Question

The molar heat of vaporization of carbon disulfide, \(\mathrm{CS}_{2},\) is \(28.4 \mathrm{kJ} / \mathrm{mol}\) at its normal boiling point of \(46^{\circ} \mathrm{C} .\) How much energy (heat) is required to vaporize \(1.0 \mathrm{g}\) of \(\mathrm{CS}_{2}\) at \(46^{\circ} \mathrm{C}\) ? How much heat is evolved when \(50 . \mathrm{g}\) of \(\mathrm{CS}_{2}\) is condensed from the vapor to the liquid form at \(46^{\circ} \mathrm{C} ?\)

Short answer

The heat required to vaporize 1.0 g of \(\mathrm{CS}_{2}\) at \(46^{\circ}\mathrm{C}\) is 0.37 kJ, and the heat evolved when 50.0 g of \(\mathrm{CS}_{2}\) is condensed at \(46^{\circ}\mathrm{C}\) is -18.6 kJ.

Step-by-step solution

Calculate moles of \(\mathrm{CS}_{2}\)

Given that 1.0 g of \(\mathrm{CS}_{2}\) needs to be vaporized, we first need to determine the number of moles in 1.0 g. To do this, we need the molar mass of \(\mathrm{CS}_{2}\). Carbon has a molar mass of 12.01 g/mol and sulfur has a molar mass of 32.07 g/mol. So, the molar mass of \(\mathrm{CS}_{2}\) is: \(\mathrm{Molar\ mass} = 12.01 + 2(32.07) = 76.15 \mathrm{\ g/mol}\) Now we can find the number of moles in 1.0 g of \(\mathrm{CS}_{2}\): \(\mathrm{Moles\ of\ CS}_2 = \cfrac{1.0\ g}{76.15\ g/mol} = 1.31 \times 10^{-2}\ \text{mol}\)

Compute the heat required to vaporize 1.0 g of \(\mathrm{CS}_{2}\)

Now that we have the number of moles in 1.0 g of \(\mathrm{CS}_{2}\), we can calculate the heat required to vaporize it using the molar heat of vaporization provided in the problem: \(\text{Heat\ required} = 1.31 \times 10^{-2}\ \text{mol} \times 28.4\ \cfrac{\mathrm{kJ}}{\mathrm{mol}} = 0.37\ \mathrm{kJ}\)

Calculate moles of \(\mathrm{CS}_{2}\) in 50.0 g

Now we need to determine the number of moles in 50.0 g of \(\mathrm{CS}_{2}\): \(\mathrm{Moles\ of\ CS}_2 = \cfrac{50.0\ g}{76.15\ g/mol} = 0.656\ \text{mol}\)

Compute the heat evolved when 50.0 g of \(\mathrm{CS}_{2}\) condenses

To find the heat evolved when the substance condenses, we multiply the moles by the negative of the molar heat of vaporization: \(\text{Heat\ evolved} = 0.656\ \text{mol} \times (-28.4)\ \cfrac{\mathrm{kJ}}{\mathrm{mol}} = -18.6\ \mathrm{kJ}\) So, the heat required to vaporize 1.0 g of \(\mathrm{CS}_{2}\) at \(46^{\circ}\mathrm{C}\) is 0.37 kJ, and the heat evolved when 50.0 g of \(\mathrm{CS}_{2}\) is condensed at \(46^{\circ}\mathrm{C}\) is -18.6 kJ (negative sign indicates heat is being evolved).

Key concepts

Enthalpy Change

Enthalpy change is a crucial concept in thermodynamics, especially when studying chemical reactions and phase changes. It refers to the heat transferred in a system at constant pressure. For phase changes like vaporization or condensation, enthalpy change helps predict the energy required to transform a substance from one phase to another.

The molar heat of vaporization is a type of enthalpy change. It signifies the amount of energy needed to vaporize one mole of a substance at its boiling point. For carbon disulfide (CS extsubscript{2}), this value is given as 28.4 kJ/mol. When vaporizing 1 mole of CS extsubscript{2}, 28.4 kJ of energy is absorbed. Conversely, when CS extsubscript{2} vapor condenses back to liquid, this amount of energy is released.

Understanding enthalpy changes can help in calculating the energy associated with these transformations, which is crucial in both academic and industrial settings. It simplifies predicting heat exchanges during reactions and underpins many processes in chemical engineering.

Phase Change

Phase changes occur when a substance transitions between solid, liquid, and gaseous states. Key phase changes include melting, freezing, vaporization, and condensation. Each phase change involves an amount of energy exchange without altering the substance’s chemical composition.

During vaporization, a liquid absorbs heat to become a gas. This process requires energy, hence the significance of the molar heat of vaporization. For carbon disulfide, vaporizing 1.0 g involves converting mass to moles and then using the enthalpy value to find heat absorbed.

Conversely, condensation happens when a gas releases heat to form a liquid. In our example, 50.0 g of CS extsubscript{2} gas condensing releases heat. The negative sign indicates an exothermic process, where heat flows out of the system.

• Vaporization: Absorbs heat
• Condensation: Releases heat

Understanding these changes is essential for mastering energy calculations in chemical processes.

Carbon Disulfide

Carbon disulfide (CS extsubscript{2}) is a chemical compound with interesting properties. It's a colorless, volatile liquid with a distinct odor. Often used in industrial applications, it serves as a building block in various chemical synthesis processes.

CS extsubscript{2} exhibits specific behaviors during phase transitions. With a normal boiling point of 46°C, it readily evaporates at room temperature, making it important to consider safety in handling due to its flammability and odor.

The molecular structure of carbon disulfide, consisting of one carbon atom and two sulfur atoms, results in a molar mass of 76.15 g/mol. Understanding its properties and how it interacts during phase changes aids in practical chemical applications and environmental considerations.

• Boiling Point: 46°C
• Molar Mass: 76.15 g/mol

Knowing CS extsubscript{2}'s characteristics is beneficial in both theoretical studies and real-world chemical engineering applications.