The standard enthalpies fo formation of `CO_(2) (g), H_(2) O (1)`, and glucose (s) at `25^(@)C` are `- 400 kJ mol^(-1), - 300 kJ mol^(-)`, and `- 1300 kJ mol^(-1)`, respectively. The standard enthalply of combustion per gram of glucose at `25^(@)C` is

Options

A. `+ 16.11 kJ`
B. `- 16.11 kJ`
C. `+ 2900 kJ`
D. `- 2900 kJ`

Explanation

To find the standard enthalpy of combustion per gram of glucose at 25°C, we will follow these steps:

### Step 1: Write the combustion reaction of glucose
The combustion reaction of glucose (C₆H₁₂O₆) can be represented as:
\[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \]

### Step 2: Use the standard enthalpy of formation values
We know the standard enthalpies of formation for the substances involved:
- \( \Delta H_f^\circ \) for \( CO_2 (g) = -400 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) for \( H_2O (l) = -300 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) for \( C_6H_{12}O_6 (s) = -1300 \, \text{kJ/mol} \)

### Step 3: Calculate the enthalpy change for the combustion reaction
Using the formula:
\[ \Delta H_{reaction} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

For the products:
- \( 6 \times \Delta H_f^\circ (CO_2) + 6 \times \Delta H_f^\circ (H_2O) \)
- \( = 6 \times (-400) + 6 \times (-300) \)
- \( = -2400 - 1800 = -4200 \, \text{kJ} \)

For the reactants:
- \( \Delta H_f^\circ (C_6H_{12}O_6) + 6 \times \Delta H_f^\circ (O_2) \)
- Since \( O_2 \) is in its standard state, its enthalpy of formation is 0.
- \( = -1300 + 0 = -1300 \, \text{kJ} \)

Now, substituting these values into the reaction enthalpy equation:
\[ \Delta H_{reaction} = -4200 - (-1300) \]
\[ = -4200 + 1300 = -2900 \, \text{kJ} \]

### Step 4: Calculate the enthalpy of combustion per gram of glucose
We know that the enthalpy change calculated is for 1 mole of glucose (180 g). To find the enthalpy of combustion per gram:
\[ \text{Enthalpy of combustion per gram} = \frac{-2900 \, \text{kJ}}{180 \, \text{g}} \]
\[ = -16.11 \, \text{kJ/g} \]

### Final Answer
The standard enthalpy of combustion per gram of glucose at 25°C is:
\[ \boxed{-16.11 \, \text{kJ/g}} \]