Question

Calculate the standard enthalpy of formation of propane if its enthalpy of combustion is

Answer 1

Given the equation C3H8+5O2⟶3CO2+4H2OCX3HX8+5OX2⟶3COX2+4HX2O and that enthalpies of formation for H2O(l)HX2O(l) is −285.3 kJ/mol−285.3 kJ/mol and CO2(g)COX2(g) is −393.5 kJ/mol−393.5 kJ/mol, and the enthalpy of combustion for the reaction is −2220.1 kJ/mol−2220.1 kJ/mol, I need to find the heat of formation of propane.

My initial idea was to use Hess's law, and I got[3(−393.5)+4(−285.3)]−[−2220.1]=−101.6 kJ[3(−393.5)+4(−285.3)]−[−2220.1]=−101.6 kJ

I then doubted myself because Hess's law gives the ΔHorxnΔHrxno which is different from the ΔHofΔHfo forC3H8CX3HX8 that we are trying to find.

So I used another approach using the individual equations for the formation of H2O,CO2HX2O,COX2, and C3H8+5H2O⟶3CO2+4H2OCX3HX8+5HX2O⟶3COX2+4HX2O and combining the equations to get the equation for the formation of C3H8CX3HX8 (which is 3C+4H2⟶C3H83C+4HX2⟶CX3HX8). After manipulating the equations, I got +2220.1−1141.2−1180.5=−101.6 kJ

Answer 2

Answer:

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