(30 Points): For each of the following materials, clearly write out an equation
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Question
(30 Points): For each of the following materials, clearly write out an equation for their specific enthalpy () and provide the numerical final result: 4. a. Ethylene Glycol@ 400 'C and 1 atm using the heat of formation approach. b. Chloroform 50 'C and 200 atm using the heat of reaction approach. c. Carbon dioxide @ -100 C and 1 atm using the heat of formation approach Note: you may need to use literature resources to estimate the numerical values. State any assumptions you made in your analysis and cite the resources you usedExplanation / Answer
a) Heat of formation of Ethylene glycol at 25 C = -460 KJ/mol
Specific heat of Ethylene glycol = 3.14 KJ/kg/K
Molecular Wt. of Ethylene Glycol = 62 g/mol
=>Specific heat of Ethylene Glycol. Cp = 3.14/1000*62=.19 KJ/mol/K
=>Heat of formation of Ethylene Glycol at 400 C and 1 atm = Heat of formation at 25C+Cp*(400-25)
=>Heat of formation of Ethylene Glycol at 400 C and 1 atm =-460+.19*(400-25)=-388.75 KJ/mol
b) CH4 + Cl2 ? CH3Cl + HCl
Heat of reaction = -99 KJ/mol
Heat of formation of Methane = -67 KJ/mol
Heat of formation of Cl2=0 (Elemental form)
Heat of formation of HCl = -92 KJ/mol
Heat of reaction = Heat of formation of (HCl+Chloform-Cl2-Methane)
=> Heat of formation of Chloroform at 25 C, 1 atm = -99--92+0+-67=-74 KJ/mol
Specific heat of chloroform = 1.05 KJ/kg/K
Molecular Wt. of Chloroform = 119 g/mol
=>Specific heat of Chloroform = 1.05/1000*119=.12 KJ/mol/K
Heat of formation of Chloroform at 50 C = Heat of formation of Chloroform at 25 C+Cp*(50-25)=-74+.12*(50-25)=-71 KJ/mol
c) Heat of formation of CO2 at 25 C = -393 KJ/mol
Specific heat of CO2 = .03 KJ/mol/K
=> Heat of formation of CO2 at -100 C = -393+.03*(-100-25)=-396 KJ/mol
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