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Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A 0.1964 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter with a heat capacity of 1.56 \(\mathrm{kJ} / \mathrm{C}\) . The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C}\) . Calculate the energy of combustion of quinone per gram and per mole.

Short Answer

Expert verified
The energy of combustion of quinone per gram is calculated as: $$ \frac{q_\text{sys}}{m} = \frac{1.56\,\text{kJ/}^\circ\text{C} \times 3.2\,^\circ\text{C}}{0.1964\,\text{g}} = 25.3\,\text{kJ/g} $$ The molar mass of quinone is calculated as: $$ M = 6 \times 12.01\,\text{g/mol} + 4 \times 1.008\,\text{g/mol} + 2 \times 16.00\,\text{g/mol} = 108.10\,\text{g/mol} $$ The energy of combustion per mole is calculated as: $$ \text{Energy of Combustion per Mole} = 25.3\,\text{kJ/g} \times 108.10\,\text{g/mol} = 2734\,\text{kJ/mol} $$ The energy of combustion of quinone per gram is \(25.3\,\text{kJ/g}\) and per mole is \(2734\,\text{kJ/mol}\).

Step by step solution

01

Calculate the Heat Gained by the System

To calculate the heat gained by the system, we need to multiply the heat capacity of the calorimeter by the temperature increase. $$ q_\text{sys} = C_\text{cal} × \Delta T $$ Here, \(q_\text{sys}\) is the heat gained by the system, \(C_\text{cal}\) is the heat capacity of the calorimeter (1.56 kJ/°C), and \(\Delta T\) is the change in temperature (3.2°C).
02

Calculate the Energy of Combustion per Gram

Now that we've calculated the heat gained by the system, we can calculate the energy of combustion per gram. To do this, divide the heat gained by the system by the mass of the quinone sample. $$ \frac{q_\text{sys}}{m} = \frac{\text{Energy of Combustion}}{\text{mass}} $$ Here, \(m\) is the mass of the quinone sample (0.1964 g).
03

Find the Molar Mass of Quinone

To calculate the energy of combustion per mole, we first need to find the molar mass of quinone by adding the atomic masses of its constituent elements. The molecular formula for quinone is \(C_6H_4O_2\), which means it contains 6 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms. The molar mass of Carbon: \(12.01 \text{ g/mol}\), The molar mass of Hydrogen: \(1.008 \text{ g/mol}\), and The molar mass of Oxygen: \(16.00 \text{ g/mol}\)
04

Calculate the Energy of Combustion per Mole

Now that we have the energy of combustion per gram and the molar mass of quinone, we can calculate the energy of combustion per mole. To do this, multiply the energy of combustion per gram by the molar mass of quinone. $$ \text{Energy of Combustion per Mole} = \frac{q_\text{sys}}{m} \times M $$ Here, \(M\) is the molar mass of quinone.

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Key Concepts

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

Quinone
Quinone is a fascinating molecule with great significance in biological processes, particularly in photosynthesis. It plays a key role in the transport of electrons in plants, enabling them to convert water, carbon dioxide, and sunlight into glucose. Quinone's structure,
  • is based on a six-carbon ring,
  • comprises four hydrogen and two oxygen atoms,
  • has a chemical formula of \(\text{C}_6\text{H}_4\text{O}_2\).
Understanding quinone involves not just chemistry, but also its biological functions and implications in the energy cycles of plants.
Heat of Combustion
The heat of combustion is a measure of the energy released when a substance burns in oxygen. This energy is crucial in determining how much heat is generated during a chemical reaction, such as burning quinone in a bomb calorimeter.
To calculate the heat of combustion:
  • Identify the calorimeter's heat capacity (\(1.56 \ \mathrm{kJ}/\mathrm{°C}\)).
  • Measure the rise in temperature (\(3.2\ \mathrm{°C}\)).
  • Calculate the heat gained using the formula: \(q_\text{sys} = C_\text{cal} \times \Delta T\).
This heat value represents the energy released per gram and can be scaled to a per mole basis for practical applications.
Molar Mass Calculation
Calculating the molar mass of a compound is a fundamental step in chemistry. It involves adding up the atomic masses of each element in the molecule. For quinone \(\left(\text{C}_6\text{H}_4\text{O}_2\right)\):
  • Carbon: \(6 \times 12.01 \, \text{g/mol} = 72.06 \, \text{g/mol}\)
  • Hydrogen: \(4 \times 1.008 \, \text{g/mol} = 4.032 \, \text{g/mol}\)
  • Oxygen: \(2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}\)
The total molar mass of quinone is the sum of these values: \(72.06 + 4.032 + 32.00 = 108.092 \, \text{g/mol}\). This value is used to convert energy per gram into energy per mole during combustion analysis.
Bomb Calorimeter
A bomb calorimeter is a device used to measure the heat of combustion of a substance. It consists of a robust container or "bomb" where the substance is burned in a controlled, oxygen-rich environment. Key features of bomb calorimeters include:
  • Insulation to prevent heat loss to surroundings.
  • A thermometer to accurately gauge temperature changes.
  • A known heat capacity to calculate the total energy change.
In our calculation, the heat capacity was given as \(1.56 \, \mathrm{kJ}/\mathrm{°C}\).
Using this apparatus ensures precise measurement of the heat produced during combustion, which is critical in energy calculations like those involving quinone.

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Most popular questions from this chapter

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing 150.0 \(\mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\) . After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C}\) . If the specific heat capacity of nickel is 0.444 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

Consider an airplane trip from Chicago, Illinois, to Denver, Colorado. List some path-dependent functions and some state functions for the plane trip.

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and 1.00 atm pressure. \(\Delta H_{\text { vap }}\) for the liquid is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C} .\)

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