Answer:
The order of energy released per mass is
CH₃OH (-2.268 × 10⁴ kJ/kg) < C₈H₁₈ (-4.826 × 10⁴ kJ/kg) < H₂ (-2.835 × 10⁵ kJ/kg)
Explanation:
In order to calculate the enthalpy of a reaction (ΔH°r) we can use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where
ΔH°f(i) are the enthalpies of formation of reactants and products
ni are the moles of reactants and products
Combustion of hydrogen
H₂(g) + 1/2 O₂(g) ⇒ H₂O(l)
ΔH°r = 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(H₂) - 1/2 mol × ΔH°f(O₂)
ΔH°r = 2 mol × (-285.8 kJ/mol) - 1 mol × 0 - 1/2 mol × 0
ΔH°r = -571.6 kJ
571.6 kJ are released when 1 mole of H₂ is burned. The amount of heat released per kilogram is:
[tex]\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}[/tex]
Combustion of methanol
CH₃OH(l) + 3/2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l)
ΔH°r = 1 mol × ΔH°f(CO₂) + 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(CH₃OH) - 3/2 mol × ΔH°f(O₂)
ΔH°r = 1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol) - 1 mol × (-238.4 kJ/mol) - 3/2 mol × 0
ΔH°r = -726.7 kJ
726.7 kJ are released when 1 mole of CH₃OH is burned. The amount of heat released per kilogram is:
[tex]\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH[/tex]
Combustion of octane
C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)
ΔH°r = 8 mol × ΔH°f(CO₂) + 9 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(C₈H₁₈) - 12.5 mol × ΔH°f(O₂)
ΔH°r = 8 mol × (-393.5 kJ/mol) + 9 mol × (-285.8 kJ/mol) - 1 mol × (-208.4 kJ/mol) - 12.5 mol × 0
ΔH°r = -5511.8 kJ
5511.8 kJ are released when 1 mole of C₈H₁₈ is burned. The amount of heat released per kilogram is:
[tex]\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}[/tex]
