SCHOOL OF CHEMICAL & BIOMEDICAL ENGINEERINGCHEMICAL & BIOMOLECULAR ENGINEERING

CHEMICAL/ BIOMOLECULAR ENGINEERING LABORATORY 2

Year 1: Material/ Energy Balance for a CombustionProcess at Steady State

Name: Muhammad Ilyas Hassan

Matriculation Number: U1322800G

Group: 6

Date of Experiment: 24th Feb 2014

1. Aim

The objective of the experiment is to derive the material and energy balance formula for a steady-state combustion process by calculating the air excess ratio, stoichiometry, higher heating value, lower heating value, etc and thus, understand the concept of a steady state combustion process and the technique of Gas Chromatography.

2. Abstract

The traditional combustion process is heavily used today for to drive many purposes and processes mainly due to its energy output. In this combustion process, a substance with high bond energy typically hydrocarbons, are used to be burnt excess air, due to its oxygen content, to produce a great deal of energy. Typical end products for combustion is carbon dioxide and water, sometimes carbon monoxide, depending on weather the supply of oxygen is limited or not. There are many methods to determine the air content to monitor the combustion process and one way is the use of Gas Chromatography. In this study, we explore the basic principles unlying the combustion process by combusting propane, C3H8 with excess air and monitoring its mass and energy balance using selected methods of calculation from experimental values and not to mention, Gas Chromatography to give us a clearer idea of combustion as a whole.

3. Principles

3.1 Stoichiometry Equation and Conditions

Just like any other chemical process/experiment, we have to understand the theoretical or stoichiometrical equation of the combustion process, which not only shows the chemical compositions of the reactants and products, but also the theoretical or ideal mole ratios.

The conditions at which the reactions take place also matter, as it affects the compositions and enthalpies of the chemical substances. In this experiment, the combustion takes place at atmospheric pressure of 1 atm and room temperature of 24.4 oC.

Under the atmospheric pressure, a fuel (propane gas) reacts with O2 (Air, O2: N2 = 1 :3.76 ) and this process is exothermic and the stoichiometric combustion formula is:C3H8+5(O2 +3.76N2) 3CO2 +4H2O+18.8N2 + H where H is enthalphy of combustionAs the Air/fuel ratio changes, the enthalpy of combustion and the composition of the product (exhaust gases) will change accordingly. However, the principles of energy and materials balance are always valid and can be used to solve the unknowns.

3.2 Mass Balance and TerminologiesMass balance is the calculation of the different amount of reactants and/or products based on given values and the stoichiometric equation above. It gives us specific figures on the amounts of input and output to the system, which in this case is the Fireboy burner, the combustion chamber where the process takes place.

Theoretical & Excess Air:Theoretical air is the quantity of air that contains theoretical oxygen (5 moles in the above example)Percent excess air: (moles air) fed (moles air)theo 100% (moles air)theoDry and Wet basis: composition on a wet basis denotes the component mole fractions of a gas that contains water; composition on a dry basis signifies the component mole fraction of the same gas without water.

3.3 Energy Balance

Similar to mass, the energy balance helps us find the energy output of the combustion process. Energy balance is based on 1st Law of Thermodynamics, law of the conservation of energy.

First laws for open system at steady state: Our system is the open system at steady state, the firstlaws can be written as:

Fig 1: First laws for open system at steady state

Where,

Fig 2: Heat component of the equation on the open system at steady state

Standard heat (enthalpy) of combustion, H c0 : the heat of reaction of the substance with oxygen to yield specific products (e.g., CO2, H2O), with both reactants and products are at the referencetemperature and pressure (25C, 1 atm).Standard heat (enthalpy) of formation, H 0f : enthalpy change associated with the formation of 1 mole compound from naturally occurring elements under reference state (25C, 1 atm ). The heat of combustion is related to heat of formation: H0 = H0 H0

Enthalpy changes for processes involving chemical reaction: Two methods exists for the calculation of the overall enthalpies changes, they are: (1) heat of reaction method (molecular species at 25C, 1atm), and (2) heat of formation method (naturally occurring elemental species at 25C, 1atm). In this experiment, the former method is used.3.4 Gas Chromatography

Gas Chromatography (GC) is a technique used to separate substances that are volatile. It makes use of an inert carrier gas to separate the compounds due to their differences in partitioning behavior between the mobile gas phase and the stationary phase in the column. Simply said, this technique helps in the analysis of the exhaust gas for the combustion process. More about GC in the Appendix A.

4. Equipment and Materials

The experiment is fairly a straight-forward one with 3 main parts; combustion, sampling and analysis.

Combustion chamber

Fireboy burners equipped with quartz tube are used as the combustion device (chambers), propane gas as the fuel.A MKS Mass gas flow controller is used to regulate the volumetric flow rate of the fuel gas. Shimadzu GC 14B is used to analyze the composition of the exhaust gases (Appendix I).Air samplingA micro-syringe is used as the sampling device.Analysis

Gas Chromatogram

Other equipments include common lab apparatus including the traditional thermometer.

5. Procedure

The experiment is fairly a straight-forward one with 3 main parts; combustion, sampling and analysis. A brief outline of the procedure is listed below.

1. The GC was started first to ensure that the base-line was stable before using later in the experiment.2. Gas fuel supply was turned on, in which at the same time the mass flow controller (MFC) was switched on and checked zero. 3. Propane flow rate was set, and the process was ignited.4. Waiting time of 3 minutes to ensure that the process has gone to steady state.5. Sample of the exhaust gas was taken using the micro-syringe and the temperature was recorded.6. Sample gas was injected into the analytical port of the GC before starting the analysis.7. Computation and calculation of results

6.Results and Calculations

6.1 GC Analysis

The Gas Chromatography (GC) technique was used to analyse and determine the content composition of the air mixture sample drawn from the outlet of the Fireboy burner, that was taken using the micro-syringe.

Table 1 below is taken straight from the GC.

GC ResultsMole fraction (dry basis)

CO20.1030

N20.8290

020.0680

Total (dry basis)1.0000

H200.1373

Total (wet basis)1.1373

Table 1: GC analysis results (dry basis)

Table 2 is derived from table one, taking into account the water vapour content of the gas.

Wet basisMole fraction (wet basis)

CO20.09056272

N20.728898007

O20.05978898

H200.120750293

Total (wet basis)1.000

Table 2: GC Analysis Results (wet basis)

6.2 Results

Using the above data from the GC, the mass and energy balances and the input molar flow rate of the propane gas, the respective values for the different components of input and output were able to be determined using calculations.

Below are the tabulated results from the calculations:

InletOutletGC analysis

Yini^Hiyini^Hi1

%mol/skJ/mol%mol/skJ/molmol frac

N276.506.665 x 10-3072.906.665 x 10-311.0800.728898007

O220.331.772 x 10-305.985.467 x 10-412.0610.05978898

C3H83.172.76042 x 10-40nilnilnil0

H2Onilnilnil12.081.104 x 10-313.1320.120750293

CO2Nilnilnil9.068.281 x 10-416.9670.09056272

COnilnilnilnilnilnil0

Data calcuTable 3: Tabulation of all calculations and results

6.3 Calculations and DerivationsBelow are the workings showing the derivation of the values tabulated above. These values are calculated in sequential order and makes use of relevant assumptions and mass and energy balances. Below is a simplified diagram of the system.

QTin = 25C Tout = 401C n propn oxy n nitro n propn oxy n nitron CO2n COn water

Fig 3: Simple diagram of the open system

6.3.1 Flow Rate of Propane Gas (Inlet)Visual Flow Rate of C3H8 = 371 cm3/ min1 mol of (any) gas has a net volume of 24dm3 = 24000cm3 at R.T.PFlow Rate of C3H8 = 371 cm3/ min x 1mol/ 24000cm3 x 1min/ 60s = 2.76042 x 10-4 mol/s

6.3.2 Flow Rate of Carbon Dioxide (Outlet)Stoichiometric Equation:C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + HFor this equation to be valid, it was assumed that complete combustion occurred and was justified with the absence of a Carbon Monoxide product of incomplete combustion.Flow Rate of CO2 = 3 x 2.76042 x 10-4 mol/s = 8.281 x 10-4 mol/s

6.3.3 Flow Rate of Water (Outlet)Stoichiometric Equation:C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + HFor this equation to be valid, it was assumed that complete combustion occurred and was justified with the absence of a Carbon Monoxide product of incomplete combustion.Flow Rate of H2O = 4 x 2.76042 x 10-4 mol/s = 1.104 x 10-3 mol/s

6.3.4 Flow Rate of Nitrogen (Outlet)Stoichiometric Equation:C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + HThe flow rate of Nitrogen and Oxygen cannot be obtained using the stoichiometric ratio with the basis of Propane gas (i.e 18.8 x 2.5347 x 10-4 mol/s). This is based on the chromatogram obtained, there are no peaks indicative of Carbon Monoxide. This implies that complete combustion has occurred yielding the Carbon Dioxide and Water. However, we can assume that since Propane gas has undergone complete combustion, the limiting reactant is thus Propane. This can inevitably mean that the air was supplied in excess to ensure that complete combustion occurs. Thus the stoichiometric values and coefficient will not give a fair and reliable representation of answer due to the fuel air ratio. It is only recommended that we check with the chromatogram and use the comparison method for the answers.Based on GC, the mol frac of CO2 = 0.090562720.09056272 = 8.281 x 10-4 mol/sBased on GC, the mol fraction of N2 = 0.728898007Flow Rate of N2 = [(8.281 x 10-4 mol/s)/ 0.09056272] x 0.728898007Flow Rate of N2 = 6.665 x 10-3 mol/s

6.3.5 Flow Rate of Oxygen (Outlet)Stoichiometric Equation:

C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + H

Based on GC, the mol frac of CO2 = 0.090562720.09056272 = 8.281 x 10-4 mol/sBased on GC, the mol fraction of O2 = 0.05978898Flow Rate of O2 = [(8.281 x 10-4 mol/s)/ 0.09056272] x 0.05978898Flow Rate of O2 = 5.467 x 10-4 mol/s

6.3.6 Flow Rate of Nitrogen (Inlet)Stoichiometric Equation:

C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + HFlow Rate of N2 (Inlet) = Flow Rate of N2 (Outlet) = 6.665 x 10-3 mol/sThis is due to the nature of inertness of Nitrogen Gas.The inlet flow rate of Nitrogen is the same to that of the outlet flow rate of Nitrogen due to its inertness and its presence in the chromatogram despite the fact that complete combustion will not yield any Nitrogen products. The presence of Nitrogen is due to its role as a constituent in air and is part of the air fuel ratio. This is because, for every 1 mole of Oxygen, there is 3.76 moles of Nitrogen. Thus the chromatogram which displays the peak for Nitrogen simply implies that this Nitrogen gas constituent is non reactive as it does not contribute to the combustion but it is present due to its function as a component of air. Hence, the number of moles of inlet Nitrogen gas is the same as the outlet.

6.3.7 Flow Rate of Oxygen (Inlet)Stoichiometric Equation:C3H8 + 5(O2 + 3.76N2) 3CO2 + 4H2O + 18.8N2 + HFor every 1 mole of Oxygen, there is 3.76 moles of NitrogenFlow Rate of N2 (Inlet) = 6.665 x 10-3 mol/sFlow Rate of O2 (Inlet) = (6.665 x 10-3 mol/s)/ 3.76 = 1.772 x 10-3 mol/s

6.3.8 Inlet Molar Flow Rate and Mole FractionInlet Molar Flow Rate = n total (C3H8 + O2 + N2)Inlet Molar Flow Rate = 6.665 x 10-3 mol/s + 1.772 x 10-3 mol/s + 2.76042 x 10-4 mol/sInlet Molar Flow Rate = 8.713 x 10-3 mol/sMole Fraction of C3H8 = (2.76042 x 10-4 mol/s)/ (8.713 x 10-3 mol/s) = 0.0317Mole Fraction of O2 = (1.772 x 10-3 mol/s)/ (8.713 x 10-3 mol/s) = 0.2033Mole Fraction of N2 = (6.665 x 10-3 mol/s)/ (8.713 x 10-3 mol/s) = 0.7650

6.3.9 Outlet Molar Flow Rate and Mole FractionOutlet Molar Flow Rate = n(CO2 + H2O + N2 + O2)Outlet Molar Flow Rate = 0.009144215Mole Fraction of CO2 = 0.09056272Mole Fraction of H2O = 0.120750293Mole Fraction of N2 = 0.728898007Mole Fraction of O2 = 0.05978898

6.3.10 Enthalpies (Hi) of Outlet StreamState Reference for enthalpies would be respective components at 25OC, 1 atmEnthalpies of Components = Carbon Dioxide (CO2):Hf of CO2 = Hf of CO2 = - [36.11 x 10-3 T + 4.233 x 10-5 ] (UL: 903.15, LL: 298.15)Hf of CO2 = + 16.96748504 kJ/mol = 16.967kJ/molWater (H2O):Hf of H2O = -285.8kJ/mol + Hf of H2O = -285.8kJ/mol + [33.46 x 10-3 T + 0.6680 x 10-5 ] (UL: 903.15, LL: 298.15)Hf of H2O = 13.13196544 kJ/mol = 13.132 kJ/mol

Nitrogen (N2):Hf of N2 = Hf of N2 = [29.00 x 10-3 T + 0.2199 x 10-5 ] (UL: 903.15, LL: 298.15)Hf of N2 = 11.08011351 kJ/mol = 11.080 kJ/mol

Oxygen (O2):Hf of O2 = 0 kJ/mol + Hf of O2 = [29.61 x 10-3 T + 1.158 x 10-5 ] (UL: 903.15, LL: 298.15)Hf of O2 = 12.06077904 kJ/mol = 12.061 kJ/mol

6.3.11 Higher Heating Value (HHV)The value is determined by bringing all the products of combustion back to the original pre-combustion temperature, and the condensation of any vapor produced. The reference temperature utilized is 25C which is the same as the thermodynamic heat of combustion; since the enthalpy change for the reaction assumes a common temperature of the compounds before and after combustion, in which case the water produced by combustion is liquid. The higher heating value also takes into account the latent heat of vaporization of water.HHV = LHV + Latent Heat of Vaporization of H2OHHV = -(Hc) HHV = 2220.0kJ/mol

6.3.12 Lower Heating Value (LHV)The value is determined by subtracting the heat of vaporization of the water vapor from the higher heating value which treats any H2O formed as a vapor. This methodology assumes that the water component of a combustion process is in vapor state at the end of combustion, as opposed to the higher heating value where all of the water in a combustion process is in a liquid state after a combustion process.LHV = -(Hc + Latent Heat of Vaporization of H2O )LHV = - (-2220.0kJ/mol + 44.013kJ/mol) = 2175.99 kJ/mol

6.3.13 Heat Rejection of Combustion Chamber (Q)First Law for Open System at Steady State:Q + Ws = H + Ek + EpQ = H = out niHi - in niHi + Hc (heat of reaction method)Q = [nCO2HCO2 + nH2OHH2O + nN2HN2 + nO2HO2] [nC3H8HC3H8 + nO2HO2 + nN2HN2] + HcQ = [(16.967 kJ/mol)(8.281 x 10-4 mol/s) + (13.132 kJ/mol)(1.104 x 10-3 mol/s) + (11.080 kJ/mol)(6.665 x 10-3 mol/s) + (12.061 kJ/mol)(5.467 x 10-4 mol/s)] [( 0 kJ/mol)(2.760 x 10-4 mol/s)] + [(-2220 kJ/mol)( 2.760 x 10-4 mol/s)]Q = -0.503816339 kJ/s = - 503.816 J/s

The negative result signifies that heat rejected from the system, releasing energy in a form of heat. The above method makes use of Hess Law which can be found in Appendix B.

6.3.14 Percentage of Air Excess Ratio (Nitrogen Gas Based)Percent Excess Air = [(moles air)fed - (moles air)theo]/[(moles air)theo] x 100%Percent Excess Air = [(8.437 x 10-3 mol/s) - (5.834 x 10-3 mol/s)]/ [(5.834 x 10-3 mol/s)] Percent Excess Air = 44.63%The Percent Excess Air obtained is 44.63%. This implies that the air to fuel ratio is sufficient for complete combustion of the propane gas. Also, based on the gas chromatography, there were no evident peaks depicting the presence of Carbon Monoxide; only to prove that total complete combustion has occurred. However, this assumes that extent of reaction is 100 percent, using the amount of reacted O2 as the theorectical moles of O2.

7.Discussion

7.1 Methodology

For this study, the fireboy burner was used as the combustion chamber and the Gas Chromatogram was used to test the outlet air sample using the aid of micro-syringe to capture the air sample. Overall, it was a relatively easy method of getting fast results to do quick and easy calculations. However, each step has its cons as well, with sources for error, leading to a less accurate study.

Fireboy Burner

The fireboy burner has a tube-like combustion chamber and is a safe and easy way of burning fuel, with the inlet components flowing in from the bottom and the residual gas leaving at the top of the tube.

Source(s) of error and recommendation:

However, the top of the tube has a sufficiently large diameter making it difficult to take a sample that represents the outlet gas as its constituents might not be homogenous and uniform throughout the area at the top of the tube, for the the sampling using the micro-syringe. Hence, a possible suggestion might actually be using a smaller diameter tube-like combustion chamber to minimise this non-homogeneity discrepancy. The air samples can also be taken at a specific spot around the top of the tube to emphasise on consistancy and accuracy of results, if more than one air samples are to be taken.

Micro-syringe air sampling

Air sampling is a fast and convenient way of collecting air for measurement of air ratios. The micro-syringe is light and features a lock to collect and seal off the air compartment effectively. However, the technique of using air samples to represent the whole of the residual gas from the combustion chamber gives room from a sources of error. Techniques of using the microsyringe also has its limitations.

Source(s) of error & recommendation:

First, the study was conducted using only one air sample to represent the outlet gas mixture. A rough estimation of after 3 mins after the combustion process started was given before it was assumed to be at steady state. Using only one air sample does not give confirmation that combustion has already reached steady state. It was still possible that the outlet air ratio could be changing when the air sample was taking. Hence, it is highly recommended that for future experiments, there should be multiple trials to monitor the state of the combustion process. Also, a larger volume of sample could be utilized, producing a more effective and reliable result.

In addition to that, due to safety reasons, the air sample was not taken directly from the outlet of the combustion chamber, but instead at a distance to prevent the risk of getting burned. Thus, this proves to be a significant source of error, simply due to the fact that the air taken was not completely from the outlet to the combustion chamber but also partially made up of generic surrounding air. This thus affects the final reading of the air mixture ratio on the GC, eventually affeting the sample calculations. Therefore, a syringe which has a longer tip and better safety feature could have been used to sample direct air from the outlet of the tube instead.

Gas Chromatography

The gas chromatogram also is a quick way of determining the air mixture ratios with the help of an inbuilt Thermal Conductivity Detector.

This detection process involves carrying the sample gas by helium or any other inert gas which does not contribute to the separation process (carrier gas). Due to the difference in molecular weight and other viable properties, the compounds will be separated but all different retention timings due to their nature of molecular structure and interaction with the column. The mechanism in which the TCD detects is via the electrical resistance of the sample passing though a wire which has a constant voltage applied. Any organic vapor of the highly heated compound that passes through the wire will change the resistance. This change in resistance is displayed in form of peak in the graph plotted according to the time when the change of resistance occurs. This thus, is an effective way in detecting and determining the different gases in the outlet gas mixture.

This GC process can be done in less than 20 mins to display not only figures of the air mixture ratio but also a curve that shows relatively the peaks of the constituent gases. Below is a picture taken of the curve from the experiment.

Fig 4: Curves computed from the GC analysis

The image above shows two curves; one in pink which represents the air generic surrounding air and black one showing the outlet gas constituents. The first peak (furthest left) shows the relative composition of CO2 gas while the second peak shows the relative compositions of O2 and N2 gases. This curve shows that O2 and N2 makes a up majority of the air mixture constituents, be it being the surrounding air or the residual gas from the combustion chamber. However, the main significance of this curve is that it shows clearly the difference for each peaks. For CO2 , there was hardly a peak for the pink curve while there is a relatively significant one for the black curve. This shows that C02 was not present in the inlet air and that was only formed after the combsution process and comprises a significant portion of the outlet residual gas. For the second peak, the black one is lower, implying that the amount of O2 has dropped, as it was one of the reactants for the combustion process. More about peaks and curves can be found in Appendix A.

Source(s) of error & Recommendations:

However, the curves also shows that the experiment has sources of error. Theoretically, the peak of O2 and N2 should be shown as two separate and distinct peaks, instead of just one large one. Also, there were some small peaks for the black curve, simply implying that there were other gas constituents apart from the ones accounted for in the calculations above. A possible reason for the source of error is the injection process of the micro-syringe to the GC, to transfer the air sample into the system to go through the Thermal Conductivity Detector. The injection process has to be a quick and swift one so that the air is properly injected and transferred. A possible recommendation would be to use a machine to transfer this gas or simply use other means of transferring the outlet gas for air mixture ratio measurements, to minimise this discrepancy.

7.2 Calculations and Results

The calculations were relatively straight forward and values were all derived simply using the mass and energy balances discussed earlier.

Assumptions used as basis for calculations

From this study, we are able to use the mass and energy balances, to derive values such as heat output, a useful method that applies not only to combustion but also other chemical processes. However, there were quite a few basis of assumptions we needed to take into account before using the mass and energy balances, which may affect the actual values of the calculations above.

The first assumption was that the initial conditions prior to this experiment was judged to be at standard state condition with reference to the rooms temperature: 25C and pressure 101.325kPa. This initial condition plays an important role as reference states of the gaseous constituents are taken at 25 OC, 1 atm. The reference states allows us to calculate the enthalpies and acts as a basis these enthalpy calculations, where at reference states, enthalpies are 0 kl/mol.

In addition, It was assumed that since generic air from the surroundings was the inlet air, meaning that the composition of air was with regards to only oxygen and nitrogen. In addition, the ratio of oxygen to nitrogen is 1 : 3.76. This thus disregards the presence of other contituents of the generic air such as water vapor and carbon dioxde, even though their respective mole fractions are relatively insignificant. The air ratio also was used to determine the mole fraction of inlet oxygen, thus affecting the amount of oxygen reacted that was calculated.

Another key assumption was that the reaction had reached steady state, meaning that there were no change in variables over time. This assumes that the composition of the air sample at the outlet was not changing with respect to time and that taking one air sample and finding its mole ratio was enough. This assumption also suggests that there was no built up in the combustion chamber. This assumptions thus allows us to simply calculate the enthalpies and moles rates easily without taking into account and variable factors.

The last key assumption was that in the first law of open system with steady state motion, it was assumed that there was negligible work, change in potential energy and kinetic energy. This means that the change in kinetic energy and potential energy can be disregarded in using the energy balance equation for open system, which also implies that heat loss or rejected is equal to enthalphy change of the chemical constituents.

Sources of error & Recommendation:These assumptions serve as sources of error. When assumptions are made, the tendency of error increases, depending on the significance of assumptions. Therefore, to improve the accuracy of the experiment, the significance of these assumptions can be tested.

9.Conclusion

In conclusion, methodology used in the experiment has been simple, effective and successful enough in yielding fast and positive results to find the air mixture ratio which eventually enabled us to calculate the many different properties of the input and output constituents and experiment. Furthermore, the material and energy balance for a steady-state combustion process has been formulated, giving a basic understanding of air excess ratio, stoichiometric, higher heating value and lower heating value. As an added benefit, the concept of the gas chromatography was better understood with practicality even though there was a significant possible source of error using the injection technique of the micro-syringe and quite a handful of assumptions have to be made. Of course, this experiment is only one of the many methods that can be used. To have a better understanding of the combustion process and deriving the calculated values, other methods such as mass spectrometry (Appendix C) can be used in conjuction with GC to further back up the assumptions and minimise uncertainty of the derived values.

10. References

[1] Mass Spectrometry , Retrieved 8 March, 2014 from http://www.premierbiosoft.com/tech_notes/mass-spectrometry.html

[2] Fundamentals of Gas Chromatography, Retrived 8 March, 2014, from http://www.chem.agilent.com/Library/usermanuals/Public/G1176-90000_034327.pdf

[3] Hess Law Conservation of energy , Retrived 8 March, 2014, from http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html

11. Appendix

11.1 Appendix A Gas Chromatography

11.11 Gas Chromatography[2]

Gas chromatography (GC) is a laboratory technique that separates mixtures into individual components. It is used to identify components and to measure their concentrations.

Rather than a physical separation (such as distillation and similar techniques), GC creates a time separation.It does this by passing the vaporized mixture (or a gas) through a tube containing a material that retards some componentsmore than others. This separates thc components in time. After detection, the result is a chromatogram (Figure 1), where each peak represents a different component of the original mixture.The appearance time can be used to identify each component; the peak size (height or area) is a measure of the amount.

A gas chromatographic system consists of: A regulated and purified carrier gas source, which moves the sample through the GC An inlet, which also acts as a vaporizer for liquid samples A column, in which the time separation occurs A detector, which responds to the components as they occur by changing its electrical output Data interpretation of some sort

Fig 5: Components of the GC[3]

11.12 Peak Measurements [2]

Two basic measurements can be made on a peak: The time after injection when the peak is detected The size of the peak

Retention timeThe appearance time, measured from injection to detection, is the sum of two parts: The plumbing timehow long it takes for the carrier gas to pass through the column. It is measured by injecting air or some other non-interacting substance.

The retention timethe additional time caused by the components interaction with the stationary phase in the column.

For most purposes, the plumbing time is ignored and theretention time is taken as the appearance time.

Peak sizeSize can be measured either as peak area or peak height, both measured relative to a constructed baseline.The baseline under the peak cannot be measured directly. It must be constructed from the baselines on either side of the peak.

This is simple with well-separated peaks. It is much more difficult when peaks are merged, on the trailing edge of a solvent peak, or otherwise less than ideal. For this reason, time spent improving the peak separation is time well spent.

Peak height

This is the simplest measurement, requiring only a ruler. It is the vertical distance from the top of the peak to the baseline.

Peak area

This is the area enclosed by the peak signal and the baselineunder it. It is best measured by electronic means.

11.2 Appendix B Hess Law [3]

Hess's lawstates that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction.

11.3 Appendix C Mass Spectrometry [1]

Mass spectrometry is a powerful analytical technique used to quantify known materials, to identify unknown compounds within a sample, and to elucidate the structure and chemical properties of different molecules. It generates multiple ions from the sample under investigation, it then separates them according to their specific mass-to-charge ratio (m/z), and then records the relative abundance of each ion type. Similar to the GC,the Mass Spectrometer identifies the different components to give a curve.

Fig 6: Components of Mass Spectrometry[1]