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A 0.692 -g sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is burned in a constant volume calorimeter. The temperature rises from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C} .\) The calorimeter contains \(575 \mathrm{g}\) of water and the bomb has a heat capacity of \(650 \mathrm{J} / \mathrm{K}\). What quantity of heat is evolved per mole of glucose?

Short Answer

Expert verified
The heat evolved per mole of glucose is approximately 2798.24 kJ/mol.

Step by step solution

01

Calculate Temperature Change

Determine the change in temperature of the system by subtracting the initial temperature from the final temperature: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 25.22^\circ C - 21.70^\circ C = 3.52^\circ C \]
02

Calculate Heat Absorbed by Water

Use the formula for heat evolved: \[ q_{\text{water}} = m \cdot c \cdot \Delta T \] where \(m\) is the mass of the water, \(c = 4.18 \, \text{J/g}^\circ \text{C}\) is the specific heat capacity of water, and \(\Delta T\) is the temperature change.Calculate: \[ q_{\text{water}} = 575 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times 3.52^\circ C = 8459.84 \, \text{J} \]
03

Calculate Heat Absorbed by Calorimeter

Use the calorimeter's heat capacity and temperature change to calculate heat absorbed by the calorimeter: \[ q_{\text{calorimeter}} = C_\text{calorimeter} \cdot \Delta T \] where \(C_\text{calorimeter} = 650 \, \text{J/K}\).Calculate: \[ q_{\text{calorimeter}} = 650 \, \text{J/K} \times 3.52 \, K = 2288 \, \text{J} \]
04

Total Heat Evolved in Reaction

Sum the heat absorbed by the water and calorimeter to find the total heat evolved:\[ q_{\text{total}} = q_{\text{water}} + q_{\text{calorimeter}} = 8459.84 \, \text{J} + 2288 \, \text{J} = 10747.84 \, \text{J} \]
05

Calculate Moles of Glucose

Calculate the amount of moles of glucose combusted using the molar mass of glucose \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\), which is 180.16 g/mol:\[ n = \frac{0.692 \, \text{g}}{180.16 \, \text{g/mol}} = 0.003841 \, \text{mol} \]
06

Calculate Quantity of Heat Evolved per Mole of Glucose

Determine the heat evolved per mole of glucose:\[ q_\text{per mole} = \frac{q_{\text{total}}}{n} = \frac{10747.84 \, \text{J}}{0.003841 \, \text{mol}} = 2798242.45 \, \text{J/mol} \]Convert to kJ/mol:\[ q_\text{per mole} = 2798.24 \, \text{kJ/mol} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Capacity
Heat capacity is a crucial property in calorimetry because it defines how much heat a substance can store as its temperature changes.
It is particularly important for understanding how a system in a calorimeter will evolve during a reaction. In calorimetry, two types of heat capacities are often noted: the specific heat capacity and the calorimeter's heat capacity.
  • Specific heat capacity, denoted as \(c\), is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
  • The calorimeter heat capacity refers to the heat needed to raise the entire calorimeter's temperature by 1 degree Celsius.
In many experiments, you’ll find the water's specific heat capacity used because water is often the medium in which reactions occur, due to its known and consistent heat capacity of \(4.18 \, \text{J/g}^\circ \text{C}\). This experiment also provides the bomb's calorimeter heat capacity, which is 650 \(\text{J/K}\).
Both water and the calorimeter contribute to the overall measurement of heat evolved during the glucose combustion process.
Glucose Combustion
Combustion refers to a chemical process that involves burning a substance in the presence of oxygen to release energy in the form of heat and light.
For glucose combustion, the chemical reaction can be represented as: \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) + 6\mathrm{O}_2(g) \rightarrow 6\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(g)\ + \, \text{energy}\] In this process, when glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) is burned, it reacts with oxygen to produce carbon dioxide, water, and releases energy.
This experiment uses a calorimeter to keep this reaction confined and measure the heat released.
Understanding glucose combustion helps us know the amount of energy in foods, as glucose is a primary energy source in biological systems.
Enthalpy Change
Enthalpy change represents the total heat content change of a system when a chemical reaction occurs at constant pressure.
In constant volume calorimetry, such as a bomb calorimeter, we instead measure the \(\Delta U\), but it closely approximates enthalpy change (\(\Delta H\)) under constant pressure conditions because no PV work is performed.Bomb calorimeters directly measure the heat evolved (or absorbed) by the reaction.
Knowing the total heat evolved helps us to understand how much energy is involved in the overall reaction.
We calculated this heat as the sum of the heat absorbed by the water in the calorimeter and the calorimeter's own absorbed heat.For the glucose combustion, the calculated total heat released translates to a calorimeter reading whereby the rise in temperature helps us compute the energy change that happened within the calorimeter's enclosed system.
Molar Mass Calculation
The molar mass of a substance is the mass of one mole of its chemical units. It is an essential concept for converting measured mass into moles, a standard unit for quantifying chemical reactions.
For glucose, with the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), its molar mass is calculated as follows:
  • 6 carbon atoms at \(12.01 \, \text{g/mol}\) = \(72.06 \, \text{g/mol}\)
  • 12 hydrogen atoms at \(1.008 \, \text{g/mol}\) = \(12.096 \, \text{g/mol}\)
  • 6 oxygen atoms at \(16.00 \, \text{g/mol}\) = \(96.00 \, \text{g/mol}\)
  • Total molar mass = \(72.06 \, + 12.096 \, + 96.00 \, = 180.16 \, \text{g/mol}\)
To find how many moles of glucose were burned, the mass of glucose (\(0.692 \, \text{g}\)) is divided by this molar mass, providing a basis to calculate energy released per mole.
The precise conversion from mass to moles allows for the comparison of energy release on a per-mole basis, aiding in comprehensive understanding across various chemical reactions.

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Most popular questions from this chapter

You add \(100.0 \mathrm{g}\) of water at \(60.0^{\circ} \mathrm{C}\) to \(100.0 \mathrm{g}\) of ice at \(0.00^{\circ} \mathrm{C} .\) Some of the ice melts and cools the water to \(0.00^{\circ} \mathrm{C} .\) When the ice and water mixture has come to a uniform temperature of \(0^{\circ} \mathrm{C},\) how much ice has melted?

The standard enthalpy of formation of solid barium oxide, \(\mathrm{BaO},\) is \(-553.5 \mathrm{kJ} / \mathrm{mol},\) and the enthalpy of formation of barium peroxide, \(\mathrm{BaO}_{2},\) is \(-634.3 \mathrm{kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic? $$\mathrm{BaO}_{2}(\mathrm{s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ (b) Draw an energy level diagram that shows the relationship between the enthalpy change of the decomposition of \(\mathrm{BaO}_{2}\) to \(\mathrm{BaO}\) and \(\mathrm{O}_{2}\) and the enthalpies of formation of \(\mathrm{BaO}(\mathrm{s})\) and \(\mathrm{BaO}_{2}(\mathrm{s})\).

For each of the following, define a system and its surroundings and give the direction of heat transfer between system and surroundings. (a) Methane is burning in a gas furnace in your home. (b) Water drops, sitting on your skin after a dip in a swimming pool, evaporate. (c) Water, at \(25^{\circ} \mathrm{C},\) is placed in the freezing compartment of a refrigerator, where it cools and eventually solidifies. (d) Aluminum and \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) are mixed in a flask sitting on a laboratory bench. A reaction occurs, and a large quantity of heat is evolved.

Calculate the quantity of heat required to convert \(60.1 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) at \(0.0^{\circ} \mathrm{C}\) to \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100.0^{\circ} \mathrm{C} .\) The heat of fusion of ice at \(0^{\circ} \mathrm{C}\) is \(333 \mathrm{J} / \mathrm{g}\); the heat of vaporization of liquid water at \(100^{\circ} \mathrm{C}\) is \(2260 \mathrm{J} / \mathrm{g}\).

Assume you mix \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{CsOH}\) with \(50.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) in a coffee-cup calorimeter. The following reaction occurs: $$\mathrm{CsOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CsCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The temperature of both solutions before mixing was \(22.50^{\circ} \mathrm{C},\) and it rises to \(24.28^{\circ} \mathrm{C}\) after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all \(1.00 \mathrm{g} / \mathrm{mL}\) and the specific heat capacities of the solutions are \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

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