Question

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete combustion of hydrogen and methanol (Identify phases) and use standard enthalpies of formation to calculate the amount of heat released per kilogram of the fuel (kJ/kg). Which fuel contains the most energy in the least mass? How does the energy of these fuels compare to that of octane (C8H18) (amount of heat released by octane (kJ/kg))?

Answer 1

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:

$\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}$

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:

$\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH$

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:

$\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}$