Chapter 5: Problem 46
A \(0.692-\mathrm{g}\) sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) was burned in a constant-volume calorimeter. The temperature rose from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C}\). The calorimeter contained \(575 \mathrm{g}\) of water, and the bomb had a heat capacity of \(650 \mathrm{J} / \mathrm{K}\). What is \(\Delta U\) per mole of glucose?
Short Answer
Step by step solution
Calculate Heat Absorbed by Water
Calculate Heat Absorbed by Calorimeter
Find Total Heat Released by the Reaction
Calculate Moles of Glucose Burned
Calculate Change in Internal Energy per Mole
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
In our exercise, glucose undergoes a combustion reaction. This process releases energy, resulting in a negative \( \Delta U \), which signifies an exothermic reaction. We calculate this by finding the total heat absorbed by both the water and the calorimeter in the system. Then, this value is adjusted according to the amount of substance that reacted, expressed per mole, so we understand the energy change for a standard quantity of glucose.
Heat Capacity
Two heat capacities are considered in the exercise: that of the water, and that of the calorimeter bomb itself. Water has a specific heat capacity of \( 4.184 \text{ J/g°C} \), a known constant used in calculations for thermal energy exchange with water.
Combustion Reaction
When glucose undergoes combustion, it reacts with oxygen to form carbon dioxide and water, releasing energy that is then absorbed by the surrounding water and the calorimeter bomb. The purpose of measuring these heat exchanges is to determine the total energy output of the reaction. By relating this to the moles of glucose, we find the standard energy change per mole, an essential step in understanding the enthalpic nature of glucose combustion.
Molar Mass
In this exercise, the molar mass of glucose (\( \text{C}_{6}\text{H}_{12}\text{O}_{6} \)) is \( 180.18 \text{ g/mol} \). Knowing the sample's mass (\( 0.692 \text{ g} \)), we can compute the number of moles of glucose involved in the reaction.