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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
<u>Combustion of hydrogen</u>
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:
<u>Combustion of methanol</u>
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:
<u>Combustion of octane</u>
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: