The Bomb Calorimeter
- Pages: 5
- Word count: 1068
- Category: Bomb
A limited time offer! Get a custom sample essay written according to your requirements urgent 3h delivery guaranteed
Order NowTo calibrate a bomb calorimeter by the combustion of benzoic acid. Then to use the calibrated calorimeter to measure the heat of combustion of naphthalene and calculate the heat of formation of naphthalene.
Theory:
q = C.ΔTBA C = q / ΔTBA
Moles = mass / Mr
ΔŪ = ΔU / moles
ΔHoc,298 = ΔŪoc,298 + PDV = ΔŪoc,298 + RTΔn
C10H8 (s) + 12O2 (g) 10CO2 (g) + 4H2O (l)
ΔHoc,298 = – 1 x ΔHof (Nap) + 10 x ΔHof (CO2) + 4 x ΔHof (H2O)
Procedure:
1. First, a pellet of approximately 1.0g of benzoic acid was weighed out using a four decimal place mass balance and the weight was recorded at 1.0022g recorded.
2. Then approximately 2400cm3 of distilled water was added to the calorimeter. A pipette was used to measure out 10cm3 of distilled water which was immediately 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 connect the two
electrodes together, to ensure that the wire was not taut.
4. The wire was then wound around the electrode to ensure a good electrical point contact was made.
5. A length of cotton was then tied around the pellet and then the cotton attached to the tungsten wire and pellet place in the metal crucible below the electrodes.
6. The benzoic acid pellet was engraved on to make it easier to knot the cotton onto it.
7. Teflon tape was then wound around the edge of the lid and then inserted into the bomb and screwed in tightly. The bomb was then filled with oxygen to a pressure of 25 atmospheres.
8. The same steps from 1-8 was repeated for Naptheline.
Calculations to determine the heat capacity of the calorimeter, C :
Using the equationq = C.ΔTBA C = q / ΔTBA
Where q is the heat liberated by the combustion of the pellet of benzoic acid.
q = (ΔUo c,298 = -26.43 kJ g-1) x 1.0022g
q = -26.48 kjTherefore,C = -26.48C = -12.6 kJ K-1
2.1
Calculation to determine the heat of combustion of naphthalene pellet:
q = C.ΔTNap
q = -12.6 kJ K-1 x 2.1
q = -26.48kJ
Calculations to determine the heat of combustion per mole of naphthalene:
Moles = mass / Mr
Mr of naphthalene (C10H8) = 128 g mol-1
Moles = 0.6497 = 5.0758×10-3mol
128
ΔŪ = ΔU / moles
ΔŪ = – 26.5 kJ = -5220.9 kJ mol-1
5.0758×10-3mol
Calculations to determine the heat of formation of naphthalene:
ΔHoc,298 = ΔŪoc,298 + PDV = ΔŪoc,298 + RTΔn
Combustion of naphthalene
C10H8 (s) + 12O2 (g) 10CO2 (g) + 4H2O (l)
| |C10H8(s) |12O2 (g) |10CO2 (g) |4H2O (l)
| |Molar ratio |1 |12 |10 |4 |
Δn = gas molecules in reactants –gas molecules in products
Δn = 12 – 10 = 2
ΔŪoc,298 = -5220.9 kJ mol-1
RTΔn = 8.314 x 298 x 2 (where R = gas constant)
RTΔn = 4955.144 J
RTΔn = 4955.144 J
1000
RTΔn = 4.955 kJ
ΔHoc,298 = ΔŪoc,298 + RTΔn
ΔHoc,298 = -5220.9 kJ mol-1 + 4.955 kJ
ΔHoc,298 = -5215.9 kJ mol-1
Using Hess’s LawΔHoc,298 = -5215.9 kJ mol-1
C10H8 (s) + 12O2 (g) 10CO2 (g) + 4H2O (l)
1 x ΔHof (Nap) 10 x ΔHof (CO2) + 4 x ΔHof (H2O)
ΔHoc,298 = – 1 x ΔHof (Nap) + 10 x ΔHof (CO2) + 4 x ΔHof (H2O)
1 x ΔHof (Nap) = – (-5215.9 – (-393.52 x 10) – (-285.83 x4))
ΔHof (Nap) = – (-5215.9) – (-5078.52)
ΔHof (Nap) = 137.38 kJ mol-1
Errors:
Errors are found in experiments due to experimental equipment, human errors and other factors that affect the way in which the experiment is carried out such as transferring different chemicals from one appliance to another. In this investigation the calculation had to found from the experimental procedure to find the standard enthalpy of formation of naphthalene. The estimated error for this calculation was +/- 10 units. As there is no theoretical value of enthalpy of formation to compare against there is a significant margin of error taken into account.
There are errors to consider for temperature which are as follows:
ΔT=(T2-T1)± T
T= [pic][( T1)2+( T2)2]
Benzoic acid:
T1=21.0 oC ±0.2
T2=23.2 oC ±0.2
ΔT=(23.2-21.0)± T=2.2± T
T =[pic][(0.2)2+(0.2)2]= [pic]2/5.
ΔT=2.1±[pic]2/5
Naptheline:
T1=22.7 oC ±0.2
T2=24.8 oC ±0.2
ΔT=(24.8-22.7)± T= 2.1± T
T =[pic][(0.2)2+(0.2)2]= [pic]2/5.
ΔT=2.1±[pic]2/5
Discussion:
The experimental results do reflect the tasks given so calibration of the bomb calorimeter was successful. The enthalpy of naphthalene could therefore be calculated using several equations and the results taken from the combustion of benzoic acid. The results taken for this calculations shows the experiment was performed to an acceptable standard taking into account the small errors caused by transfers of solutions and the mass readings.
There were improvements that could have been made to further increase the accuracy of recordings obtained such as a more accurate thermometer could have been used that would gave readings to more decimal places with more graduations. Also an exact amount of distilled water could have been used for the bomb calorimeter rather that an approximation. This could also be applied for the mass of benzoic acid and naphthalene used. These adjustments would have only made a slight improvement on recordings already6 collated but it would have been an improvement nonetheless.
Tongs and gloves had to be used to prevent contamination on the pellet if those was not used then oil from skin could contaminate the pellet which could of resulted to different weight readings.
Lab coats, goggles and gloves was worn at all time during this experiment. Lab coats was worn to prevent contamination of the chemicals on skin and clothing witch could have resulted to corrosion of the clothes and irritation on the skin.
The experiment was carried out to a high degree of care and all procedure points were followed out carefully which proved successful as the graphs and recordings came out as expected.
ΔT =2.1oC = 2.1K
ΔT =2.1oC = 2.1K