The combustion of glucose (C6H12O6)

Question

The combustion of glucose (C₆H₁₂O₆), producing carbon dioxide and water vapor (gas) has a Δ𝐻ΔH of -2540 kJ per mole of C₆H₁₂O₆.

Calculate the energy involved in making 50.0 g of glucose from carbon dioxide and liquid water. Remember from the video that enthalpy change for converting water from a gas to a liquid is -44 kJ/mol.

Answer 1

This is a problem involving Hess's Law.

Given:

eq.1: C₆H₁₂O₆(s) + 6O₂(g) ==> 6CO₂(g) + 6H₂O(g) .. balanced equation .. ∆H = -2540 kJ / mol

eq.2: H2O(g) ==> H2O(l) .. ∆H = -44 kJ / mol

Asked for:

6CO₂(g) + 6H₂O(l) ==> C₆H₁₂O₆(s) + 6O₂ .. ∆H = ?

Procedure:

50.0 g C₆H₁₂O₆ x 1 mol / 180 g = 0.278 mols C₆H₁₂O₆

reverse eq.1: 6CO₂(g) + 6H₂O(g) <==> C₆H₁₂O₆(s) + 6O₂(g) .. ∆H = +2540 kJ / mol x 0.278 mol = +706 kJ

reverse eq.2 and x 6: 6H₂O(l) <==> 6H₂O(g) .. ∆H = 6x-44 kJ/mol = +264 kJ

Add together and combine/cancel like terms to end up with:

6CO₂(g) + 6H₂O(l) <==> C₆H₁₂O₆(s) + 6O₂(g) .. asked for equation

∆H = 706 kJ + 264 kJ

∆Hrxn = +970 kJ