Calculate the enthalpy for the decomposition of glucose given the following reactions: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) ΔH = ? C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O (l) ΔH = -2820kJ

Answer 1

This is Hess' Law. You rearrange the equations with the given values of ∆H in such a way that when you add them together, you get the equation of interest (target equation).

Target Equation: C6H12O6 ==> 2C2H5OH + 2CO2 ... ∆H = ?

Eq.1: C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O (l) ... ΔH = -2820kJ

Eq. 2: C2H5OH(l) + 3O2(g) → 2CO2 (g) + 3H2O (l) ... ΔH = -1368 kJ

Copy eq. 1: C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O (l) ... ΔH = -2820kJ

Reverse eq 2 and x2: 4CO2 + 6H2O --> 2C2H5OH + 6O2 ... ∆H = +2736 kJ

Add them up: C6H12O6(s) + 6O2 + 4CO2 + 6H2O ==> 6CO2(g) + 6H2O + 2C2H5OH + 6O2

Combine and cancel where appropriate to end up with...

C6H12O6 ==> 2CO2 + 2C2H5OH TARGET EQUATION

∆H = -2820 kJ + 2736 kJ

∆H = -84 kJ