Acetone (C))3H_6O (l)) bums in oxygen according to the equation: 4 O_2 (g) + C_3
ID: 535118 • Letter: A
Question
Acetone (C))3H_6O (l)) bums in oxygen according to the equation: 4 O_2 (g) + C_3 H_6 O(l) = 3 CO_2 (g) + 3H_2 O(l) At 25 degree C the enthalpy change for this reaction is -1789.9 kJ The standard enthalpy of formation Delta_f H degree of CO_2 (g) and H_2O(l) are -393.5 1 and -285.83 kJ/mol, respectively Determine the enthalpy of formation of C_3H_6O(l) at 25 degree C. Please enter your answer (unit: kJ/mol) with 2 decimal as 101.23. Please pay attention to the sign of your answer. Using the result from last question to calculate the reaction enthalpy Delta_r H at 0 degree C (instead of 25 degree C. as given in the last question), if the heat capacities at constant pressure (C_p) of C_3 H_6 O(l), O_2 (g), CO_2 (g) and H_2 O(l)are 124.7 J K^-1 mol^-1, 29.36 J K^-1 mol^-1, 29.10 J K^-1 mol^-1, and 75.29 J K^-1 mol^-1, respectively. Please enter your answer with 2 decimals as 23.22. Using KJ/mol as the unit Pay attention to the sign Continue with the first question Calculate reaction Internal energy Delta_r U degree at 25 degree C Please enter with 2 decimals ns 1123.23 with unit of kJ/mol.Explanation / Answer
Enthalpy of reaction= sum of enthalpy of products-sum of enthalpy of reactants
=3* enthalpy of formation of CO2(g)+3* enthalpy of formation of H2O- { 4* enthalpy of formation of O2+ enthalpy of formation of C3H6O}
3,3,4 and 1 are coefficients of CO2, H2O, O2 and C3H6O respectively.
Since enthalpy of formation of O2=0
=-1789.9= 3*(-393.51)+3*(-285.83)- {4*0+enthalpy of formation of C3H6O}
Enthalpy of formation of C3H6O= -2038.02+1789.9=-248.12 Kj/mole
Enthalpy change of the reaction
H= U+PV
deltaH= deltaU+delta(PV)
deltaH= deltaU+ RT*deltan
deltan= change in no of moles of gases= 3-4=-1
deltaH= deltaU-1*8.314*(25+273)
-1789.3*1000 = deltaU-2478
deltaU= -1789.3*1000+2478=-1786822 J/mole= -1786.2*1000 J/mole