Answer and Explanation:
The 3 reactions represented are
C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O
C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O
C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O
ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol
ΔH°f = ΔH°(products) - ΔH°(reactants)
For reaction 1,
ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol
ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol
ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol
For reaction 2,
ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol
ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol
ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol
For reaction 3,
ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol
ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol
ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol
b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.
Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol
Number of moles = 254/296 = 0.858 mole
heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol
c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole
5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole
The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole
q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ
QED!!!
