Calculate the reaction enthalpy ΔrH when 1 mol of glucose is oxidized to acetic acid at 80°C. 2O2 (g) + C6H12O6 (s) →2CH3COOH (l) + 2CO2 (g) + 2H2O (l) Express your answer to four significant figures and include the appropriate units.

The reaction enthalpy for the oxidation of 1 mole of glucose is - 1,055 kJ.

Explanation:

The standard enthalpy of a reaction can be calculated from the standard enthalpies of formation of each substance, using the following expression:

ΔH°r = ∑n (p).ΔH°f (p) - ∑n (p).ΔH°f (p)

where,

ΔH°r is the standard enthalpy of the reaction

n (i) is the number of moles of reactants and products in the balanced equation

ΔH°f (i) is the standard enthalpy of formation of reactants and products, which are tabulated.

Considering the balanced equation:

2 O₂ (g) + C₆H₁₂O₆ (s) → 2 CH₃COOH (l) + 2 CO₂ (g) + 2 H₂O (l)

The standard enthalpy of reaction is:

ΔH°r = [2 x ΔH°f (CH₃COOH (l)) + 2 x ΔH°f (CO₂ (g)) + 2 x ΔH°f (H₂O (l)) ] - [2 x ΔH°f (O₂ (g)) + 1 x ΔH°f (C₆H₁₂O₆ (s)) ]

ΔH°r = [2 mol x (-483.5 kJ/mol) + 2 mol x (-393.5 kJ/mol) + 2 mol x (-285.8 kJ/mol) ] - [2 mol x 0 kJ/mol + 1mol x (-1,271 kJ/mol) ]

ΔH°r = - 1,055 kJ

In the balanced equation there is 1 mole of glucose. So, the enthalpy of reaction for such amount is - 1,055 kJ (exothermic).