Problem 104 - Chemistry: The Central Science

Question

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is a toxic, highly flam mable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(l)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) : \begin{tabular}{lrl} \hline & \(\Delta H_{f}^{\circ}(\mathbf{k J} / \mathrm{mol})\) & \(\Delta G_{f}^{0}(\mathbf{k J} / \mathrm{mol})\) \\ \hline \(\mathrm{CS}_{2}(l)\) & \(89.7\) & \(65.3\) \\ \(\mathrm{CS}_{2}(g)\) & \(117.4\) & \(67.2\) \\ \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? (b) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) bums 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 K? (e) Use the data in the preceding table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(l)\). Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the preceding table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(\mathrm{l})\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(1 \mathrm{~atm}\) ?

Step-by-Step Solution

Answer

The CS2 molecule has a linear structure with C-S bond order of 2. The combustion reaction of liquid CS2 in O2 is: \(CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\). This reaction is exothermic and spontaneous at 298 K due to negative ΔH° and ΔG° values. The ΔS° for vaporization of CS2(l) is positive, as expected. The boiling point of CS2(l) is approximately 319.8 K, indicating that CS2 will be a liquid at 298 K and 1 atm.

Step 1: Start by counting the total number of valence electrons in the CS2 molecule. Carbon has 4 valence electrons, and each sulfur atom has 6, resulting in a total of 16 valence electrons: \[C: 4 + S: 6 \times 2 = 16\] Place the carbon atom in the center since it has the lowest electronegativity and connect it to the sulfur atoms. Then, arrange the remaining electrons in pairs and distribute them to fulfill the octet rule. They will form double bonds between the carbon and the sulfur atoms. The bond order of the C-S bonds is 2 since there are double bonds between C and S atoms. #a_Predicting Structure with VSEPR Method#

Count the electron groups around the central carbon atom. Each double bond is considered as one electron group. In the CS2 molecule, there are 2 electron groups. According to the VSEPR method, two electron groups result in a linear arrangement, which leads to a bond angle of 180°. Therefore, the structure of the CS2 molecule is linear. #a_Balancing the Combustion Equation#

Step 2: When CS2(l) combusts in O2, it forms CO2(g) and SO2(g). Write a balanced equation for this reaction: \[CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\] #a_Calculating ΔH° and ΔG° for the Reaction#

Use the available thermodynamic data for CS2(l), CS2(g), CO2(g), and SO2(g) with the following relationships: ΔH° = sum of ΔHf°(products) - sum of ΔHf°(reactants) ΔG° = sum of ΔGf°(products) - sum of ΔGf°(reactants) ΔH° = (1 × ΔHf°(CO2) + 2 × ΔHf°(SO2)) - (1 × ΔHf°(CS2) + 3 × ΔHf°(O2)) ΔG° = (1 × ΔGf°(CO2) + 2 × ΔGf°(SO2)) - (1 × ΔGf°(CS2) + 3 × ΔGf°(O2)) Note that ΔHf°(O2) and ΔGf°(O2) are both zero since O2 is a reference state. Use the given values for other species: ΔH° = (1 × (-393.5) + 2 × (-296.8)) - (1 × 89.7) ≈ -1073.9 kJ/mol ΔG° = (1 × (-394.4) + 2 × (-300.4)) - (1 × 65.3) ≈ -1024.9 kJ/mol The reaction is exothermic since ΔH° is negative. The reaction is also spontaneous at 298 K since ΔG° is negative. #a_Calculating ΔS° for Vaporization of CS2(l)#

Step 3: We can use the given thermodynamic data and the relationship between ΔH°, ΔS°, and ΔG°: ΔG° = ΔH° - TΔS° Rearrange the equation and solve for ΔS°: ΔS° = (ΔH° - ΔG°) / T Use the data for CS2(l) and CS2(g) at 298 K: ΔH° = ΔHf°(CS2(g)) - ΔHf°(CS2(l)) = 117.4 - 89.7 = 27.7 kJ/mol ΔG° = ΔGf°(CS2(g)) - ΔGf°(CS2(l)) = 67.2 - 65.3 = 1.9 kJ/mol T = 298 K ΔS° = (27.7 - 1.9) / 298 ≈ 0.086 kJ/mol K The sign of ΔS° is positive, which is expected for a vaporization process. #a_Estimating the Boiling Point of CS2(l)#

Use the Clausius-Clapeyron equation to estimate the boiling point at 1 atm: \[\ln{\frac{P_2}{P_1}} = \frac{\Delta H_\text{vap}}{R}(\frac{1}{T_1} - \frac{1}{T_2})\] Assume that the vapor pressure is 1 atm at the boiling point (T2). At 298 K (T1), the vapor pressure (P1) can be calculated using ΔG° and the relationship: ΔG° = -RT ln(P1) Rearrange and solve for P1: \[P_1 = e^\frac{-\Delta G^\circ}{RT}\] Use the given ΔG° of 1.9 kJ/mol at 298 K: \[P_1 = e^\frac{-(1.9 \times 10^3)}{(8.314)(298)} \approx 0.883\, \text{atm}\] Now, apply the Clausius-Clapeyron equation: \[\ln\frac{1}{0.883} = \frac{(27.7 \times 10^3)}{(8.314)}\left(\frac{1}{298} - \frac{1}{T_2}\right)\] Rearrange and solve for T2 (the boiling point): \[T_2 = \frac{1}{\frac{1}{298} - \frac{(8.314)(\ln\frac{1}{0.883})}{(27.7)(10^3)}} \approx 319.8\, \text{K}\] The boiling point of CS2(l) is estimated to be 319.8 K, and since this is higher than 298 K, it is predicted that CS2 will be a liquid at 298 K and 1 atm.

Key Concepts

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is crucial for predicting the shapes of molecules. When looking at the CS extsubscript{2} molecule, using VSEPR theory means understanding how electron pairs around the central carbon atom influence its geometric shape.
In CS extsubscript{2}, there are two double bonds between the carbon and sulfur atoms. Each double bond is considered a single electron group in VSEPR theory.
With two electron groups present, the theory predicts a linear structure for the molecule. This results in bond angles of 180°, making the CS extsubscript{2} molecule straight.
This clarity in structure helps us understand how CS extsubscript{2} interacts with other molecules and predicts physical and chemical properties.

Lewis Structures

Drawing the Lewis structure of a molecule like CS extsubscript{2} helps visualize how atoms are bonded and their electron connectivity. Start by calculating valence electrons for carbon and sulfur.
Carbon has 4, while each sulfur atom has 6. This totals 16 valence electrons, crucial for completing the Lewis structure.
Carbon, being less electronegative, sits at the center with sulfur atoms at both ends. To satisfy the octet rule, carbon forms double bonds with each sulfur, a stable configuration with two bonding pairs.
The correct Lewis structure is both a guide and a tool, illustrating the predicted chemical behavior and reactivity of the molecule.

Exothermic Reactions

Exothermic reactions release energy, typically in heat form, when bonds form in the product molecules. In this case, consider the combustion of CS extsubscript{2} in oxygen to form CO extsubscript{2} and SO extsubscript{2}.
This reaction is specifically exothermic since the computed ΔH° is negative (-1073.9 kJ/mol).
The energy released signifies stronger bonds in the products than in the initial reactants.
Recognizing exothermic reactions is essential; such reactions provide energy in various industrial and environmental processes.

Spontaneous Reactions

A reaction is spontaneous if it occurs without continuous external energy input. This is determined by analyzing the Gibbs free energy change, ΔG°.
In our example, the ΔG° of -1024.9 kJ/mol for CS extsubscript{2}'s combustion indicates spontaneity at 298 K.
Negative ΔG° implies that the process naturally progresses towards product formation without intervention.
Understanding spontaneity helps predict whether a reaction will proceed under given conditions, critical for practical and theoretical chemistry.

Entropy Change

Entropy concerns the disorder or randomness of a system. The change in entropy, ΔS°, is calculated when understanding processes like vaporization.
During vaporization of CS extsubscript{2}, the structure shifts from liquid to gas, increasing molecular disorder.
This is reflected in a positive ΔS° value, computed as 0.086 kJ/mol·K, aligning with increased entropy expectations for transitioning to a gaseous state.
Recognizing ΔS° values aids in assessing whether processes favor increased disorganization, a key principle in thermodynamics and chemistry.