Determination Of The Heat Of Combustion And The Standard Enthalpy Of Formation Of Glucose

Categories: ChemistryPhysics

Abstract:

In this experiment, the primary objective is to determine the heat of combustion of glucose using a constant-volume bomb calorimeter. Subsequently, this value will be utilized to calculate both the enthalpy of combustion and the standard enthalpy of formation of glucose.

The initial step involves using benzoic pellets as a standard for determining the calorimeter's heat capacity. Additionally, benzoic acid is employed along with glucose pellets as a spiking material to facilitate the ignition of glucose, which is challenging to ignite directly.

The results obtained indicate percentage errors of 0.0985% for the heat of combustion and 0.6897% for the heat of formation of glucose. These errors, although small, could potentially arise from various factors discussed in the subsequent Discussion section.

Introduction:

In this section, we will outline the objectives of the experiment in our own words, avoiding direct copying of theories provided in handouts. We will clearly define what we aim to measure and describe how we intend to obtain the desired quantity from experimental measurements.

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The primary goal of this experiment is to determine the heat of combustion of glucose. Subsequently, we will utilize this value to calculate both the enthalpy of combustion and the standard enthalpy of formation of glucose. These calculations will be performed using a constant-volume bomb calorimeter.

It is important to note that the process ideally should be isothermal, but due to practical limitations, the temperature will not remain constant throughout. Thus, the isothermal process is carried out in two steps.

Firstly, we conduct an adiabatic process in which we obtain our products (A (T₀) + B (T₀) + S (T₀) → C (T₁) + D (T₁) + S (T₁)).

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Subsequently, we bring the products back to their initial temperature by either adding or removing heat from the system (C (T₁) + D (T₁) + S (T₁) → C (T₀) + D (T₀) + S (T₀)).

It is important to account for the formation of nitrogen dioxide (NO₂) gas in air, which will react with oxygen to produce nitric acid (HNO₃). To quantify this reaction, we employ sodium carbonate for titration to determine the volume of the resulting acid.

During calculations, we will determine the heat capacity of the calorimeter by employing a standard benzoic acid pellet. Subsequently, we will calculate the heat of combustion of glucose and, based on this value, derive the enthalpy of combustion and the enthalpy of formation of glucose.

Experimental:

The experimental procedure outlined in the Lab Manual was meticulously followed, with one exception: sodium carbonate was used for titration instead of sodium hydroxide. The apparatus employed for this experiment was a constant-volume bomb calorimeter, and titration was conducted to determine the quantity of nitric acid formed.

To prepare the mixed pellets, the following proportions of glucose and benzoic acid were used:

  • Pellet 1: 0.7904g of glucose and 0.2101g of benzoic acid.
  • Pellet 2: 0.7906g of glucose and 0.2101g of benzoic acid.

Data:

Table 1: Given values to be used

Quantity Value
∆H combustion of glucose -2808 kJ/mol
∆H0f CO2 -393.5 kJ/mol
∆H0f H2O -285.83 kJ/mol
∆U wire combustion -1400 cal/g
∆U nitric acid -14.1 cal/g
∆U Benzoic acid -6318 cal/g
Normality of Na2CO3 0.07163N

Table 2: Data for benzoic acid (standardization pellets)

Standard Pellet T1 T2
Mass of pellet (benzoic acid) (±0.0001g) 1.0033 0.9880
Mass of the wire (±0.0001g) 0.0173 0.0168
Mass of the wire remaining (±0.0001g) 0.0037 0.0031
Mass of burned wire (±0.0001g) 0.0136 0.0137
Initial temperature T0 (±0.0001oC) 19.4489 21.1457
Rise in temperature ∆T (±0.0001oC) 2.6128 2.6292
Volume of Na2CO3 (±0.02mL) 0.20 0.10

Table 3: Data for mixed benzoic acid and glucose pellets

Mixed Pellet T3 T4
Mass of glucose ±0.0001g 0.7904 0.7906
Mass of benzoic acid ±0.0001g 0.2101 0.2101
Mass of pellet ±0.0001g 0.9963 0.9931
Mass of the wire (±0.0001g) 0.0157 0.0156
Mass of the remaining wire (±0.0001g) 0.0023 0.0011
Mass of burned wire (±0.0001g) 0.0133 0.0145
Initial temperature T0 (±0.0001oC) 23.2998 24.5531
Rise in temperature ∆T (±0.0001oC) 1.7725 1.7711
Volume of Na2CO3 (±0.02mL) 1.60 0.60

Results:

Calculations:

Heat capacity (Cv) of the calorimeter from standardization calculation:

Standardization 1:

  • ΔU benzoic acid = ΔU benzoic acid * m benzoic acid
  • ΔU benzoic acid = -6318 cal/g * 1.0033g = -6338.8494 cal
  • ΔU wire = ΔU wire * m burned wire
  • ΔU wire = -1400 cal/g * 0.0136g = -19.04 cal
  • ΔU nitric acid = Δ nitric acid * m eq
  • At equivalence: nNa2CO3= nHNO3
  • nHNO3= (0.07163)*(0.20 mL) = 0.0143 meq
  • ΔU nitric acid = ΔU nitric acid * m nitric acid = - 14.1 cal/g * 0.0143 meq = -0.2016 cal
  • ΔU isothermal = (∆U benzoic acid + ∆U wire + ∆U nitric acid)
  • ΔU isothermal = -6338.8494 cal - 19.04 cal - 0.2016 cal = -6358.1 cal
  • ΔU isothermal = -Cv*(T1 – T0)
  • Cv = - (ΔU isothermal) / ∆T
  • Cv = - (-6358.1 cal) / (2.6128) = 2433.44 cal/°C

Standardization 2:

  • ΔU benzoic acid = ΔU benzoic acid * m benzoic acid
  • ΔU benzoic acid = -6318 cal/g * 0.9880g = -6242.184 cal
  • ΔU wire = ΔU wire * m burned wire
  • ΔU wire = -1400 cal/g × 0.0137g = -19.18 cal
  • ΔU nitric acid = Δ nitric acid * m eq
  • At equivalence: nNa2CO3= nHNO3 (titration by Na2CO3)
  • nHNO3 =0.07163*0.10 mL = 0.0072 meq
  • ΔU nitric acid = ΔU nitric acid * m nitric acid = - 14.1 cal/g * 0.0072 meq = -0.1 cal
  • ΔU isothermal = (∆U benzoic acid + ∆U wire + ∆U nitric acid)
  • ΔU isothermal = -6242.184 cal - 19.18 cal - 0.1 cal = -6261.464 cal
  • ΔU isothermal = -Cv*(T1 – T0)
  • Cv = - (ΔU isothermal) / ∆T
  • Cv = - (-6261.464) / (2.6292) = 2381.147 cal/°C

Cv (average) = (2433.44 + 2381.147) / 2 = 2407.2935 cal/°C

Heat of combustion of glucose (ΔU) calculation:

Mixed pellet 1:

  • % mpure benzoic acid = (mpure benzoic acid / mtotal) * 100
  • % mpure benzoic acid = (0.2101 / (0.7904 + 0.2101)) * 100 = 21%
  • mbenzoic acid after mixing = (% mpure benzoic acid * mtotal) / 100 = (21 * 0.9963) / 100 = 0.2092g
  • % mglucose = 100 - % mbenzoic acid = 100 - 21 = 79%
  • mglucose after mixing = (% mglucose * mtotal) / 100 = (79 * 0.9963) / 100 = 0.7871g
  • At equivalence: nNa2CO3 = nHNO3 (titration by Na2CO3)
  • nHNO3 = 0.07163 * 1.60 = 0.1146 mol
  • ΔUisothermal = mglucoseΔUglucose + mbenzoic acidΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire
  • mglucose ΔUglucose = ΔUisothermal - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)
  • ΔUglucose = [ΔUisothermal - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)] / mglucose
  • ΔUglucose = [-Cvs ΔT - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)] / mglucose = [-2407.2935 * 1.7725 - (0.2092 * -6318 + 0.1146 * -14.1 + 0.0133 * -1400)] / 0.7871 = -3716.1304 cal/g

Mixed pellet 2:

  • % mpure benzoic acid = (mpure benzoic acid / mtotal) * 100
  • % mpure benzoic acid = (0.2101 / (0.7906 + 0.2101)) * 100 = 21%
  • mbenzoic acid after mixing = (% mpure benzoic acid * mtotal) / 100 = (21 * 0.9931) / 100 = 0.2085g
  • % mglucose = 100 - % mbenzoic acid = 100 - 21 = 79%
  • mglucose after mixing = (% mglucose * mtotal) / 100 = (79 * 0.9931) / 100 = 0.7845g
  • At equivalence: nNa2CO3 = nHNO3 (titration by Na2CO3)
  • nHNO3 = 0.07163 * 0.60 = 0.0430 mol
  • ΔUisothermal = mglucoseΔUglucose + mbenzoic acidΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire
  • mglucose ΔUglucose = ΔUisothermal - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)
  • ΔUglucose = [ΔUisothermal - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)] / mglucose
  • ΔUglucose = [-Cvs ΔT - (mbenzoic acid ΔUbenzoic acid + meqΔUHNO3 + mwireΔUwire)] / mglucose = [-2407.2935 * 1.7711 - (0.2085 * -6318 + 0.0430 * -14.1 + 0.0145 * -1400)] / 0.7845 = -3728.9334 cal/g

ΔUglucose (average) = (-3716.1304 -3728.9334) / 2 = -3722.5319 cal/g

Enthalpy of combustion of glucose (ΔH) calculation:

Combustion of Glucose:

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)

  • ΔH = ΔU + ΔPV = ΔU + ΔngasRT
  • ΔngasRT = 0 because Δngas = 6 - 6 = 0
  • Molar mass of glucose is 180.1559 g/mol [2]
  • ΔH = ΔUglucose * Mglucose = -3728.9334 * 180.15589 = -671789.3154 cal/mol = -2810.7665 kJ/mol

ΔHcombustion of glucose = -2810.7665 kJ/mol

Heat of formation of glucose (ΔHf) calculation:

Hess’s law: ΔH∘r= ΔH∘fproducts−ΔH∘freactants [3]

  • ΔH∘r = 6 (ΔH∘CO2 + ΔH∘H2O) – (ΔH∘glucose)
  • ΔH∘glucose = 6 (ΔH∘CO2 + ΔH∘H2O) - ΔH∘r = 6(-285.83 - 393.5) - (-2810.7665) = -1265.2135 kJ/mol
  • ΔHf∘ = -1265.2135 kJ/mol

Percentage Error = [(exp. - theoretical) / theoretical] * 100

Sample Calculation: % Error ΔHcombustion = [(-2810.7665 + 2808) / -2808] * 100 = 0.0712%

Discussion:

In reality, measuring ∆H directly in an isothermal process is not feasible due to temperature fluctuations caused by heat transfer. Therefore, an adiabatic process is employed, which consists of two steps:

  • In the calorimeter, products are obtained at a new temperature through an adiabatic change of state: A (To) + B (To) + S (To) → C (T1) + D (T1) + S(T1)
  • Subsequently, the products are returned to the initial temperature by adding or removing heat: C (T1) + D (T1) + S(T1) → C (T0) + D (T0) + S(T0)

For the isothermal process, we have:

ΔUisothermal = ΔUcal + ΔUimg

ΔUcal is zero because it's adiabatic and occurs at a constant volume. Therefore:

ΔUisothermal = ΔUT1→To = Cv (To - T1) = -Cv ΔT

where Cv = Cv (H2O) + Cv (CO2) + Cv (s)

It's worth noting that Cv (H2O) and Cv (CO2) are negligible in comparison to Cv (s) due to their much smaller masses.

In this experiment, we made several assumptions:

  • We assumed that the gases behave ideally, following the ideal gas law (PV = nRT).
  • We also assumed that the heat capacities of CO2 and H2O are negligible compared to the system because of their much smaller masses.
  • Additionally, we assumed that the heat capacities (Cv) remain constant over small temperature changes.

Table 4: Experimental and Real ΔHcombustion & ΔHf0 of Glucose Comparison

Experimental Result Real Value Percentage Error
ΔHcombustion (kJ/mol) -2810.7665 -2808 0.0985%
ΔHf0 (kJ/mol) -1265.2135 -1274 [4] 0.6897%

Corrections:

We considered that some air might still be present in the bomb, leading to the possibility of nitrogen gas reacting to form HNO3. To account for this, we used sodium bicarbonate for titration to determine the heat of combustion.

We also weighed the wire before and after combustion to include its heat of combustion in our calculations.

Benzoic acid served as a standard due to its ability to facilitate the combustion of glucose, which is not easily ignited.

The percentage errors were found to be 0.0985% and 0.6897%, both of which are considered small and acceptable.

Possible sources of errors include systematic errors from the machine or the digital balance, slight losses of glucose and benzoic acids during the preparation of mixed pellets, incorrect volume readings on the burette during titration, and adding excess sodium carbonate beyond the equivalence point.

Cellular Respiration:

Cellular respiration is a process that releases energy in the form of ATP by breaking down food. It has two types: aerobic and anaerobic respiration, with both using glucose as a reactant.

Table 5: Comparison between Aerobic and Anaerobic Respiration

Aerobic Respiration Anaerobic Respiration
Occurrence Takes place more in animals and plants Takes place more in microorganisms
Oxygen Takes place in the presence of Oxygen Takes place in the absence of Oxygen
Place Only takes place inside the cell Takes place anywhere
Adenosine Triphosphate Gives 36 ATPs per glucose molecule. Gives 2 ATPs per glucose molecule. In animals: lactic acid. In Bacteria: methane and hydrogen sulfide.
End Products End products: CO2 & H2O End products: CH3CH2OH & CO2. In animals: lactic acid. In Bacteria: methane and hydrogen sulfide.
Oxidation of Substrate Substrate is oxidized completely into carbon dioxide and water Substrate is oxidized incompletely
Chemical Reaction C6H12O6 + 6O2 → 6CO2 + 6H2O + 2,900 kJ/mol C6H12O6 → 2C2H5OH + 2CO2 + 118 kJ/mol

Strain Energy of Cyclopropane:

Some molecular geometries require molecules to bend or stretch their bonds from their normal state, resulting in strain energy (S). Cyclopropane is a well-known molecule with strain energy, as its carbon-carbon bond angles are constrained to be 60º.

Strain Energy Calculation:

S = ∆HII - ∆HI

Therefore, bomb calorimetry is used to determine the enthalpies of these two compounds, and their difference yields the strain energy of cyclopropane (S).

Resonance Energy of Benzene (R):

Benzene exhibits two equivalent structures, known as Kekule Structures:

C6H6

However, these are resonance structures where pi electrons are delocalized among all the carbon atoms. This equal sharing of electrons between all carbons contributes to benzene's enhanced stability.

To determine the resonance energy of benzene, we can compare the enthalpies of the real benzene molecule, which exhibits resonance, with one of the Kekule benzene structures (cyclohexatriene). The difference between these two enthalpies represents the remaining energy, known as the resonance energy.

These two molecules can be TTCC and Cyclohexane:

TTCC Cyclohexane Benzene

By employing bomb calorimetry, we can determine the enthalpies of TTCC and cyclohexane and then subtract their values. This calculation will yield the imaginary enthalpy of the Kekule benzene structure.

Subsequently, using bomb calorimetry again, we would combust benzene and calculate its enthalpy.

Finally, by subtracting the enthalpy of benzene from the enthalpy of the calculated Kekule structure, we can determine the resonance energy of benzene:

R = ∆Hbenzene - ∆Hkekule benzene

Conclusion:

In this experiment, our primary objective was to calculate the enthalpy of combustion and the standard enthalpy of formation of glucose using a bomb calorimeter.

The experimental enthalpy of combustion was found to be -2810.7665 kJ/mol with a percentage error of 0.0985%. The enthalpy of formation was determined to be -1265.2135 kJ/mol with a percentage error of 0.6897%. These errors, although present, are relatively small and may be attributed to various sources of error discussed earlier.

It is important to note that Kekule structures do not exist in reality. Therefore, we need to explore other molecules with similar structures.

While there are no exact matches, we can consider using two molecules with differing structures (bonds and atoms) that, when combined, would theoretically yield the benzene structure. This approach assumes that these two molecules have no other source of energy, such as strain energy.

References:

  1. Halaoui, L. Experiment 1: Bomb Calorimetry: enthalpy of combustion and formation of glucose. PowerPoint Presentation.
  2. Molar mass of glucose. (n.d.). Retrieved from https://www.webqc.org/molecular-weight-ofglucose.html
  3. Libretexts. (2018, November 26). Hess's Law. Retrieved from https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/Hesss_Law
  4. Standard enthalpy of formation at 298 K: ΔHf0 (kJ mol-1). (n.d.). Retrieved from https://scilearn.sydney.edu.au/fychemistry/Questions/Thermodata.htm
  5. Panawala, L. Difference Between Aerobic and Anaerobic Respiration. ResearchGate. 2017.
  6. Garland, C. W. Experiments in Physical Chemistry; Nibler, J. W., Shoemaker, D. P.; McGraw Hill, Boston, 8th edition, 2009; p 159.
  7. Halpern, A. M. Experimental Physical Chemistry, second ed.; Prentice Hall: Upper Saddle River, NJ, 1997.
  8. Halaoui, L. Handout: Bomb Calorimetry: Enthalpy of Combustion and Formation of Glucose.
Updated: Jan 11, 2024
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Determination Of The Heat Of Combustion And The Standard Enthalpy Of Formation Of Glucose. (2024, Jan 11). Retrieved from https://studymoose.com/document/determination-of-the-heat-of-combustion-and-the-standard-enthalpy-of-formation-of-glucose

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