Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining process. If all the capturing of energy from the reaction is done through electrical process (non P-V work), then calculate, maximum available energy which can be captured by combustion of 34.2 g of sucrose :
(Given : `DeltaH_("combustion")("sucrose")=-6000kJmol^(-1)`
`DeltaS_("combustion")=180j//K-mol` and bodyntemperature is 300 K)

Options

A. 600 kJ
B. `594.6` kJ
C. `5.4` kJ
D. 605.4 kJ

Explanation

To calculate the maximum available energy that can be captured by the combustion of 34.2 g of sucrose, we can follow these steps:

### Step 1: Calculate the number of moles of sucrose
The molar mass of sucrose (C₁₂H₂₂O₁₁) is calculated as follows:
- Carbon (C): 12 x 12.01 g/mol = 144.12 g/mol
- Hydrogen (H): 22 x 1.008 g/mol = 22.176 g/mol
- Oxygen (O): 11 x 16.00 g/mol = 176.00 g/mol

Total molar mass of sucrose = 144.12 + 22.176 + 176.00 = 342.296 g/mol (approximately 342 g/mol)

Now, we can calculate the number of moles of sucrose in 34.2 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{34.2 \, \text{g}}{342 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

### Step 2: Use the Gibbs free energy equation
The maximum available energy (useful work) can be calculated using the Gibbs free energy change (ΔG) equation:
\[ \Delta G = \Delta H - T \Delta S \]
Where:
- ΔH = -6000 kJ/mol (given)
- T = 300 K (given)
- ΔS = 180 J/K·mol = 0.180 kJ/K·mol (convert J to kJ)

### Step 3: Substitute the values into the equation
Substituting the values into the Gibbs free energy equation:
\[ \Delta G = -6000 \, \text{kJ/mol} - (300 \, \text{K} \times 0.180 \, \text{kJ/K·mol}) \]
Calculating the second term:
\[ 300 \times 0.180 = 54 \, \text{kJ/mol} \]
Now substituting this back:
\[ \Delta G = -6000 - 54 = -6054 \, \text{kJ/mol} \]

### Step 4: Calculate the energy for the given moles of sucrose
Since we calculated ΔG for 1 mole of sucrose, we need to find the energy for 0.1 moles:
\[ \text{Maximum available energy} = \Delta G \times \text{number of moles} = -6054 \, \text{kJ/mol} \times 0.1 \, \text{mol} = -605.4 \, \text{kJ} \]

### Final Answer
The maximum available energy that can be captured by the combustion of 34.2 g of sucrose is approximately **605.4 kJ**.

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