1. Solid urea, (NH 2 ) 2 CO, burns to give CO 2 ,N 2 , and liquid H 2 O. Its hea
ID: 902144 • Letter: 1
Question
1. Solid urea, (NH2)2CO, burns to give CO2,N2, and liquid H2O. Its heat of combustion is -632.2 kJ/mol. Balance the equation for combustion of urea, and calculate the heat generated per mole of H2O formed. (Enter in kJ.)
1B.Using this heat of combustion and the appropriate thermodynamic data, determine the heat of formation of urea. (Enter in kJ.)
2. Given that the Horxn = -2842.0 kJ, for the following reaction:
2C3H5(NO3)3(l) 6CO2(g) + 1/2O2(g) + 3N2(g) + 5H2O(g)
determine the standard enthalpy of formation, Hof, for: C3H5(NO3)3(l)(nitroglycerin). Enter in kJ/mol.
3a. Use standard enthalpies of formation to determine Horxn for:
2SO2(g) + O2(g) 2SO3(g)
Enter in kJ.
3b. Use standard enthalpies of formation to determine Horxn for:
2NO2(g) N2O4(g)
Enter in kJ.
Explanation / Answer
You can get the necessary values of standard formation of enthalpies of compounds from the ' Table of thermodynamic values'. Just google it.
1. The balanced equation is : 2 (NH2)2CO + 3 O2 = 2 CO2 + 2 N2 + 4 H2O
Enthalpy of combustion of reaction = -632.2 *2 = -1264.4 kJ
1.B. Enthalpy of reaction/combustion = formation enthalpy of products - formation enthalpy of reactants
or, -1264.4 = [{2*393.509 + 2*0 + 4* (-241.818)} - { 2*Hf,urea + 3*0}]
or, -1264.4 = -180.254 - 2*Hf,urea
or, Hf,urea = 542.073 kJ/mol
2. 2C3H5(NO3)3(l) 6CO2(g) + 1/2O2(g) + 3N2(g) + 5H2O(g)
Horxn = -2842.0 kJ =
= [ 6*393.509 + 0.5*0 + 3*0 + 5*(-241.818)] - 2*Hf.2C3H5(NO3)3
=1151.964 - 2C3H5(NO3)3
or, Hf. 2C3H5(NO3)3 = 1996.982kJ/mol
3a. 2SO2(g) + O2(g) 2SO3(g)
Enthalpy of reaction = formation enthalpy of products - formation enthalpy of reactants
= [ 2*( -395.72) ] - [ 2*( -296.83) + 0]
= - 197.78 kJ
3b. 2NO2(g) N2O4(g)
Enthalpy of reaction = formation enthalpy of products - formation enthalpy of reactants
=9.16 - 2* 33.18
= -57.2 kJ
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