When 0.36 g of glucose was burned in a bomb calorimeter ( Heat capacity `600 JK^(-1)`)the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion (MJ/mole).

Explanation

To solve the problem of calculating the standard molar enthalpy of combustion of glucose when 0.36 g is burned in a bomb calorimeter, we can follow these steps:

### Step 1: Calculate the number of moles of glucose
We know that the molar mass of glucose (C6H12O6) is 180 g/mol. We can calculate the number of moles of glucose using the formula:

\[ \text{Number of moles} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} \]

Substituting the values:

\[ \text{Number of moles} = \frac{0.36 \, \text{g}}{180 \, \text{g/mol}} = 0.002 \, \text{moles} \]

### Step 2: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter can be calculated using the formula:

\[ q = C \times \Delta T \]

where:
- \( q \) is the heat absorbed,
- \( C \) is the heat capacity of the calorimeter (600 J/K),
- \( \Delta T \) is the temperature rise (10 K).

Substituting the values:

\[ q = 600 \, \text{J/K} \times 10 \, \text{K} = 6000 \, \text{J} \]

### Step 3: Calculate the molar enthalpy of combustion
The molar enthalpy of combustion (\( \Delta H_c \)) can be calculated using the formula:

\[ \Delta H_c = \frac{q}{\text{number of moles}} \]

Substituting the values we found:

\[ \Delta H_c = \frac{6000 \, \text{J}}{0.002 \, \text{moles}} = 3000000 \, \text{J/mol} \]

### Step 4: Convert to MJ/mol
To convert joules to megajoules, we divide by \( 10^6 \):

\[ \Delta H_c = \frac{3000000 \, \text{J/mol}}{10^6} = 3 \, \text{MJ/mol} \]

### Final Answer
The standard molar enthalpy of combustion of glucose is \( 3 \, \text{MJ/mol} \).

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