Standard enthalpies of formation are obtained from thermodynamic tables as: C2H5
ID: 703227 • Letter: S
Question
Standard enthalpies of formation are obtained from thermodynamic tables as: C2H5OH(l) -224 kJ/mol CO2(g) -398 kJ/mol H2O(l) -281 kJ/mol. Calculate the enthalpy change of the reaction C2H5OH(l) + 3 O2 ? 2 CO2(g) + 3 H2O Follow the procedures based on Hess's Law: First write down the reactions corresponding to the enthalpies of formation you have been given, reverse the equations if necessary, remembering to change the sign of ?Ho, then combine the equations to give the required process, and evaluate the associated ?Ho State the answer to four significant figures without decimal point, ±xxxxStandard enthalpies of formation are obtained from thermodynamic tables as: C2H5OH(l) -224 kJ/mol CO2(g) -398 kJ/mol H2O(l) -281 kJ/mol. Calculate the enthalpy change of the reaction C2H5OH(l) + 3 O2 ? 2 CO2(g) + 3 H2O Follow the procedures based on Hess's Law: First write down the reactions corresponding to the enthalpies of formation you have been given, reverse the equations if necessary, remembering to change the sign of ?Ho, then combine the equations to give the required process, and evaluate the associated ?Ho State the answer to four significant figures without decimal point, ±xxxx
Standard enthalpies of formation are obtained from thermodynamic tables as: C2H5OH(l) -224 kJ/mol CO2(g) -398 kJ/mol H2O(l) -281 kJ/mol. Calculate the enthalpy change of the reaction C2H5OH(l) + 3 O2 ? 2 CO2(g) + 3 H2O Follow the procedures based on Hess's Law: First write down the reactions corresponding to the enthalpies of formation you have been given, reverse the equations if necessary, remembering to change the sign of ?Ho, then combine the equations to give the required process, and evaluate the associated ?Ho State the answer to four significant figures without decimal point, ±xxxx
Explanation / Answer
Formation of C2H5OH from its elements
2C + 3H2 + 1/2O2 = C2H5OH
Hf1 = - 224 kJ/mol
Reverse the equation
C2H5OH = 2C + 3H2 + 1/2O2.......... Eq1
H1 = 224 kJ/mol
Formation of CO2 from its elements
C + O2 = CO2
Hf2 = - 398 kJ/mol
Multiply by 2
2C + 2O2 = 2CO2.........eq2
H2 = - 2 x 398 = - 796 kJ/mol
Formation of H2O from its elements
H2 + 1/2O2 = H2O
Hf3 = - 281 kJ/mol
Multiply by 3
3H2 + 3/2O2 = 3H2O............eq3
H3 = - 3 x 281 = - 843 kJ/mol
Now add the above equations
Eq1 + Eq2 + Eq3
We get
C2H5OH + 3O2 = 2 CO2 + 3 H2O
From the Hess law
enthalpy change of the reaction
H = H1 + H2 + H3
= 224 - 796 - 843
= - 1415 kJ/mol
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