Enthalpy of Combustion

• Saran Singh Sound

Aim: To determine the enthalpy change of combustion of Ethanol (C₂H₅OH), Propanol (C₃H₇OH) and Butanol (C₄H₉OH).

Chemicals –

• Ethanol (C₂H₅OH)
• Propanol (C₃H₇OH)
• Butanol (C₄H₉OH)

Data Collection –

Table 1: List of apparatus and Least Count and Uncertainties of Measuring Instruments Used

Quantitative Data

• Temperature in degrees Celsius ± 0.1°C
• Time in seconds ± 0.01s

Qualitative Observation:

• It was observed that the flame on the spirit burner swayed due to wind interference. Also, the flame went off at few instances. When the flame was initially burnt its intensity was much greater than when it had burnt for a few seconds.

Section A – Finding the water equivalent of calorimeter

Table 2.2: Collected Data

Formulas:

Mass = No. of Moles x Molar Mass

M_w x C (T_hot – T_final) = M_w x C (T_final – T_cold) + Q (T_final – T_cold)

â-²T = Initial Temperature – Extrapolation Temperature

Enthalpy Change = Mass x Specific Heat Capacity x â-²T

Molar Mass of Ethanol = 46.068g

Molar Mass of Propanol = 60.095g

Molar Mass of Butanol = 74.122g

Percentage Deviation = x100

Uncertainty Formulas:

Error in mass of Water = x 100

Error in temperature = x 100

Error in time = x 100

Avg. Uncertainty =

Trial 1:

Uncertainty of temperature is small and hence error bars cannot be seen on the graph

The graph is used to estimate the water equivalent of calorimeter. The gradient of the best fit line shows the rate of decrease of temperature.

Uncertainties:-

Error in mass of Water = x 100 = 1%

Error in Temperature of cold water/T_c = x 100 = 0.4%

Error in Temperature of hot water/T_h = x 100 = 0.2%

Error in Extrapolated Temperature/T_f = x 100 = 0.2%

Calculation:-

M_w x C (T_hot – T_final) = M_w x C (T_final – T_cold) + Q (T_final – T_cold)

.^.. 50.0 x 4.18 (57.1 – 40.4) = 50.0 x 4.18 (40.4 – 27.6) + Q (40.4 – 27.6)

So, Q = 63.8g

Percentage Uncertainty for Q= 1% + (0.2% – 0.2%) = 1% + (0.2% -0.4%) + (0.2% – 0.4%)

.^.. 1% + 0.6% = 1.6% Uncertainty

Absolute Uncertainty = 1.6% x 63.8 = ± 1g

So, Water equivalent of calorimeter = (64 ± 1)g

Note : There were three assumptions made during this experiment.

1. The Specific Heat Cpacity of the calorimeter is same as water
2. No Heat is lost to the surrounding
3. Mass of water is equal to volume of water

Trial 2:

Uncertainty of temperature is small and hence error bars cannot be seen on the graph

The graph is used to estimate the water equivalent of calorimeter. The gradient of the best fit line shows the rate of decrease of temperature.

Uncertainties:-

Error in mass of Water = x 100 = 1%

Error in Temperature of cold water/T_c = x 100 = 0.3%

Error in Temperature of hot water/T_h = x 100 = 0.2%

Error in Extrapolated Temperature/T_f = x 100 = 0.2%

Calculation:-

M_w x C (T_hot – T_final) = M_w x C (T_final – T_cold) + Q (T_final – T_cold)

.^.. 50.0 x 4.18 (63.6 – 45.6) = 50.0 x 4.18 (45.6 – 30.1) + Q (45.6 – 30.1)

So, Q = 33.7g

Percentage Uncertainty for Q= 1% + (0.2% – 0.2%) = 1% + (0.2% – 0.3%) + (0.2% – 0.3%)

.^.. 1% + 0.8% = 1.8% Uncertainty

Absolute Uncertainty = 1.8% x 33.7 = ± 0.6 g

So, Water equivalent of calorimeter = (33.7 ± 0.6)g

Note : There were three assumptions made during this experiment.

1. The Specific Heat Cpacity of the calorimeter is same as water
2. No Heat is lost to the surrounding
3. Mass of water is equal to volume of water

Trial 3:

Uncertainty of temperature is small and hence error bars cannot be seen on the graph

The graph is used to estimate the change in temperature of the displacement reaction between CuSO4 solution and Zinc (s) powder. The gradient of the best fit line shows the rate of decrease of temperature.

Uncertainties:-

Error in mass of Water = x 100 = 1%

Error in Temperature of cold water/T_c = x 100 = 0.3%

Error in Temperature of hot water/T_h = x 100 = 0.2%

Error in Extrapolated Temperature/T_f = x 100 = 0.2%

Calculation:-

M_w x C (T_hot – T_final) = M_w x C (T_final – T_cold) + Q (T_final – T_cold)

.^.. 50.0 x 4.18 (62.6 – 44.6) = 50.0 x 4.18 (44.6 – 29.8) + Q (44.6 – 29.8)

So, Q = 45.2g

Percentage Uncertainty for Q= 1% + (0.2% – 0.2%) = 1% + (0.2% – 0.3%) + (0.2% – 0.3%)

.^.. 1% + 0.8% = 1.8% Uncertainty

Absolute Uncertainty = 1.8% x 45.2 = ± 0.8 g

So, Water equivalent of calorimeter = (45.2 ± 0.8)g

Note : There were three assumptions made during this experiment.

1. The Specific Heat Cpacity of the calorimeter is same as water
2. No Heat is lost to the surrounding
3. Mass of water is equal to volume of water

Average of all Trials:

Average Water Equivalent of calorimeter/Q = = 47.6 g

Uncertainty for Average Water Equivalent of calorimeter/Q = = ± 15g

Therefore, Average Water Equivalent of calorimeter/Q = (48 ± 15)g

Section B – Finding the enthalpy of combustion of three alcohols.

Table 1.1: Collected Data

Note: Mass of Water was kept constant at 50.0 ± 0.5g

Enthalpy of combustion (ΔH^Ø_C) of ethanol :

Trial 1:

Mass of Ethanol lost = 136.832 – 136.269 = (0.563 ± 0.002)g

Temperature Change/ΔT = 28.9 – 46.1 = (-17.2 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004)mol

ΔH^Ø_C = = -442040 J.mol^-1 = = -442.040 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1.2%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2.2% + 32.5%) + 4% = 38.7%

Absolute Uncertainty = 38.7% x -442.040 = ± 171 kJ.mol^-1

So, Enthalpy of Combustion of ethanol = (-442 ± 171 ) kJ.mol^-1

Trial 2:

Mass of Ethanol lost = 136.269 – 135.934 = (0.335 ± 0.002)g

Temperature Change/ΔT = 28.9 – 47.7 = (-18.7 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004)mol

ΔH^Ø_C = = -480590 J.mol^-1 = = -480.590 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1.2%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2.2% + 32.5% + 4% = 38.7%

Absolute Uncertainty = 38.7% x -480.590 = ± 186 kJ.mol^-1

So, Enthalpy of Combustion of ethanol = (-481 ± 186) kJ.mol^-1

Trial 3:

Mass of Ethanol lost = 135.934 – 135.558 = (0.376 ± 0.002)g

Temperature Change/ΔT = 28.9 – 50.9 = (-21.7 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004) mol

ΔH^Ø_C = = -557690 J.mol^-1 = = -557.690 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2% + 32.3% + 4% = 38.3%

Absolute Uncertainty = 38.3% x -557.690 = ± 214 kJ.mol^-1

So, Enthalpy of Combustion of ethanol = (-558 ± 214) kJ.mol^-1

Average of all Trials:

Average Enthalpy of Combustion of ethanol = = -494 kJ.mol^-1

Uncertainty in Average Enthalpy of Combustion of ethanol = = ± 58 kJ.mol^-1

So, Average Enthalpy of Combustion of ethanol = (-494 ± 58) kJ.mol^-1

Literature Value = -1368 kJ.mol ^-1

Therefore, Percentage Deviation = 63.9%

Enthalpy of combustion (ΔH^Ø_C) of propanol :

Trial 1:

Mass of Propanol lost = 138.554 – 138.087 = (0.467 ± 0.002)g

Temperature Change/ΔT = 28.9 – 44.9 = (-15.7 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004) mol

ΔH^Ø_C = = -403490 J.mol^-1 = = -403.490 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1.2%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2.2% + 32.5% + 4% = 38.7%

Absolute Uncertainty = 38.7% x -403.490 = ± 156 kJ.mol^-1

So, Enthalpy of Combustion of propanol = (-403 ± 156) kJ.mol^-1

Trial 2:

Mass of Propanol lost = 138.087 – 137.812 = (0.275 ± 0.002)g

Temperature Change/ΔT = 28.9 – 45.9 = (-16.9 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004)mol

ΔH^Ø_C = = -434330 J.mol^-1 = = = -434.330 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1.2%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2.2% + 32.5% + 4% = 38.7%

Absolute Uncertainty = 38.7% x -434.330 = ± 168 kJ.mol^-1

So, Enthalpy of Combustion of propanol = (-434 ± 168) kJ.mol^-1

Trial 3:

Mass of Propanol lost = 137.812 – 137.547 = (0.265 ± 0.002)g

Temperature Change/ΔT = 28.9 – 46.9 = (-17.9 ± 0.2)°C

No of moles = = (0.0100 ± 0.0004)mol

ΔH^Ø_C = = -460030 J.mol^-1 = = -460.030 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 4%

Error in Temperature Change/ ΔT = = 1.2%

Error in Water equivalent of Calorimeter/Q = = 31.3%

Percentage Uncertainty in ΔH^Ø_C = 2.2% + 32.5% + 4% = 38.7%

Absolute Uncertainty = 38.7% x -460.030 = ± 178 kJ.mol^-1

So, Enthalpy of Combustion of propanol = (-460 ± 178) kJ.mol^-1

Average of all Trials:

Average Enthalpy of Combustion of propanol = = -432 kJ.mol^-1

Uncertainty in Average Enthalpy of Combustion of propanol = = ± 29 kJ.mol^-1

So, Average Enthalpy of Combustion of propanol = (-432 ± 29) kJ.mol^-1

Literature Value = -2021 kJ.mol ^-1

Therefore, Percentage Deviation = 78.5%

Enthalpy of combustion (ΔH^Ø_C) of butanol :

Trial 1:

Mass of butanol lost = 137.844- 137.667 = (0.177 ± 0.002)g

Temperature Change/ΔT = 28.9 – 38.9 = (-10.0 ± 0.2)°C

No of moles = = (0.0020 ± 0.0004)mol

ΔH^Ø_C = = -1285000 J.mol^-1 = = -1285.000 kJ.mol^-1

Uncertainties:

Error in mass of Water = x 100 = 1%

Error in No of Moles = = 0.2%

Error in Temperature Change/ ΔT = Share this: Facebook Twitter Reddit LinkedIn WhatsApp