/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 A sample of benzene, \(\mathrm{C... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), weighing \(3.51 \mathrm{~g}\) was burned in an excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rose from \(25.00^{\circ} \mathrm{C}\) to \(37.18^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter and contents was \(12.05\) \(\mathrm{kJ} /{ }^{\circ} \mathrm{C}\), what is the value of \(q\) for burning \(1 \mathrm{~mol}\) of benzene at constant volume and \(25.00^{\circ} \mathrm{C}\) ? The reaction is $$\mathrm{C}_{6} \mathrm{H}_{6}(l)+\frac{15}{2} \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$$

Short Answer

Expert verified
The heat released for burning 1 mol of benzene is approximately \(3268.64\, \mathrm{kJ/mol}\).

Step by step solution

01

Calculate the Temperature Change

Find the difference in temperature recorded by the calorimeter: \[\Delta T = T_{final} - T_{initial} = 37.18^{\circ}\mathrm{C} - 25.00^{\circ}\mathrm{C} = 12.18^{\circ}\mathrm{C}\]
02

Determine the Heat Absorbed by Calorimeter

Use the formula for heat absorbed, which is the product of heat capacity and temperature change: \[q = C \times \Delta T = 12.05\, \mathrm{kJ}/^{\circ}\mathrm{C} \times 12.18^{\circ}\mathrm{C}\]\[q = 146.769\, \mathrm{kJ}\]
03

Calculate Moles of Benzene Burned

Find the number of moles of benzene by dividing the mass by its molar mass. The molar mass of benzene (\(\mathrm{C}_6\mathrm{H}_6\)) is \(78.11\, \mathrm{g/mol}\):\[n = \frac{3.51\, \mathrm{g}}{78.11\, \mathrm{g/mol}} = 0.0449\, \mathrm{mol}\]
04

Calculate Heat Released per Mole

Determine the heat released per mole by dividing the calorimeter heat by the number of moles:\[q_{mol} = \frac{q}{n} = \frac{146.769\, \mathrm{kJ}}{0.0449\, \mathrm{mol}}\]\[q_{mol} \approx 3268.64\, \mathrm{kJ/mol}\]

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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 key concept in understanding how substances react to temperature changes. It is the amount of heat required to change the temperature of a system by one degree Celsius. In the context of our experiment, the bomb calorimeter has a specific heat capacity of 12.05 kJ/°C. This means that for every 1°C that the calorimeter's temperature increases, 12.05 kJ of heat is absorbed by the calorimeter and its contents.

Consider what happens when we ignite benzene inside a calorimeter. As combustion occurs, heat energy is released, raising the temperature of the calorimeter. The calorimeter's heat capacity allows us to quantify exactly how much energy has been absorbed. In this case, the temperature rose by 12.18°C, and by multiplying this temperature change by the calorimeter's heat capacity, we can calculate the total heat absorbed in the process.
Bomb Calorimeter
A bomb calorimeter is an essential tool for measuring the heat of combustion of a substance. It is a closed system that prevents any gas exchange with the surrounding environment, making it ideal for measuring energy changes at constant volume.

In our exercise, benzene is burned in a bomb calorimeter, which ensures that the entire heat generated is transferred to the calorimeter itself, and none is lost to the environment. This setup allows us to precisely measure the increase in temperature due solely to the combustion of the substance. Bomb calorimeters are designed to withstand the high pressure and temperature that result from a combustive reaction, providing reliable data for energy calculations.
Molar Mass Calculation
Calculating the molar mass of a compound is crucial for determining how much of a substance is involved in a reaction. Molar mass is the weight of one mole of a substance in grams/mol.

For benzene, which has the formula \( \mathrm{C}_{6} \mathrm{H}_{6} \), the molar mass is obtained by summing the atomic masses of all the constituent atoms. Each carbon atom contributes about 12.01 g/mol, and each hydrogen atom contributes about 1.008 g/mol. With six carbon atoms and six hydrogen atoms, the molar mass of benzene comes out to be 78.11 g/mol.

This calculation allows us to convert from grams to moles, letting us determine how many moles of benzene were combusted. In the exercise, by dividing the mass of benzene burned (3.51 g) by its molar mass (78.11 g/mol), we find the number of moles burned.
Combustion Reaction
A combustion reaction is a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. For organic compounds like benzene, the products of complete combustion are carbon dioxide and water.

The given reaction \( \mathrm{C}_{6} \mathrm{H}_{6}(l) + \frac{15}{2} \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l) \) exemplifies this process. In our case, benzene reacts with oxygen, and the balanced reaction tells us the stoichiometric amounts of reactants and products.

Understanding the combustion reaction is essential because it explains how heat is produced in the calorimeter. By knowing the exact amount of heat released per mole of benzene, we can calculate energy released in kilojoules per mole, providing a deeper insight into the energy content of the substance.

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Most popular questions from this chapter

A bullet weighing 235 grains is moving at a speed of \(2.52 \times 10^{3} \mathrm{ft} / \mathrm{s}\). Calculate the kinetic energy of the bullet in joules and in calories. One grain equals \(0.0648 \mathrm{~g}\).

A \(19.6-\mathrm{g}\) sample of a metal was heated to \(61.67^{\circ} \mathrm{C}\). When the metal was placed into \(26.7 \mathrm{~g}\) of water in a calorimeter, the temperature of the water increased from \(25.00^{\circ} \mathrm{C}\) to \(30.00^{\circ} \mathrm{C}\). What is the specific heat of the metal?

Hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), is produced during decomposition of organic matter. When \(0.5000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{~S}\) burns to produce \(\mathrm{SO}_{2}\left(\mathrm{~g}\right.\) ) and \(\mathrm{H}_{2} \mathrm{O}(l),-281.0 \mathrm{~kJ}\) of heat is released. What is this heat in kilocalories?

An industrial process for manufacturing sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), uses hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), from the purification of natural gas. In the first step of this process, the hydrogen sulfide is burned to obtain sulfur dioxide, \(\mathrm{SO}_{2}\). $$\begin{gathered}2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}l)+2\mathrm{SO}_{2}(g) ; \\ \Delta H^{\circ}=-1124 \mathrm{~kJ}\end{gathered}$$ The density of sulfur dioxide at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is \(2.62 \mathrm{~g} /\mathrm{L}\), and the molar heat capacity is \(30.2 \mathrm{~J} /\left(\mathrm{mol} \cdot{ }^{\circ} \mathrm{C}\right) .\) (a) How much heat would be evolved in producing \(1.00 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? (b) Suppose heat from this reaction is used to heat \(1.00 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) from \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) to \(500^{\circ} \mathrm{C}\) for its use in the next step of the process. What percentage of the heat evolved is required for this?

What is the original definition of the calorie? What is the present definition?

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