Standard Enthalpies of formation of CO2(g), H2O(l), and glucose(solid) at 25°C are -400, -300 and -1300 KJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 25°C is ?

To determine the standard enthalpy of combustion per gram of glucose, we first need to understand the combustion reaction of glucose and how to calculate its enthalpy change. The combustion of glucose (C6H12O6) in the presence of oxygen produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is:

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

Next, we can use the standard enthalpies of formation for each of the substances involved in the reaction to find the enthalpy change for the combustion of glucose. The standard enthalpy change of a reaction can be calculated using the formula:

ΔH° = ΣΔH°f (products) - ΣΔH°f (reactants)

Where ΔH°f represents the standard enthalpy of formation of each compound. Now, let’s break it down step by step:

• Products: For the products, we have 6 moles of CO2 and 6 moles of H2O.
• ΔH°f for CO2(g) = -400 kJ/mol
• ΔH°f for H2O(l) = -300 kJ/mol
• So, the total enthalpy of formation for the products is:
• 6 × (-400 kJ/mol) + 6 × (-300 kJ/mol) = -2400 kJ + -1800 kJ = -4200 kJ
• Reactants: For the reactants, we have 1 mole of glucose.
• ΔH°f for glucose(s) = -1300 kJ/mol
• Thus, the total enthalpy of formation for the reactants is:
• 1 × (-1300 kJ/mol) = -1300 kJ

Now, substituting these values into our equation gives:

ΔH° = (-4200 kJ) - (-1300 kJ) = -2900 kJ

This value, -2900 kJ, represents the enthalpy change for the combustion of 1 mole of glucose. To find the enthalpy of combustion per gram, we need to convert this value into a per gram basis. The molar mass of glucose (C6H12O6) is:

• C: 6 × 12.01 g/mol = 72.06 g/mol
• H: 12 × 1.008 g/mol = 12.096 g/mol
• O: 6 × 16.00 g/mol = 96.00 g/mol
• Total: 72.06 + 12.096 + 96.00 = 180.156 g/mol

Now, we can calculate the enthalpy of combustion per gram of glucose:

Enthalpy of combustion per gram = ΔH° / Molar mass

Substituting the values we have:

Enthalpy of combustion per gram = -2900 kJ / 180.156 g/mol ≈ -16.1 kJ/g

Therefore, the standard enthalpy of combustion per gram of glucose at 25°C is approximately -16.1 kJ/g. This value indicates the amount of energy released when one gram of glucose is completely combusted in oxygen, which is significant for understanding energy production in biological systems and various applications in chemistry and energy science.