To find the standard enthalpy of combustion of glucose, we will first write the balanced chemical equation for the combustion of glucose. The chemical formula for glucose is $ C_6H_{12}O_6 $. Combustion involves the reaction of a substance with oxygen to produce carbon dioxide (CO₂) and water (H₂O) typically. The reaction for the combustion of glucose can be represented as:

$$ C_6H_{12}O_6 (s) + 6O_2 (g) \rightarrow 6CO_2 (g) + 6H_2O (l) $$

Each molecule of glucose (C6H12O6) produces six molecules of CO₂ and six molecules of H₂O. Now, we use the given standard enthalpies of formation (ΔH_f°) for these substances to find the enthalpy of the reaction. The enthalpy of a reaction can be calculated using the formula:

$$ \Delta H_{\text{reaction}} = \sum \Delta H_{f,\text{products}} - \sum \Delta H_{f,\text{reactants}} $$

For the given reaction:

• ΔH_f° of CO₂(g) = –400 kJ/mol
• ΔH_f° of H₂O(l) = –300 kJ/mol
• ΔH_f° of glucose(s) = –1300 kJ/mol

The sum of the standard enthalpies of formation of the products:

$$ [6 \times (-400) \text{ kJ/mol}] + [6 \times (-300) \text{ kJ/mol}] = (-2400 + -1800) \text{ kJ} = -4200 \text{ kJ} $$

The sum of the standard enthalpies of formation of the reactants (glucose as only part contributing with enthalpy value other than zero):

$$ [-1300 \text{ kJ/mol}] + [6 \times 0 \text{ kJ/mol for } O_2] $$

Therefore, the standard enthalpy of combustion of glucose is:

$$ \Delta H_{\text{reaction}} = (-4200 \text{ kJ}) - (-1300 \text{ kJ}) = -2900 \text{ kJ} $$

So, -2900 kJ of energy is released per mole of glucose combusted at standard conditions. The standard enthalpy of combustion per gram of glucose can be calculated using the molar mass of glucose. Glucose has a molar mass of approximately 180 g/mol.

$$ \Delta H_{\text{combustion per gram}} = \frac{-2900 \text{ kJ/mol}}{180 \text{ g/mol}} = -16.11 \text{ kJ/g} $$

Thus, the correct answer is -16.11 kJ per gram of glucose, which corresponds to option C –16.11 kJ.