Compute the molar enthalpy of combustion of glucose. Given that combustion of 0.305 g of glucose caused a rise in temperature of 6.30 °C of a calorimeter with total heat capacity C... Compute the molar enthalpy of combustion of glucose. Given that combustion of 0.305 g of glucose caused a rise in temperature of 6.30 °C of a calorimeter with total heat capacity C=755 J/°C. Note that what is given is the heat capacity of the calorimeter, not the specific heat capacity of the substance inside the calorimeter. The molecular weight of glucose is 180.2 amu. Read more

Steps to Solve

1. Calculate the heat absorbed by the calorimeter

To compute the heat absorbed by the calorimeter, we can use the formula:

$$ q = C \times \Delta T $$

where ( C ) is the heat capacity of the calorimeter and ( \Delta T ) is the temperature change.

Given:

• ( C = 755 , \text{J/°C} )
• ( \Delta T = 6.30 , \text{°C} )

So,

$$ q = 755 , \text{J/°C} \times 6.30 , \text{°C} = 4,756.5 , \text{J} $$

2. Determine moles of glucose combusted

Next, we need to calculate the number of moles of glucose that were combusted using the formula:

$$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $$

Given:

• Mass of glucose = 0.305 g
• Molar mass of glucose ( C_6H_{12}O_6 ) = 180.2 g/mol

So,

$$ \text{moles} = \frac{0.305 , \text{g}}{180.2 , \text{g/mol}} = 0.00169 , \text{mol} $$

3. Calculate the molar enthalpy of combustion

Now, we find the molar enthalpy of combustion by dividing the total heat produced by the number of moles of glucose combusted:

$$ \Delta H_{combustion} = \frac{q}{\text{moles}} $$

Substituting the values we found:

$$ \Delta H_{combustion} = \frac{4,756.5 , \text{J}}{0.00169 , \text{mol}} \approx 2,814,222.5 , \text{J/mol} $$

Converting to kilojoules per mole:

$$ \Delta H_{combustion} \approx 2814.2 , \text{kJ/mol} $$

4. Determine the negative sign

Since combustion is an exothermic reaction, we express enthalpy as a negative value:

$$ \Delta H_{combustion} \approx -2814.2 , \text{kJ/mol} $$