To graduate a bomb calorimeter by the burning of benzoic acid. Then to utilize the graduated calorimeter to mensurate the heat of burning of naphthalene and cipher the heat of formation of naphthalene.Theory:Q = C. ?TBA C = Q / ?TBAGram molecules = mass / Mr?U = ?U / moles?Hoc.
298 = ?Uoc. 298 + PDV = ?Uoc. 298 + RT?nC10H8 ( s ) + 12O2 ( g ) 10CO2 ( g ) + 4H2O ( cubic decimeter )?Hoc. 298 = – 1 x ?Hof ( Nap ) + 10 ten ?Hof ( CO2 ) + 4 ten ?Hof ( H2O )Procedure:1. First.
a pellet of about 1. 0g of benzoic acid was weighed out utilizing a four denary topographic point mass balance and the weight was recorded at 1. 0022g recorded.2.
Then about 2400cm3 of distilled H2O was added to the calorimeter. A pipette was used to mensurate out 10cm3 of distilled H2O which was instantly added to a clean bomb.3. The top of the bomb electrode was held so that a length of tungsten wire was passed through the hole in each electrode to link the twoelectrodes together. to guarantee that the wire was non tight.4.
The wire was so wound around the electrode to guarantee a good electrical point contact was made.5. A length of cotton was so tied around the pellet and so the cotton attached to the wolfram wire and pellet topographic point in the metal crucible below the electrodes.6.
The benzoic acerb pellet was engraved on to do it easier to knot the cotton onto it.7. Polytetrafluoroethylene tape was so wound around the border of the palpebra and so inserted into the bomb and screwed in tightly. The bomb was so filled with O to a force per unit area of 25 ambiances.8.
The same stairss from 1-8 was repeated for Naptheline.Calculations to find the heat capacity of the calorimeter. Degree centigrade:Using the equationq = C. ?TBA C = Q / ?TBAWhere Q is the heat liberated by the burning of the pellet of benzoic acid.Q = ( ?Uo c.
298 = -26. 43 kJ g-1 ) x 1. 0022gQ = -26. 48 kjTherefore. C = -26. 48C = -12. 6 kJ K-12. 1Calculation to find the heat of burning of naphthalene pellet:Q = C.
?TNapQ = -12. 6 kJ K-1 ten 2. 1Q = -26. 48kJCalculations to find the heat of burning per mole of naphthalene:Gram molecules = mass / MrMr of naphthalene ( C10H8 ) = 128 g mol-1Gram molecules = 0. 6497 = 5. 0758?10-3mol128?U = ?U / moles?U = – 26. 5 kJ = -5220. 9 kJ mol-15.
0758?10-3molCalculations to find the heat of formation of naphthalene:?Hoc. 298 = ?Uoc. 298 + PDV = ?Uoc. 298 + RT?nCombustion of naphthaleneC10H8 ( s ) + 12O2 ( g ) 10CO2 ( g ) + 4H2O ( cubic decimeter )| |C10H8 ( s ) |12O2 ( g ) |10CO2 ( g ) |4H2O ( cubic decimeter )| |Molar ratio |1 |12 |10 |4 |?n = gas molecules in reactants –gas molecules in merchandises?n = 12 – 10 = 2?Uoc. 298 = -5220. 9 kJ mol-1RT?n = 8. 314 ten 298 ten 2 ( where R = gas invariable )RT?n = 4955.
144 JouleRT?n = 4955. 144 Joule1000RT?n = 4. 955 kJ?Hoc. 298 = ?Uoc. 298 + RT?n?Hoc. 298 = -5220. 9 kJ mol-1 + 4.
955 kJ?Hoc. 298 = -5215. 9 kJ mol-1Using Hess’s Law?Hoc. 298 = -5215.
9 kJ mol-1C10H8 ( s ) + 12O2 ( g ) 10CO2 ( g ) + 4H2O ( cubic decimeter )1 ten ?Hof ( Nap ) 10 ten ?Hof ( CO2 ) + 4 ten ?Hof ( H2O )?Hoc. 298 = – 1 x ?Hof ( Nap ) + 10 ten ?Hof ( CO2 ) + 4 ten ?Hof ( H2O )1 ten ?Hof ( Nap ) = – ( -5215. 9 – ( -393. 52 ten 10 ) – ( -285. 83 x4 ) )?Hof ( Nap ) = – ( -5215.
9 ) – ( -5078. 52 )?Hof ( Nap ) = 137. 38 kJ mol-1Mistakes:Mistakes are found in experiments due to experimental equipment. human mistakes and other factors that affect the manner in which the experiment is carried out such as reassigning different chemicals from one contraption to another. In this probe the computation had to found from the experimental process to happen the standard heat content of formation of naphthalene. The estimated mistake for this computation was +/- 10 units. As there is no theoretical value of heat content of formation to compare against there is a important border of mistake taken into history.There are mistakes to see for temperature which are as follows:?T= ( T2-T1 ) ± TT= [ movie ] [ ( T1 ) 2+ ( T2 ) 2 ]Benzoic acid:T1=21.
0 oC ±0. 2T2=23. 2 oC ±0. 2?T= ( 23. 2-21. 0 ) ± T=2.
2± ThymineT = [ movie ] [ ( 0. 2 ) 2+ ( 0. 2 ) 2 ] = [ movie ] 2/5.
?T=2. 1± [ movie ] 2/5Naptheline:T1=22. 7 oC ±0. 2T2=24. 8 oC ±0. 2?T= ( 24.
8-22. 7 ) ± T= 2. 1± ThymineT = [ movie ] [ ( 0. 2 ) 2+ ( 0.
2 ) 2 ] = [ movie ] 2/5.?T=2. 1± [ movie ] 2/5Discussion:The experimental consequences do reflect the undertakings given so standardization of the bomb calorimeter was successful. The heat content of naphthalene could hence be calculated utilizing several equations and the consequences taken from the burning of benzoic acid. The consequences taken for this computations shows the experiment was performed to an acceptable criterion taking into history the little mistakes caused by transportations of solutions and the mass readings.There were betterments that could hold been made to farther increase the truth of recordings obtained such as a more accurate thermometer could hold been used that would gave readings to more denary topographic points with more graduations. Besides an exact sum of distilled H2O could hold been used for the bomb calorimeter instead that an estimate.
This could besides be applied for the mass of benzoic acid and naphthalene used. These accommodations would hold merely made a little betterment on recordings already6 collated but it would hold been an betterment nonetheless.Tongss and baseball mitts had to be used to forestall taint on the pellet if those was non used so oil from tegument could pollute the pellet which could of resulted to different weight readings.Lab coats. goggles and baseball mitts was worn at all clip during this experiment.
Lab coats was worn to forestall taint of the chemicals on tegument and vesture enchantress could hold resulted to corrosion of the apparels and annoyance on the tegument.The experiment was carried out to a high grade of attention and all process points were followed out carefully which proved successful as the graphs and recordings came out as expected.?T =2.
1oC = 2. 1K?T =2. 1oC = 2. 1K