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Chapter 6: Problem 71

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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Most popular questions from this chapter

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{kJ}, w=+88 \mathrm{kJ}\) b. \(q=+82 \mathrm{kJ}, w=-47 \mathrm{kJ}\) c. \(q=+47 \mathrm{kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{kJ}\) . If $2.50 \times 10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\( at \)100 .^{\circ} \mathrm{C}$ and 3.50 atm and \(4.50 \times\) \(10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at $100 .^{\circ} \mathrm{C}$ and 3.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

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.

In a coffee-cup calorimeter, 1.60 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with 75.0 \(\mathrm{g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\) . After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\) . Assuming the solution has a heat capacity of 4.18 $\mathrm{J} / \mathrm{C} \cdot \mathrm{g}$ and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of kJ/mol.

A biology experiment requires the preparation of a water bath at $37.0^{\circ} \mathrm{C}$ (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with 90.0 \(\mathrm{g}\) cold water, what mass of hot water must be added to reach $37.0^{\circ} \mathrm{C} ?$
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