The user wants to find the standard enthalpy of combustion per gram of glucose at 25°C.

Step 1: Write the balanced chemical equation for the combustion of glucose.

Glucose ($C_6H_{12}O_6$) is a solid that combusts in the presence of oxygen ($O_2$) gas to produce carbon dioxide ($CO_2$) gas and liquid water ($H_2O$) at standard conditions (25°C).

The unbalanced equation is: $C_6H_{12}O_6(s) + O_2(g) \rightarrow CO_2(g) + H_2O(l)$

To balance the equation:

1. Balance Carbon (C): There are 6 C atoms on the left, so we need 6 $CO_2$ molecules on the right. $C_6H_{12}O_6(s) + O_2(g) \rightarrow 6CO_2(g) + H_2O(l)$
2. Balance Hydrogen (H): There are 12 H atoms on the left, so we need 6 $H_2O$ molecules on the right. $C_6H_{12}O_6(s) + O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$
3. Balance Oxygen (O): On the right, there are (6 × 2) + (6 × 1) = 18 O atoms. On the left, there are 6 O atoms in glucose. We need 12 more O atoms from $O_2$. This requires 6 $O_2$ molecules.

The balanced equation is: $C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)$

Step 2: Calculate the standard enthalpy of combustion (\\Delta H_c^\\circ) for one mole of glucose.

The standard enthalpy of a reaction can be calculated from the standard enthalpies of formation (\\Delta H_f^\\circ) of the products and reactants using the following formula: $\Delta H_{rxn}^\circ = \sum (n \cdot \Delta H_f^\circ)_{\text{products}} - \sum (m \cdot \Delta H_f^\circ)_{\text{reactants}}$ where 'n' and 'm' are the stoichiometric coefficients.

Given data:

• $\Delta H_f^\circ(CO_2(g)) = -400$ kJ/mol
• $\Delta H_f^\circ(H_2O(l)) = -300$ kJ/mol
• $\Delta H_f^\circ(C_6H_{12}O_6(s)) = -1300$ kJ/mol
• The standard enthalpy of formation of an element in its standard state is zero, so $\Delta H_f^\circ(O_2(g)) = 0$ kJ/mol.

Now, substitute the values into the formula for the combustion reaction: $\Delta H_c^\circ = [6 \times \Delta H_f^\circ(CO_2(g)) + 6 \times \Delta H_f^\circ(H_2O(l))] - [1 \times \Delta H_f^\circ(C_6H_{12}O_6(s)) + 6 \times \Delta H_f^\circ(O_2(g))]$ $\Delta H_c^\circ = [6 \times (-400) + 6 \times (-300)] - [1 \times (-1300) + 6 \times 0]$ $\Delta H_c^\circ = [-2400 - 1800] - [-1300]$ $\Delta H_c^\circ = -4200 + 1300$ $\Delta H_c^\circ = -2900 \text{ kJ/mol}$ This is the standard enthalpy of combustion for one mole of glucose.

Step 3: Calculate the standard enthalpy of combustion per gram of glucose.

To find the enthalpy per gram, we need to divide the molar enthalpy by the molar mass of glucose.

Calculate the molar mass of glucose ($C_6H_{12}O_6$):

• Atomic mass of C = 12 g/mol
• Atomic mass of H = 1 g/mol
• Atomic mass of O = 16 g/mol Molar mass = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol.

Now, calculate the enthalpy of combustion per gram: $\text{Enthalpy per gram} = \frac{\Delta H_c^\circ}{\text{Molar mass}} = \frac{-2900 \text{ kJ/mol}}{180 \text{ g/mol}}$ $\text{Enthalpy per gram} = -16.111... \text{ kJ/g}$ Rounding to two decimal places, we get $-16.11$ kJ/g.

Step 4: Compare the result with the given options.

• A: +2900 kJ (Incorrect value and sign, units are per mole)
• B: – 2900 kJ (Correct value per mole, but the question asks per gram)
• C: –16.11 kJ (Matches our calculation)
• D: +16.11 kJ (Incorrect sign; combustion is an exothermic process, so $\Delta H$ must be negative)

The correct option is C.