Question

Compare the energy available from the combustion of a given volume of methane and the same volume of hydrogen at the same temperature and pressure. (Standard enthalpy of formation of methane is -75 kJ/mol)

Answer 1

Ans:

We know that equal moles of gases occupy equal volume.

So, let us consider the combustion of 1 mol of methane (CH₄) and 1 mol of hydrogen (H₂).

ΔHf = enthalpy of formation

ΔHf for elements is zero {ΔHf of O₂, H₂ = 0}

ΔHf (CO₂) = -393.5 kJ/mol & ΔHf (H₂O) = -285.8 kJ/mol {data collected}

ΔHreaction = enthalpy of reaction

Combustion of Methane;

1 CH₄ (g) + 2 O₂(g) ---> 1 CO₂ (g) + 2 H₂O(l); ΔHreaction = ?

ΔHreaction = ∑ΔHf{products}−∑ΔHf{reactants}

ΔHreaction = {ΔHf (CO₂) + (2 x ΔHf (H₂O))} – { ΔHf (CH₄) + (2 x ΔHf (O₂))}

ΔHreaction = {(-393.5) + (2 x (-285.8))} – {-75 + (2 x 0)} = -890.1 kJ

{negative sign indicates that energy is released}

Combustion of 1 mol of methane releases 890.1 kJ of energy.

Combustion of Hydrogen;

1 H₂ (g) + ½ O₂(g) ---> 1 H₂O(l); ΔHreaction = ?

ΔHreaction = ∑ΔHf{products}−∑ΔHf{reactants}

ΔHreaction = {ΔHf (H₂O)} – { ΔHf (H₂) + ( ½ x ΔHf (O₂))}

ΔHreaction = {(-285.8)} – {0 + ( ½ x 0} = -285.8 kJ

{negative sign indicates that energy is released}

Combustion of 1 mol of hydrogen releases 285.8 kJ of energy.

Conclusion: 1 mol of methane release more energy than 1 mol of hydrogen