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Chapter 6: Problem 114

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 \(1.56 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). 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 is \(25.4 \, \frac{\mathrm{kJ}}{\mathrm{g}}\) and \(2744.8 \, \frac{\mathrm{kJ}}{\mathrm{mol}}\).

Step by step solution

01

Calculate the heat released

To calculate the heat released during the combustion of quinone, we can use the formula: Heat released (q) = Heat capacity of calorimeter x Change in temperature Given heat capacity of the calorimeter = 1.56 kJ/°C and change in temperature = 3.2 °C, we can calculate the heat released using the formula. \[q = (1.56 \, \mathrm{kJ/^\circ C}) \times (3.2 \, \mathrm{^\circ C}) = 4.992 \, \mathrm{kJ}\]
02

Calculate the energy of combustion per gram

Now we can calculate the energy of combustion per gram using the formula: Energy of combustion per gram = Heat released (q) / Mass of sample From Step 1, we have calculated the heat released and we are given the mass of the sample as 0.1964 g. So, \[Energy \, of \, combustion \, per \, gram = \frac{4.992 \, \mathrm{kJ}}{0.1964 \, \mathrm{g}} = 25.4 \, \frac{\mathrm{kJ}}{\mathrm{g}}\]
03

Calculating the molar mass of quinone

To calculate the energy of combustion per mole, we need to determine the molar mass of quinone using the formula: Molar mass = (number of carbon atoms x atomic mass of carbon) + (number of hydrogen atoms x atomic mass of hydrogen) + (number of oxygen atoms x atomic mass of oxygen) Given chemical formula of quinone: \(C_6H_4O_2\) Atomic mass of carbon (C) = 12.01 g/mol Atomic mass of hydrogen (H) = 1.01 g/mol Atomic mass of oxygen (O) = 16.00 g/mol Molar mass of quinone = (6 x 12.01 g/mol) + (4 x 1.01 g/mol) + (2 x 16.00 g/mol) = 108.06 g/mol
04

Calculate the energy of combustion per mole

Finally, we can calculate the energy of combustion per mole using the formula: Energy of combustion per mole = Energy of combustion per gram x Molar mass of quinone Using the values from Steps 2 and 3, we can calculate the energy of combustion per mole: \[Energy \, of \, combustion \, per \, mole = (25.4 \, \frac{\mathrm{kJ}}{\mathrm{g}}) \times (108.06 \, \mathrm{g/mol}) = 2744.8 \, \frac{\mathrm{kJ}}{\mathrm{mol}}\] Hence, the energy of combustion of quinone is 25.4 kJ/g and 2744.8 kJ/mol.

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Most popular questions from this chapter

The specific heat capacity of silver is \(0.24 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{~g}\) Ag from \(273 \mathrm{~K}\) to \(298 \mathrm{~K}\). b. Calculate the energy required to raise the temperature of \(1.0\) mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{~kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample of silver.

The bomb calorimeter in Exercise 106 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A 1.056-g sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right.\) ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Which has the greater kinetic energy, an object with a mass of \(2.0 \mathrm{~kg}\) and a velocity of \(1.0 \mathrm{~m} / \mathrm{s}\) or an object with a mass of \(1.0 \mathrm{~kg}\) and a velocity of \(2.0 \mathrm{~m} / \mathrm{s} ?\)

What is the difference between \(\Delta H\) and \(\Delta E\) ?
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