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.
Understand the Problem
The question is asking to compute the molar enthalpy of combustion of glucose based on the given data of heat capacity, mass of glucose, and the observed temperature change in a calorimeter.
Answer
The molar enthalpy of combustion of glucose is approximately \(-2814.2 \, \text{kJ/mol}\).
Answer for screen readers
The molar enthalpy of combustion of glucose is approximately (-2814.2 , \text{kJ/mol}).
Steps to Solve
- 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,
$$ q = 755 , \text{J/°C} \times 6.30 , \text{°C} = 4,756.5 , \text{J} $$
- 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,
$$ \text{moles} = \frac{0.305 , \text{g}}{180.2 , \text{g/mol}} = 0.00169 , \text{mol} $$
- 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:
$$ \Delta H_{combustion} \approx 2814.2 , \text{kJ/mol} $$
- Determine the negative sign
Since combustion is an exothermic reaction, we express enthalpy as a negative value:
$$ \Delta H_{combustion} \approx -2814.2 , \text{kJ/mol} $$
The molar enthalpy of combustion of glucose is approximately (-2814.2 , \text{kJ/mol}).
More Information
The combustion of glucose is a critical reaction for cellular respiration, where glucose is oxidized to provide energy in biological systems. This value illustrates the energy available from glucose when used as fuel.
Tips
- Forgetting to convert the final answer into the appropriate units (e.g., from J to kJ).
- Not considering the negative sign for exothermic reactions.
AI-generated content may contain errors. Please verify critical information
