Thermodynamics: Enthelpy of Sucrose

In summary: Delta U is the energy from the problem statement because in the calorimeter the combustion is taking place at constant volume. From the problem statement, we know that the heat of combustion of sucrose is 12 mol O2+ 1 mol Sucrose = 12 mol CO2 + H2O. So Delta H is in terms of a constant volume measurement, which apparently isn't what we're supposed to use, because Delta H is usually calculated via constant pressure. The molar value of Enthalpy is 79314.6 kJ/mol, which even for a big molecule, seems absurd compared to the value in the table for the next part of the problem. I think the big part is that I'm assuming Ideal gas
  • #1
Thyferra2680
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Homework Statement



A sample of sucrose, C12H22O11, with mass 0.1265 g, is burned in a bomb calorimeter initially at 298 K. The temperature rises by 1.743 K. To produce the same temperature increase with an electrical heater in this apparatus, it is found to require 2.0823 kJ of energy.

(1) Determine Δ H0 (298) for combustion of sucrose.

(2) Use data in Table 19.2 to calculate Δ H0 (298) for combustion of sucrose, and compare
your answer to (1).

(Table 19.2 states that Sucrose has a Molar Enthalpy of formation of -2220 kJ/Mol)

Homework Equations



DeltaH = DeltaU + Delta n R T

DeltaH = DeltaU + P DeltaV

U = Heat(constant v)

H = Heat(constant p)

Molar mass of Sucrose
Personal Assumption (Ideal gas?)

The Attempt at a Solution


So, I've tried a few things.

First I tried saying that the combustion of Sucrose is from 12 mol O2+ 1 mol Sucrose = 12 mol CO2 + H2O, thus delta n = 11 moles. Delta U is the energy from the problem statement because in the calorimeter the combustion is taking place at constant volume.

But if I use this, then my Delta H is in terms of a constant volume measurement, which apparently isn't what we're supposed to use, because Delta H is usually calculated via constant pressure. Even then, the molar value of Enthalpy is 79314.6 kJ/mol, which even for a big molecule, seems absurd compared to the value in the table for the next part of the problem.

I think the big part is that I'm assuming Ideal gas for the product gases. Is there any way out of this mess?
 
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  • #2
Start by calculating the heat of combustion of sucrose in the bomb. How many moles of sucrose were burned? How much energy was released? What is the energy per mole?


In your expression for Delta U, what is the value of the term 'P Delta V' in that expression?
 
  • #3


I would approach this problem by first clarifying the assumptions and conditions given in the problem statement. It seems that the combustion of sucrose is taking place in a bomb calorimeter, which means it is occurring at constant volume. The bomb calorimeter is also initially at 298 K, so we can assume that the heat capacity of the calorimeter is constant at this temperature.

To calculate the enthalpy change of combustion, we can use the equation ΔH = ΔU + PΔV. Since the volume is constant, ΔV = 0 and we are left with ΔH = ΔU. We are also given the change in temperature (ΔT = 1.743 K) and the amount of energy required to produce the same temperature change with an electrical heater (2.0823 kJ). We can use the heat capacity of the calorimeter to calculate the change in internal energy (ΔU = CΔT).

Now, to determine the enthalpy change of combustion, we can use the relationship ΔH = ΔU + PΔV = ΔU = CΔT. We can rearrange this equation to solve for ΔH. Plugging in the values given in the problem statement, we get ΔH = 2.0823 kJ - PΔV.

To calculate PΔV, we can use the ideal gas law, PV = nRT. We know the initial volume (V), temperature (T), and moles of gas (n) for the products (12 mol CO2 and 1 mol H2O). We can also calculate the pressure (P) using the ideal gas law and the volume and temperature of the calorimeter. With these values, we can solve for PΔV and plug it into the equation for ΔH.

For the second part of the problem, we can use the table to calculate the enthalpy of formation of sucrose. The enthalpy of formation (ΔHf) is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. In this case, we are given the molar enthalpy of formation for sucrose, which is -2220 kJ/mol.

To calculate the enthalpy change of combustion using the enthalpy of formation, we can use the equation ΔH = ΣΔ
 

Related to Thermodynamics: Enthelpy of Sucrose

1. What is the definition of enthalpy?

Enthalpy is a thermodynamic property that represents the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

2. How is enthalpy of sucrose measured?

The enthalpy of sucrose is typically measured using calorimetry, which involves measuring the heat exchanged between a substance and its surroundings. In the case of sucrose, the sample is burned in a controlled environment and the heat released is measured.

3. What factors affect the enthalpy of sucrose?

The enthalpy of sucrose is affected by the temperature, pressure, and physical state of the substance. Changes in these factors can result in a change in the heat content of the system and therefore the enthalpy.

4. How is the enthalpy of sucrose used in chemical reactions?

The enthalpy of sucrose is used in chemical reactions as a measure of the energy released or absorbed during the reaction. This allows scientists to predict the direction and extent of a reaction, as well as the amount of energy that will be released or required.

5. Can the enthalpy of sucrose be negative?

Yes, the enthalpy of sucrose can be negative if the reaction involving sucrose releases more energy than is required to break the bonds in the sucrose molecule. This is known as an exothermic reaction and is often accompanied by a decrease in temperature.

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