Question

Use the enthalpy data below to determine the standard reaction enthalpy ∆h⁰_r for the reaction:
C6H14 (g) + 9.5 O2 (g) = 6 CO2 (g) + 7 H2O (g)
Data:
C6H14 (l) + 9.5 O2 (g) = 6 CO2 (g) + 7 H2O (l): ∆h0r = −1791 BTU / lb-mol
Justify your answer clearly.

Answer 1

Solution :

here,

the standard reaction enthalpy ∆h0r = standard heat of formation of products - standard heat of formation of reactants

   ∆h0r = (∆h0f)prod - (∆h0f)react.........................(1)

now, heat of formation data :

heat of formation of CO2 = -393.5 KJ/mol

heat of formation of H2O = -285.82 KJ/mol

heat of formation of C6H14 = -198.7 KJ/mol

heat of formation O2 = 0 KJ/mol

from equation (1),

∆h0r = (6*(-393.5) + 7*(-285.82)) - (1*(-198.7) + 9.5*0)

∆h0r = -4163.04 KJ/mol

now for unit conversion,  1 BTU / lb-mol = 2.326 KJ/mol

so, ∆h0r = -4163.04/2.326 BTU / lb-mol

∆h0r = -1789.785 BTU / lb-mol, which is nearer the answer.