Photosysthesis si a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosysthesis of glucose can be represented as:
`6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i)`
The energy of one mole of a photon of wave lenght is known as one Einstein.
Heats of combustion of graphite and hydrogen at 298K are `-393.5 KJ"mol"^(-1)` and `-285.8 KJ"mol"^(-1)` respectively. If on combustion 1 g of glucose releases 15.58 KJ of energy , Calcualate the heat of formation of glucose at 298 K.

Answer

To calculate the heat of formation of glucose at 298 K, we can follow these steps:

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

### Step 2: Identify the heat of combustion of glucose
We know that the combustion of 1 gram of glucose releases 15.58 kJ of energy. To find the heat of combustion for 1 mole of glucose, we need to convert grams to moles. The molar mass of glucose \( C_6H_{12}O_6 \) is approximately 180 g/mol.

Thus, the heat of combustion for 1 mole of glucose is:
\[ \text{Heat of combustion} = 15.58 \, \text{kJ/g} \times 180 \, \text{g/mol} = 2804.4 \, \text{kJ/mol} \]

### Step 3: Use Hess's Law to find the heat of formation
According to Hess's Law, the change in enthalpy for a reaction is equal to the sum of the changes in enthalpy for the individual steps of the reaction. The heat of formation (\( \Delta H_f \)) can be calculated using the formula:
\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

For the combustion reaction, we can express it as:
\[ \Delta H = \Delta H_f \text{(products)} - \Delta H_f \text{(glucose)} - \Delta H_f \text{(oxygen)} \]
Since the heat of formation of elemental oxygen is zero, we can simplify this to:
\[ \Delta H = \Delta H_f \text{(products)} - \Delta H_f \text{(glucose)} \]

### Step 4: Substitute known values
The heats of formation for the products are:
- For \( CO_2(g) \): \( \Delta H_f = -393.5 \, \text{kJ/mol} \)
- For \( H_2O(g) \): \( \Delta H_f = -285.8 \, \text{kJ/mol} \)

Thus, the total heat of formation for the products is:
\[ \Delta H_f \text{(products)} = 6 \times (-393.5) + 6 \times (-285.8) \]
Calculating this gives:
\[ = -2361 - 1714.8 = -4075.8 \, \text{kJ} \]

### Step 5: Set up the equation
Now, we can set up the equation:
\[ -2804.4 \, \text{kJ} = -4075.8 \, \text{kJ} - \Delta H_f \text{(glucose)} \]

### Step 6: Solve for the heat of formation of glucose
Rearranging gives:
\[ \Delta H_f \text{(glucose)} = -4075.8 + 2804.4 \]
Calculating this gives:
\[ \Delta H_f \text{(glucose)} = -1271.4 \, \text{kJ/mol} \]

### Final Answer
The heat of formation of glucose at 298 K is:
\[ \Delta H_f \text{(glucose)} = -1271.4 \, \text{kJ/mol} \]