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Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & \(\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(I) .\) Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

Short Answer

Expert verified
The Lewis structure of the \(CS_2\) molecule is linear with double bonds between carbon and the two sulfur atoms. The structure of the \(CS_2\) molecule is linear with an angle of 180° between the C-S bonds. The balanced reaction of \(CS_2(l)\) burning in \(O_2(g)\) is given by \(CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\). The reaction is exothermic with \(\Delta H^{\circ} = -1263.7\, kJ/mol\) and spontaneous at 298 K with \(\Delta G^{\circ} = -1060.5 \, kJ/mol\). The \(\Delta S^{\circ}\) for vaporization of \(CS_2(l)\) is positive, 0.082 \(J/(mol\cdot K)\). The boiling point of \(CS_2\) is estimated to be 1094 K, so it will be in the liquid state at 298 K and 101.3 kPa.

Step by step solution

01

(a) Lewis structure and the bond order of C-S bonds

\( Carbon (C) has 4 valence electrons and Sulfur (S) has 6 valence electrons. Since one molecule CS_2 contains 2 sulfur atoms, the total number of valence electrons in CS_2 molecule is \(4 + 2*6 = 16\). To reach an octet, carbon shares 8 electrons with two sulfur atoms. The Lewis structure of the \(CS_2\) molecule is: C = S = C Each C-S bond in \(CS_2\) has a double bond, so the bond order of the C-S bonds is 2.
02

(b) VSEPR Prediction of Molecular Structure

\( According to VSEPR theory, molecular structures minimize electron pair repulsion. In \(CS_2\), Carbon has two electron domains (two double bonds with sulfur atoms). Since there are no lone pairs on carbon, the molecule will have a linear structure. Therefore, the structure of the \(CS_2\) molecule is linear with an angle of 180° between the C-S bonds.
03

(c) Balanced equation with O2

\( \(CS_2\) reacts with \(O_2\) to form CO_2 and SO_2. To balance the equation, we adjust the stoichiometric coefficients: \(CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\)
04

(d) \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) calculation

\( To calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction, we will use the given thermodynamic data for \(CS_2\) and data from Appendix C for the other substances. We will apply the following equations: \(\Delta H^{\circ} = \sum H_{products} - \sum H_{reactants}\) \(\Delta G^{\circ} = \sum G_{products} - \sum G_{reactants}\) Using the data given, and the values for \(O_2(g), CO_2(g),\) and \(SO_2(g)\) from Appendix C: \(\Delta H^{\circ} = [1(-393.5) + 2(-296.8)] - [1(89.7) + 3(0)] = -1174 - 89.7 = -1263.7 \, kJ/mol\) \(\Delta G^{\circ} = [1(-394.4) + 2(-300.4)] - [1(65.3) + 3(0)] = -995.2 - 65.3 = -1060.5 \, kJ/mol\) The reaction is exothermic since \(\Delta H^{\circ} < 0\), and it is spontaneous at 298 K since \(\Delta G^{\circ} < 0\).
05

(e) \(\Delta S^{\circ}\) for vaporization

\( To calculate \(\Delta S^{\circ}\) for vaporization at 298 K, we can use the following equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) Solving for \(\Delta S^{\circ}\): \(\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}\) Using the given data for \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the vaporization of \(CS_2(l)\): \(\Delta S^{\circ} = \frac{89.7 - 65.3}{298} = \frac{24.4}{298} = 0.082 \, J/(mol\cdot K)\) The positive sign for \(\Delta S^{\circ}\) is expected for a vaporization process, as it indicates an increase in entropy.
06

(f) Boiling point estimation and state at 298K and 101.3 kPa

\( To estimate the boiling point of \(CS_2\), we can use the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) At the boiling point, \(\Delta G^{\circ} = 0\), so: \(0 = \Delta H^{\circ} - T\Delta S^{\circ}\) Solving for T: \(T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{89.7}{0.082} = 1094 \, K\) Since the boiling point of \(CS_2\) is much higher than 298 K, we can predict that the substance will be in the liquid state at 298 K and 101.3 kPa.

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Key Concepts

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

Lewis Structure
In chemistry, the Lewis structure is a visual representation that shows how atoms are bonded within a molecule. It's a blueprint of molecular structure highlighting shared electron pairs, also known as covalent bonds, and lone pairs of electrons. Understanding Lewis structures helps us see how atoms achieve a full valence shell, akin to the noble gases.

For carbon disulfide (\(CS_2\)), the process begins by adding up the valence electrons. Carbon (C) has 4 valence electrons, and each sulfur (S) has 6. In a single \(CS_2\) molecule, the total count is \(4 + 2 \times 6 = 16\) valence electrons. This guide directs us to construct the structure where carbon is central, forming a double bond with each sulfur atom, resulting in the linear Lewis structure C=S=C.

Double bonds mean two pairs of shared electrons between atoms. Therefore, each C-S bond is a double bond, giving each C-S a bond order of 2. The bond order refers to the number of chemical bonds between a pair of atoms, impacting stability and bond strength.
VSEPR Theory
The VSEPR (Valence Shell Electron Pair Repulsion) theory is a straightforward approach to predict the geometric layout of a molecule from its Lewis structure. According to this theory, the shape of electrons around a central atom is organized so that electron-pair repulsion is minimized.

When examining carbon disulfide (\(CS_2\)), the VSEPR theory plays a crucial role. The central carbon atom in \(CS_2\) forms two electron domains due to the two double bonds with sulfur atoms. Since there are no lone pairs of electrons on carbon in this scenario, the electron repulsions are minimized by positioning the sulfur atoms at the opposite ends. This results in a linear geometry for \(CS_2\), with a bond angle of 180°, ideal for reducing any repulsive forces.

This prediction aligns tightly with the molecular structure depicted through the Lewis structure, reinforcing the dual insights provided by these chemistry tools.
Enthalpy
Enthalpy is a measure of the total energy within a chemical system at constant pressure. It's frequently represented by \(\Delta H\), where a negative value signifies an exothermic process (energy is released), and a positive value indicates an endothermic process (energy is absorbed).

When considering chemical reactions, enthalpy changes are vital to determining whether reactions are energy-releasing or energy-absorbing. In the scenario of the combustion of liquid \(CS_2\), it reacts with oxygen to form carbon dioxide \((CO_2)\) and sulfur dioxide \((SO_2)\). From the provided data, we calculated \(\Delta H^{\circ}\) for the reaction as -1263.7 \(kJ/mol\). The negative sign indicates this reaction is exothermic, which aligns with the observation that burning \(CS_2\) produces a blue flame, a hallmark of energy release.

Enthalpy changes not only indicate the heat flow but help predict the spontaneous nature of reactions through the Gibbs free energy equation. This relates directly to the overall energy considerations being crucial for evaluating reaction viability and conditions.

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Most popular questions from this chapter

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\), (b) \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)

Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy can only be produced but cannot not be destroyed. (b) In a certain process the entropy of the system changes by \(1.2 \mathrm{~J} / \mathrm{K}\) (increase) and the entropy of the surroundings changes by \(-1.2 \mathrm{~J} / \mathrm{K}\) (decrease). Thus, this process must be spontaneous. (c) In a certain process the entropy of the system changes by \(1.3 \mathrm{~J} / \mathrm{K}\) (increase) and the entropy of the surroundings changes by \(-1.2 \mathrm{~J} / \mathrm{K}\) (decrease). Thus, this process must be reversible.

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathbf{b})\) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K} ?\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

Does the entropy of the system increase, decrease, or stay the same when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution?

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

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