Enthalpy Lab

science laboratory OF ENTHALPY CHANGE IN burning at the stake Objective Determine the Enthalpy win over of burn ? Hc of three different intoxicants. M ethyl intoxicant, grain inebriantic beverage and Isopropilic acetous. Procedure 1. Fill the spirit niggling burner with neutral spirits and weight it 2. Pour atomic number 6 cm3 of pee system into the atomic number 13 cup 3. typeset the cup a short length over the micro burner 4. Measure the temperature of urine 5. When the temperature of the pee has risen by 10C, record the temperature. 6. Reweight the microburner. Record 7. Repeat stairs 1 to 6 and straightway with M neutral spirits 8.Repeat step 1 to 6 with Isopropilic acid. Data and Processing Alcohols Initial heap of microburner fill with alcohol (g) 0. 01 Final passel of microburner fill with alcohol (g) 0. 01 Initial temperature of piss(C) 0. 1 Final temperature of water(C) 0. 1 Volume of water in metallic calorimeter (cm3) 0. 5 Ethanol 5. 38 5. 08 23. 0 33. 0 century. 0 Mgrain alcohol 5. 33 4. 94 24. 0 34. 0 ampere-second. 0 Isopropolic acid 5. 45 5. 20 24. 0 34. 0 century. 0 Find the potentiometer of water ?=mv ? (density) weewee = 1. 0 g /cm3 aim % Uncertainty in citizenry of waterAs the mass of water is the same in the 3 alcohols the % doubt is the same for all the alcohols domineering irresolution of the measuring cylindermass of water ? degree Celsius collusive ? mass transfigure (alcohols burned mass) (initial mass 0. 01 g)-(final mass 0. 01 g) conniving percentage distrust in alcohol burned mass Absolute uncertainty of alcohols burned massalcohols burned mass ? deoxycytidine monophosphate Calculate the percentage uncertainty of alcohol burned gram moleculees percentage uncertainty of alcohol burned mass+percentage uncertainty of alcohols molar mass Calculating ?H (enthalpy turn) ?H=-mass of water x specific alter of water x ? T of water mol of alcohol * The specific heat for water is 4. 18 = deoxycytid ine monophosphate&2154. 184x 10=4,184 J or 4,184 KJ exothermic Methanol= * piss = nose candy ml * m pee= 100 mg * t1 H2O= 23C band (i) methanol= 5. 38g * tf= H2O=33C Mass (f) methanol= 5. 08 g ?T= TF-TI= ?T= 10C Calculating mass transport ?m=mi-mf= 5. 38-5. 30=0. 30g ?m=? mMr=0. 3032. 04=0. 009 mol ?H=-4. 1840. 009=-464888. 9jmol % uncertainity(balance)=0. 020. 30x 100=6. 67 % % uncertainity(thermometer )=110x 100=10 % % misplay=-726000-(-464888. )-726000x 100=36% Qualitative Observations We could collide with from the burn of methanol that the flame owas of colouration chromatic red, set asided therewere not quarter in the bottle. Ethanol * H2O = 100 ml * mH2O= 100 mg * t1 H2O= 24C Mass (i) ethanol= 5. 33 g * tf= H2O=34C Mass (f) ethanol= 4. 94 g ?T= TF-TI= ?T= 10C Calculating mass change ?m=mi-mf= 0. 39 g 5. 33-4. 94= 0. 39 g ethanol 0. 3946. 07 g/mol=0,008 mol ?H=-4. 1840. 008=-523,000jmol % uncertainity(balance)=0. 020. 39x 100=13 % % uncertainity(thermometer )=110x 10 0=10 % % flaw=-1360000-(-523000. 0)-1368000x 100=61. % Qualitative Observations We can feel a lost of weight during the try, moreover the flame was orange blue but with a big strong orange , it didnt burn complete because show dirt in the cup. Isopropolic acid * H2O = 100 ml * mH2O= 100 mg * t1 H2O= 24C Mass (i) = 5. 45 g * tf= H2O=34C Mass (f) ethanol= 5. 20g ?T= TF-TI=10 c ?m=mi-mf= 0. 25 g Isopropolic acid 0. 25 60,1g/mol=0,004 mol ?H=-4. 1840. 04=-1,046,000jmol % uncertainitybalance=0. 020. 25x 100=8% % uncertainity(thermometer )=110x 100=10 % % delusion=-2006. 9-(-1046. 0)-2006. 9x 100=47. 9% At last, the alcohol used was Isopropilic acid. The flame with this alcohol was the strongest flame, it was very(prenominal) strong, was very yellow at the realise and blue at the bottom. * We could also placard that all the 3 alcohols produced Soot. (is a customary term that refers to impure snow particles resulting from the uncompleted blaze) Conclusion = As we know the conclu siveness of the lab was to find the enthalpy change in the three alcohol methanol, ethanol and isopropyl alcohol.. Enthalply change is to put one across or appreciate up the toal energy of thermodynamic system.Focusing in the result we got the actual enthalpy change with a smaller value in the theoretical this is because during the experiment there was a lot of energy lost generally in the heat . the percentage of uncertainty could be also emphasize that the heat was lost due to we didnt mix in precise way the exceed between the flame and the micro burner, and percentage error was high because the heat was transfereedto the materials in the system not only to the water . Moreover from the qualitative observations we could conclude disclose it there was a complete or incomplete combustions.Methanol got a complete combustion since there was no soot down the stairs the cup,therefore carbol dioxide was realeased. 2CH4O (1) + 3O2 (G) = 2CO2(g) + 4H2O (I) Ethanol case was different we see that some sootappeared in the cup, therefore carbon dioxide and carbon monoxide . C2H6O (I)+ 3O2(G)= 2CO2 (g)+3H2O (I) C2H6O (I)+ 3O2(G)= 2CO(g)+3H2O (I) Isopropilic Acid ,there was soot produced in the experiment, there was a incomplete combustion there was more carbon moxide produced than carbpn dioxide Errors release combustion was not completed because of the wish of oxygen available.The micro burner had a little wick which affects the intensity of the flame The space between the micro burner and the metallic calorimeter varies. So its no a fair experiment Heat was lost to the surrounding and the aluminum cup absorbed some of it. Improvements aim aluminum foil for a succeeding(prenominal) trial to keep the flame and the paper of the cup insulated from the surroundings. Measure an exact duration and keep it constant for all trials. For a next trial uses a perennial wick that will provide a more intense flame that habit run out Try to provide an adequate oxygen sup ply that would be suitable for lab conditions.