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Thermodynamics: Enthalpy of Sucrose

by Thyferra2680
Tags: enthalpy, sucrose, thermodynamics
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Feb8-09, 03:16 PM
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1. The problem statement, all variables and given/known data

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)

2. Relevant 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?)

3. 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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Feb9-09, 03:32 PM
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P: 3,724
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?

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