated by optical radiation, and the radiation source could be an

element-specific line radiation source or a continuum radiation source. The radiation then

passes through a monochromator in order to separate the element-specific radiation from any

other radiation emitted by the radiation source, which is finally measured by a detector.
http://elchem.kaist.ac.kr/vt/chem-ed/quantum/at-lvls.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://en.wikipedia.org/wiki/Monochromatorhttp://en.wikipedia.org/wiki/Monochromatorhttp://elchem.kaist.ac.kr/vt/chem-ed/data/wcurve.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htmhttp://elchem.kaist.ac.kr/vt/chem-ed/quantum/at-lvls.htm


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3. After running series of standard chromium solution and the unknown sample, the following datais collected. You are preparing a callibation curve by plotting absorbance versus concentration

for standard solutions. How do you calculate the concentration of Cr in mud, if unknown sample

is obtained from 2 gram of dry mud after you dried it in oven at 105 C? it is dissolved in 100ml

water

chromium (ppm) absorption

0.1 0.006

0.5 0.033

1 0.064

2 0.121

3 0.174

4 0.232

Answer:

unknown sample absorption = 0.105

0.105 = x 0.0572 + 0.039
http://elchem.kaist.ac.kr/vt/chem-ed/spec/atomic/graphics/aa-expt.gif


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x = 1.77 ppm/100ml

Assignment 3:

1. How do you describe the fundamental theory and instrumentation for IR Spectroscopy analysis?Answer:- Theory

IR radiation does not have enough energy to induce electronic transitions as seen with

UV. Absorption of IR is restricted to compounds with small energy differences in the

possible vibrational and rotational states.

For a molecule to absorb IR, the vibrations or rotations within a molecule must cause a

net change in the dipole moment of the molecule. The alternating electrical field of the

radiation (remember that electromagnetic radation consists of an oscillating electrical

field and an oscillating magnetic field, perpendicular to each other) interacts with

fluctuations in the dipole moment of the molecule. If the frequency of the radiationmatches the vibrational frequency of the molecule then radiation will be absorbed,

causing a change in the amplitude of molecular vibration.

----

y = 0.0572x + 0.0039

0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5

absorption

absorption

Linear (absorption)


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- InstrumentationSources

An inert solid is electrically heated to a temperature in the range 1500-2000 K. The

heated material will then emit infra red radiation.

The Nernst gloweris a cylinder (1-2 mm diameter, approximately 20 mm long) of rare

earth oxides. Platinum wires are sealed to the ends, and a current passed through the

cylinder. The Nernst glower can reach temperatures of 2200 K.

The Globar source is a silicon carbide rod (5mm diameter, 50mm long) which is

electrically heated to about 1500 K. Water cooling of the electrical contacts is needed to

prevent arcing. The spectral output is comparable with the Nernst glower, execept at

short wavelengths (less than 5 mm) where it's output becomes larger.

The incandescent wire source is a tightly wound coil of nichrome wire, electrically

heated to 1100 K. It produces a lower intensity of radiation than the Nernst or Globar

sources, but has a longer working life.

Detectors

There are three catagories of detector;

Thermal

Pyroelectric

Photoconducting

2. What is the characteristic of functional group absorption peaks of phenol spectra. If you havepropanoic acid with OH and CO functional groups will the IR spectra be different from IR spectra

of phenol?

Answer:

Phenol, also known as carbolic acid, phenic acid, is an organic compound with the

chemical formula C6H5OH. It is a white crystalline solid. The molecule consists of a phenyl (-

C6H5), bonded to a hydroxyl (-OH) group.

Propanoic acid (from 'propane', and also known as propionic acid) is a naturallyoccurring carboxylic acid conneted to (-OH) group with chemical formula CH3CH2COOH.

Alcohol/Phenol O-H Stretch 3550 - 3200 (broad, s)

Carboxylic Acid O-H Stretch 3000 - 2500 (broad, v)


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3. If you have phenol in mud, what sample preparation techniques can be applied for IRspectroscopy analysis?

Answer:

First you have to get the phenol in the mud as pure, which you can achieve by the

Purification Method, and then you can get the pure evaporated phenol.

Another method would be the other way round, just to freeze the mud, because of the different

Melting Points; there will be substances that will freeze earlier and some that will freeze later, in

this steps we can separate the Phenol from the mud and so have the prepared.

If you have solids you have heat it up with a heat of 1500K-2000K, this melting then is ready to

be analyzed.

4. The following is the IR spectra of compound with the molecular formula of C6H12O, can yougive prediction what compound is? Give explanation based on functional group absorption

peaks.

Answer:

3342 is O-H as the functional Group Alcohol

2935 and 2858 are double peaks which combine =C-H as the functional Group Alkana

1453 isC-H as the functional Group Alkane

1070 is C-O as the functional Group alcohol

971 is =C-H as the functional Group Alkene


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REFERENCES

http://www.wpi.edu/Academics/Depts/Chemistry/Courses/General/infrared.html

http://www.atsdr.cdc.gov/toxprofiles/tp17-c7.pdf

http://orgchem.colorado.edu/hndbksupport/irtutor/analyzeir.htmlhttp://www.chem.ucla.edu/~webspectra/irtable.html

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