March 29, 2024, 04:38:04 AM
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Topic: Thermodynamics: Enthalpy of Combustion to Standard Enthalpy of Formation  (Read 1835 times)

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Offline adschaff

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Hi all,

New here, came across the forums while trying to understand a section of data analysis for my Lab report, hoping somebody around here can explain some things to me, as I am completely stuck and in a situation without knowledgeable resources.

The experiment was using bomb calorimetry to determine the standard enthalpy of formation for Naphthalene. I'm stuck at adjusting my experimental enthalpy of combustion to the standard enthalpy of formation. Namely with the equations provided to adjust for the pressure difference, as the combustion took place in 30 bar of oxygen and the standard is 1 bar.

This is from my experiment handout instructions:

" The second enthalpy correction brings the solid reactant and liquid product (water) from their high pressure condition in the bomb (~30 bar) to the standard state (1 bar). The needed expression, (dH/dP)t, can be derived from the fundamental equation for enthalpy, dH = TdS + VdP and the Maxwell relation (dS/dP)t = -(dV/dT)p providing the following expression:

(dH/dP)t = - TVα + V = V( 1- αT)

where the coefficient of thermal expansion is defined as α = (1/V)(dV/dT)p. Integration of this equation between the pressure of the bomb (~30 bar) and 1 bar can be carried out by assuming that these materials are incompressible. This integration yields ΔH2 "


We weren't given the volume of the bomb (oxygen pressurized chamber), which seems pretty imperative to the problem, unless I'm integrating it wrong, or incorrect in assuming (dV/dT)p = 0. Anyway I come up with 29*V, and I'm not sure how to apply that to my enthalpy.

Any thoughts or explanations would be appreciated greatly, Also original experiment handout attached, if desired.

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