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Chapter 19: Problem 108

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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Most popular questions from this chapter

Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

A standard air conditioner involves a \(r\) frigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at $31^{\circ} \mathrm{C}$. Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous of nonspontaneous?

Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

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(a) Can endothermic chemical reactions be spontaneous? (b) Can a process be spontaneous at one temperature and nonspontaneous at a different temperature? (c) Water can be decomposed to form hydrogen and oxygen, and the hydrogen and oxygen can be recombined to form water. Does this mean that the processes are thermodynamically reversible? (d) Does the amount of work that a system can doon its Id on the nath of the nrocese?
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