Standard Enthalpy of Combustion per Gram of Glucose

To determine the standard enthalpy of combustion per gram of glucose, we first write the balanced chemical equation for the complete combustion of one mole of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$):

$\text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \longrightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l)$

The standard enthalpy of combustion per mole of glucose ($\Delta_c H^\circ$) is given by the difference between the sum of the standard enthalpies of formation of the products and the reactants:

$\Delta_c H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})$

$\Delta_c H^\circ = \left[ 6 \times \Delta_f H^\circ(\text{CO}_2, g) + 6 \times \Delta_f H^\circ(\text{H}_2\text{O}, l) \right] - \left[ \Delta_f H^\circ(\text{C}_6\text{H}_{12}\text{O}_6, s) + 6 \times \Delta_f H^\circ(\text{O}_2, g) \right]$

Given:

• $\Delta_f H^\circ(\text{CO}_2, g) = -400\text{ kJ/mol}$
• $\Delta_f H^\circ(\text{H}_2\text{O}, l) = -300\text{ kJ/mol}$
• $\Delta_f H^\circ(\text{C}_6\text{H}_{12}\text{O}_6, s) = -1300\text{ kJ/mol}$
• $\Delta_f H^\circ(\text{O}_2, g) = 0\text{ kJ/mol}$ (by definition for an element in its standard state)

Substitute these values into the expression:

$\Delta_c H^\circ = [6(-400) + 6(-300)] - [-1300 + 0]$ $\Delta_c H^\circ = [-2400 - 1800] - [-1300]$ $\Delta_c H^\circ = -4200 + 1300 = -2900\text{ kJ/mol}$

Next, we calculate the molar mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$): $\text{Molar mass} = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180\text{ g/mol}$

The standard enthalpy of combustion per gram of glucose is therefore:

$\text{Enthalpy of combustion per gram} = \frac{\Delta_c H^\circ}{\text{Molar mass of glucose}} = \frac{-2900\text{ kJ}}{180\text{ g}} \approx -16.11\text{ kJ/g}$

Thus, the correct option is (C).